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CBSE Sample Papers for Class 10 Maths Paper 1 is part of CBSE Sample Papers for Class 10 Maths Here we have given CBSE Sample Papers for Class 10 Maths Paper 1. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Sample Papers for Class 10 Maths Paper 1

Board	CBSE
Class	X
Subject	Maths
Sample Paper Set	Paper 1
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 1 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.
Given that LCM (91, 26) = 182, find HCF (91, 26).

Question 2.
Does the rational number \(\frac { 441 }{ { 2 }^{ 2 }.{ 5 }^{ 7 }.{ 7 }^{ 2 } } \) has a terminating or a non-terminating decimal representation?

Question 3.
If 1 is a zero of the polynomial p(x) = ax² – 3 (a – 1) x – 1, then find the value of a.

Question 4.
The nth turn of an AP is 7 – 4n. Find its common difference.

Question 5.
If adjoining figure, DE || BC and AD = 1 cm, BD = 2 cm. What is the ratio of the area of ∆ABC to the area of ∆ADE?

Question 6.
In adjoining figure, PA and PB are tangents to the circle drawn an external point P. CD is a third tangent touching the circle at PB = 10 cm and CQ = 2 cm, what is the length of PC?

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
Determine the values of m and n, so that the following system of linear equations has infinite number of solutions:
(2m – 1)x + 3y – 5; 3x + (n – 1)y – 2 = 0.

Question 8.
Find the roots of the quadratic equation: 4 √3 x² + 5x – 2 √3 =0.

Question 9.
A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag. Find the probability of getting:
(i) a white ball or a green ball.
(ii) neither a green ball nor a red ball.

Question 10.
Without using trigonometric tables, find the value of
\(\frac { cos{ 70 }^{ o } }{ sin{ 20 }^{ o } } \) + cos 57° . cosec 33° – 2 cos 60°.

Question 11.
In the adjoining figure, E is a point on the side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC,
prove that: AB x EF = AD x EC.

Question 12.
Two circles touch externally. The sum of their areas is 58π cm² and the distance between their centres is 10 cm. Find the radii of the two circles.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
Prove that

Question 14.
Prove that: (1 + cot A – cosec A) (1 + tan A + sec A) = 2.

Question 15.
Solve the following pair of linear equations:

OR
The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator both increased by 3, they are in the ratio 2 : 3. Determine the fraction.

Question 16.
Show that \(\frac { 1 }{ 2 }\) and \(\frac { -3 }{ 2 }\) are the zeroes of the polynomial 4x² + 4x – 3 and verify the relationship between zeroes and coefficients of the polynomial.
OR
Quadratic polynomial 2x² – 3x + 1 has zeroes as α and β. Form a quadratic polynomial whose zeroes are 3α and 3β.

Question 17.
Construct a ∆ABC in which CA = 6 cm, AB = 5 cm and ∠BAC = 45°, then construct a triangle similar to the given triangle whose sides are \(\frac { 6 }{ 5 }\) of the corresponding sides of the ∆ABC.

Question 18.
Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear. .
OR
If P(9a – 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.

Question 19.
Prove that the points A (-3, 0), B (1, – 3) and C(4, 1) are the vertices of an isosceles right triangle.

Question 20.
Cards marked with all 2-digit numbers are placed in a box and are mixed thoroughly. One card is drawn at random. Find the probability that the number on the card is
(i) divisible by 10
(ii) a perfect square number
(iii) a prime number less than 25.
OR
The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of
(i) hearts
(ii) queen
(iii) clubs.

Question 21.
Find the median of the following data:

Question 22.
The following frequency distribution shows the number of rims scored by some batsmen of India in one-day cricket matches:

Find the mode for the above data.

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
Prove that n³ – n is divisible by 6 for every positive integer n.

Question 24.
A sum of Rs 1600 is to be used to give 10 cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
What is the importance of an academic prize in student’s life?

Question 25.
A motor boat whose speed is 20 km/h in still water takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Find the speed of the stream.
OR
A shopkeeper buys some books for Rs 80. If he had bought 4 more books for the same amount, each book would have cost Rs 1 less. Find the number o.f books he bought.

Question 26.
Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Question 27.
In the adjoining figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersects XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

OR
The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.

Question 28.
The angles of elevation of the top of a tower from two points P and Q at distances of a and b respectively from the base and in the same straight line with it are complementary. Prove that the height of the tower is √ab .
OR
The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

Question 29.
In the adjoining figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find
(i) the area of the shaded region.
(ii) the perimeter of the shaded region.

Question 30.
A building is in the form of cylinder surmounted by a hemispherical dome (shown in the adjoining figure). The base diameter of the dome is equal to \(\frac { 2 }{ 3 }\) of the total height of the building. Find the height of the building if it contains \(67\frac { 1 }{ 21 }\) m³ of air.

Answers

Answer 1.
∵ HCF x LCM = 91 x 26
⇒HCF x 182 = 91 x 26
⇒HCF = \(\frac { 91X26 }{ 182 }\) = 13.

Answer 2.

Now numerator and denominator are co-prime and denominator is of the form 2^m x 5ⁿ. Hence, given rational number is terminating.

Answer 3.
Given 1 is a zero of p(x) = ax² – 3 (a – 1) x – 1
⇒ p(1) = 0 ⇒ a x 1² – 3(a – 1) x 1 – 1 = 0
⇒ a – 3a + 3 – 1 = 0
⇒ – 2a = – 2 ⇒ a = 1

Answer 4.
Given a_n = 7 – 4n ⇒a₂ = 7 – 4 x 2 = -1 and a₁ = 7 – 4 x 1 = 3
∴ common difference d = a₂ – a₁ = -1 – 3 = -4

Answer 5.

Area of ∆ABC : Area of ∆ADE = 9:1

Answer 6.
PC = PA – CA = PB – CQ = (10 – 2) cm = 8 cm (∵ PA = PB and CA = CQ)

Answer 7.
For infinite number of solutions
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)

Answer 8.
4√3 x² + 5x – 2√3 = 0
=> 4√3 x² + 8x – 3x – 2√3 = 0
=> 4x(√3x + 2) – √3(√3x + 2) = 0
=> (√3x + 2) (4x – √3 ) = 0
=> √3x + 2 = 0 or 4x – √3 = 0
\(x=-\frac { 2 }{ \sqrt { 3 } } or\frac { \sqrt { 3 } }{ 4 } \)

Answer 9.
Total number of possible outcomes = 5 + 8 + 7 = 20
(i) Number of favourable outcomes = number of white balls or number of green balls
= 7 + 8 = 15
∴ P (a white or a green ball) = \(\frac { 15 }{ 20 } =\frac { 3 }{ 4 } \)
(ii) Number of favourable outcomes = number of white balls = 7
∴ P (neither green nor red ball) = \(\frac { 7 }{ 20 }\)

Answer 10.
\(\frac { cos{ 70 }^{ o } }{ sin{ 20 }^{ o } } \) + cos 57° . cosec 33° – 2 cos 60°.

Answer 11.
Given, ∆ABC is an isosceles triangle with AB = AC.
⇒ ∠C = ∠B (angles opp. equal sides of a ∆ are equal)
In ∆ABD and ∆ECF,
∠ABD = ∠ECF (∵ ∠B = ∠C, proved above)
and ∠ADB = ∠EFC (each = 90°)
∴ ∆ABD ~ ∆ECF (by AA similarity criterion)
∴ \(\frac { AB }{ EC } =\frac { AD }{ EF } \) ⇒ AB x EF = AD x EC, as required.

Answer 12.
Let r₁ cm and r₂ cm be the radii of two circles, then
r₁ + r₂ = 10 ….(i)

If r₁ = 7, then r₂ = 3 and if r₁ = 3, then r₂ = 7.
Hence, the radii of two circles are 7 cm and 3 cm.

Answer 13.

Answer 14.
(1 + cot A – cosec A) (1 + tan A + sec A)

Answer 15.
Given

Answer 16.
Let f(x) = 4x² + 4x – 3, then

Answer 17.
Steps of construction:
1. Draw AB = 5 cm.
2. At the point A draw ∠BAX = 45°.
3. From AX cut off AC = 6 cm.
4. Join BC. ∆ABC is formed with given data.
5. Draw \(\overrightarrow { AL } \) making acute angle with AB as shown in the figure.
6. Draw 6 arcs at P1 P2, P3, P4, P5 and P6 such that AP1 = P1P2 = P2P3 = P3P4 = P4P5 = P5P6.
7. Join BP5.
8. Draw B’P6 || BP5 meeting AB produced at B’.
9. From B’, draw B’C’ || BC meeting AX at C’.
∆AB’C’ ~ ∆ABC.

Answer 18.
Since the given points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear, the area of triangle formed by them is zero

Answer 19.

∴AB² + BC² = AC² and AB = BC
Hence ABC is a right angled isosceles triangle , right angled at B

Answer 20.
Cards have numbers 10 to 99 (both inclusive) i.e. 10, 11, 12, … 99.
∴ Total number of cards = 90 (99 – 9 = 90)
(i) The numbers from 10 to 99 which are divisible by 10 are 10, 20, 30, …, 90
The number of’such numbers = 9.
∴ Required probability = \(\frac { 9 }{ 90 } =\frac { 1 }{ 10 } \)
(ii) The numbers from 10 to 99 which are perfect squares are 16, 25, 36, 49, 64, 81.
The number of such numbers = 6.
∴ Required probability = \(\frac { 6 }{ 90 } =\frac { 1 }{ 15 } \)
(iii) The numbers from 10 to 99 which are prime numbers less than 25 are 11, 13, 17, 19, 23.
The number of such numbers = 5.
∴ Required probability = \(\frac { 5 }{ 90 } =\frac { 1 }{ 18 } \)
OR
After removing king, queen and jack of clubs, the number of cards left = 49
(i) The number of cards of hearts left in the remaining cards = 13
∴ P (a card of hearts) = \(\frac { 13 }{ 49 }\)
(ii) The number of cards of queens left in the remaining cards = 13
∴ P (a card of queen) = \(\frac { 3 }{ 49 }\)
(iii) The number of cards of clubs left in the remaining cards = 10
∴ P (a card of clubs) = \(\frac { 10 }{ 49 }\)

Answer 21.

Answer 22.
The maximum class frequency is 10 and the class corresponding to this frequency is 6000 – 8000. So, the modal class is 6000 – 8000.
Here, l = 6000, h = 2000, f₁ = 10, f₀ = 8 and f₂ = 2

∴ Mode = 6400

Answer 23.
Given n is any positive integer. Applying Euclid’s lemma with divisor = 6, we get
n = 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4 or 6q + 5, where q is some whole number.
Six cases arise.
Case I.
If n = 6q, then
n³ – n = n(n – 1) (n + 1) = 6q(6q – 1 )(6q + 1), which is divisible by 6.

Case II.
If n = 6q + 1, then
n³ – n = n(n – 1 )(n + 1) = (6q + 1)(6q)(6q + 2)
= 12q(6q + 1)(3q + 1), which is divisible by 6.

Case III.
If n = 6q + 2, then
n³ – n – n(n – 1 )(n + 1) = (6q + 2)(6q + 1)(6q + 3)
= 6(3q + 1)(6q + 1)(2q + 1), which is divisible by 6.

Case IV.
If n – 6q + 3, then
n³ – n = n(n – 1 )(n + 1) = (6q + 3)(6q + 2)(6q + 4)
= 12(2q + 1)(3q + 1)(3q + 2), which is divisible by 6.

Case V.
If n = 6q + 4, then
n³ – n = n(n – 1 )(n + 1) = (6q + 4)(6q + 3)(6q + 5)
= 6(3q + 2)(2q + 1)(6q + 5), which is divisible by 6.

Case VI.
If n = 6q + 5, then
n³ – n = n(n – 1 )(n + 1) = (6q + 5)(6q + 4)(6q + 6)
= 12(69 + 5)(3q + 2)(q + 1), which is divisible by 6.
Thus, in all cases, n³ – n is divisible by 6.
Hence, for every positive integer n, n³ – n is divisible by 6.

Answer 24.
Let the first cash prize be Rs x.
It is given that each prize is Rs 20 less than its preceding prize.
These cash prizes form an AP with a = x, d = – 20, S_n = 1600 and n = 10.

⇒ 320 = 2x – 180
⇒ 2x = 500
⇒ x = 250.
Therefore, the prizes are 250, 250 – 20, 250 – 2 × 20, 250 – 3 × 20, 250 – 4 × 20, 250 – 5 × 20,
i.e., 250, 230, 210, 190, 170, 150, 130, 110, 90, 70
Hence, the values of cash prizes are
Rs 250, Rs 230, Rs 210, Rs 190, Rs 170, Rs 150, Rs 130, Rs 110, Rs 90 and Rs 70.
Academic prize generates spirit of competition. One starts doing hard work which leads to success.

Answer 25.
Let the speed of the stream be x km/h.
Given, speed of the motor boat in still water = 20 km/h.
Then, the speed of the boat upstream = (20 – x) km/h
and the speed of the bpat downstream = (20 + x) km/h
Time taken by the motor boat to cover 48 km upstream = \(\frac { 48 }{ 20-x }\) hours,
time taken by the motor boat to cover 48 km downstream = \(\frac { 48 }{ 20+x }\) hours.
According to given, \(\frac { 48 }{ 20-x }\) – \(\frac { 48 }{ 20+x }\) = 1
⇒ 48(20 + x) – 48(20 – x) = (20 – x) (20 + x) ⇒ 48(2x) = 20² – x²
⇒ x² + 96x – 400 = 0
⇒ (x – 4) (x + 100) = 0
⇒ x = A or x = – 100 but x being the speed of the stream cannot be negative
⇒ x = 4
Hence, the speed of the stream is 4 km/h.
OR
Let the number of books the shopkeeper bought be x.
Given, total cost of books = Rs 80
Then cost of one book = \(\frac { 80 }{ x }\)
When he had bought 4 more books for same amount i.e. Rs 80, then cost of one book = \(\frac { 80 }{ x+4 }\)
According to given \(\frac { 80 }{ x }\) – \(\frac { 80 }{ x+4 }\) = 1
⇒ 80(x + 4) – 80x = x(x + 4)
⇒ 320 = x² + 4x
⇒ x² + 4x – 320 = 0
⇒(x + 20) (x- 16) = 0
⇒ x = 16 or x = – 20 but x being the number of books cannot be negative
⇒ x = 16
Hence, the number of books he bought = 16.

Answer 26.
Given. ABC is a right triangle right angled at A so that BC is its hypotenuse.
To prove. BC² = AB² + AC².
Construction. From A, draw AD ⊥ BC.
Proof. In ∆DBA and ∆ABC,

∠ABD = ∠ABC (same angle)
and ∠ADB = ∠BAC (each = 90°)
∴ ∆DBA ~ ∆ABC (AA similarity criterion)
\(\frac { AB }{ BC } =\frac { BD }{ AB } \) => AB² = BD x BC …(i)
In ∆DAC and ∆ABC,
∠ACD = ∠ACB (same angle)
and ∠ADC = ∠BAC (each = 90°)
∴ ∆DAC ~ ∆ABC (AA similarity criterion)
\(\frac { AC }{ BC } =\frac { DC }{ AC } \) => AC² = DC x BC …(ii)
On adding (i) and (ii), we get
AB² + AC² = BD x BC + DC x BC
= (BD + DC) x BC = BC x BC
=> AB² + AC² = BC².
Hence, BC² = AB² + AC².

Answer 27.
Join OP, OQ and OC. Mark the angles as shown in the figure.

In ∆OAC and ∆OAP,
OA = OA (common)
OC = OP (radii of some circle)
AC = AP (tangent drawn from A)
∴ ∆OAC ≅ ∆OAP (by SSS rule of congruency)
∴∠1 = ∠2 => ∠PAC = 2∠1 …(i)
Similarly,
∴∆OBC ≅ ∆OBQ,
∴ ∠3 = ∠4 => ∠QBC = 2∠3 …(ii)
As XY || X’Y’ and AB is a transversal,
∠PAC + ∠QBC = 180° (sum of co-int ∠S)
=> 2∠1 + 2∠3 = 180° [using (i) and (ii)]
=> ∠1 + ∠3 = 90° …(iii)
In ∆OAB, ∠AOB + ∠1 + ∠3 = 180° (sum of angles of a ∆)
=> ∠AOB + 90° = 180° (using (iii))
=> ∠AOB = 90°.
OR

Let the line BD meet the bigger circle at E. Join AE. Let O be the centre of two concentric circles.
As AB is a diameter of the bigger circle, O is mid-point of AB.
BD is tangent to the smaller circle and OD is radius of smaller circle through the point of contact D, OD ⊥ BE.
Since BE is a chord of the bigger circle and OD ⊥ BE,
BD = DE
(∵ Perpendicular from the centre to a chord bisects it)
=> D is mid-point of BE
∴ OD = \(\frac { 1 }{ 2 }AE\)
(∵ Segment joining the mid-points of any two sides of a triangle is half of the third side)
=> AE = 2 OD => AE = (2 x 8) cm = 16 cm (∵ OD = 8 cm)
In ∆OBD; ∠ODB = 90°, by Pythagoras theorem, we get
OB² = BD² + OD² => 13² = BD² + 8² (∵ OB is radius of bigger circle, so OB = 13 cm)
=> BD² = 169 – 64 = 105 => BD = √105 cm
But DE = BD => DE = √105 cm
Now ∠AEB = 90° (∵ angle in a semicircle = 90°)
=> ∠AED = 90°
In ∆ADE; ∠AED = 90°, by Pythagoras theorem, we get
AD² = AE² + DE² => AD² = 16² + (√105)² = 256 + 105 = 361
=> AD = √361 cm => AD = 19 cm

Answer 28.
Let AB be the tower of height h (units).
AP = a, QA = b.
As the angles of elevation are complementary, if ∠APB = θ° then ∠AQB = 90° – θ°

Let CD be a vertical tower of height h metres. From a point A on the ground, the angle of elevation of the top is 60°. B is another point 10 m vertically above A and angle of elevation of the top of the tower from B is 45°.
Let AC = d metres
From B, draw BE ⊥ CD.
Then BE = AC and EC = BA = 10 m.
From right angled ∆ADC, we get

Answer 29.
(i) Radius of the circle = 3.5 cm = \(\frac { 7 }{ 2 }\)
∴ Area of the quadrant OACB = \(\frac { 1 }{ 4 }\) x πr²

Answer 30.
Let the radius of the spherical dome be r metres and the total height of the building be h metres.
Since the base diameter of the dome is \(\frac { 2 }{ 3 }\) of the total height of the building, therefore,

Hence , the height of the building = 6 m

We hope the CBSE Sample Papers for Class 10 Maths Paper 1 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 1, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 6 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 6. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Sample Papers for Class 10 Maths Paper 6

Board	CBSE
Class	X
Subject	Maths
Sample Paper Set	Paper 6
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 6 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.
If \(\frac { p }{ q }\) is a rational number (q ≠ 0), what is the condition on q so that the decimal representation of \(\frac { p }{ q }\) is terminating?

Question 2.
Find the value of k so that the following system has no solution:
3x – y – 5 = 0; 6x – 2y – k = 0.

Question 3.
Find the 10th term of the AP √2, √8, √18 , … .

Question 4.
In the adjoining figure, if ∠ATO = 40°, find ∠AOB.

Question 5.
If cos A = \(\frac { 3 }{ 5 }\), find 9 cot² A – 1.

Question 6.
In the adjoining figure, ∠M = ∠N = 46°. Express x in terms of a, b and c, where a, b and c are lengths of LM, MN and NK respectively.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
For any natural number n check whether 6ⁿ end with digit 0.

Question 8.
Simplify:

Question 9.
If A, B and C are the interior angles of a triangle ABC, show that:

Question 10.
The angles of a cyclic quadrilateral ABCD are ∠A = (6x + 30)°, ∠B = (5x)°, ∠C = (x + 10)° and ∠D = (3y – 10)°. Find x and y.

Question 11.
Find the middle term(s) of the following AP:
213, 205, 197, …, 37.

Question 12.
The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10 √2 units.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
If the roots of the equation (a – b) x² + (b – c) x + (c – a) = 0 are equal, prove that 2a = b + c.

Question 14.
The sum of first 7 terms of an AP is 63 and sum of its next 7 terms is 161. Find 28th term of AP.
OR
Find the sum of all multiples of 8 lying between 201 and 950.

Question 15.
Show that the square of any positive integer cannot be of the form 5m + 2 or 5m + 3 for some integer m.

Question 16.
In the adjoining figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE, and hence find the lengths of AE and DE.

Question 17.
In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC² + BD² = AD² + BC².

Question 18.
The average score of boys in the examination of a school is 71 and that of the girls is 73. The average score of the boys and girls in the examination is 71.8. Find the ratio of number of boys to the number of girls who appeared in the examination.
OR
Given below is the distribution of weekly pocket money received by students of a class. Calculate the pocket money that is received by most of the students.

Question 19.
If sec θ+ tan θ = p, prove that sin θ = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)
OR
Evaluate the following:
sec 41° sin 49° + cos 49° cosec 41° – \(\frac { 2 }{ \sqrt { 3 } } \) tan 20° . tan 60° . tan 70° – 3(cos² 45° – sin² 90°)

Question 20.
In the adjoining figure, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14)

Question 21.
In the adjoining figure, find the area of the shaded region, . enclosed between two concentric circles of radii 7 cm and 14 cm, where ∠AOC = 40°.

OR
In the adjoining figure, APB and AQO are semicircles and OA = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region.

Question 22.
Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle whose sides are \(\frac { 3 }{ 5 }\) times the corresponding sides of the given triangle. 5

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
In the adjoining figure, the vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that \(\frac { AD }{ AB } =\frac { AE }{ AC } =\frac { 1 }{ 3 } \). Calculate the area of ∆ADE and compare it with area of ∆ABC.

OR
Find the point on the x-axis which is equidistant from the points (5, 4) and (-2, 3). Also find the area of the triangle formed by these points.

Question 24.
If α and β are the zeroes of the polynomial p(x) = 2x² + 5x + k satisfying the relation α² + β² + αβ = \(\frac { 21 }{ 4 }\), then find the value of k. Also find the zeroes of the polynomial p(x).
OR
If the remainder on division of x³ + 2x² + kx + 3 by x – 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x³ + 2x² + kx – 18.

Question 25.
A number x is selected from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is
(i) less than 16
(ii) not less than 16.

Question 26.
Due to heavy floods in a State, thousands were rendered homeless. 50 schools collectively offered to the State Government to provide place and canvas for 1500 tents to be fixed by the Government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs 120 per sq. m, find the amount shared by each school to set up the tents.
What value is generated by the above problem?

Question 27.
The median of the following data is 525. Find the values of x and y, if the total frequency is 100.

Question 28.
A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m (shown in the adjoining figure). Find the dimensions of the rectangular park.

Question 29.
In the adjoining figure, two equal circles with centres O and O’, touch each other at X. OO’ produced meets the circle with O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of \(\frac { DO’ }{ CO }\)

OR
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Question 30.
A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird. (Take √3 = 1.732).

Answers

Answer 1.
q is of the form 2^m x 5ⁿ, where m, n are non-negative integers, where p, q are coprime.

Answer 2.
Given: 3x – y – 5 = 0
6x – 2y – k – 0
For no solution, we have

Answer 3.
Given AP is √2, √8, √18 ,……..
i.e. √2, 2√2, 3√2,……
∴ a = √2, d = √2
T₁₀ = a + (10 – 1 )d
= √2 + 9√2 = 10√2 = √200
Hence, 10th term of given AP is √200 .

Answer 4.
Given ∠ATO = 40°
and ∠OAT = 90° (tangent is perpendicular to the radius)
∴ ∠AOT = 180° – (90° + 40°)
= 50°
∠AOB = 2∠AOT = 2 x 50° = 100°

Answer 5.
∵ cos A = \(\frac { 3 }{ 5 }\)

Answer 6.
In ∆KPN and ∆KLM,
∠N = ∠M = 46° (given)
∠K = ∠K (common)
∴ ∆KPN ~ ∆KLM

Answer 7.
6ⁿ can end with digit 0 only if 6ⁿ is divisible by 2 and 5 both. .
But prime factors of 6ⁿ = 2ⁿ x 3ⁿ, so 6ⁿ is not divisible by 5.
∴ By Fundamental Theorem of Arithmetic, there is no natural number n for which 6ⁿ ends with digit zero.
Hence, 6ⁿ does not end with digit zero

Answer 8.

Answer 9.
As A, B and C are the interior angles of ∆ABC, A + B + C = 180

Answer 10.
In cyclic quadrilateral ABCD,
∠A + ∠C = 180°
=> (6x + 30)° + (x + 10)° = 180°
=> 7x = 140 => x = 20 …(i)
and ∠B + ∠D = 180°
=> (5x)° + (3y – 10)° = 180°
=> 5x + 3y = 190
=> 5 x 20 + 3y = 190 => 3y = 90 => y = 30 (using (i))
x = 20, y = 30.

Answer 11.
Given AP is 213, 205, 197, … 37.
Here, a = 213, d = -8, l = 37
∵ l = a + (n – 1 )d
=> 37 = 213 + (n – 1) (-8) => n – 1 = \(\frac { -176 }{ -8 }\) => n – 1 = 22
=> n = 23
Total number of terms = 23

Answer 12.
Radius of circle = \(\frac { 1 }{ 2 }\) x 10√2 = 5√2 .
The centre of the circle is C(2a, a – 7) and it passes through the point P(11, -9), so CP = radius of circle

Answer 13.
The given equation is (a – b)x² + (b – c) x + (c – a) = 0.
Comparing it with Ax² + Bx + C = 0, we get
A = a – b, B = b – c, C = c – a.
Discriminant = B² – 4AC = (b – c)² – 4(a – b) (c – a).
For equal roots, discriminant = 0
=> (b – c)² – 4(a – b) (c – a) = 0
=> b² – 2bc + c² – 4(ca – a² – bc + ab) = 0
=> 4a² + b² + c² – 4ab – 4ca + 2bc = 0
=> (2a – b – c)² = 0
=>2a – b – c = 0
=> 2a = b + c.

Answer 14.
Let the first term and common difference of AP be a and d respectively.
Given, S₇ = 63 => \(\frac { 7 }{ 2 }\) [2a + (7 – 1 )d] = 63
=> 2a + 6d = 18 …(i)
and sum of next 7 terms = 161
i.e. S₁₄ – S₇ = 161
=> S₁₄ = 161 + 63 => S14 = 224
=> \(\frac { 14 }{ 2 }\) [2a + (14 – 1 )d] = 224
=> 2 a + 13d = 32 … (ii)
Subtracting (i) from (ii), we get
2a + 13d – 2a – 6d = 32 – 18 => 7d = 14 => d = 2.
Substituting d = 2 in equation (i), we get
2A + 6 x 2 = 18 => 2a – 6 => a = 3.
Now, T₂₈ = 3 + (28 – 1) x 2 = 3 + 54 = 57.
Hence, 28th term of AP is 57.
OR
The multiples of 8 lying between 201 and 950 are 208, 216, 224, …, 944.
These numbers form are AP with a = 208, d = 8 and l = 944.
Let the number of these numbers be n, then
944 = 208 + (n – 1) x 8 => 736 = 8 (n – 1)
=> 92 = n – 1 => n = 93.
Sum of these integers = \(\frac { 93 }{ 2 }\) (208 + 944) = \(\frac { 93 }{ 2 }\) x 1152
= 93 x 576 = 53568.

Answer 15.
Let n be any positive integer. Applying Euclid’s division lemma with divisor = 5, we get
n = 5q, 5q + 1, 5q + 2, 5q + 3 or 5q + 4, where q is some whole number.
Now (5q)² – 25q² = 5 m, where m = 5q², which is an integer;
(5q + 1)² = 25q² + 10q + 1 = 5(5q² + 2q) + 1 = 5m + 1, where
m = 5q² + 2q, which is an integer;
(5q + 2)² = 25q² + 20q + 4 = 5(5q² + 4q) + 4 = 5m + 4, where
m = 5q² + 4q, which is an integer;
(5q + 3)² = 25q² + 30q + 9 = 5(5q² + 6q + 1) + 4 = 5m + 4, where
m = 5q² + 6q + 1, which is an integer;
(5q + 4)² = 25q² + 40q + 16 = 5(5q² + 8q + 3) + 1 = 5m + 1, where
m = 5q² + 8q + 3, which is an integer.
Thus, the square of any positive integer is of the form 5m, 5m + 1 or 5m + 4 for some integer m.
It follows that the square of any positive integer cannot be of the form 5m + 2 or 5m + 3 for some integer m.

Answer 16.
In ∆ABC and ∆ADE
∠BAC = ∠EAD (same angle)
∠ACB = ∠AED (each 90°)
∴ ∆ABC ~ ∆ADE (AA Similarity criterion)

Answer 17.

Given, in quadrilateral ABCD,
∠A + ∠D = 90° …(i)
Produce AB and DC to meet at the point E.
In ∆AED, ∠A + ∠D + ∠E = 180°
=> 90° + ∠E = 180° (using (i))
=> ∠E = 90°.
In ∆AED, ∠AED = 90°
∴ AD² = AE² + ED² …(ii)
In ∆BEC, ∠BEC = 90°
∴ BC² – BE² + EC² …(iii)
In ∆AEC, ∠AEC = 90°
∴ AC² = AE² + EC² …(iv)
In ∆BED, ∠BED = 90°
∴ BD² = BE² + ED² …(v)
On adding (iv) and (v), we get
AC² + BD² = (AE² + ED²) + (BE² + EC²)
=> AC² + BD² = AD² + BC² (using (ii) and (iii))

Answer 18.
Let the number of boys who appeared in the examination be x and that of the girls be y, so the total number of students who appeared in the examination = x + y.
Since the average score is boys in the examination is 71 and that of the girls is 73, therefore, the sum of scores of boys = 71x and the sum of scores of girls = 73y.
∴ The total sum of scores of boys and girls = 71x + 73y.
The average score of the boys and girls

Hence, the pocket money received by most of the students is Rs 86.32

Answer 19.
Given sec θ + tan θ = p

Answer 20.
In ∆ABC, ∠ACB = 90°. By Pythagoras theorem,
AB² = AC² + BC²
=> 13² = 12² + BC² (given AB = 13 cm and AC = 12 cm)
=> BC² = 25 => BC = 5 cm

Answer 21.
Let R and r be the radii of the outer and inner concentric circles then R = 14 cm and r = 7 cm.
The central angle made by the major sector of the outer and inner circles θ = 360° – 40° = 320°.
∴ Area of shaded region = area of major sector OAC – area of major sector OBD

Answer 22.
Steps of construction:
1. Draw AB = 6.5 cm.
2. With A as centre and radius 5.5 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 a triangle with AB = 6.5 cm, AC = 5.5 cm and BC = 5 cm.
5. Draw any ray AX making an acute angle with AB on the side opposite to the vertex C.
6. Locate 5 points A1, A2, A3, A4 and A5 on AX such that AA1 = A1 A2 = A2A3 = A3A4 = A4A5.
7. Join A5B. Through A3 draw a line parallel to A5B to intersect AB at B’.
8. Through B’ draw a line parallel to BC to intersect the AC at C’. Then AB’C’ is the required triangle.

Answer 23.
Given the vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2).

Answer 24.
Given α and β are the zeroes of the polynomial p(x) = 2x² + 5x + k.
∴ Sum of the zeroes = α + β = \(\frac { -5 }{ 2 }\) …(i)
and product of the zeroes = αβ = \(\frac { k }{ 2 }\) …(ii)

Answer 25.
For the product of outcomes, we construct a table as under:

Total number of outcomes are 4 x 4 i.e. 16 which are all equally likely.
(i) Table shows that the outcomes favourable to the event ‘product xy less than 16’ are (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (3, 1), (3, 4) and (4, 1) i.e. products as 1, 2, 3, 4, 4, 8, 9, 12 which are 8 in numbers.
∴ Required probability = \(\frac { 8 }{ 16 }\) = \(\frac { 1 }{ 2 }\)
(ii) P(Product of x and y is not less than 16)
= 1 – P(Product of x and y is less than 16).
= \(1-\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\).

Answer 26.
Let r and h be the radius and height of the cylindrical
part and l be the slant height of conical part.
Given r = 2.8 m, l = 3.5 m, height of cone = 2.1 m

Answer 27.
Construct the cumulative frequency distribution table as under:

It is given n = 100, so 76 + x + y = 100 => x + y = 24 …(i)
As the median is 525, which lies in the class 500 – 600, so the median class is 500 – 600.
∴I = 500, f = 20, c.f. = 36 + x and h = 100

=> 5(14 – x) = 25 => 14 – x = 5 =>x = 9
∴ From (i), 9 + y = 24 => y = 15.
Hence, x = 9 and y = 15.

Answer 28.

Let the breadth of the rectangular park be x metres, then its length = (x + 3) metres.
∴ Area of rectangular park = (x + 3) x m²
Base of isosceles triangle = x metres and its height = 12 m.
∴ Area of isosceles triangle = \(\left( \frac { 1 }{ 2 } \times x\times 12 \right) \) m² = 6x m².
According to given, (x + 3) x = 6x + 4
=> x² + 3x = 6x + 4 => x² – 3x – 4 = 0
=> x² – 4x + x – 4 = 0 => x(x – 4) + 1 (x – 4) = 0
=> (x – 4) (x + 1) = 0 =>x – 4 = 0 or x + 1 = 0
=> x = 4 or x = -1.
Since x is the breadth of the rectangular park, it cannot be negative.
Hence, the breadth of the park is 4 metres and its length = (x + 3) metres
= (4 + 3) metres = 7 metres.

Answer 29.
As AC is tangent to the circle with centre O at the point C and OC is radius,
OC ⊥ AC => ∠ACO = 90°.

But these are corresponding angles, therefore, AB || PQ.
Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Answer 30.
Let AB be the tree and B be the position of bird.
From a point C on the ground the angle of elevation of bird is 45°. Bird flies away horizontally and reach the point D in 2 seconds. The angle of elevation from point to the bird is now 30°.
Let AC = x m and AE = y m.

We hope the CBSE Sample Papers for Class 10 Maths Paper 6 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 6, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 10 Maths Paper 7 is part of CBSE Sample Papers for Class 10 Maths . Here we have given CBSE Sample Papers for Class 10 Maths Paper 7. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

CBSE Sample Papers for Class 10 Maths Paper 7

Board	CBSE
Class	X
Subject	Maths
Sample Paper Set	Paper 7
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 7 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.
If the sum of zeroes of the quadratic polynomial 3x² – kx + 6 is 3, then find the value of k.

Question 2.
If 1 is a root of both the equations ay² + ay + 3 = 0 and y² + y + b = 0, then find the value of ab.

Question 3.
Consider the following distribution, find the frequency of class 30-40.

Question 4.
Cards marked with number 3, 4, 5, ..50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.

Question 5.
In which quadrant the point P that divides the line segment joining the points A(2, -5) and B(5, 2) in the ratio 2 : 3 lies?

Question 6.
If sec 2A = cosec (A – 27°), where 2A is an acute angle, find the measure of ∠A.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
Find whether the following pair of linear equations is consistent or inconsistent:
2x – 3y = 8; 4x – by = 9.

Question 8.
The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, -5) and R(-3, 6), find the coordinates of P.

Question 9.
In the adjoining figure, AP and BP are tangents to a circle with centre O, such that AP = 5 cm and ∠APB = 60°, find the length of chord AB.

Question 10.
How many terms of the AP 18, 16, 14, … be taken so that their sum is zero?

Question 11.
Show that the mode of the sequences obtained by combining the two sequences S1 and S2 given below is different from that of S1 and S2 taken separately:
S1 : 3, 5, 8, 8, 9, 12, 13, 9, 9
S2 : 7, 4, 7, 8, 7, 8, 13.

Question 12.
In the adjoining figure, ABC and DBC are two right triangles. Prove that AP x PC = BP x PD.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
Solve the following pair of equations
49x + 51y = 499
51x + 49y = 501.
OR
Sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number.

Question 14.
Divide 56 in four parts in AP such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5 : 6.

Question 15.
D and E are points on the sides AB and AC respectively of ∆ ABC such that DE is parallel to BC, and AD : DB = 4:5. CD and BE intersect each other at F. Find the ratio of the areas of ∆DEF and ∆CBF.

Question 16.
Find the area of ∆PQR with Q(3, 2) and the mid-points of the sides through Q being (2,-1) and (1, 2).
OR
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

Question 17.
Prove the following identity:

Question 18.
Three alarm clocks ring at intervals of 4, 12 and 20 minutes respectively. If they start ringing together, after how much time will they next ring together?

Question 19.
Prove that 3 + 2√3 is an irrational number.

Question 20.
In the adjoining figure, ABCD is a square of side 14 cm. Semicircles are drawn with side of square as diameter. Find the area of the shaded region.

Question 21.
In the adjoining figure shows two arcs PAQ and PBQ. Arc PAQ is a part of a circle with centre O and radius OP while arc PBQ is a semicircle drawn on PQ as diameter. If OP = PQ = 10 cm, show that the area of the shaded region is 25 (√3 – \(\frac { \pi }{ 6 } \))cm².

OR
Sides of a triangular field are 15 m, 16 m, 17 m. With the three comers of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

Question 22.
Two different dice are tossed together. Find the probability that the product of the two numbers on the top of dice is
(i) 6
(ii) a perfect square number.
OR
A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a trader, will only accept the shirts which are good, but Kewal, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that
(i) it is acceptable to Ramesh?
(ii) it is acceptable to Kewal?

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
If α and β are the zeroes of the polynomial p(x) = 3x² + 2x + 1, find the polynomial whose zeroes are

Question 24.
If sec θ – tan θ = x, show that sec θ + tan θ = \(\frac { 1 }{ x }\) and hence, find the values of cos θ and sin θ.
OR
If tan (A + B) = √3 , tan (A – B) = \(\frac { 1 }{ \sqrt { 3 } } \), 0° < A + B < 90°, A > B, find A and B. Also calculate tan A sin (A + B) + cos A tan (A – B).

Question 25.
From the top of a tower h metres high, the angles of depression of two objects, which are in line with the foot of tower are α and β (β>α). Find the distance between the two objects.

Question 26.
Sushant has a vessel of the form of an inverted cone, open at the top, of height 11 cm and radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are put in the vessel due to which \(\frac { 2 }{ 5 }\) th of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds.
What value has been shown by Sushant?

Question 27.
A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m away from the bottom of a pillar, a snake is coming to its hole at the base of a pillar, seeing the snake, the peacock pounces on it. If their speeds are equal, at what distance from the hole is the snake caught?
OR
Find the value(s) of p for which the quadratic equation
(2p + 1) x² – (7p + 2) x + (7p – 3) = 0 has equal roots. Also find these roots.

Question 28.
Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on the outer circle, construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculations.

Question 29.
In the adjoining figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS.

OR
In the adjoining figure, O is the centre of the circle. Determine ∠ACB, if PA and PB are tangents and ∠APB = 50°.

Question 30.
The following table gives the daily income of 50 workers of a factory. Draw both types (less than type and greater than type) ogives. Hence, obtain the median income.

Answers

Answer 1.
Given sum of zeroes of the polynomial 3x² – kx + 6 is 3
⇒ 3 = \(\frac { -(-k) }{ 3 }\) ⇒ k =9

Answer 2.
Given 1 is a root of both the equations => ay² + ay + 3 = 0 and y² + y + b =0
=> a.1² + a.1 + 3 = 0 and 1² + 1 + b = 0
=> a + a + 3 = 0 and 1 + 1 + b = 0
=> 2a + 3 = 0 and 2 + b = 0

Hence the values of ab is 3

Answer 3.

Hence, the frequency of class 30-40 is 3

Answer 4.
Total number of cards = 48
Perfect square number cards from 3 to 50 are 4, 9, 16, 25, 36 and 49. These are 6 in number.
∴ P(Perfect square number) = \(\frac { 6 }{ 48 }\) = \(\frac { 1 }{ 8 }\)

Answer 5.

Answer 6.
sec 2A = cosec (A – 27°) => cosec (90° – 2A) = cosec(A – 27°)
=> 90° – 2A = A – 27° => 3A = 117°
=> A = \(\frac { { 117 }^{ o } }{ 3 } \) => A = 39°
Hence, ∠A = 39°.

Answer 7.
The given pair of linear equations can be written as 2x – 3y – 8 = 0 and 4x – 6y – 9 = 0.

Answer 8.
Given the coordinates of Q(2, -5) and R(-3, 6).
Let the coordinates of P be (2y, y).
According to given, PQ = PR
=> (PQ)² = (PR)²
=> (2 – 2y)² + (- 5 – y)² = (- 3 – 2y)² + (6 – y)²
=> 4 – 8y + 4y² + 25 + 10y + y² – 9 + 12y + 4y² + 36 – 12y + y²
=> 2y + 29 = 45 => 2y = 16 => y = 8
Hence, coordinates of P are (16, 8).

Answer 9.
Given ∠APB = 60° and AP = 5 cm
In ∆APB, PA = PB (tangents drawn from an external point are equal)
∴∠PAB = ∠PBA (angles opposite to equal sides are equal)
Now, ∠PAB + ∠PBA + ∠APB = 180° (angle sum property of a triangle)
=> ∠PAB + ∠PAB + 60° = 180°
=> 2∠PAB = 120° => ∠PAB = 60°
=> ∠PAB = ∠PBA = ∠APB = 60°
∴∆APB is an equilateral triangle.
So, AB = AP = 5 cm.

Answer 10.
Given AP is 18, 16, 14, …
Here, a – 18, d = -2
If the sum of n terms of the given AP is zero, their S_n = 0

But n cannot be zero.
Hence, the required number of terms is 19.

Answer 11.
Given S1 : 3, 5, 8, 8, 9, 12, 13, 9, 9
S2 : 7, 4, 7, 8, 7, 8, 13
In S1 : Number 9 occurs maximum number of times i.e. 3 times
Hence, the mode of sequence S1 = 9
In S2 : Number 7 occurs maximum number of times i.e. 3 times
Hence, the mode of sequence S2 = 7
After combining sequences S1 and S2, we have
3, 4, 5, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 12, 13, 13
Here number 8 occurs maximum number of times i.e. 4 times.
Hence, the mode of combined sequence = 8
Hence, mode of S1 and S2 taken combined is different from that of S1 and S2 taken separately.

Answer 12.
In ∆APB and ∆DPC,
∠BAP = ∠CDP (each 90°)
∠APB = ∠DPC (vert. opp. ∠s)
∴∆APB ~ ∆DPC (by AA similarity criterion)

Answer 13.
Given 49x + 51 y = 499
and 51x + 49y = 501
Adding (i) and (ii), we get
100x + 100y – 1000
=> x + y = 10
Subtracting (ii) from (i), we get
-2x + 2y = -2
=> -x + y = -1
Now, adding (iii) and (iv), we get
2y = 9 => y = \(\frac { 9 }{ 2 }\)
Putting y = \(\frac { 9 }{ 2 }\) in (iii), we get

OR
Let the two digit number be 10x + y, then
according to given, x + y = 8
and | 10x + y – (10y + x) | = 18
=> |9x – 9y| = 18 => |x – y| = 2 =>x – y = ±2
When x + y = 8 and x – y = 2, then
on adding these equations, we get 2x = 10 => x = 5.
Putting x = 5 in x + y = 8, we get y = 3
∴ The original number is 10 x 5 + 3 i.e. 53
When x + y = 8 and x – y = -2, then
on adding these equations, we get 2x – 6 => x = 3
Putting x = 3 in x + y = 8, we get y = 5
∴ The original number is 10 x 3 + 5 i.e. 35
Hence, the original number is 53 or 35.

Answer 14.
Let the four numbers in AP be
a – 3d, a – d, a + d and a + 3d, then
according to given,
a – 3d + a – d + a + d + a + 3d = 56
4a = 56 => a = 14

Answer 15.
Given AD : DB = 4 : 5

Answer 16.
Let D and E are the mid-points of the sides through Q.
Let the coordinates of P and R be (x, y) and (u, v) respectively
∵D is the mid-point of QP

Answer 17.

Answer 18.
To find the time when the clocks will next ring together, we have to find the LCM of 4, 12 and 20.
Prime factorisation 4, 12 and 20
4 = 2 x 2
12 = 2 x 2 x 3
20 = 2 x 2 x 5
∴ LCM of 4, 12 and 20 = 2 x 2 x 3 x 5 = 60
Hence, the alarm clocks will ring together again after 60 minutes i.e. one hour.

Answer 19.
Let us assume that 3 + 2√3 is a rational number, say r.

Answer 20.

Radius of each semicircle = \(\frac { 14 }{ 2 }\) cm = 7 cm
Mark the shaded regions I, II, III and IV (as shown in adjoining figure)
Area of region I + area of region III
= area of square ABCD – area of two semicircles each of radius 7 cm = (14 x 14 – 2.\(\frac { 1 }{ 2 }\) π x 72) cm²
= (196 – \(\frac { 22 }{ 7 }\) x 49) cm²
= (196 – 154) cm² = 42 cm²
Similarly, area of region II + area of region IV = 42 cm²
Required area = area of shaded regions I, II, III and IV
= (2 x 42) cm² = 84 cm²

Answer 21.
OQ = OP (radii of same circle)
Given OP = PQ = 10 cm => OP = OQ = PQ = 10 cm.
Thus, OPQ is an equilateral triangle of side 10 cm.
Also ∠POQ = 60° (angle of an equilateral triangle)
Area of shaded region = area of a semicircle with PQ as diameter i.e. radius 5 cm + area of equilateral AOPQ of side 10 cm – area of sector of a circle of radius 10 cm and central angle 60°

The region of the field left ungrazed by the three animals is shown shaded. Sum of areas of three sectors of radius 7 m (each) and having central angles as ∠A, ∠B and ∠C
= area of a sector of radius 7 m and having central angle as sum of ∠A, ∠B and ∠C
= area of a sector of radius 7 m and having central angle as ∠A + ∠B + ∠C i.e. an angle of 180°

Answer 22.
When two different dice are tossed together, the total number of outcomes is 36 and all the outcomes are equally likely.
(i) The outcomes favourable to the event ‘the product of two numbers on the top of two dice is 6’ are (1, 6), (6, 1), (2, 3) and (3, 2). These are 4 in number.
∴ P (product of two numbers is 6) = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\).
(ii) The outcomes favourable to the event ‘the product of two numbers on the top of dice is a perfect square number’ are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 4), (4, 1).
These are 8 in number.
∴ P (product of two numbers is a perfect square number) = \(\frac { 8 }{ 36 }\) = \(\frac { 2 }{ 9 }\).
OR
One shirt is drawn at random from a carton containing 100 shirts means all shirts are equally likely to be drawn. So, the sample space of the experiment has 100 equally likely outcomes.
(i) Since Ramesh accepts only good shirts and there are 88 good shirts in the carton. Therefore, the number of favourable (i.e. acceptable) outcomes to Ramesh = 88.
∴ P (acceptable to Ramesh) = \(\frac { 88 }{ 100 }\) = 0.88
(ii) As Kewal rejects the shirts which have major defects and the number of such shirts is 4, so the number of shirts acceptable to Kewal = 100 – 4 = 96.
Therefore, the number of favourable (acceptable) outcomes to Kewal = 96.
∴ P (acceptable to Kewal) = \(\frac { 96 }{ 100 }\) = 0.96

Answer 23.
Given a and (3 are the zeroes of the polynomial p(x) = 3x² + 2x + 1

Answer 24.
Given sec θ – tan θ = x
∵ sec² θ – tan² θ = 1
=> (sec θ tan θ) (sec θ + tan θ) = 1
x . (sec θ + tan θ) = 1
=> sec θ + tan θ = \(\frac { 1 }{ x }\) …(ii)
Adding (i) and (ii), we get
sec θ – tan θ + sec θ + tan θ = x + \(\frac { 1 }{ x }\)

Answer 25.
Let P be the top of the tower MP of height h metres, and A, B be the two objects. Let BM = x metres and AB = d metres. The angles of depression of the two objects are shown in the adjoining figure, then
∠MAP = α and ∠MBP = β.
From right angled ∆MBP, we get

Answer 26.
Volume of water in the cone
= Volume of cone of height 11 cm and radius 2.5 cm

Hence, the number of metallic balls put into the vessel = 440
Sushant’s love for plants, nature and environment has been depicted.

Answer 27.

Let CD be the pillar and initially the snake be at the point A and the peacock is sitting at the top of pillar i.e. at the point D. Then CD = 9 m and AC = 27 m.
Let the snake be caught at the point B. As the snake and peacock have equal speeds, they cover equal distance in same time.
Let AB = x metres, then BD = AB = x metres and BC = AC – AB = (27 – x) metres.
In ∆BCD, ∠C = 90°.
By Pythagoras theorem, we get
BD² = BC² + CD²
=> x² = (27 – x)² + 9² => x² = 729 – 54x + x² + 81
=> 54x = 810 => x = 15.
∴ BC = (27 – x) metres = (27 – 15) metres = 12 metres.
Hence, the snake is caught at a distance of 12 metres from its hole.
OR
The given equation is (2p + 1) x² – (7p + 2) x + (7p – 3) = 0 …(i)
Comparing it with ax² + bx + c = 0, we get
a = 2p + 1, b = – (7p + 2), c = 7p – 3.
Discriminant = b² – 4ac = (-(7p + 2))² – 4 (2p + 1) (7p – 3)
= 49p² + 28p + 4 – 4 (14p² – 6p + 7p – 3)
= 49p² + 28p + 4 – 56p² – 4p + 12
= -7p² + 24p + 16.
For equal roots, discriminant = 0
=> -7p² + 24p + 16 = 0 => 7p² – 24p – 16 = 0
=> 7p² – 28p + 4p – 16 = 0 => 7p (p – 4) + 4 (p – 4) = 0
=> (p – 4) (7p + 4) = 0 => p – 4 = 0 or 7p + 4 = 0

Answer 28.
Steps of construction:
1. Draw two concentric circles of radii 3 cm and 5 cm with point O as their centre.
2. Let P be a point on the outer circle. Join OP and draw its perpendicular bisector to meet OP at M.
3. Taking M as centre and OM (or MP) as radius, draw a circle. Let this circle intersect the smaller circle i.e. circle of radius 3 cm at points A and B.
4. Join PA and PB. Then PA and PB are the required tangents on measuring PA (or PB), we find that PA = 4 cm.

Calculation of length PA:
Join OA.
In ∆OAP, ∠OAP = 90° (angle in a semicircle)
By Pythagoras theorem, we get
PA² = OP² – OA² = 5² – 3² = 25 – 9 = 16
=> PA = 4 cm.

Answer 29.

Join QO and produce it to meet SR at M.
In ∆PRQ, PR = PQ (lengths of tangents)
=> ∠PRQ = ∠PQR …(i)
∠PQR + ∠PRQ + ∠RPQ = 180°
=> ∠PQR + ∠PQR + 30° = 180° (using (i))
=> 2∠PQR = 150° => ∠PQR = 75°.
As RS || PQ and RQ is transversal,
∠SRQ = ∠PQR
=> ∠SRQ = 75°
Since PQ is tangent to the circle with centre O at the point Q and OQ is radius, OQ ⊥ PQ.
Also RS || PQ (given), so QM ⊥ RS i.e. OM ⊥ RS
=> MR = MS (∵ perpendicular from centre to a chord bisects it)
In ∆QRM and ∆QSM,
MR = MS (proved above)
∠QMR = ∠QMS (each = 90°, as QM ⊥ RS)
QM = QM
∴∆QRM ≅ ∆QSM
∴ ∠MSQ = ∠MRQ (c.p.c.t.)
=> ∠RSQ = ∠SRQ.
In ∆QRS, ∠RQS + ∠SRQ + ∠RSQ = 180° => ∠RSQ + ∠SRQ + ∠SRQ = 180°
=> ∠RQS + 2 x 75° = 180° => ∠RQS = 30°.
OR

As PA is tangent to the circle at A and OA is radius, OA ⊥ AP
i.e. ∠OAP = 90°
Similarly, ∠OBP = 90°
In quadrilateral OAPB,
∠AOB + ∠OAP + ∠APB + ∠OBP = 360°
=> ∠AOB + 90° + 50° + 90° = 360°
=> ∠AOB = 130°
Reflex ∠AOB = 360° – 130° = 230°
∠ACB = \(\frac { 1 }{ 2 }\) of reflex ∠AOB
(∵ angle at the centre = double the angle at the remaining part of circle)
=> ∠ACB = \(\frac { 1 }{ 2 }\) x 230° = 115°

Answer 30.
For more than ogive:

Take 1 cm along x-axis = Rs 20 and 1 cm along y-axis = 10 workers.
Plot the points (100, 50), (120, 38), (140, 24), (160, 16), (180, 10) and (200, 0).
Join these points by a free-hand drawing. More than type ogive is drawn on the graph sheet. For less than type ogive:

Choose the same scale (as above). Plot the points (100, 0), (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50). Join these points by a free hand drawing. Less than type ogive is drawn on the same graph sheet (with same axes).

The abscissa of point of intersection of more than ogive and less than ogive represents 138 Hence, the median daily income = Rs 138.

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