Category: CBSE Class 10

CBSE Sample Papers for Class 10 Maths Paper 3 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 3. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Sample Papers for Class 10 Maths Paper 3

Board	CBSE
Class	X
Subject	Maths
Sample Paper Set	Paper 3
Category	CBSE Sample Papers

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 3 of Solved CBSE Sample Papers for Class 10 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

GENERAL INSTRUCTIONS:

• All questions are compulsory.
• This question paper consists of 30 questions divided into four sections A, B, C and D.
• Section A comprises of 6 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 8 questions 1 of 4 marks each.
• There is no overall choice. However, internal choice has been provided in one question of 2 marks, 1 three questions of 3 marks each and two questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• In question of construction, drawings shall be neat and exactly as per the given measurements.
• Use of calculators is not permitted. However, you may ask for mathematical tables.

SECTION A

Question numbers 1 to 6 carry 1 mark each.

Question 1.
Write the polynomial whose zeroes are 2 + √3 and 2 – √3.

Question 2.
If the quadratic equation px² – 2√5 px + 15 = 0 has two equal real roots, then find the value of p.

Question 3.
Which term of the arithmetic progression 4, 9, 14, 19……is 109?

Question 4.
Explain, why 17 x 5 x 11 x 3 x 2 + 2 x 11 is a composite number?

Question 5.
If the adjoining figure is a sector of a circle of radius 10.5 cm, find the perimeter of the sector. [Take π=\(\frac { 22 }{ 7 }\)]

Question 6
Find the value of a, if the distance between the points A(-3, -14) and B (a, -5) is 9 units.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7. Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

Question 8.
The coordinates of the mid-point of the line joining the points (3p, 4) and (-2, 2q) are (5, p). Find the values of p and q.

Question 9.
If the total surface area of a solid hemisphere is 462 cm², find its volume.[Take π=\(\frac { 22 }{ 7 }\)]

Question 10.
If 4 cos θ = 11 sin θ, find the value of

Question 11.
In an equilateral triangle ABC, AD is drawn perpendicular to BC meeting BC in D. Prove that AD² = 3BD².

Question 12.
The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7 respectively.

Question 14.
Solve graphically the pair of linear equations 2x + 3y = 11 and 2x – 4y = – 24. Hence, find the value of m, given that the line represented by y = mx + 3 passes through the intersection of the given pair.

Question 15.
Find the value of k for which the following pair of linear equations has no solution : x + 2y = 3; (k – 1)x + (k + 1)y = (k + 2).

Question 16.
Find the sum of all two digit numbers which when divided by 3 yield 1 as remainder.
OR
The sum of 4th and 8th terms of an AP is 24 and the sum of 6th and 10th terms is 44. Find the AP.

Question 17.
Find the mean of the following frequency distribution :

Question 18.
Find the area of the quadrilateral whose vertices taken in order are A(-5, -3), B(-4, -6), C(2, -1) and D(1, 2).
OR
Prove that the points A(2, 3), B(-2, 2), C(-1, -2) and D(3, -1) are the vertices of a square ABCD.

Question 19.
Draw a line segment of length 7.8 cm and divide it in the ratio 5 : 8. Measure the two parts. Also, justify your construction.

Question 20.
Prove the following identities:

OR
Prove that \((cosec A – sin A) (sec A – cos A) =\frac { 1 }{ tanA+cotA }\)

Question 21.
In ∆PQR, right angled at Q, PQ = 5 cm and PR + QR = 25 cm. Find the values of sin P, cos P and tan P.

Question 22.
A toy is in the form of a cone of base radius 3.5 cm mounted on a hemisphere of base diameter 7 cm. If the total height of the toy is 15.5 cm, find the total surface area of the toy.
OR
A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
Ram’s mother has given him money to buy some boxes from the market at the rate of 4x² + 3x – 2. The total amount of money given by his mother is represented by 8x⁴ + 14x³ – 2x² + 7x – 8. Out of this money he donated some amount of money to a child who was studying in the light of street lamp. Find how much amount of money he donated after purchasing the maximum number of boxes from the market? Write the values shown by Ram.

Question 24.
A student scored a total of 32 marks in class tests in mathematics and science. Had he scored 2 marks less in science and 4 more in mathematics, the product of his marks would have been 253. Find his marks in two subjects.
OR
If -4 is a root of the equation x² + 2x + 4p = 0, find the values of k for which the equation x² + p(1 + 3k)x + 7(3 + 2k) = 0 has equal roots.

Question 25.
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC at L and AD produced in E. Prove that EL = 2BL.
OR
In a ∆PQR, N is a point on PR such that QN ⊥ PR. If PN x NR = QN², prove that ∠PQR = 90°.

Question 26.
In the adjoining figure, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm, PT and QT are tangents to the circle at points P and Q respectively. OT intersects the circle at the point E. If AB is tangent to the circle at E, find the length of AB.

Question 27.
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are α and β respectively. Prove that the height of the tower is \(\frac { h\quad tan\alpha }{ tan\beta -tan\alpha } \)

Question 28.
PS is a diameter of a circle of radius 6 cm. Q and R are points on the diameter that PQ, QR and RS are equal. Semicircles are drawn with PQ and QS as diameters, as shown in the figure. Find the perimeter of the shaded region.
Also find the area of the shaded region. (Use π = 3.14)

Question 29.
A die has six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded.
(i) How many different scores are possible?
(ii) What is the probability of getting a total of 7?
OR
Two customers Shyam and Ekta visit a particular shop in the same week from Tuesday to Saturday. Each is equally likely to visit the shop on any day. Find the probability that they visit the shop
(i) on the same day
(ii) on consecutive day
(iii) on different days
(iv) neither on same day nor on consecutive days.

Question 30.
The following distribution gives the daily wages of 50 workers of a factory:

Convert the distribution to a less than type cumulative frequency distribution, and draw its ogive. Hence, obtain the median wages from the graph.

Answers

Answer 1.
Required polynomial = x² – (Sum of the zeroes) x + Product of the zeroes
= x² – (2 + √3 + 2 – √3)x + (2 + √3)(2 – √3)
= x² – 4x + 1.

Answer 2.
Given px² – 2√5 px + 15 = 0
For equal roots, D = (-2√5 p)² – 4 x p x 15 = 0
=> 20p² – 60p = 0 => 20p(p – 3) = 0
=> p = 0 or p = 3 but p ≠ 0 => p = 3.

Answer 3.
Given 4, 9, 14, 19 … 109
Here, a = 4, d = 5, l = 109
Now, l = a + (n – 1)d
109 = 4 + (n – 1) x 5 => 5(n – 1) = 105 => n – 1 = 21 => n = 22.
Hence, 22nd term is 109.

Answer 4.
Given number = 17 x 5 x 11 x 3 x 2 + 2 x 11 = 11 x 2 x (17 x 5 x 3 + 1)
which is product of more than one prime number, therefore, the given number is prime.

Answer 5.
Radius = 10.5 cm = \(\frac { 21 }{ 2 }\) cm

Perimeter = AC + BC + AB = (10.5 + 10.5 + 11) cm = 32 cm.

Answer 6.
AB = 9

Answer 7.
Let a be any positive odd integer.
Applying Euclid’s division lemma with divisor 4, we get
a = 4q, 4q + 1, 4q + 2 or 4q + 3 where q is some whole number.
But 4q and 4q + 2 are even integers and a is odd integer, therefore, a cannot be of the form 4q or 4q + 2.
Hence, a is of the form 4q + 1 or 4q + 3, where q is some whole number.

Answer 8.
The mid-point of the line segment joining the points (3p, 4) and (-2, 2q) is

and 2 + q = p => 2 + q = 4 => q = 2.
Hence, p = 4 and q = 2.

Answer 9.
Given, total surface area of hemisphere = 462 cm²
=> 3πr² = 462

Answer 10.
Given 4 cos θ = 11 sin θ
⇒\(\frac { sin\theta }{ cos\theta } =\frac { 4 }{ 11 } \) ….(i)

Answer 11.
In ∆ABD right angled at D.
By Pythagoras theorem,
AB² = AD² + BD²
=> BC² = AD² + BD² (∵ AB = BC = CA)
=> (2BD)² = AD² + BD²
(∵in an equilateral ∆ perpendicular is the median , ∴\(BD=\frac { 1 }{ 2 }BC\))
=> 4BD² = AD² + BD²
=> AD² = 3BD²

Answer 12.
Given: A circle inscribed in an isosceles ∆ABC with AB = AC, touching the sides BC, CA and AB at D, E and F respectively.
To prove: BD = DC
Proof: ∵ Length of tangents from an external point to the circle are equal

∴ AF = AE …(i)
BF = BD …(ii)
and CD = CE …(iii)
∴ AB = AC
=> AF + BF = AE + EC
=> BF = EC (using (i))
=> BD = CD (using (ii) and.(iii))

Answer 13.
Given that required number when divides 2053 and 967 leaves a remainder 5 and 7 respectively.
∴ The required number is HCF of 2053 – 5 = 2048 and 967 – 7 = 960

∴HCF of 2048 and 960 = 64
Hence, the required largest number is 64.

Answer 14.
The given equations can be written as

The given lines intersect at the point P(-2, 5). Therefore, the solution of the given pair of linear equation is x = -2, y = 5.
Given that the line y = mx + 3 passes through the intersection of the given pair i.e. (-2, 5)
∴5 = m x (-2) + 3 => -2m – 2 => m = -1.

Answer 15.
The given pair of linear equations can be written as
x + 2y – 3 = 0 and (k – 1)x + (k + 1)y – (k + 2) = 0
Here, a1 = 1,b1 = 2, c1 = -3 and a2 = k – 1, b2 = k + 1, c2 = -(k + 2).
Now, the given pair of linear equations has no solution, if

=> 2k – 2 = k + 1 and 2k + 4 ≠ 3k + 3
=> k = 3 and k ≠ 1.
Hence, the given pair of linear equations will have no solution when k = 3.

Answer 16.
Two digit numbers which leaves remainder 1 when divided by 3 are 10, 13, 16, 19 …, 97.
These numbers form an AP with a = 10, d = 3 and l = 97.
∵ l = a + (n – 1 )d
=> 97 = 10 + (n – 1) x 3 => 3(n – 1) = 87 => n – 1 = 29
=> n = 30
Now, the sum of all such numbers

= 15 x 107 = 1605.
Hence, the required sum is 1605.
OR
Let the first term and common difference of AP be a and d respectively.
Given: T4 + T8 = 24 and T6 + T10 = 44
=> a + 3d + a + 7d = 24 and a + 5d + a + 9d = 44
=> 2a + 10d = 24 …(i)
and 2a + 14d = 44 …(ii)
Subtracting (i) from (ii), we get
4d = 20 => d = 5.
Putting d = 5 in (i), we get
2a + 10 x 5 = 24
=> 2a = -26 => a = -13.
Hence, the required AP is
-13, -13 + 5, -13 + 2 x 5, -13 + 3 x 5, ……
i.e. -13, -8, -3, 2, …

Answer 17.
We shall use step-deviation method. Taking assumed mean a = 25, construct the table as under. Here, h = 10.

Answer 18.
Join AC.
Now,

∵AB = BC = CD = DA i.e. all 4 sides are equal
and AC = BD i.e. diagonals are equal.
∴ ABCD is a square.

Answer 19.
Steps of construction:
1. Draw a line segment AB of length 7.8 cm by using ruler.
2. Draw any ray AX making an acute angle with AB.
3. Locate 13 (5 + 8) points, A1, A2, A3, …, A13 on AX such that AA1 = A1A2 = A2A3 = …… = A12A13.
4. Join A13B.
5. Through A5, draw a line parallel to A13B (making an angle equal to ∠AA13B) to intersect AB at C. Then C is the required point which divides AB in the ratio 5 : 8.
On measuring, we find that AC = 3 cm and CB = 4.8 cm.

Answer 20.

Answer 21.
Given PR + QR = 25 cm => QR = (25 – PR) cm.
In ∆PQR, ∠Q = 90°. By Pythagoras Theorem,
PR² = PQ² + QR²
=> PR² = 5² + (25 – PR)²
=> PR² = 25 + 625 – 50 PR + PR²
=> 50 PR = 650 => PR = 13 cm
∴ QR = (25 – 13) cm = 12 cm.

Answer 22.
Given radius of cone = 3.5 cm
Diameter of hemisphere = 7 cm

Answer 23.
Total amount of money p(x) = 8x⁴ + 14x³ – 2x² + 7x – 8.
Cost of each box g(x) = 4x² + 3x – 2
Maximum number of boxes purchased = q(x)
Amount of money donated to the child = r(x)
We need to find q(x) and r(x).
On dividing p(x) by g(x), we have

Hence, the amount of money he donated to the child = 14x – 10
and maximum number of boxes, he purchased = 2x² + 2x – 1.
The values shown by Ram are kindness, helpfulness, well wisher of society, care for children and promoting education.

Answer 24.
Let the student’s marks in mathematics be x.
As the total marks in mathematics and science is 32,
so the student’s marks in science = 32 – x.
Given if he scores 2 less marks in science and 4 more marks in mathematics, then his new marks in science = 32 – x – 2 = 30 – x
and his new marks in mathematics = x + 4.
According to given, (x + 4) (30 – x) = 253
=> 30x + 120 – x² – 4x = 253
=> x² – 26x + 133 = 0
=> x² – 19x – 7x + 133 = 0
=> (x – 7) (x – 19) = 0 => x = 7 or x = 19.
When x = 7, 32 – x = 25; when x = 19, 32 – x = 13.
Hence, student’s marks in mathematics = 7 and in science 25
or student’s marks in mathematics = 19 and in science = 13.
OR
Given x = -4 is a root of the equation x² + 2x + 4p = 0
(-4)² + 2 x (-4) + 4p = 0
=> 16 – 8 + 4p = 0 => 4p = -8 => p = -2.
Now, the equation x² + p(1 + 3k)x + 7(3 + 2k) = 0 has equal roots
D = 0 => {p( 1 + 3k)}² – 4 x 1 x 7(3 + 2k) = 0
=> p²(1 + 6k + 9k²) – 28(3 + 2k) = 0
=> (-2)² (1 + 6k + 9k²) – 84 – 56k = 0 (∵ p = -2)
=> 4 + 24k + 36k² – 84 – 56k = 0
=> 36k² – 32k – 80 = 0
=> 9k² – 8k – 20 = 0
=> (9k + 10) (k – 2) = 0
=> \(k=-\frac { 10 }{ 9 }\) or k = 2.

Answer 25.
In ∆BMC and ∆EMD,

MC = MD (∵ M is mid-point of CD)
∠CBM = ∠DEM (alt. ∠s, BC || AE)
∠BMC = ∠EMD (vert. opp. ∠s)
∴ ∆BMC ≅ ∆EMD (AAS criterion of congruence)
=> BC = DE (c.p.c.t.)
Also AD = BC (opp sides of ||gm ABCD)
∴ AE = AD + DE = BC + BC => AE = 2BC …(i)
In AAEL and ACBL,
∠AEL = ∠CBL
∠ALE = ∠CLB
∴ ∆AEL ~ ∆CBL

Answer 26.
Since tangent is perpendicular to radius, OP ⊥ PT.
In ∆OPT, ∠OPT = 90°. By Pythagoras theorem,
PT² = OA² – OP² = 13² – 5² = 144 => PT = 12 cm
As the lengths of tangent drawn from a point to a circle are equal,
AP = AE = x cm (say)
=> AT = PT – AP = (12 – x) cm
ET = OT – OE = 13 cm – 5 cm = 8 cm.
Given, AB is tangent to the circle at E, OE ⊥ AB => ∠AET = 90°.
In ∆AET, ∠AET = 90°. By Pythagoras theorem,
AT² = AE² + ET² => (12 – x)² = x² + 8²
=> 144 – 24x + x² = x² + 64

Answer 27.
Let AB be a vertical tower and BP be a flagstaff of height h surmounted on the tower AB, and O be the point of observation on the plane
(ground), then BP = h.
Let OA = d and AB = H.
Given, ∠BOA = α and ∠POA = β.

Answer 28.
PS = 2 x radius = (2×6) cm = 12 cm.
Given PQ = QR = RS
=> PQ = \(\frac { 1 }{ 3 }\) of PS = \(\frac { 1 }{ 3 }\) of 12 cm = 4 cm
and QS = 8 cm.

Answer 29.
(i) For the sum of outcomes on the top faces of two dice i.e. the total score obtained, we construct the table as under:

The different scores obtained are 0, 1, 2, 6, 7, 12. They are six in number.
Hence, the number of different possible scores are 6 (in number).
(ii) Total number of outcomes are 6×6 i.e. 36, which are all equally likely.
The table shows that we get score 7 twelve times.
So, the number of outcomes favourable to the event ‘getting a total of 7’ is 12.
∴ P(getting a total of 7) = \(\frac { 12 }{ 36 } =\frac { 1 }{ 3 } \)
OR
Let T, W, Th, F and S represent Tuesday, Wednesday, Thursday, Friday and Saturday respectively. Since Shyam and Ekta visit a particular shop in the same week from Tuesday to Saturday, the outcomes are:
(T, T), (T, W), (T, Th), (T, F), (T, S)
(W, T), (W, W), (W, Th), (W, F), (W, S)
(Th, T), (Th, W), (Th, Th), (Th, F), (Th, S)
(F, T), (F, W), (F, Th), (F, F), (F, S)
(S, T), (S, W), (S, Th), (S, F), (S, S)
Total number of outcomes = 5 x 5 = 25 and all outcomes are equally likely.
(i) The outcomes favourable to the event ‘visit on the same day’ are (T, T), (W, W), (Th, Th), (F, F), (S, S); which are 5 in number.
∴ P (visit on the same day) = \(\frac { 5 }{ 25 } =\frac { 1 }{ 5 } \)
(ii) The outcomes favourable to the event ‘visit on consecutive days’ are (T, W), (W, T), (W, Th), (Th, W), (Th, F), (F, Th), (F, S), (S, F); which are 8 in number.
∴ P (visit on consecutive days) = \(\frac { 8 }{ 25 }\).
(iii) P (visit on different days) = 1 – P (visit on same day)
= \(1-\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
(iv) The outcomes favourable to the event ‘visit neither on same day nor on consecutive days’ are (T, Th), (T, F), (T, S), (W, F), (W, S), (Th, T), (Th, S), (F, T), (F, W), (S, T), (S, W), (S, Th); which are 12 in number.
∴P (visit neither on same day nor on consecutive days) = \(\frac { 12 }{ 25 }\)

Answer 30.
From the given data, we construct the table for cumulative frequency distribution as under:

Take 1 cm along x-axis = Rs 20 and 1 cm along y-axis = 10 workers.
Plot the points (120, 12), (140, 26), (160, 34), (180, 40), (200, 50) and (100, 0). Join these points by a free hand drawing. The required ogive (less than type) is drawn on the graph sheet given below.

Since the scale on x-axis starts at 100, a kink (break) is shown on the x-axis near the origin.
To find median:
Let A be the point on y-axis representing frequency = \(\frac { n }{ 2 } =\frac { 50 }{ 2 } \) = 25.
Through A, draw a horizontal line to meet the ogive at P. Through ,P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 138.5
Hence, the median daily wages = Rs 138.50

We hope the CBSE Sample Papers for Class 10 Maths Paper 3 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 3, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 5 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 5. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Sample Papers for Class 10 Maths Paper 5

Board	CBSE
Class	X
Subject	Maths
Sample Paper Set	Paper 5
Category	CBSE Sample Papers

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 5 of Solved CBSE Sample Papers for Class 10 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

GENERAL INSTRUCTIONS:

• All questions are compulsory.
• This question paper consists of 30 questions divided into four sections A, B, C and D.
• Section A comprises of 6 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 8 questions 1 of 4 marks each.
• There is no overall choice. However, internal choice has been provided in one question of 2 marks, 1 three questions of 3 marks each and two questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• In question of construction, drawings shall be neat and exactly as per the given measurements.
• Use of calculators is not permitted. However, you may ask for mathematical tables.

SECTION A

Question numbers 1 to 6 carry 1 mark each.

Question 1.
Which term of the sequence 114, 109, 104,…… is the first negative term?

Question 2.
Find the value of k for which the following are the consecutive terms of an AP: k, 2k – 1, 2k + 1.

Question 3.
Find the distance between the points [\(\frac { -8 }{ 5 }\),2] and [\(\frac { 2 }{ 5 }\),2]

Question 4.
Which measure of central tendency is given by the x-coordinate of the point the of intersection of “more than ogive” and “less than ogive”?

Question 5.
What is the distance between two parallel tangents of a circle of radius 4 cm?

Question 6.
In the adjoining figure, O is the centre of a circle. The area of sector OAPB is \(\frac { 5 }{ 18 }\) of the area of the circle. Find x.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
Using Euclid’s algorithm, find the HCF of 1656 and 4025.

Question 8.
Find the two numbers whose sum is 75 and difference is 15.

Question 9.
If m and n are the zeroes of the polynomial 3x² + 11x – 4, find the value of \(\frac { m }{ n } +\frac { n }{ m } \) .

Question 10.
If the distances of P(x, y) from the points A(3, 6) and B(-3, 4) are equal, prove that 3x + y = 5.

Question 11.

Question 12.
Three metallic solid cubes whose edges are 3 cm, 4 cm and 5 cm are melted and formed into a jingle cube. Find the edge of the cube so formed

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
In the adjoining figure, \(\frac { XP }{ PY } =\frac { QX }{ QZ } \) = 3. If the area of ∆XYZ is 32 cm², then find the area of the quadrilateral PYZQ.

Question 14.
In an equilateral triangle ABC, a point D is taken on base BC such that BD : DC = 2:1. Prove that 9 AD² = 7AB².

Question 15.
Find the values of a and b so that 8x⁴ + 14x³ – 2x² + ax + b is exactly divisible by 4x² + 3x – 2.
OR
If one zero of the polynomial 3x² – 8x – (2k + 1) is seven times the other, find both zeroes of the polynomial and the value of k.

Question 16.
Solve for x and y:
(a – b)x + (a + b)y = a² – 2ab – b²
(a + b) (x + y) = a² + b².

Question 17.
Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are A(2, 1), B (4, 3) and C(2, 5).
OR
Show that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a square.

Question 18.
There are 100 cards in a box on which numbers from 1 to 100 are written. A card is taken out from the box at random. Find the probability that the number on the selected card is
(i) divisible by 9 and is a perfect square
(ii) a prime number greater than 80.

Question 19.
In a single throw of a pair of different dice, what is the probability of getting
(i) a prime number of each dice?
(ii) a total of 9 or 11?

Question 20.
A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket.
OR
A solid right-circular cone of height 60 cm and radius 30 cm is dropped in a right-circular cylinder full of water of height 180 cm and radius 60 cm. Find the volume of water left in the cylinder in cubic metre.

Question 21.
Prove the following: sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ = 1.
OR
Prove that

Question 22.
If cos θ + sin θ = √2 cos θ, show that cos θ – sin θ = √2 .

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
If the roots of the quadratic equation x² + 2px + mn = 0 are real and equal, show that the roots of the quadratic equation x² – 2(m + n) x + (m² + n² + 2p²) = 0 are also real and equal.

Question 24.
The sum of three numbers in AP is 12 and sum of their cubes is 288. Find the numbers.
OR
The sums of first n terms of three arithmetic progressions are S₁, S₂ and S₃ respectively. The first term of each AP is 1 and their common differences are 1, 2 and 3 respectively. Prove that S₁ + S₃ = 2S₂.

Question 25.
Dudhnath has two vessels containing 720 mL and 405 mL of milk. Milk from these containers is poured into glasses of equal capacity to their brim. Find the minimum number of glasses that can be filled.
OR
Show that the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.

Question 26.
The following table gives production yield per hectare of wheat of 100 farms of a village.

Change the distribution to more than type, and draw its ogive. Using the ogive, find the median of the given data.
What is the value of proper knowledge of farming?

Question 27.
In the adjoining figure, AB is a chord of length 16 cm of a circle with centre O and of radius 10 cm. The tangents at A and B intersect at the point P. Find the length of PA.

OR
Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Question 28.
The angle of elevation of a cloud from a point h metres above the surface of a lake is θ and the angle of depression of its reflection in the lake is φ. Prove that the height of the cloud above the

Question 29.
Draw a circle of radius 3.5 cm. From a point P Outside the circle at a distance of 6 cm from the centre of the circle, draw two tangents to the circle. Also measure their lengths.

Question 30.
In the adjoining figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If the arcs of equal radii 7 cm with centres A, B, C and D have drawn, then find the area of the shaded region.

Answers

Answer 1.
The given sequence is an AP with first term a = 114 and common difference d = 5.
∴ a_n = a + (n – 1)d < 0
⇒114 + (n – 1) (-5) < 0
⇒ -5n + 119 < 0 ⇒ 5n > 119

∴ n = 24
Hence, 24th term is 1st negative term.

Answer 2.
Given k, 2k – 1, 2k + 1 are in AP
⇒ 2(2k – 1) = k + 2k + 1
⇒ 4k – 2 = 3k + 1
⇒ 4k – 3k = 2 + 1
⇒ k = 3.

Answer 3.

Answer 4.
Median

Answer 5.
Distance between parallel tangents = (4 + 4) cm = 8 cm.

Answer 6.
Area of sector OAPB = \(\frac { x }{ 360 } \times \pi { r }^{ 2 }\), where r is the radius of the circle.

Hence x = 100

Answer 7.
By Euclid’s division algorithm, we have
4025 = 1656 × 2 + 713; 1656 = 713 × 2 + 230
713 = 230 × 3 + 23; 230 = 23 × 10 + 0
Thus, last non-zero remainder = 23
Hence, HCF of 1656 and 4025 is 23.

Answer 8.
Let the two numbers be x and y (x > y), then according to given
x + y = 75 …(i)
and x – y = 15 …(ii)
Adding (i) and (ii), we get
2x = 90 => x = 45
Putting x = 45 in (i), we get
45 + y = 75 => y = 30
Hence, two numbers are 45 and 30.

Answer 9.
Given, m and n are zero’s of the polynomial 3x² + 11x – 4.

Answer 10.
Given PA = PB => (PA)² = (PB)²
=> (x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²
=> x² – 6x + 9 + y² – 12y + 36 = x² + 6x + 9 + y² – 8y + 16
=> 6x + 6x – 8y + 12y = 9 + 36 – 9 – 16
=> 12x + 4y = 20
=> 3x + y = 5

Answer 11.

Answer 12.
Let the edge of the cube formed by x cm, then
volume of the cube formed = sum of volumes of three cubes of edges 3 cm, 4 cm and 5 cm
=> x³ = 3³ + 4³ + 5³
=> x³ = 27 + 64 + 125 = 216 = 6³
=> x = 6
Hence, the edge of the cube so formed is 6 cm.

Answer 13.
In ∆XYZ, \(\frac { XP }{ PY } =\frac { QX }{ QZ } \) (given)
∴PQ || YZ (converse of Thales theorem)
⇒ ∠XPQ = ∠XYZ and ∠XQP = ∠XZY (corres. ∠s)
∆XPQ ~ ∆XYZ.

Answer 14.

As ∆ABC is equilateral, BC = AB.
Given BD : DC = 2 : 1 => BD = \(\frac { 2 }{ 3 }\)BC = \(\frac { 2 }{ 3 }\)AB.
Draw AE ⊥ BC, then E is mid-point of BC,
so BE = \(\frac { 1 }{ 2 }\) BC = \(\frac { 1 }{ 2 }\) AB.
From fig., ED = BD – BE = \(\frac { 2 }{ 3 }\)AB – \(\frac { 1 }{ 2 }\)AB = \(\frac { 1 }{ 6 }\)AB.
In ∆ABE, ∠AEB = 90°
∴ AB² = AE² + BE² …(i)
In ∆AED, ∠AED = 90°
∴ AD² = AE² + ED² …(ii)
Subtracting (ii) from (i), we get

⇒ 9 AD² = 7AB².

Answer 15.
Dividing 8x⁴ + 14x³ – 2x² + ax + b by 4x² + 3x – 2, we get

Answer 16.
The given equations are:
(a – b)x + (a + b)y = a² – 2ab – b²
(a + b) (x + y) = a² + b²
Equation (ii) can be written as
(a + b)x + (a + b)y = a² + b²
Subtracting equation (i) from equation (iii), we get
2bx = 2ab + 2b² => x = a + b.
Substituting x = a + b in equation (iii), we get
(a + b) (a + b) + (a + b)y = a² + b²
=> a² + b² + 2ab + (a + b)y = a² + b²

Answer 17.
Let D, E and F be the mid-points of the sides BC,
CA and AB respectively

Answer 18.
Total number of cards in the bag = 100
By saying that a card is selected at random means all cards are equally likely to be selected.
So, the sample space of the experiment has 100 equally likely outcomes.
(i) Numbers divisible by 3 and perfect square are 9, 36, 81.
∴ The number of outcomes favourable to the event ‘divisible by 3 and is a perfect square’ is 3.
Required probability = \(\frac { 3 }{ 100 }\).
(ii) Prime numbers greater than 80 but less than or equal to 100 are 83, 89, 97.
∴ The number of outcomes favourable to the event ‘prime number less than 80’ is 3.
∴Required probability = \(\frac { 3 }{ 100 }\)

Answer 19.
When two different dice are thrown, the total number of outcomes is 36 and all the outcomes are equally likely.
(i) The outcomes favourable to the event ‘a prime number on each dice’ are (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5).
These are 9 in number.
∴ P (a prime number on each die) = \(\frac { 9 }{ 36 }\) = \(\frac { 1 }{ 4 }\).
(ii) The outcomes favourable to the event ‘a total of 9 or 11’ are (3, 6), (6, 3), (4, 5), (5, 4), (5, 6), (6, 5). These are 6 in number.
∴ P (a total of 9 or 11) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\).

Answer 20.
Let h cm be height of the bucket.
Here, radius of top = R = 28 cm,
radius of bottom = r = 21 cm.
Volume of bucket = 28.490 litres = (28.490 x 1000) cm³
= 28490 cm³

Answer 21.
LHS = sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ
= (sin² θ)³ + (cos² θ)³ + 3 sin² θ cos² θ
= (sin² θ + cos² θ)³ – 3 sin² θ cos² θ (sin² θ + cos² θ) + 3 sin² θ cos² θ
(∵ a³ + b³ = (a + b)³ – 3ab (a + b))
= (1)³ – 3 sin² θ cos² θ (1) + 3 sin² θ cos² θ
= 1 – 3 sin² θ cos² θ + 3 sin² θ cos² θ = 1 = RHS
OR

Answer 22.
Given cos θ + sin θ = √2cos θ ⇒sin θ = (√2 – 1) cos θ

Answer 23.
Given that the roots of the quadratic equation x² + 2px + mn = 0 are real and equal
=> its discriminant = 0
=> (2p)² – 4 x 1 x mn = 0 => p² = mn …(i)
Now, the roots of the quadratic equation x² – 2(m + n) x + (m² + n² + 2p²) = 0 are real and equal if its discriminant = 0
i.e. if (-2 (m + n))² – 4 x 1 x (m² + n² + 2p²) = 0
i.e. if 4 (m² + n² + 2mn) – 4 (m² + n² + 2p²) = 0
i.e. if 8mn – 8p² = 0 i.e. if p² = mn, which is true from (i).

Answer 24.
Let the three numbers are in AP be a – d, a, a + d, then according to given,
a – d + a + a + d = 12
3a = 12 => a = 4 …(i)
and (a – d)³ + a³ + (a + d)³ – 288
=> a³ – 3a²d + 3ad² – d³ + a³ + a³ + 3a²d + 3ad² + d³ = 288
=> 3a³ + 6ad² = 288
=> 3(4)³ + 6 x 4 x d² = 288 [using (i)]
=> 192 + 24d² = 288
=> 24d² = 96 => d² = 4
d = ±2
When a = 4, d = 2, the numbers are 4 – 2, 4, 4 + 2 i.e. 2, 4, 6
and when a = 4, d = -2, the numbers are 4 – (-2), 4, 4 + (-2) i.e. 6, 4, 2
Hence, the three numbers are 2, 4, 6.
OR

Answer 25.
Two vessels contain 720 mL and 405 mL of milk respectively. Since we need the minimum number of glasses of equal capacity, so the capacity of each glass should be maximum. Therefore, we have to find the HCF of 720 mL and 405 mL.
720 = 2 x 2 x 2 x 2 x 3 x 3 x 5 = 2⁴ x 3² x 5¹ and
405 = 3 x 3 x 3 x 3 x 5 = 3⁴ x 5¹
∴ HCF (720, 405) = 3² x 5¹ = 9 x 5 = 45.
∴Maximum capacity of each glass = 45 mL.

Answer 26.
Construct the table for cumulative frequency distribution of more than type as under:

Take 1 cm along x-axis 5 kg of production/ha and 1 cm along y-axis = 10 farms.
Plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16) and (80, 0). Join these points by a free hand drawing. The required ogive (more than type) is drawn on the graph sheet given below. Since the scale on x-axis starts at 50, a kink (break) is shown on the x-axis near the origin.

To find median:
Let A be the point on y-axis representing frequency = \(\frac { n }{ 2 }\) = \(\frac { 100 }{ 2 }\) = 50.
Through A, draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet
the x-axis at M. The abscissa of the point M represents 70.5 kg.
Hence, the median = 70.5 kg of wheat/ha.
The values of proper knowledge of farming are:
Arrangement of irrigation, testing of soil, rotation of crops, quality and quantity of manure and
pesticides will help in increasing the yield per hectare.

Answer 27.

Given, AB = 16 cm, so AL = BL = 8 cm
(∵ OP is the perpendicular bisector of AB)
In ∆OLB, ∠OLB = 90° (∵ OP ⊥ AB)
By Pythagoras theorem,
OL² = OB² – BL² = 10² – 8² = 100 – 64 = 36
=> OL = 6 cm.
Let LP = x cm and BP = y cm, then OP = OL + LP = (6 + x) cm.
Since tangent is perpendicular to radius, OB ⊥ PB.
In ∆OPB, ∠OBP = 90°. By Pythagoras theorem,
OP² = BP² + OB²
=> (x + 6)² = y² + 10²
=> x² + 12x + 36 = y² + 100

Answer 28.
Let O be the point of observation h metres above the lake and let the cloud be at the point P. If Q is the reflection of the cloud in the lake then BQ = BP.
Let the height of the cloud above the lake be x metres. From O, draw OA ⊥ PQ. Then, AB = OM = h metres.
AP = BP – BA = (x – h) metres and
AQ = AB + BQ = (x + h) metres.
Let OA = d metres.
From right angled ∆OAP, we get

Answer 29.
Steps of construction:
1. Mark a point O. With O as centre and radius 3.5 cm, draw a circle.
2. Take a point P at a distance of 6 cm from O. P lies outside the circle.
3. Join OP and draw its perpendicular bisector to meet OP at M.
4. Taking M as centre and OM (or MP) as radius, draw a circle. Let this circle intersect the previous circle at points A and B.
5. Join PA and PB. Then PA and PB are the required tangents. On measuring, we find that AP = BP = 4.9 cm (approximately).

Justification:
Join OA, then ∠OAP = 90° (angle in a semicircle = 90°)
As OA is radius and ∠OAP = 90°, so PA has to be tangent to the circle.

Answer 30.
Area of trapezium ABCD

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CBSE Sample Papers for Class 10 Maths Paper 4 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 4. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Sample Papers for Class 10 Maths Paper 4

Board	CBSE
Class	X
Subject	Maths
Sample Paper Set	Paper 4
Category	CBSE Sample Papers

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 4 of Solved CBSE Sample Papers for Class 10 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

GENERAL INSTRUCTIONS:

• All questions are compulsory.
• This question paper consists of 30 questions divided into four sections A, B, C and D.
• Section A comprises of 6 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 8 questions 1 of 4 marks each.
• There is no overall choice. However, internal choice has been provided in one question of 2 marks, 1 three questions of 3 marks each and two questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• In question of construction, drawings shall be neat and exactly as per the given measurements.
• Use of calculators is not permitted. However, you may ask for mathematical tables.

SECTION A

Question numbers 1 to 6 carry 1 mark each.

Question 1.
The decimal expansion of the rational number \(\frac { 11 }{ { 2 }^{ 3 }\times { 5 }^{ 2 } } \) will terminate after how many places of decimal?

Question 2.
Is x = -2 a solution of the equation x² – 2x + 8 = 0?

Question 3.
In the adjoining figure, P and Q are points on the sides of AB and AC respectively of ∆ABC such that AP = 3.5 cm, PB = 7 cm, QC = 6 cm. Prove that PQ || BC.a

Question 4.
If tan θ = cot (30° + θ), find the value of θ.

Question 5.
A cylinder and a cone are of same base radius and c the ratio of volume of cylinder to that of the cone.

Question 6.
Two coins are tossed simultaneously. Find the probability of getting exactly one head.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
Complete the following factor tree and find the numbers x, y and z.

Question 8.
If α and β are zeroes of the polynomial f(x) = x² – x – k such that α – β = 9, find k.

Question 9.
The coordinates of the vertices of ∆ABC are A(4, 1), B(-3, 2) and C(0, k). Given that the area of ∆ABC is 12 unit², find the value of k. .

Question 10.
Without using trigonometric tables, evaluate the following:

Question 11.
Prove that

Question 12.
Area of a sector of a circle of radius 36 cm is 54π cm². Find the length of the corresponding arc of the sector.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
The HCF of 65 and 117 is expressible in the form 65m – 117. Find the value of m. Also find the LCM of 65 and 117 using prime factorization

Question 14.
Solve the following pair of linear equations: px + qy = p – q;qx – py = p + q.
OR
In the given figure, ABCDE is a pentagon with BE || CD and BC || ED. BC is perpendicular to CD. If the perimeter of pentagon ABCDE is 21 cm, find the values of x and y.

Question 15.
Find the roots of the following equation

Question 16.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work for 30 days?
OR
Jaspal Singh repays his total loan of Rs 118000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, what amount will be paid by him in the 30th instalment? What amount of loan does he still to pay after the 30th instalment?

Question 17.
Prove that : sin θ(1 + tan θ) + cos θ(1 + cot θ) = sec θ + cosec θ.

Question 18.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².
OR
In the adjoining figure, DB ⊥ BC, DE ⊥ AB
and AC ⊥ BC. Prove that \(\frac { BE }{ DE } =\frac { AC }{ BC } \)

Question 19.
In the adjoining figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 4 cm and 3 cm respectively. If area of ∆ABC = 21 cm², then find the lengths of sides AB and AC.

Question 20.
In the adjoining figure, APB and CQD are semicircles of diameter 7 cm each, while ARC and BSD are semicircles of diameter 14 cm each. Find the perimeter of the shaded region.

Question 21.
Two different dice are rolled together. Find the probability of getting:
(i) the sum of numbers on two dice as 5.
(ii) even numbers on both dice.
OR
The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is \(\frac { 1 }{ 4 }\). The probability of selecting a blue ball at random from the same jar is \(\frac { 1 }{ 3 }\). If the jar contains 10 orange balls, find the total number of balls in the jar.

Question 22.
Find the value of p for the following distribution whose mean is 10:

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
If pth, qth and rth terms of an AP are a, b and c respectively, then show that (a – b) r + (b – c) p + (c – a) q = 0.

Question 24.
An icecream seller sells his icecreams in two ways:
(A) In a cone of r = 5 cm, h – 8 cm with hemispherical top.
(B) In a cup in shape of cylinder with r – 5 cm, h – 8 cm.
He charges the same price both but prefers to sell his icecream in a cone.
(a) Find the volume of the cone and the cup.
(b) Which out of the two has more capacity?
(c) By choosing a cone, which value is not followed by the icecream seller?

Question 25.
Solve the following pair of linear equations graphically:
x + 3y = 6; 2x – 3y = 12
Also find the area of the triangle formed by the lines representing the given equations with y-axis.

Question 26.
If the centroid of ∆ABC, in which A(a, b), B(b, c), C(c, a) is at the origin, then calculate the value of (a³ + b³ + c³).
OR
The three vertices of a parallelogram ABCD are A(3, -4), B(-1, -3) and C(-6, 2). Find the coordinates of vertex D and find the area of parallelogram ABCD.

Question 27.
From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars. (Use √3 = 1.73)
OR
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Question 28.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
OR
Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O’P are tangents to the two circles. Find the length of the common chord PQ.

Question 29.
Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ∆ABC in which PQ = 8 cm. Also justify the construction.

Question 30.
Size of agriculture holding in a survey of 200 families is given in the following table:

Compute the median and mode size of holding.

Answers

Answer 1.

Hence, the decimal expansion of given rational number will terminate after 3 places of decimal.

Answer 2.
(-2)² – 2 x (-2) + 8 = 4 + 4 + 8 = 16≠0
∴ x = -2 is not a solution of the equation x² – 2x + 8 = 0.

Answer 3.

∴ By the converse of B.P.T., PQ || BC.

Answer 4.
tan θ = cot (30° + θ)
= tan [90° – (30° + θ)]
= tan (60° – θ)
θ = 60° – θ => 2θ = 60° => θ = 30°.

Answer 5.
Let r be the base radius of cylinder and cone and h be their heights.

Hence, the ratio of volume of cylinder to that of cone = 3:1.

Answer 6.
When two coins are tossed simultaneously, then the outcomes are HH, HT, TH, TT.
So total number of outcomes = 4.
The outcomes favourable to the event ‘getting exactly one head’ are HT and TH.
∴The number of outcomes favourable to the event = 2.
∴ P(exactly one head) = \(\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)

Answer 7.
x = 3381 x 2 = 6762
y = 161 x 7 = 1127
and z = \(\frac { 161 }{ 7 }\) = 23.

Answer 8.
Given f(x) = x² – x – k

Answer 9.
Area of ∆ABC = \(\frac { 1 }{ 2 }\)| 14(2 – k) – 3(k – 1) + 0(1 – 2) |
=> 12 = \(\frac { 1 }{ 2 }\)|8 – 4k – 3k + 3|
=> 24 = |11 – 7k|
=> 11 – 7k = ±24
=> 11 – 7k = 24 or 11 – 7k = -24
=> 7k = -13 or 7k = 35
=> k = \(-\frac { 13 }{ 7 }\) or k = 5
Hence, k = \(-\frac { 13 }{ 7 }\) or k = 5

Answer 10.

Answer 11.

Answer 12.
Let the central angle (in degrees) be θ, then

Hence, the length of the corresponding arc of the sector = 3π cm.

Answer 13.
By Euclid’s division algorithm, we have
117 = 65 x 1 + 52; 65 = 52 x 1 + 13; 52 = 13 x 4 + 0
∴ HCF of 65 and 117 = 13
∴ 65m – 117 = 13 => 65m = 130 => m = 2.
Prime factorisation of 65 and 117 are as follows:
65 = 5 x 13 and 117 = 3 x 3 x 13
∴ LCM of 65 and 117 = 3 x 3 x 5 x 13 = 585.

Answer 14.
Given equations are
px + qy = p – q …(i)
qx – py = p + q …(ii)
Multiplying (i) by p and (ii) by q, we get
p²x + pqy = p² – pq …(iii)
and q²x – pqy – pq + q² …(iv)
On adding (iii) and (iv), we have

Putting x = 1 in equation (i), we have
p x 1 + qy = p – q => qy = -q => y = -1
∴ x = 1 ,y = -1.
OR
Given BE || CD and BC || ED, also BC ⊥ CD
=> BCDE is a rectangle
CD = BE
=> x + y = 5
Again, given perimeter of pentagon ABCDE = 21 cm
=> AB + BC + CD + DE + EA = 21 cm
=> 3 + x – y + x + y + BC + 3 = 21cm
=> 2x + x – y = 21 – 6
=> 3x – y = 15
Adding (i) and (ii), we get
4x = 20 => x = \(\frac { 20 }{ 4 }\) => x = 5
Putting x = 5 in (i), we get
5 + y = 5 =>y = 0
Hence, x = 5, y = 0.

Answer 15.
Given

(x + 4) (x – 7) = -30 => x² – 7x + 4x – 28 = -30
=> x² – 3x + 2 = 0 => x² – 2x – x + 2 = 0
=> x(x – 2) – 1(x – 2) = 0 => (x – 2) (x – 1) = 0
=> x – 2 = 0 or x – 1 = 0 => x = 2 or x = 1.
Hence, the roots of the given equation are 2 and 1.

Answer 16.
200, 250, 300, … is in AP.
whose first term = 200 and common difference = 50.
Let the number of days for which work is delayed be n, then

=> n[200 + 25 (n- 1)] = 27750
=> 25n² + 175n = 27750
=> n² + 7n – 1110 = 0
=> n² + 37n – 30n – 1110 = 0
=> n(n + 37) – 30(n + 37) = 0
=> (n + 37) (n – 30) = 0
=> n = -37 or n = 30
But n cannot be negative
∴ n = 30
Hence, the work was delayed by 30 days.
OR
Jaspal Singh starts repaying loan with first instalment of Rs 1000 and increases the instalment every month by Rs 100, so the numbers involved in instalments are 1000, 1100, 1200, …
These numbers form an AP with a = 1000 and d = 100.
30th term = a + 29d = 1000 + 29 x 100 = 3900.
Hence, his 30th instalment = Rs 3900
Sum of 30 terms = \(\frac { 30 }{ 2 }\) (100 + 3900)
= 15 x 4900 = 73500.
∴ The total amount of loan repaid = Rs 73500.
∴ Amount of loan still to be paid = Amount of total loan – amount of loan repaid
= Rs 118000 – Rs 73500 = Rs 44500.

Answer 17.

Answer 18.

In ∆ABC, ∠C = 90°
∴ AB² = AC² + BC²
In AECD, ∠ECD = 90°
∴ DE² = CD² + EC²
In ∆AEC, ∠ACE = 90°
∴ AE² = AC² + EC²
In ∆BCD, ∠BCD = 90°
∴ BD² = BC² + CD²
Adding (iii) and (iv), we get
AE² + BD² = (AC² + BC²) + (CD² + EC²)
= AB² + DE²
OR

Given, DB ⊥ BC, AC ⊥ BC
=> ∠DBC = ∠ACB = 90°
∴ AC || DB(sum of interior ∠s on the same side of BC is 180°)
∠ABD = ∠BAC (alt ∠s)
Now, in ABED and AACB,
∠EBD = ∠ABD = ∠BAC
∠C = ∠E (each = 90°)
∴ ∆BED ~ ∆ACB (AA similarity)

Answer 19.
Since tangents drawn from an external point to a circle are equal.
∴BF = BD = 4 cm (given)
and CE = CD = 3 cm (given)
Let AF = AE = x cm

Now, area (∆ABC) = area (∆AOB) + area (∆BOC) + area (∆AOC)
21 = \(\frac { 1 }{ 2 }\)AB x OF + \(\frac { 1 }{ 2 }\)BC x OD + \(\frac { 1 }{ 2 }\)AC x OE
=> 21 = \(\frac { 1 }{ 2 }\)(AF + BF) x 2 + \(\frac { 1 }{ 2 }\)(BD + CD) x 2 + \(\frac { 1 }{ 2 }\)(AE + CE) x 2
=> 21 = (x + 4) + (4 + 3) + (x + 3)
=> 21 = 2x + 14 => 2x = 7 => x = \(\frac { 7 }{ 2 }\) =>x = 3.5
AB = x + 4 = (3.5 + 4) cm = 7.5 cm
and AC = x + 3 = (3.5 + 3) cm = 6.5 cm.

Answer 20.
Given BC = R = 7 cm
and AE = CF = \(\frac { 7 }{ 2 }\) cm = r (say)
The perimeter of the shaded region
= perimeter of semicircle APB + perimeter of semicircle BSD + perimeter of semicircle DQC + perimeter of semicircle CRA

Answer 21.
Total possible outcomes = 6 x 6 = 36.
(i) The possible outcomes favourable to the event ‘sum of numbers on two dice is 5’ are (2, 3), (3, 2), (1, 4), (4, 1) i.e. 4 in number.
∴ Required probability = \(\frac { 4 }{ 36 } =\frac { 1 }{ 9 } \)
(ii) The possible outcomes favourable to the event ‘even numbers on both dice’ are (2, 2), (2, 4), – (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6) i.e. 9 in number.
∴ Required probability = \(\frac { 9 }{ 36 } =\frac { 1 }{ 4 } \)
We know that the sum of probabilities of all elementary events = 1
=> P(red ball) + P(blue ball) + P(orange ball) = 1

=> x = 24.
Hence, the total number of balls in the jar = 24.

Answer 22.

Answer 23.
Let A be the first term and D be the common difference of the AP.
Given,
a = pth term => a = A + (p – 1)D ….(i)
b = qth term => b = A +(q – 1)D ….(ii)
c = rth term => c = A + (r – 1)D ….(iii)
Subtracting (ii) from (i), we get
a – b = (p – q) D ….(iv)
Similarly, b – c = (q – r) D ….(v)
and c – a = (r – p) D ….(vi)
∴ (a – b) r + (b – c) p + (c – a) q = [(p – q) r + (q – r) p + (r – p)q] D
= (pr – qr + pq + qr – pq) D
= 0 x D = 0.

Answer 24.
Volume of type A = Volume of cone + Volume of hemisphere

Answer 25.
The given equations can be written as

Select the coordinate axes. Take 1 cm = 1 unit on both axes.
Plot the points (0, 2) and (3, 1) on the graph paper and join these points by a straight line. Plot the points (3, -2) and (6, 0) on the same graph paper and join these points by a straight line.
The given lines intersect at the point P(6, 0). Therefore, the solution of the given pair of linear equations is x = 6, y = 0.
The area bounded by the given lines and y-axis has been shaded.

From the figure AB = 6 units and OP = 6 units
The area of shaded region = area of ∆ABP
= \(\frac { 1 }{ 2 }\) x AB x OP = \(\frac { 1 }{ 2 }\) x 6 x 6
= 18 sq. units

Answer 26.
Given vertices of ∆ABC as A (a, b), B (b, c), C (c, a) and the centroid of ∆ABC is G(0, 0).
As we know that centroid of A whose vertices are (x1 y1), (x2, y2) and (x3, y3)

Answer 27.
Let AB be a tower 100 m high. P and Q be the position of two cars.
Angles of depressions are shown in adjoining figure.
∴∠APB = 90° and ∠AQB = 45°.
In ∆APB, ∠ABP = 90°
30° = \(\frac { AB }{ PB }\)

Answer 28.
Given: P is an external point to a circle with centre O. PA and PB are two tangents drawn from P to the circle, A and B being points of contact.
To prove: PA = PB
Construction: Join OA, OB and OP.
Proof: Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.
∠OAP = 90° and ∠OBP = 90°
Now, in ∆OAP and ∆OBP,
OA = OB (radii of same circle)
OP = OP (common)
∠OAP = ∠OBP (each 90°)

Answer 29.
As ∆PQR ~ ∆ABC, so the side PQ of ∆PQR is the corresponding side AB of ∆ABC.
Given AB = 6 cm and we need PQ to be 8 cm, so \(\frac { PQ }{ AB } =\frac { 8 }{ 6 } =\frac { 4 }{ 3 } \)
Thus, we are required to construct ∆PQR similar to ∆ABC such that sides of ∆PQR are \(\frac { 4 }{ 3 }\) times of the corresponding sides of ∆ABC.
Steps of construction:
1. Draw AB = 6 cm.
2. With A as centre and radius 6 cm, draw an arc.
3. With B as centre and radius 5 cm, draw an arc to meet the previous arc at C.
4. Join AC and BC, then ABC is an isosceles triangle with AB = AC = 6 cm and BC = 5 cm.
5. Take points A and P same.
6. Draw any ray AX making an acute angle with AB on the side opposite to the vertex C.

7. Locate 4 (the greater of 4 and 3) points A1, A2, A3 and A4 on AX such that AA1 = A1A2 = A2A3 = A3A4.
8. Join A3B.
9. Through A4, draw a line parallel to A3B (making an angle equal to ∠AA3B) to intersect the extended line segment AB at Q.
10. Through Q, draw a line parallel to BC (making an angle equal to ∠ABC) to intersect the extended line segment AC at R.
Then AQR i.e. PQR is the required triangle.
Justification:

Answer 30.
Construct the cumulative frequency distribution table as under:

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NCERT Exemplar Solutions for Class 10 Science Chapter 9 Heredity and Evolution

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 9 Heredity and Evolution

NCERT Exemplar Solutions for Class 10 Science Chapter 9 Short Answer Questions

Question 1.
How is a sex of new born determined in humans ?
Answer:
Natural selection is preferential survival and differential reproduction of individuals of a population which possess variations that provide better adaptability to the existing environment. Peppered Moth of England has two forms, light grey and dark grey. Prior to industrial revolution, tree trunks in the forests around.
Manchester were light greyish green due to presence of lichens. Most of the Peppered Moth found in the area were light coloured with dark spots which could not be spotted easily by predator birds. During 1845 to 1890, air pollution killed the lichen flora. The deposition of soot changed the colour of the tree trunks to blackish one. Peppered Moth of the area also exhibited switch over to melanic or blackish form.
It provided better survival value against dark background. The light grey form largely disappeared as it could be easily spotted by predator birds and devoured.

More Resources

• NCERT Exemplar Solutions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Does genetic combination of mothers play a significant role in determining the sex of a new born ?
Answer:
No. Mothers have no role in determining the sex of the new born. Mothers are homogametic, that is, they produce only one type of ova (22 + X). Fathers are heterogametic, that is, they produce two types of sperms, gynosperms (22 + X ) and androsperms (22 + Y). If gynosperm (22 + X) fertilises the ovum (22 + X), the sex of new born will be female (44 + 2X). If androsperm (22 + Y) fuses with ovum (22 + X), the new born will be boy (44 + XY).

Question 3.
Mention three important features of fossils which help in the study of evolution.
Answer:
Fossils are remains or impressions of past organisms that are found in the rocks. Fossils of lower strata belong to early periods while those of upper strata are of later periods. Arranging the fossils stratumwise will indicate the occurrence of different forms of life at different times. It is found that the early fossils generally belong to simple organisms. Complexity and elaboration increased gradually with evolution. Evolution has never been linear or straight. A number of variants or branches appeared, some of which were more complex while others were less complex.

1. Fossils indicate the path of evolution of different groups.
2. They can indicate the phylogeny of some organisms, e.g, Horse, Elephant.
3. Some fossils have characteristics intermediate between two groups,
   e.g., toothed bird Archaeopteryx. They indicate how one group has evolved from another.

Question 4.
Why do all the gametes formed in human females have an X-chromosome ?
Answer:
Human females are homogametic. Their generic constitution is 44 + XX The two sex chromosomes are similar. Their ova which are produced after meiosis carry a gametic constitution of 22 + X No other combination is possible.

Question 5.
In human beings the satistical possibility of getting either a male or female child is 50 : 50. Give suitable explanation.
Answer:
Human females (44 + XX) are homogametic, that is, they produce only one type of ova (22 + X). Human males are heterogametic. They produce two types of sperms (22 + X and 22 + Y) in equal proportion, that is, 50 : 50 ratio. The chance of male or female child is also 50 : 50, as there is equal chance of androsperm (22 + Y) or gynosperm (22 + X) fertilizing an ovum.

Question 6.
A very small population of a species faces a greater threat of extinction than a larger population. Provide a suitable genetic explanation.
Answer:
A small population is always at a risk of degeneration and extinction due to

1. Excessive inbreeding that brings about inbreeding depression or degeneration,
2. Fewer recombinations and variations which are otherwise essential for maintaining vitality and vigour of the species.
3. Lesser adaptability to changes in the environment,
4. Increased threat to survival due to poaching, habitat destruction and environmental change.

Question 7.
What are homologous structures ? Give an example. Is it necessary that homologous structures always have a common ancestor ?
Answer:
Homologous structures or organs are those structures which have similar origin, similar development, similar internal structure and similar basic plan but show different external form and function, e.g., forelimbs of amphibians, reptiles, mammals and birds. Homologous structures are always have a common ancestry because otherwise there cannot be any similarity in basic plan, internal structure, development or origin. Modifications have occurred in them due to varied adaptations.

Question 8.
Does the occurrence of diversity of animals on earth suggest their diverse ancestry also ? Discuss this point in the light of evolution.
Answer:
Diversity of animals does not mean that they have diverse ancestry. Animals can be grouped into distinct lineages (e.g., mammals, birds, reptiles, annelids). Many of the lineages further show some similarities in basic traits indicating a common ancestry, e.g.,
vertebrates. Therefore, animals having a common ancestor in the remote past have successively developed newer and newer traits forming various groups of animals.

Question 9.
Give the pair of contrasting traits of the following characters in Pea plant and mention which is dominant and recessive

1. Yellow seed
2. Round seed.

Answer:

1. Seed Colour: Yellow (dominant), green (recessive).
2. Seed Shape: Round (dominant), wrinkled (recessive).

Question 10.
Why did Mendel choose Pea plant for his experiments ?
Answer:
Mendel chose Garden Pea or Edible Pea (Pisum sativum) as his experimental material because of

1. Easy availability of detectable contrasting traits of several characters,
2. Flower structure normally allows self fertilization but cross fertilization can also be carried out easily.
3. Annual nature of the plant,
4. Formation of a large number of seeds by each plant,
5. Requirement of little care except at the time of cross breeding.

Question 11.
A woman has only daughters. Analyse the situation genetically and provide a suitable, explanation.
Answer:
A woman produces only one type of ova (22 + X) while her husband produces two types of sperms, gynosperms (22 + X) and androsperms (22 + Y) in equal proportion. It is a chance that each time the woman conceived, only the gynosperm fertilised the egg so that only daughters were born.

NCERT Exemplar Solutions for Class 10 Science Chapter 9 Long Answer Questions

Question 12.
Does geographical isolation of individuals of a species lead to formation of a new species ? Provide a suitable explanation.
Answer:
Yes. Geographical isolation of a population will lead to genetic drift as there will be no gene flow between it and the parent species. Inbreeding in small population will result in fixation of certain alleles and elimination of others. There will be change in gene frequency. Mutations will produce new alleles and hence a new gene pool. Accumulation of new alleles and hence new variations over several generations will ultimately lead to the formation of new species.

Question 13.
Bacteria have a simpler body plan when compared with human beings. Does it mean that human beings are more evolved than bacteria. Provide a suitable explanation.
Answer:
Both bacteria and human beings perform all the activities of life and live comfortably in their environments. They, therefore, seem
to be equally evolved. However, human beings have a far more complex organisation and differentiation which are absent in bacteria. Since complex organisation and differentiation develop only through evolution, humans are far more evolved than bacteria.

Question 14.
All the human races like Africans, Asians, Europeans, Americans and others might have evolved from a common ancestor. Provide a few evidences in support of this view.
Answer:
All the human races have evolved from a common ancestor because they possess

1. Common body plan
2. Common structure
3. Similar physiology
4. Similar metabolism
5. Similar chromosome number
6. Common genes or genetic blue print
7. Free interbreeding.

Question 15.
Differentiate between inherited and acquired characters. Give one example of each type.
Answer:

Acquired Traits	Inherited Traits
1.     Development. The traits develop during life time of an individual. 2.     Nature. They are somatic variations. 3.     Cause. Acquired traits develop due to direct effect of environment, use and disuse and conscious efforts. 4.     Fate. They die with the death of the individual. Ex. Muscular body of a wrestler.	The traits are obtained from the parents. They are genetic variations. The traits develop due to mutations and reshuffling of genetic material. They are passed on to the next generation. Ex. Fused and free ear lobes.

Question 16.
Give reasons why acquired characters are not inherited.
Answer:
Many of the variations have no immediate benefit to the species. They function as preadaptations which can be beneficial under certain environmental conditions like heat tolerance variation if the temperature of the area rises.

Question 17.
Evolution has exhibited greater stability of molecular structures when compared with morphological structures. Comment on the statement and justify your opinion.
Answer:
Life is a highly organised system of biochemicals and their reactions. It is expressed inside the cells. They are the same right from bacteria to human beings. However, the method of procurement of the raw materials for forming bio-chemicals is not the same in various organisms. It has helped in avoiding cut throat competition amongst the living beings. They developed different morphological features and modes of obtaining nutrients. As a result a huge diversity has appeared in the living world.

Question 18.
In the following crosses, write the characteristics of the progency

Answer:
(a) All Round Yellow
(b) Round Yellow, Round Green, Wrinkled Yellow and Wrinkled Green in the ratio of 9 : 3 : 3 : 1.
(c) All Wrinkled Green (et) All Round Yellow (Rr Yy).

Question 19.
Study the following cross showing self pollination in F₁. Fill in the blank and answer the question that follows :

Answer:
Rr Yy (Round Yellow ).

Question 20.
In the above question, what are combination of characters in F₂ progeny ? What are their ratios ?
Answer:
Round Yellow – 9,
Round Green – 3,
Wrinkled Yellow – 3,
Wrinkled Green -1
i.e., 9 : 3 : 3 : 1.

Question 21.
Give the basic features of the mechanism of inheritance.
Answer:
Basic Features of Inheritance

1. Unit Characters: An organism is made of a large number of characters, each of which behaves as a unit.
2. Genes: Characters are controlled by genes.
3. One Gene-One Character: A single gene generally controls one character.
4. Location of Genes: Genes are located on chromosomes.
5. Alleles: A gene may have two or more forms called alleles. They represent different traits of a character.
6. Paired Alleles: An individual possesses two alleles of every gene. The two may be similar or dissimilar.
7. Dominance: Where there are two different forms or alleles of the gene, generally one expresses its effect. It is called dominant allele. The other which does not express its effect in presence of dominant allele, is called recessive allele.
8. Segregation: The two alleles separate at the time of gamete formation. A gamete has only one allele or form of the gene.
9. Independent Assortment: The alleles of different genes located on separate chromosomes behave independent of one another.
10. Pairing: Fusion of gametes during fertilization, brings together the two forms of a gene in the zygote.

Question 22.
Give reasons for the appearance of new combination of characters in the F₂ progeny.
Answer:
Independent Assortment: The two forms of a gene separate and pair independent of the two forms of other genes during gametogenesis and fertilisation. It causes new combination of characters, e.g.;

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HOTS Questions for Class 10 Science Chapter 9 Heredity and Evolution

These Solutions are part of HOTS Questions for Class 10 Science. Here we have given HOTS Questions for Class 10 Science Chapter 9 Heredity and Evolution

Question 1.
Observe the diagram carefully. What does it depict ? Identify the parts shown by lines.

Answer:
Reconstruction of fossil bird Archaeopteryx.

1. Claw
2. Free fingers
3. Beak
4. Teeth
5. Tail
6. Feathers

More Resources

• HOTS Questions for Class 10 Science
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• Value Based Questions in Science for Class 10
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Name the organism in which feathers appeared for the first time.
Answer:
Members of Dromaesaur family which were small dinosaurs.

Question 3.
Which one is the edible part in Kale, Kohlrabi, Broccoli, Brussel’s Sprout, Cabbage and Cauliflower ?
Answer:
Kale — Leaves,
Kohlrabi — Swollen stem,
Broccoli — Immature green flowers,
Brussel’s Sprout — Axillary buds,
Cabbage — Terminal bud,
Cauliflower — Immature inflorescence of sterile flowers.

Question 4.
Give an example where temperature determines the sex of the new bom.
Answer:

1. Chrysema picta (a turtle). Temperature above 33°C produces females and below 28°C males.
2. Agama agama (a lizard). High temperature produces males.

Question 5.
Name a recessive trait which is quite common in human beings.
Answer:
Blood group O (I°I°).

Question 6.
Why is variation beneficial for the species but not necessarily for the individual ? (CBSE Foreign 2010)
Answer:
Preadaptation is a variation which under normal conditions is of no advantage to the individual bearing it. However, it becomes highly useful in survival under changed environment, e.g, heat wave in temperate environment, insecticide or antibiotic resistance.

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Value Based Questions in Science for Class 10 Chapter 15 Our Environment

These Solutions are part of Value Based Questions in Science for Class 10. Here we have given Value Based Questions in Science for Class 10 Chapter 15 Our Environment

Question 1.
Why have the railway vendors been switching over from re-usable glasses to plastic cups, then clay cups and now paper cups while we stress greater role for renewable articles.
Answer:
There is no denying the fact that reusable articles are ecofriendly. However, they are also liable to get contaminated, especially when railway vendors do not have time and resources to sterilise them. Therefore, railways switched over to disposable articles. Reusable glasses were replaced with disposable plastic cups. However, plastic cups are not biodegradable. Their disposal posed a big problem. They could not be burnt as burning produced toxic gases. The dumping place became sites of breeding flies, mosquitoes and others. The practice was, therefore, discontinued and replaced by clay cups or kulhads. However, making lakhs of kulhads daily removed top fertile soil from several hundred acres of land. The clay cups also required dumping places. Ultimately, disposable paper cups have come to stay with the vendors. The waste cups are recyclable as well as can be burnt without causing problem of dumping.

More Resources

• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Hospital waste is not allowed to be mixed with municipal trash and garbage. Neither it is dumped separately. How is it disposed off and why ?
Answer:
Hospital waste is highly contaminated and can be a source of several diseases wherever it is dumped or disposed off. It also contains infected, cancerous and other waste tissues that can be source of several diseases. Therefore, hospital wastes are collected in separate bags and taken to incinerators for disposal. In incinerators the articles are burnt at 850°C under aerobic conditions. Smoke and grit are precipitated to avoid air pollution. Only ash and unburnt metallic materials are left. They are disposed off in land fills.

Question 3.
Most Australians develop skin problems as they grow old. Why ?
Answer:
Australia lies near Antarctica in the zone where ozone hole appears every year. More high energy ultraviolet rays (UV – B) reach the earth in the area of ozone hole. UV radiations cause skin cancers, mutations and cataracts besides reducing immunity. Therefore, most Australians come to have skin problems as they grow old.

Question 4.
Mahi’s father is a manufacturer who purchases raw materials from only one firm and sells the finished product to another single firm. While studying the difference between food chain and food web Mahi noted that his father is following a faulty practice which must be stopped. He also tried to convince his father about it.
What is the fault in the business of Mahi’s father and how has Mahi got this idea ?
Answer:
In a food chain, members of one trophic level are eaten by members of higher trophic level. A disease or scarcity due to drought that kills members of one trophic level, affects the population of both lower trophic level (very high number, glut) as well as higher trophic level (starvation and death). By anology, Mahi found that any problem in the supplier firm or purchaser firm will ruin his father’s business. Therefore, like a food web, there should be a few alternative sources of raw material supply as well as sale of finished products.
You plan to organise a compaign on “Harmful effects of smoking on human health” in your neighbourhood areas and guide them.

Question 5.
(a) List any three reasons that you will give to convince the people about harmful effects of smoking on human health.
(b) List any three values that are indicated with such approach. (CCE 2014)
Answer:
(a)

1. Smoking causes cancer of oral cavity and lungs, bronchitis, emphysema, gastric ulcers and heart disease.
2. Tobacco smoke is equally harmful to nonsmokers who live along with the smoker. They develop respiratory problems and CO toxicity.
3. Smoking is a wasteful, rather harmful drain on the resources of the family.
4. Smoke is irritant to nonsmokers.

(b) Values:

1. To protect the residents from the harmful effect of smoking .
2. It will educate your classmates and others in the compaign not to indulge in smoking despite peer pressure.
3. Participants in the compaign will be able to spread the message to their families and acquaintances. Ecoclub of your school is organising a debate on the topic “Nature’s fury unleashed by human greed (Uttarakhand disaster) “.

Question 6.
(a) List three arguments that you will use to convince the audience that humans are responsible for this disaster.
(b) List three values that will be inculcated with this debate. (CCE 2014)
Answer:
(a) Nature’s Fury: Humans are responsible for excessive landslides and floods in the hilly areas due to following reasons :

1. Building dams to store water and generate electricity. However, the geology of hills is usually unstable.
   Seepage from dams make the hills prone to excessive landslides and earthquakes.
2. Hills attract a large number of tourists. Roads are build up and a great deal of construction activity is carried out along the roads to accomodate tourists.
3. Trees are cut off for construction activity and roads. Debris of construction sites is often thrown into gorges and channels near the roads and buildings.
4. A lot of garbage and other wastes are generated by tourists and the locals. However, there is no system for their proper disposal.

(b) Values:

1. It is important to know the features of our environment and surroundings.
2. No human activity should disturb the balance of nature.
3. Trees are life line against disturbances in nature. They should not be felled unless protective measures have been taken.
4. There should be proper and scientific method for disposal of wastes.
   You have been asked to talk on “Ozone layer and its protection” in the school assembly on ‘Envrionment Day’.

Question 7.
(a) Why should ozone layer be protected to save the environment ?
(b) List any two ways that you would stress in your talk to bring in awareness amongst your fellow friends that would also help in protection of ozone layer as well as envronment. (CBSE Delhi 2017)
Answer:
Ozone layer is present in the stratosphere roughtly 23-25 km above the equator and 10-16 km above the poles. It protects the earth by filtering out the harmful UV radiations,. However, due to release of ozone depleting substances (ODS), ozone layer has thinned out by 8% over the equator. A big hole appears every year in August-September over Antarctica. This has increased the level of UV-B radiations reaching the earth by 15-20%. These radiations are causing increased number of skin cancers, cataracts and reduced immunity in human beings. There is also increased incidence of blinding of animals, death of young ones, reduced photosynthesis, higher number of mutations and damage to articles.
The remedy lies in banning the use of ozone depleting substances like halons and chlorofluorocarbons (CFCs).
All the classmates and the public in general should be made aware of the consequences of depletion of ozone layer, causes of depletion as well other aspects of degradation of environment. This could be done by

1. Arranging poster making competition as well as exhibition highlighting the effects of ozone layer depletion,
2. Conducting street plays about the harmful effects of our tampering with envronment and how to remedy the situation.

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HOTS Questions for Class 10 Science Chapter 15 Our Environment

These Solutions are part of HOTS Questions for Class 10 Science. Here we have given HOTS Questions for Class 10 Science Chapter 15 Our Environment

Question 1.
Study the figure. What does it depict ?

Answer:
It is representation of movement of energy and matter in the biosphere.
Energy flow is unidirectional while matter is repeatedly recycled.

More Resources

• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Which trophic level eats nothing and which one is not eaten.
Answer:
Eats Nothing: Producers.
Not Eaten: Top carnivores.

Question 3.
What is the reason that a food chain consists of only 3-5 steps ? (CCE 2011)
Answer:
As per 10% law of Lindeman (1942), the energy available decreases by 90% with the rise of trophic level. 2000 J of energy available at the producer or T₁ level will provide only 2 J of energy to second order carnivores (T₄).
Therefore, an ecosystem cannot have food chains of several steps.

Question 4.
Describe how decomposers facilitate recycling of matter in order to maintain balance in the ecosystem. (CBSE Foreign 2010)
Answer:
Decomposers act on organic or biodegradable wastes by secreting digestive enzymes over them. Organic waste is broken down into soluble simpler substances. Decomposers pick up the simple organic substances for their own use leaving the inorganic substances. The phenomenon is called mineralisation. Minerals released from decaying organic matter become available to plants for reuse. Decomposers, therefore, help in recycling of minerals and maintain the balance in the ecosystem.

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NCERT Exemplar Solutions for Class 10 Science Chapter 7 Control and Coordination

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 7 Control and Coordination

NCERT Exemplar Solutions for Class 10 Science Chapter 7 Short Answer Questions

Question 1.

1. Label the parts a, b, c and d and show the direction of electrical signals in the figure. (CBSE 2010)
   
2. Draw the figure shown here and label motor neuron, relay neuron and spinal cord. What is the name of this connection ?
   (CCE 2011)

More Resources

• NCERT Exemplar Solutions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Answer:

1. a – Sensory neuron.
   b – Spinal cord (CNS).
   c – Mortor neuron.
   d – Effector (muscle in arm)
2. Name — Reflex arc.
   Direction :
   1. Receptor (hand) to spinal cord through sensory neuron,
   2. Sensory neuron to motor neuron through Spinal cord to effector through motor neuron.

Question 2.
Name the plant hormone responsible for the following :
(a) Elongation of cells
(b) Growth of stem
(c) Promotion of cell division
(d) Falling of senescent leaves.
Answer:
(a) Elongation of Cells. Auxin.
(b) Growth of Stem. Gibberellin.
(c) Promotion of Cell Division. Cytokinin.
(d) Falling of Senescent Leaves. Abscisic acid.

Question 3.
Label the endocrine glands in the figure

Answer:
a – Pineal gland.
b – Pituitary gland.
c – Thyroid
d – Thymus.

Question 4.
In the figures (a), (b) and (c) which appears more accurate and why ?

Answer:
Figure (a) is more accurate as it shows appropriate response of both shoot and root to the vector of gravity. Shoot is negatively geotropic. It, therefore, bends upwardly. Root is positively geotropic. It, therefore, bends downwardly.

Question 5.
Label the parts of a neuron shown in the figure.

Answer:
a – Dendrite.
b – Cyton (Cell body).
c – Axon.
d – Axon terminal (bouton).

Question 6.
Match the terms of column A with those of column B.

A	B
(a)   Olfactory receptors (b)  Thermoreceptors (temperature receptors) (c)  Gustatoreceptors (d)  Photoreceptors	(I) Tongue (ii) Eye (iiî) Nose (iv) Skin

 

Answer:
a – IIi,
b — iv,
c — i,
d — ii.

Question 7.
What is a tropic movement ? Explain with an example. (CCE 2011)
Answer:
Tropic Movement: It is directional paratonie growth movement of curvature in which direction of movement is determined by the direction of the stimulus. They mostly occur in cylindrical organs. Growth response occurs due to differences in the distribution of auxin.
Example. On unidirectional exposure to light, shoots bend towards the source of light (positive phototropism) while roots may bend away from the source of light (negative phototropism).

Question 8.
What will happen if intake of iodine in our diet is low ?
Answer:

1. Low intake of iodine will affect production of thyroxine. Reduced thyroxine reduces metabolism of protein, carbohydrate and fat. Physical activity and consumption of energy are impaired.
2. With continued reduced intake of iodine, thyroid gland enlarges (in order to produce more thyroxine). It results in swelling of neck called goitre.

Question 9.
What happens at the synapse between the neurons ?

                                                   Or

State the events in sequence that take place when an electrical impulse travels from a dendritic tip of a nerve cell to another nerve cell. (CCE 2011, 2012, 2013)
Answer:
At synapse, axon terminal is expanded to form presynaptic knob. The dendrite terminal that lies close to it is slightly broadened and depreseed to form post-synaptic depression. A fluid filled narrow space, called synaptic cleft, occurs between the two. When an impulse reaches the presynaptic knob, it stimulates the release of neurotransmitter into synaptic cleft. Neurotransmitter comes in contact with chemoreceptor sites of the membrane of postsynaptic depression. This generates an electrochemical signal or impulse in the dendrite part of second neuron.

Question 10.
Answer the following :
(a) Which hormone is responsible for the changes noticed in female at puberty ?
(b) Dwarfism results due to deficiency of which hormone ?
(c) Blood sugar level rises due to deficiency of which hormone ?
(d) Iodine is necessary for synthesis of which hormone ?
Answer:
(a) Puberty Changes in Females. Oestrogen.
(b) Dwarfism. Growth hormone.
(c) Blood Sugar. Insulin.
(d) Iodine Hormone. Thyroxine

Question 11.
Answer the following :
(a) Name the endocrine gland associated with brain.
(b) Which gland secrets digestive enzymes as well as hormones ?
(c) Name the endocrine gland associated with kidney.
(d) Which endocrine gland is present in males but not in females ?
Answer:
(a) Endocrine Gland Associated with Brain. Pituitary.
(b) Gland with Digestive Enzymes and Hormones. Pancreas.
(c) Endocrine Gland Associated with Kidneys. Adrenal glands.
(d) Endocrine Gland in Males only. Testis.

NCERT Exemplar Solutions for Class 10 Science Chapter 7 Long Answer Questions

Question 12.
Draw the structure of a neuron and explain its function. (CCE 2011)
Answer:

Functions:

1. Dendrites: Picking up sensations and transmitting the same to cell body.
2. Cell Body:
   1. Sustaining structure and function of dendrites and axon,
   2. Functioning as passage¬way for transmission of sensation or impulse to axon.
3. Axon: Carrying impulse to another neuron, muscle, gland and organ. A single impulse can be transmitted to several structures with the help of axon terminals.

Question 13.
What are the major parts of the brain ? Mention the function of different parts.
Answer:
Major Parts of Brain
Fore-Brain: Olfactory lobes (2), Cerebral hemispheres (2) or cerebrum, Diencephalon.
Mid-Brain: Cerebral peduncles, Corpora quadrigemina.
Hind Brain: Pons, Cerebellum, Medulla oblongata.
Functions

1. Olfactory Lobes: Receive sensation of smell from olfactory epithelium and relay the same to temporal lobes of cerebral hemispheres.
2. Cerebral Hemispheres:
   1. Frontal Lobes: Centres of intelligence, control of movements and facial muscles, speech.
   2. Parietal Lobes: Taste and cutaneous sensations with some components of speech.
   3. Temporal Lobes: Hearing, smell, recall of audiovisual memory, some components of speech.
   4. Occipital Lobes: Sight.
3. Diencephalon: Hypothalamus part regulates activity of smooth muscles and functions as control centre for hunger, thirst, fatigue, sleep, sweating, temperature and emotions. Pituitary gland and pineal gland are components of diencephalon.
4. Corpora Quadrigemina: Superior – sight reflexes. Inferior – auditory reflexes.
5. Cerebellum: Controls posture and equilibrium.
6. Pons: Part of respiratory centre.
7. Medulla Oblongata: Centre for reflexes, blood pressure, heart beat, breathing.

Question 14.
What constitutes the central and peripheral nervous systems ? How are the components of central nervous system protected?
(CCE 2012)
Answer:
Central Nervous System (CNS): It is hollowed part of nervous system that lies along the mid-dorsal part of the body. It has two parts, brain (located in the head) and spinal cord (located in the trunk).
Peripheral Nervous System (PNS): It is soild lateral part of nervous system that develops from CNS and connects different parts of the body with CNS. Peripheral nervous system has two components, voluntary and involuntary. Voluntary peripheral nerous system is under the control of will. It consists of cranial nerves from brain and spinal nerves from spinal cord. Involuntary peripheral nervous system works independent of will. It develops from some cranial and spinal nerves. Involuntary peripheral nervous system is also called autonomic nervous system. It has two parts, sympathetic (for emergency) and parasympathetic (for moderation). They control the functioning of various internal body parts.
Protection of Brain: Brain box or Cranium, meninges and cerebrospinal fluid.
Protection of Spinal Cord: Vertebral Column, meninges and cerebrospinal fluid.

Question 15.
Mention one function for each of these hormones.
(a) Thyroxine
(b) Insulin
(c) Adrenaline
(d) Growth hormone
(e) Testosterone. (CCE 2010, 2013, 2015)
Answer:
(a) Thyroxine: Regulates metabolism of carbohydrates, fats and proteins, release of energy and body activity.
(b) Insulin: Regulates blood glucose by its absorption in liver, muscles (for formation of glycogen) and individual cells (for metabolic activity).
(c) Adrenaline: For meeting an emergency by increasing heart rate and supply of blood to various concerned organs.
(d) Growth Hormone (GH) or Somatotrophic Hormone: Regulates growth and development of the body through anabolic activity for growth of bones, muscles and visceral organs.
(e) Testosterone (Male Sex Hormone): Controls changes in body associated with puberty in males (beard, moustache, low pitch voice, increased growth of bones and muscles).

Question 16.
Name the various plant hormones. Also give their physiological effects on plant growth and development. (CCE 2016)
Answer:
Five types — auxin, gibberelin, cytokinin, ethylene and abscisic acid.
Auxin: Cell enlargement, root formation, apical dominance, inhibition of abscission, fruit growth.
Gibberellin: Growth in stem and leaves, higher fruit yield, overcoming dormancy.
Cytokinin: Essential for cell division, differentiation, prevention of senescence and overcoming apical dominance.
Ethylene: Promotes transverse growth, fruit ripening and overcoming dormancy of some parts.
Abscisic Acid (ABA): Induces dormancy, senescence and abscission, checking excessive activity of growth promoting hormones, closure of stomata under water stress.

Question 17.
What are reflex actions ? Give two examples. Explain a reflex arc.
Answer:
Definition: Reflex actions seem to have evolved quite early in the physiology of animals when complex neuron network for processing had not been evolved. Even after evolution of processing centres, reflex actions have continued to persist because of their more efficiency for quick responses.
Two Examples:

1. On being pricked or coming in contact with hot surface, hand is withdrawn even before pain is perceived (by brain),
2. Wider openin0 of pupil in dim light and its narrowing in strong light.
   

Question 18.
“Nervous and hormonal systems together perform the function of control and coordination in human being.” Justify the statement.
Answer:
Both nervous system and hormonal (or endocrine) system are involved in control, regulation and coordination of body parts. Nervous system is connected to receptors of all senses. Information obtained from sensory organs is passed rapidly to CNS for interpretation.
On the basis of interpretation, a message is sent to effector organ or organs (muscles, glands, etc.). The rate of information or impulse transfer is very high, some 100m/sec. Every action and activity is well coordinated as the information is sent to all the regions required for that action. The action can be voluntary (under the will) or involuntary. Passage of food in the alimentary canal is due to an involuntary movement of alimentary canal called peristalsis. Picking up food and placing it inside the mouth is a voluntary movement. Both these movements are highly coordinated by nervous system.
Nervous system is also connected with the functioning of endocrine or hormonal system. The endocrine system functions at the biochemical level while nervous system functions at the physical and physiological level. For their functioning endocrine glands pour their secretion into blood which transports them to all parts of the body. Target cells have receptors for picking up the hormones and working as per hormonal stimulus. There is a feedback system which determines the requirement of hormones and the activity of endocrine glands. Other stimuli are also involved for coordinated functioning. Presence of food in the stomach stimulates its wall to secrete hormone gastrin. Gastrin stimulates gastric glands to pour gastric juice over the food. As the partially digested food passes into duodenum, the latter forms hormone secretin which induces passage of bile and pancreatic juice into duodenum.
Joint working of both the systems is observed during an emergency. Both sympathetic nervous system and adrenal glands prepare the body for meeting the emergency. There is higher rate of heart beat, more blood supply to cardiac and skeletal muscles and higher rate of breathing for quicker oxygenation of blood.

Question 19.
How does chemical co-ordination take place in animals ?
Answer:
In animals, chemical coordination is achieved through the agency of hormones which function as chemical messengers or informational molecules. Hormones are secreted by ductless glands in response to specific conditions or nervous stimulation. Timing and amount of a hormone released are regulated by feed-back mechanism. After a meal, sugar level of blood rises. It is detected by pancreas. Pancreas responds by producing hormone insulin from (3-cells of islets of Langerhans. Insulin causes glucose to be absorbed by all cells as well as get stored in liver and muscles in the form of glycogen. As the level of glucose falls in blood, insulin secretion is reduced.

Question 20.
Why the flow of signals in a synapse is from axonal end of one neuron to dendrite end of another neuron but not the reverse ?
Answer:
As the electrical impulse reaches the axon terminal in the region of axon-dendrite synapse, it stimulates the exocytosis of vesicles containing neurotransmitter (e.g. acetylcholine). Neurotransmitter attaches to the chemoreceptor sites of the membrane covering the dendrite end of synapse. It creates a new impulse that travels through cell body and axon of the second neuron. At synapse, the axon end does not contain any chemoreceptor sites so that reverse flow of electrochemical impulses is not possible.

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 7 Control and Coordination are helpful to complete your science homework.

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NCERT Exemplar Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce?

These Solutions are part of NCERT Exemplar Solutions for Class 10 Science. Here we have given NCERT Exemplar Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce?

NCERT Exemplar Solutions for Class 10 Science Chapter 8 Short Answer Questions

Question 1.
In a bisexual flower, inspite of the young stamens being removed artificially, the flower produces fruit. Provide a suitable explanation for the above situation.
Answer:
The open flowers are generally cross pollinated. Removal of stamens of a bisexual flower will not affect pollination of its intact pistil and formation of fruit.

More Resources

• NCERT Exemplar Solutions for Class 10 Science
• NCERT Solutions for Class 10 Science
• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Can you consider cell division as a type of reproduction in unicellular organisms ? Give one reason.
Answer:
Yes. Cell division in a unicellular organism which produces two d&ughter cells, actually forms two daughter individuals.

Question 3.
What is a clone ? Why do offspring formed by asexual reproduction exhibit remarkable similarity ?
Answer:
Clone is an exact genetic replica of another individual. All the offspring formed from a parent through asexual reproduction are clones of one another as well as their parent. The remarkable similarity of asexually produced daughter individuals is due to genetic similarity as they possess exact copies of DNA of their parent.

Question 4.
Explain how, offspring and parents of organisms reproducing sexually have the same number of chromosomes/DNAs.
(CCE 2014, CBSE Delhi 2017)
Answer:
The parents are diploid (2n) as each of them has two sets of chromosomes (DNAs). They form haploid (In) male and female gametes through the process of meiosis. The haploid gametes have one set of chromosomes (DNAs). During fertilization, one male gamete fuses with one female gamete. It restores the diploid (2n) chromosome/DNA number in the offspring that is formed from fusion product or zygote (2n).

Question 5.
Colonies of yeast fail to multiply in water but multiply in sugar solution. Give one reason for this.
Answer:
Yeast is heterotrophic. It obtains its nourishment from outside. Plain water cannot provide nourishment to yeast while sugar solution can do so. Therefore, Yeast multiplies in sugar solution and not in plain water.

Question 6.
Why does Bread Mould grow profusely on a moist slice of bread rather than on a dry slice of bread ?
Answer:
Growth of Bread Mould requires both water and nutrients. Dry slice of bread contains nutrients but no water. Therefore, Bread Mould does not grow over it. Moist slice possesses both water and nutrients. Hence, Bread Mould grows over it.

Question 7.
Give two reasons for the appearance of variations among the progeny formed by sexual reproduction.
Answer:
Variations appear in the progeny of sexually reproducing organisms due to

1. Random separation and coming together of chromosomes during gamete formation and gamete fusion
2. Crossing over and mutations.
3. Coming together of chromosomes of the parents.

Question 8.
Would a Planaria cut vertically into two halves regenerate into two individuals ?
Answer:
Yes. Each piece of Planaria is able to grow the missing parts and form the complete organism.

Question 9.
Correlate the number of chromosomes with the size of the organism and answer the following questions :
(a) Do larger organisms have more number of chromosomes/cell ?
(b) Can organism with fewer chromosomes reproduce more easily than organism with more number of chromosomes ?
(c) More the number of chromosomes/cell, greater is the DNA content. Justify.
Answer:
(a) No. There is no correlation between chromosome number and size of the individual.
(b) No. Chromosome number does not affect reproduction.
(c) Yes. Each chromosome represents a single duplex of DNA. More chromosomes means more DNA.

Question 10.
In Tobacco, male gametes have 24 chromosomes.
(a) What is the number of chromosomes in the female gamete ?
(b) What is the number of chromosomes in the zygote ?
Answer:
(a) 24
(b) 48.

Question 11.
Why cannot fertilization take place in flowers if pollination does not occur ?
Answer:
Pollination is essential for bringing the male gametes. In the absence of pollination, there will be no male gamete to perform fertilization.

Question 12.
Is the chromosome number of zygote, emboryonal cells and adult of a particular organism always constant ? How is the constancy maintained in these three stages ?
Answer:
Zygote is the first diploid structure. It undergoes mitotic divisions to form embryonal cells. Embryonal cells further divide by mitosis to form the adult. Mitosis maintains the same chromosome number in the daughter cells.

Question 13.
Where is the zygote located in the flower after fertilization ?
Answer:
Zygote is the fertilized oosphere which occurs in the embryo sac present inside an ovule located in the ovary part of the pistil.

Question 14.
Reproduction is linked to stability of population of a species. Justify the statement.
Answer:
Reproduction maintains the size and composition of population of a species by regularly adding new individuals for replacing the ones getting killed due to ageing, disease or predation. It also adds variations that allow the individuals to adapt to changing environment.

Question 15.
How are general growth and sexual maturation different from each other ?
Answer:
General growth is the growth of an individual in size, height, shape, weight, etc. Sexual maturation is a set of changes in body of an individual that enable one to take part in reproduction, e.g., maturation of gonads, genitalia and other accessory structures. In human beings, beginning of sexual maturation is indicated by puberty changes like cracking of voice, new hair pattern, development of breast in female, etc.

Question 16.
Trace the path of sperm during ejaculation and mention the glands and their function associated with the male reproductive system.
Answer:
Ejaculated sperms are the ones which are stored in epididymes. They are formed regularly in seminiferous tubules from where they pass through vasa efferentia into epididymes. At the time of ejaculation, the sperms are first pushed through vasa deferentia, enter ejaculatory duct where they receive secretion of seminal vesicles and then urinogenital duct where the secretion of prostate gland is poured to form semen. The urinogenital duct is lubricated by secretion of a pair of Cowper’s glands (bulbourethral glands).
Secretion of Seminal Vesicles. Nourishment, activation and providing fluid medium for sperm transport.
Secretion of Prostates Gland. Motility of Sperms.

Question 17.
What changes are observed in the uterus if fertilization does not occur ? (CBSE Delhi 2017)
Answer:
Glands present in the mucosa of uterine endometrium begin to degenerate. This sloughs off the endometrial lining, releasing a lot of mucus and blood. It is called menstruation. Menstruation lasts for 3-5 days.

Question 18.
What changes are observed in the uterus subsequent to implantation of young embryo ? (CBSE Delhi 2017)
Answer:
Endometrium thickens, becomes glandular and highly vascular. The contact region between embryo and uterine wall grows into placenta. Placenta helps the embryo in obtaining nourishment and oxygen from blood sinuses of the uterus. It also helps in transfer of wastes from embryo to the blood of the mother.

Question 19.
What are the benefits of using mechanical barriers during sexual act ?
Answer:

1. Prevention of Pregnancy : Mechanical barriers (e.g., Condoms, diaphragms) prevent the passage of sperms into the genital tract of the female. Consequently, pregnancy does not occur.
2. Non-transmission of Infections: There is no transfer of venereal diseases from the infected partner to the non-infected partner.

Question 20.
In the female reproductive system, name the parts that are involved in
(a) Production of egg.
(b) Site of fertilization
(c) Site of implantation
(d) Entry of sperms.
Answer:
(a) Production of Egg: Ovary
(b) Site of Fertilization: Ampulla part of fallopian tube.
(c) Site of Implantation: Uterus.
(d) Entry of Sperms: Vagina.

Question 21.
What would be the ratio of chromosome number between an egg and its zygote ? How is the sperm genetically different from the egg ?
Answer:
Chromosome Number in Egg and Zygote. 1 : 2
Genetic Difference between Sperm and Egg. Sperms are genetically of two types, X-containing and Y-containing. Eggs are always of one type, X-containing.

NCERT Exemplar Solutions for Class 10 Science Chapter 8 Long Answer Questions

Question 22.
Why are budding, fragmentation and regeneration all considered as asexual types of reproduction ? With neat diagrams, explain the process of regeneration in Planaria.
Answer:
(a) Asexual Reproduction: Budding, fragmentation and regeneration are all considered to be modes of asexual reproduction as they involve formation of new individuals from single parents without fertilization or fusion of gametes.
(b) Regeneration in Planaria:

Question 23.
Write two points of differences between asexual and sexual types of reproduction. Describe why variations are observed in the offspring formed by sexual reproduction.
Answer:
(a) Differences between Asexual and Sexual Reproduction

Asexual Reproduction	Sexual Reproduction
1. Parents: It is monoparental.	It is generally biparental
2. Meiosis: Meiosis does not occur during asexual reproduction. There are no gametes and no fertilization.	Meiosis occurs. Gametes are formed and fertilization occurs in sexual reproduction.
Ex. Binary fission in Amoeba, Budding in Yeast	Ex. Offspring in Wheat, Human.

(b) Development of Variation in Sexual Reproduction:
Sexual reproduction produces a number of variations in the population due to

1. Chance separation of chromosomes at the time of meiosis.
2. Crossing over during meiosis.
3. Chance combination of chromosomes during fertilization. It produces a unique combination of variations already accumulated by each individual in its DNA obtained from its parents,
4. Mutations or defective DNA replication.

Question 24.
Distinguish between pollination and fertilization. Mention the site and product of fertilization in a flower. Draw a neat, labelled diagram of a pistil showing pollen tube growth and its entry into the ovule.
Answer:
(a) Differences between Pollination and Fertilization

Pollination	Fertilization
1. Definition: It is transfer of pollen grains from anther to the stigma of a flower. 2. Step: Pollination precedes fertilization. 3. Purpose: It carries the male gamete producing pollen grains to the female sex organ. 4.  Process: Pollination is a physical process. 5. Occurrence: It occurs only in seed plants.	It is the fusion of male and female gametes. Fertilization occurs only after pollination when the pollen grain has germinated and male gametes are carried into ovule. It actually brings about fusion of gametes. Fertilization is a physico-chemical (biological) process. It occurs in both plants and animals of various types.

(b)

1. Site of Fertilization. Embryo sac in ovule.
2. Product of Fertilization. Zygote, Primary endosperm cell.

(c)

Question 25.
Distinguish between a gamete and zygote. Explain their roles in sexual reproduction.
Answer:
Differences between Gamete and Zygote

Gamete	Zygote
1. Nature: It is sex or germ cell that takes part in fertilization.	It is a product of fertilization.
2. Types: There are two types of gametes, male and female.	Zygote is of one type.
3. Chromosome Number: A gamete has haploid or In chromosome number.	Zygote has diploid or 2n chromosome number.
4. Characteristics: A gamete carries characteristics of only one parent.	It carries characteristics of both the parents.
5. Generation: Gamete is the last cell of its generation.	It is the first cell of new generation.

Role of Gamete. Gamete is sex or germ cell which is specialized to take part in sexual reproduction. There are two types of gametes, male and female. Each of them carries one set of chromosomes which are randomly obtained from its parent. Fusion of male gamete with a female gamete produces a zygote.
Role of Zygote. It is the first’cell of new generation. Zygote develops into embryo that later forms the new individual.

Question 26.
Draw the diagram of a flower and label the four whorls. Write the names of gamete producing organs in the flower.
Answer:
(a)

(b) Male Gamete Forming Part. Pollen grain, produced in anther of stamen.
Female Gamete Forming Part. Embryo sac developed inside ovule of ovary part of pistil.

Question 27.
What is placenta ? Mention its role during pregnancy. (CCE 2011, 2014)
Answer:
Placenta: It is a special double layered, spongy tissue connection between the foetus and uternine wall found in pregnant females. It has finger-like outgrowths or villi which are in contact with blood sinuses present in the uterine wall.
Role:

1. Attachment: Placenta attaches the foetus to uterine wall.
2. Villi: Placenta has finger-like outgrowths or villi which develop a large surface area for fixation and absorption.
3. Nutrients: Placenta picks up nutrients from mother’s blood and passes it to the blood of the foetus.
4. Waste Products: Waste products produced by the foetus passes out through the placenta into mother’s blood.
5. Gases: Foetus obtains oxygen supply from mother’s blood and eliminates carbon dioxide through placenta.

Question 28.
What are the various ways to avoid pregnancy ? Elaborate any one method.
Answer:
(a) Methods to Avoid Pregnancy: Prevention of pregnancy is called contraception. The techniques used in preventing the occurrence of pregnancy are called contraceptive devices.
(b) Mechanical Barrier Methods of Contraception:

1. Mechanical Barriers like condoms, cervical cap, diaphragm.
2. Oral Contraceptives or oral pills like Mala D, Saheli
3. Intrauterine Contraceptive Devices (IUCD) like loop, bow, Cu-T.
4. Surgical Methods like vasectomy in males and tubectomy in females.

Question 29.
Flow does fertilization take place ? Fertilization occurs once in a month. Comment.
Answer:
In human beings, fertilization is internal. It can occur within 2-3 days of ovulation (between 14th and 16th day of menstrual cycle). Fertilization generally takes place in ampulla (ampulla-isthmus border) part of fallopian tube where ovum rests for several hours after entry into oviduct. The sperms have to reach there. Sperms deposited in the vagina reach there partly by their own movement and partly by local movements of reproductive tract. Both sperms and eggs are viable for 24-48 hours. A sperm reaching the surface of ovum, dissolves its covering and passes its head and middle part into it. It results in fertilization and formation of zygote.
Fertilization only once in a Month. A single ovum is released by female roughly in the middle of menstrual cycle which has a span of 4 weeks. Therefore, fertilization can occur only once in a month.

Question 30.
Reproduction is essentially a phenomenon that is not for survival of individual but for the stability of a species. Justify. Reproduction is not essential for survival of the individual.
Answer:
Survival of the individual depends upon input of nutrients and energy and elimination of wastes. Reproduction has ho role in these. It is, however, essential for the stability and survival of the species. Reproduction takes part in
(a) Perpetuation of Species: Perpetuation or continuation of species.
(b) Replacement: Replacement of dead individuals and maintaining the organisation in population.
(c) Population Characteristics:

1. Education: There is an inverse ratio between education and population growth.
2. Population Education: It is imparting knowledge to public about the effects of excessive population, advantages of small families and means to achieve it.
3. Marriageable Age: Number of births is reduced if young persons marry late. Gainful employment and higher social status of women also reduce birth rate.

(d) Variations:
Variations are differences found in morphological, physiological and other traits of individuals belonging to the same organism, race or family. They develop due to

1. Faulty DNA replication
2. Crossing over
3. Chance separation of chromosomes during meiosis and chance combination during fertilization. Variations are highly important.

(e) Absence of Variations: Due to absence of variations, asexually reproduced organisms are unable to adapt themselves to changes in the environment.

Question 31.
Describe sexually transmitted diseases and mention the ways to prevent them.
Answer:
It is a group of infections caused by different types of pathogens that are transmitted by sexual contact between a healthy person and an infected person. The sexually transmitted diseases are also called venereal diseases (VDs). Some 30 different types of STDs are known.

1. It is caused by bacterium Neisseria gonorrhoeae. Gonorrhoea spreads through sexual contact, common toilets and under clothes. Incubation period is 2-5 days. Bacterium resides in genital tubes. It causes pus containing discharge, pain around genitalia and burning sensation during urination. Effective medicine is ampicillin.
2. The disease is caused by corkscrew like bacterium Treponema pallidum. It spreads through sexual contact and from mother to infants. Incubation periods is 3-5 weeks, Painless ulcer on genitalia and swelling of lymph glands occur in first stage. In second stage skin lesions, rashes and hair loss occur. Tertiary stage is characterised by chronic ulcers and damage to vital organs. Effective medicine is tetracycline.
3. Genital Warts: The disease is due to Human Papilloma virus (HPV). Hard benign outgrowths called warts appear over external genitalia and perianal area. In women infection may enter vagina and cervix causing acute pain. Cryosurgery and Podophyllum preparations are effective.
4. AIDS (Acquired Immune Deficiency Syndrome): It is caused by human immunodeficiency virus or HIV. The virus is transmitted through sexual contact, blood contact (as using common needles, syringes, razors, transfusion) and placenta (mother to foetus). Incubation period is generally 27-28 months but symptoms may appear early. Count of T-helper cells becomes low, 200/ml or less. There is headache, rashes, nausea, pharyngitis and fever. Immunity is drastically reduced so that many infections (opportunistic infections) begin to appear. A proper treatment has not yet been discovered. ART (antiretrovirus treatment) is given to patients to reduce the effect of infection.

Hope given NCERT Exemplar Solutions for Class 10 Science Chapter 8 How do Organisms Reproduce? are helpful to complete your science homework.

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NCERT Solutions for Class 10 Hindi are the part of NCERT Solutions for Class 10. Here we have given Class 10 Hindi NCERT Solutions.

NCERT Solutions for Class 10 Hindi Course A and B are designed by our experienced subject teachers according to the CBSE Latest Hindi Syllabus and guidelines. These solutions are prepared in a conceptual manner for easy understanding of the subject concepts. Students can learn all chapters from NCERT Class 10 Hindi Solutions and perform well in their final exams.

Students will understand how to answer a particular question in the exam after preparing the concepts from NCERT Solutions for Grade 10 Hindi. You can cover Hindi Course-A Kshitij Bhag 2 (17 chapters), Kritika Bhag 2 (5 chapters), and Hindi Course-B Sparsh Bhag 2(17 chapters), Sanchayan Bhag 2 (3 chapters) in the NCERT Solutions PDF links available here. So download them and ace up your preparation.

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Each and every chapter in the NCERT textbook for Class 10 Hindi are explained in a simple language for students to understand easily and prepare all answers properly. These NCERT 10th Hindi Textbook Solutions help you to practice all exercise questions and score well in the main examinations. Overall, Class 10 Hindi NCERT Solutions PDF for all chapters clear all your doubts and make you learn each and every point perfectly.

So, download the best study resources ie., NCERT Solutions for Class Hindi Course A and B in PDF format by tapping on the respective links provided here for future reference.

NCERT Solutions for Class 10 Hindi – A

NCERT Solutions for Class 10 Hindi Kshitij Bhag 2 क्षितिज भाग 2

काव्य – खंड

• Chapter 1 पद
• Chapter 2 राम-लक्ष्मण-परशुराम संवाद
• Chapter 3 सवैया और कवित्त
• Chapter 4 आत्मकथ्य
• Chapter 5 उत्साह और अट नहीं रही
• Chapter 6 यह दंतुरहित मुस्कान और फसल
• Chapter 7 छाया मत छूना
• Chapter 8 कन्यादान
• Chapter 9 संगतकार

गद्य – खंड

• Chapter 10 नेताजी का चश्मा
• Chapter 11 बालगोबिन भगत
• Chapter 12 लखनवी अंदाज़
• Chapter 13 मानवीय करुणा की दिव्या चमक
• Chapter 14 एक कहानी यह भी
• Chapter 15 स्त्री शिक्षा के विरोधी कुतर्कों का खंडन
• Chapter 16 नौबतखाने में इबादत
• Chapter 17 संस्कृति

NCERT Solutions for Class 10 Hindi Kritika Bhag 2 कृतिका भाग 2

• Chapter 1 माता का आँचल
• Chapter 2 जॉर्ज पंचम की नाक
• Chapter 3 साना-साना हाथ जोड़ि
• Chapter 4 एही ठैयाँ झुलनी हेरानी हो रामा!
• Chapter 5 मैं क्यों लिखता हूँ?

NCERT Solutions for Class 10 Hindi – B

NCERT Solutions for Class 10 Hindi Sparsh Bhag 2 स्पर्श भाग 2

काव्य – खंड

• Chapter 1 साखी
• Chapter 2 पद
• Chapter 3 दोहे
• Chapter 4 मनुष्यता
• Chapter 5 पर्वत प्रदेश में पावस
• Chapter 6 मधुर-मधुर मेरे दीपक जल
• Chapter 7 तोप
• Chapter 8 कर चले हम फ़िदा
• Chapter 9 आत्मत्राण

गद्य – खंड

• Chapter 10 बड़े भाई साहब
• Chapter 11 डायरी का एक पन्ना
• Chapter 12 तताँरा-वामीरो कथा
• Chapter 13 तीसरी कसम के शिल्पकार शैलेंद्र
• Chapter 14 गिरगिट
• Chapter 15 अब कहाँ दूसरे के दुख से दुखी होने वाले
• Chapter 16 पतझर में टूटी पत्तियाँ
• Chapter 17 कारतूस

NCERT Solutions for Class 10 Hindi Sanchayan Bhag संचयन भाग 2

• Chapter 1 हरिहर काका
• Chapter 2 सपनों के-से दिन
• Chapter 3 टोपी शुक्ला

Benefits of Studying NCERT Class 10 Hindi Solutions Book PDF

• Candidates of class 10 can find these solutions book very handy at the time of revision.
• Students can identify and examine their performance by using the NCERT Solutions Study Book.
• Also, they can spend the required amount of time for the weak areas and enhance their subject knowledge.
• All main concepts and poetry lessons can easily understandable by the students while studying from NCERT Solutions PDF.
•  You can even complete your given Homework and assignments based on NCERT textbook solutions.

FAQs on NCERT Solutions for Class 10 Hindi PDF

1. How helpful are the NCERT Solutions for Class 10 Hindi?

NCERT Solution book is very helpful and essential for students’ exam preparation. With NCERT Class 10 Hindi Solutions, you can easily learn and understand all chapters quickly & easily. These solutions are written by subject experts as per the latest CBSE 10th Hindi syllabus. Also, you can find many benefits while preparing from NCERT Solutions Class 10 Hindi.

2. How many chapters are there in Class 10 Hindi?

There are 42 chapters total in both Class X Hindi Course A and B. In Hindi Course-A, Kshitij Bhag 2 (17 chapters), Kritika Bhag 2 (5 chapters). In Hindi Course-B, Sparsh Bhag 2(17 chapters), Sanchayan Bhag 2 (3 chapters).

3. How to download NCERT Solutions of grade 10 Hindi in PDF?

All students of class 10 can download 10th standard Hindi course A & B NCERT Solutions in PDF format by clicking on the available direct links over here. You can access these PDF links free of cost so download and prepare well.

Conclusion

We believe the details shared about the NCERT Solutions for Class 10 Hindi PDF will benefit you a lot. If you need any assistance to download NCERT 10th Hindi Solutions PDF, feel free to share it with us via the comment section. We’ll revert back to you soon with the best possible solutions meanwhile, visit our site learninsta.com and get updated info on NCERT Solutions for Class 10 subjects.

Value Based Questions in Science for Class 10 Chapter 8 How do Organisms Reproduce?

These Solutions are part of Value Based Questions in Science for Class 10. Here we have given Value Based Questions in Science for Class 10 Chapter 8 How do Organisms Reproduce?

Question 1.
A classmate of yours has a pus containing discharge, pain around genitalia and burning sensation during urination. What disease he is suffering from. How has he caught the disease ? How will you help your classmate ? What precautions are required not to get the repeat of the disease ?
Gonorrhoea. Gonorrhoea is commonly a sexually transmitted disease. My classmate has caught the disease despite his being not involved sexually with any one. The other possibility is that he is using common underwears. It is also not possible as he is the lone kid of the family. No body else in his family has this disease. The last possibility is that he has caught the disease by using toilets at his relative’s home or in the school.
Clean and hygienic toilets can prevent the occurrence of this and many other diseases.

More Resources

• Value Based Questions in Science for Class 10
• HOTS Questions for Class 10 Science
• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
Surbhi and Mukesh are a young couple married only four years ago. They have two children. How will you convince the couple not to have any more child ? What guidelines will you give to the couple ?
Answer:
Conviction: I will tell the couple that :

1. A safe, satisfying reproductive health is possible only when the number of children is small.
2. Small number of children helps in enjoying the family life better both at home and outside.
3. The children can be reared with better resources. They can be properly educated and settled in life.
4. The couple can also save and plan properly for their post retirement life.

Guidelines: The couple should start using contraceptive device(s) that suit them best, may be oral pills, IUCD or condom.

Question 3.
Seeing the different colours of Bougainvillea growing in the corner of school campus the Principal suggested that the boundary of the whole campus be decorated with the plant.
How you and your classmates will perform this task ?
Answer:

1. The first task is to select the colour scheme that will best suit the campus boundary wall,
2. How much space is to be given to each colour,
3. Digging and properly watering the soil along the boundary wall but leaving a space of 25-30 cm from the wall for preventing any damage to it and keeping a narrow pathway for movement of the gardener,
4. Preparing stem cuttings of 25-30 cm length from one year old branches.
5. Fixing the cuttings in the soil in their natural position at a distance of about 30 cm from one another in 3-4 rows as per plan,
6. Regular but well spaced watering of the cuttings,
7. The cuttings will become independent plants within 2-3 months.

Question 4.
You have read in newspapers that sex ratio in many parts of the country has gone down to less than 900:1000. What does this mean ? What is the reason behind it ? How can you contribute in retrieving the situation ?
Answer:
The ratio 900 : 1000 is called sex ratio. It indicates that there are 900 females to 1000 males. This is not a healthy sign for the society. A number of males will not be able to find suitable brides leading to several types of social problems.
The reason behind this decline in sex ratio is the availability of sex determining devices for the foetus and carrying out abortion in case of female foetus.
Retrieving Situation :

1. Organising rallies against female foeticide
2. Organising vigilance group to find out the centres where sex of the foetus is being determined and reporting the same to the health authorities.

Our government lauches compaigns to provide information about AIDS prevention, testing and treatment by putting posters, conducting radio shows and using other agencies of advertisements.

Question 5.
(a) To which category of diseases AIDS belong ? Name its caustive organism.
(b) Which kind of value is government trying to develop in the citizens by conducting the above kind of programmes. (Sample Paper 2017-18)
Answer:
(a) AIDS is a sexually transmitted disease (STD) which is also contracted through blood (using common syringes, needles, razors) and placenta.
Causative Agent: Virus— Human Immuno deficiency Virus or HIV.
(b) Values:

1. Government is trying to save the citizens from the debilitating and fatal disease by telling every citizen through all means of communication how the disease spreads and what are the ways to prevent it.
2. Government is also informing the citizens about the institutes where free testing is available and asking the patients to get registered with the treatment centres for free medicines.
3. The public is also informed not to shun the AIDS patient as the disease does not spread by mere contact. It will reduce the trauma being suffered by the patient.

Hope given Value Based Questions in Science for Class 10 Chapter 8 How do Organisms Reproduce? are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.