## NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3.

 Board CBSE Textbook NCERT Class Class 7 Subject Maths Chapter Chapter 8 Chapter Name Comparing Quantities Exercise Ex 8.3 Number of Questions Solved 11 Category NCERT Solutions

## NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

Question 1.
Tell what is the profit or loss in the following transactions. Also find profit percent or loss per cent in each case.

(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
(b) A refrigerater bought for ₹ 12,000 and sold at ₹ 13,500.
(c) Acupboard bought for ₹ 2,500 and sold at ₹ 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.

Solution:

Question 2.
Convert each part of the ratio to percentage

(a) 3 : 1
(b) 2 : 3 : 5
(c) 1 : 4
(d) 1 : 2 : 5

Solution:

Question 3.
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:

Question 4.
Arun bought a car for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of the price increase?
Solution:

Question 5.
1 buy a T.V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Solution:
C.P. = ₹ 10,000
Profit = 20% of C.P.
= 20% of ₹ 10,000
= ₹ $$\frac { 20 }{ 100 }$$ × 10,000 = ₹ 2,000
∴ S.P. = C.P. + Profit
= ₹ 10,000 + ₹ 2,000 = ₹ 12,000
Hence, I get ₹ 12,000 for it.

Question 6.
Juki sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she brought it?
Solution:

Question 7.
(i) Chalk contains calcium, carbon, and oxygen in the ratio of 10 : 3 : 12. Find the percentage of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3 g, what is the weight of the chalk stick?
Solution:

Question 8.
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Solution:
C.P. = ₹ 275
Loss = 15% of C.P = $$\frac { 15 }{ 100 }$$ × C.P
= $$\frac { 15 }{ 100 }$$ × 275
= ₹ 41.25
S.P. = C.P. – Loss = 275 – 41.25 = ₹ 233.75
hence, she se11 it for ₹ 233.75.

Question 9.
Find the amount to be paid at the end of 3 years in each case:

(a) Principal = ₹ 1,200 at 12% p.a.
(b) Principal = ₹ 7,500 aat 5% p.a.

Solution:

Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?
Solution:

Question 11.
If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3, drop a comment below and we will get back to you at the earliest.

## NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2.

 Board CBSE Textbook NCERT Class Class 7 Subject Maths Chapter Chapter 8 Chapter Name Comparing Quantities Exercise Ex 8.2 Number of Questions Solved 10 Category NCERT Solutions

## NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

EXERCISE 8.2

Question 1.
Convert the given fractional numbers to per cents.

(a) $$\frac { 1 }{ 8 }$$
(b) $$\frac { 5 }{ 4 }$$
(c) $$\frac { 3 }{ 40 }$$
(d) $$\frac { 2 }{ 7 }$$

Solution:

Question 2.
Convert the given decimal fractions to per cents.

(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35

Solution:

Question 3.
Estimate what part of the figures is coloured and hence find the percent which is coloured.

Solution:

Question 4.
Find:

(a) 15% of 250
(b) 1% of 1 hour
(c) 20% of ₹ 2500
(d) 75% of 1 kg.

Solution:

Question 5.
Find the whole quantity if

(a) 5% of it is 600.
(b) 12% of it is ₹ 1080.
(c) 40% of it is 500 km.
(d) 70% of it is 14 minutes.
(e) 8% of it is 40 liters.

Solution:

Question 6.
Convert given percents to decimal fractions and also to fractions in the simplest forms.

(a) 25%
(b) 150%
(c) 20%
(d) 5%

Solution:

Question 7.
In a city, 30% are females, 40% are males, and the remaining are children. What percent are children?
Solution:
[100% (30% + 40%)] are children
⇒ [100% – 70% ] are children
⇒ 30% are children.

Question 8.
Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Solution:
Percentage of voters who voted = 60%
Percentage of voters who did not vote = (100 – 60)% = 40%
Total number of voters = 15000
Number of voters who did not vote = 40% of 15000
= ( $$\frac { 40 }{ 100 }$$ × 15000 )
= 6000

Question 9.
Meeta saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?
Solution:
Let her salary be ₹ P. Then, 10% of P = 400
⇒ $$\frac { 10 }{ 100 }$$ × p = 400
⇒ $$\frac { P }{ 10 }$$ = 400
⇒ P = 400 × 10
⇒ P = 4000
Hence, her salary is ₹ 4000.

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Solution:
Out of 100, 25% matches are won. Then, out of 20, the number of matches that the team won
= $$\frac { 25 }{ 100 }$$ × 20
= $$\frac { 1 }{ 4 }$$ × 20
= 5

We hope the NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2, drop a comment below and we will get back to you at the earliest.

## NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2.

 Board CBSE Textbook NCERT Class Class 7 Subject Maths Chapter Chapter 7 Chapter Name Congruence of Triangles Exercise Ex 7.2 Number of Questions Solved 10 Category NCERT Solutions

## NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

Question 1.
Which congruence criterion do you use in the following?

Given:
So.
AC = DF
AB = DE
BC = EF
so ∆ ABC = ∆ DEF

Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ∆ PQR ≅ ∆ XYZ

Given : ∠MLN = ∠ FGH
∠NML = ∠GFH
ML = GF
So, ∆ LMN ≅ ∆ GFH

Given : EB = DB
AE = BC
∠A = ∠C = 90°
So, ∆ ABE ≅ ∆ CDB
Solution:

(a) SSS congruence criterion
(b) SAS congruence criterion
(c) ASA congruence criterion
(d) RHS congruence criterion.

Question 2.
You want to show that ∆ ART ≅ ∆ PEN,

(а) If you have to use SSS criterion, then you need to show
(i)AR = (ii) RT = (iii) AT =

(b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
(i) RT = and (ii) PN =

(c) If it is given that AT = PN and you are to use ASA criterion, you need to have
(i) ? (ii) ?
Solution:

Question 3.
You have to show that ∆ AMP = ∆ AMQ.
In the following proof, supply the missing reasons.

Solution:

Question 4.
In ∆ ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ∆ PQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ∆ ABC = ∆ PQR? by AAA congruence criterion. Is he justified’? Why or why not?
Solution:
No! he is not justified because AAA is not a criterion for congruence of triangles.

Question 5.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ∆ RAT = ?

Solution:
∆ RAT ≅ ∆ WON

Question 6.
Complete the congruence statement:

Solution:
∆ BCA = ∆ BTA
∆ QRS = ∆ TPQ

Question 7.
In a squared sheet, draw two triangles of equal areas such that

1. the triangles are congruent
2. the triangles are not congruent. What can you say about their perimeters?

Solution:

Question 8.
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Solution:

In ∆ ABC and ∆ DEF,
AB = DF (= 2 cm)
BC = ED (= 4 cm)
CA = EF (= 3 cm)
∠BAC = ∠EDF
∠ABC = ∠DEF
But ∆ ABC is not congruent to ∆ DEF.

Question 9.
If ∆ ABC and ∆ PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Solution:
BC = RQ by ASA congruence rule.

Question 10.
Explain why ∆ ABC ≅ ∆ FED

Solution:
∠ABC = ∠FED (= 90°) BC = ED
∠ACB = ∠FDE
∵ The sum of the measures of the three angles of a triangle is 180°.
∆ ABC ≅ ∆ FED (By SAS congruence criterion)

We hope the NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2, drop a comment below and we will get back to you at the earliest.

## NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5.

 Board CBSE Textbook NCERT Class Class 7 Subject Maths Chapter Chapter 6 Chapter Name The Triangle and its Properties Exercise Ex 6.5 Number of Questions Solved 8 Category NCERT Solutions

## NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5

Question 1.
PQR is a triangle right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Solution:
QR2 = 102 + 242 By Pythagoras Property
⇒ = 100 + 576 = 676
⇒ QR = 26 cm.

Question 2.
ABC is a triangle right-angled at C. If AB – 25 cm and AC = 7 cm, find BC.

Solution:
AC2 + BC2 = AB2 By Pythagoras Property
⇒ 72 + BC2 = 252
⇒ 49 + BC2 = 625
⇒ BC2 = 625 – 49
⇒ BC2 = 576
⇒ BC = 24 cm.

Question 3.
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Solution:
Let the distance of the foot of the ladder from the wall be a m. Then,

Hence, the distance of the foot of the ladder from the wall is 9 m.

Question 4.
Which of the following can be the sides of a right triangle ?

1. 2.5 cm, 6.5 cm, 6 cm.
2. 2 cm, 2 cm, 5 cm.
3. 1.5 cm, 2 cm, 2.5 cm.

In the case of right-angled triangles, identify the right angles.
Solution:
1. 2.5 cm, 6.5 cm, 6 cm We see that
(2.5)2 + 62 = 6.25 + 36 = 42.25 = (6.5)2
Therefore, the given lengths can be the sides of a right triangle. Also, the angle between the lengths, 2.5 cm and 6 cm is a right angle.

2. 2 cm, 2 cm, 5 cm
∵ 2 + 2 = 4 $$\ngtr$$ 5
∴ The given lengths cannot be the sides of a triangle
The sum of the lengths of any two sides of a triangle is greater than the third side

3. 1.5 cm, 2 cm, 2.5 cm We find that
1.52 + 22 = 2.25 + 4 = 6.25 = 2.52
Therefore, the given lengths can be the sides of a right triangle.
Also, the angle between the lengths 1.5 cm and 2 cm is a right angle.

Question 5.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Solution:
AC = CD Given
In right angled triangle DBC, DC2 = BC2 + BD2
by Pythagoras Property = 52 + 122 = 25 + 144 = 169

⇒ DC = 13 ⇒ AC = 13
⇒ AB = AC + BC = 13 + 5 = 18
Therefore, the original height of the tree = 18 m.

Question 6.
Angles Q and R of a ∆ PQR are 25° and 65°. Write which of the following is true:

(i) PQ2 + QR2 = RP2
(ii) PQ2 + RP2 = QR2
(iii) RP2 + QR2 = PQ2
Solution:
(ii) PQ2 + RP2 = QR2 is true.

Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Solution:
In right-angled triangle DAB, AB2 + AD2 = BD2
⇒ 402 + AD2 = 412 ⇒ AD2 = 412 – 402
⇒ AD2 = 1681 – 1600
∴ Perimeter of the rectangle = 2(AB + AD) = 2(40 + 9) = 2(49) = 98 cm
Hence, the perimeter of the rectangle is 98 cm.

Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Solution:
Let ABCD be a rhombus whose diagonals BD and AC are of lengths 16 cm and 30 cm respectively.
Let the diagonals BD and AC intersect each other at O.
Since the diagonals of a rhombus bisect each other at right angles. Therefore
BO = OD = 8 cm,
AO = OC = 15 cm,
∠AOB = ∠BOC
= ∠COD = ∠DOA = 90°
In right-angled triangle AOB.
AB2 = OA2 + OB2
By Pythagoras Property
⇒ AB2 = 152 + 82
⇒ AB2 = 225 + 64
⇒ AB2 = 289
⇒ AB = 17cm
Therefore, perimeter of the rhombus ABCD = 4 side = 4 AB = 4 × 17 cm = 68 cm
Hence, the perimeter of the rhombus is 68 cm.

We hope the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5, drop a comment below and we will get back to you at the earliest.