Category: CBSE Class 7

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.2
• Algebraic Expressions Class 7 Ex 12.3
• Algebraic Expressions Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 12
Chapter Name	Algebraic Expressions
Exercise	Ex 12.4
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4

Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.

If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern.
How many segments are required to form 5, 10, 100 digits of the kind

Solution:

Let the number of digits formed be n, Then, the number of segments required to form n digits is given by the algebraic expression 5n + 1.
So,

1. the number of segments required to form 5 digits of this kind = 5 × 5 + 1 = 25 + 1 = 26
2. the number of segments required to form 10 digits of this kind = 5 × 10 + 1 = 50 + 1 = 51
3. the number of segments required to form 100 digits of this kind = 5 × 100 + 1 = 500 + 1 = 501.

Let the number of digits formed be n. Then, the number of segments required to form n digits is given by the algebraic expression 3n + 1.
So,

1. the number of segments required to form 5 digits of this kind = 3 × 5 + 1 = 15 + 1 = 16
2. the number of segments required to form 10 digits of this kind = 3 × 10 + 1 = 30 + 1 = 31
3. The number of segments required to form 100 digits of this kind = 3 × 100 + 1 = 300 + 301.

Let the number of digits formed be n. Then, the number of segments required to form n digits is given by the algebraic expression 5n + 2.
So,

1. the number of segments required to form 5 digits of this kind = 5 × 5 + 2 = 25 + 2 = 27
2. the number of segments required to form 10 digits of this kind = 5 × 10 + 2 = 50 + 2 = 52
3. the number of segments required to form loo digits of this kind = 5 × 100 + 2 = 500 + 2 = 502.

Question 2.
Use the given algebraic expression to complete the table of number patterns.

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.2
• Algebraic Expressions Class 7 Ex 12.4
• Algebraic Expressions Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 12
Chapter Name	Algebraic Expressions
Exercise	Ex 12.3
Number of Questions Solved	10
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

Question 1.
If m = 2, find the value of :

1. m – 2
2. 3m – 5
3. 9 – 5m
4. 3m² – 2m – 7
5. \(\frac { 5m }{ 2 } \) – 4

Solution:

Question 2.
If p = – 2, find the value of :

1. 4p + 7
2. – 3p² + 4p + 7
3. – 2p³ – 3p² + 4p + 7.

Solution:

Question 3.
Find the value of the following expressions, when x = – 1 :

1. 2x – 7
2. – x + 2
3. x² + 2x + 1
4. 2x² – x – 2.

Solution:

Question 4.
If a = 2, b = – 2, find the value of :

1. a² + b²
2. a² + ab + b²
3. a² – b².

Solution:

Question 5.
When a = 0, b = – 1, find the value of the given expressions :

1. 2a + 2b
2. 2a² + b² + 1
3. 2a²b + 2ab² + ab
4. a² + ab + 2.

Solution:

Question 6.
Simplify the expressions and find the value ifx is equal to 2.

1. x + 7 + 4 (x – 5)
2. 3 (x + 2) + 5x – 7
3. 6x + 5 (x – 2)
4. 4 (2x – 1) + 3x + 11.

Solution:

Question 7.
Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.

1. 3x – 5 – x + 9
2. 2 – 8x + 4x + 4
3. 3a + 5 – 8a + 1
4. 10 – 3b – 4 – 5b
5. 2a – 2b – 4 – 5 + a.

Solution:

Question 8.
(i) If z = 10, find the value of z³ – 3(z – 10).
(ii) If p = -10, find the value of p² – 2p – 100.
Solution:

Question 9.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0 ?
Solution:

Question 10.
Simplify the expression and find its value when a = 5 and b = – 3 2(a² + ab) + 3 – ab.
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.3
• Algebraic Expressions Class 7 Ex 12.4
• Algebraic Expressions Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 12
Chapter Name	Algebraic Expressions
Exercise	Ex 12.2
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify combining like terms:

1. 21b – 32 + 7b – 20b
2. – z² + 13z² – 5z + 7z³ – 15z
3. p – (p – q) – q – (q – p)
4. 3a – 2b – ab – (a – b + ab) + 3ab + b – a
5. 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
6. (3y² + 5y – 4) – (8y – y² – 4).

Solution:

Question 2.
Add:

Solution:

Question 3.
Subtract:

1. -5y² from y²
2. 6xy from – 12xy
3. (a – b) from (a + b)
4. a (b – 5) from b (5 – a)
5. -m² + 5mn from 4m² – 3mn + 8
6. -x² + 10x – 5 from 5x – 10
7. 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
8. 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq.

Solution:

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get -3a + 76 + 16?
Solution:

Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20?
Solution:

Question 6.
(a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2.

• Perimeter and Area Class 7 Ex 11.1
• Perimeter and Area Class 7 Ex 11.3
• Perimeter and Area Class 7 Ex 11.4
• Perimeter and Area Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 11
Chapter Name	Perimeter and Area
Exercise	Ex 11.2
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

Question 1.
Find the area of the following parallelograms:


Solution:

Question 2.
Find the area of each of the following triangles:


Solution:

Question 3.
Find the missing values:

Solution:

Question 4.
Find the missing values:

Solution:

Question 5.
PQRS is a parallelogram (in Figure). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:
(a) the area of the parallelogram PQRS
(b) QN, if PS = 8 cm.

Solution:
(a) Area of the parallelogram PQRS = base × height = SR × QM = 12 × 7.6 cm² = 91.2 cm²
(b) Area of the parallelogram PQRS = base × height = PS × QN
⇒ 91.2 = 8 × QN
⇒ QN = \(\frac { 912 }{ 8 } \) cm
⇒ QN = 11.4 cm.

Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (in Figure). If the area of a parallelogram is 1470 cm2, AB = 35 cm, and AD = 49 cm, find the length of BM and DL.

Solution:

Question 7.
∆ ABC is right-angled at A (in Figure). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of ∆ ABC. Also, find the length of AD.

Solution:

Question 8.
∆ ABC is isosceles with AB = AC = 7.5 cm, and BC = 9 cm (in Figure). The height of AD from A to BC is 6 cm. Find the area of ∆ ABC. What will be the height from C to AB i.e., CE?

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.3
• Practical Geometry Class 7 Ex 10.4
• Practical Geometry Class 7 Ex 10.5

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 10
Chapter Name	Practical Geometry
Exercise	Ex 10.2
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

Question 1.
Construct A XYZ in which XY = 4.5 cm, YZ = 5 cm and, ZX = 6 cm.
Solution:
Steps of Construction

1. Draw a line segment YZ of length 5 cm.
   
2. With Y as centre, draw an arc of radius 4.5 cm.
3. With Z as centre, draw an arc of radius 6 cm,
4. Mark the point of intersection of arcs as X.
5. Join XY and XZ. ∆ XYZ is now ready.

Question 2.
Construct an equilateral triangle of side 5.5 cm.
Solution:
Steps of Construction:

1. Draw a line segment BC of length 5.5 cm.
2. With B as centre, draw an arc of radius 5.5 cm.
   
3. With C as centre, draw an arc of radius 5.5 cm.
4. Mark the point of intersection of arcs as A.
5. Join AB and AC. Equilateral ∆ ABC is now ready.

Question 3.
Draw ∆ PQR with PQ = 4 cm, QR =3.5 cm and PR = 4 cm. What type of triangle is this?
Solution:
Steps of Construction:

1. Draw a line segment QR of length 3.5 cm.
2. With Q as centre, draw an arc of radius 4 cm.
3. With R as centre, draw an arc of radius 4 cm.
   
4. Mark the point of intersection of arcs as P.
5. Join PQ and PR.

∆ PQR is now ready,
∵ PQ = PR
∴ ∆ PQR is isosceles.

Question 4.
Construct ∆ ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Solution:
Steps of Construction

1. Draw a line segment BC of length 6 cm.
   
2. With B as centre, draw an arc of radius 2.5 cm.
3. With C as centre, draw an arc of a radius of 6.5 cm.
4. Mark the point of intersection of arcs as A.
5. Join AB and AC.
6. ∆ ABC is now ready. On measurement, ∠B = 90°.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.2
• Practical Geometry Class 7 Ex 10.4
• Practical Geometry Class 7 Ex 10.5

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 10
Chapter Name	Practical Geometry
Exercise	Ex 10.3
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

Question 1.
Construct ADEF such that DE = 5 cm, DF 3 cm, and m ∠EDF = 90°.
Solution:
Steps of Construction:

1. Draw a line segment DE = 5cm.
2. Draw ∠EDX = 90°.
3. With centre D and radius = 3 cm, draw an arc to intersect DX at F.
4. Join EF to obtain the required triangle DBF.

Question 2.
Construct an isosceles triangle in which the length of each of its equal sides is 6.5 cm and the angle between them is 110°.
Solution:
Steps of Construction

1. Draw a line segment QR of length 6.5 cm.
2. At Q, draw QX making 110° with QR, using a protractor.
   
3. With Q as centre, draw an arc of a radius of 6.5 cm. It cuts QX at P.
4. Join PR. ∆ PQR is now obtained.

Question 3.
Construct ∆ ABC with BC = 7.5 cm, AC = 5 cm and m ∠C = 60°.
Solution:
Steps of Construction:

1. Draw a line segment BC = 7.5 cm.
2. Draw ∠BCX = 60°.
3. With C as centre and radius = 5 cm, draw an arc intersecting CX at A.
4. Join AB to obtain the required ∆ABC.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.2
• Practical Geometry Class 7 Ex 10.3
• Practical Geometry Class 7 Ex 10.5

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 10
Chapter Name	Practical Geometry
Exercise	Ex 10.4
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

Question 1.
Construct ∆ ABC, given m ∠A = 60°, m ∠B = 30° and AB = 5.8 cm.
Solution:
Steps of Construction

1. Draw AB of length 5.8 cm.
2. At A, draw a ray AP making an angle of 60° with AB.
3. At B, draw a ray BQ making an angle of 30° with BA.
4. Mark the point of intersection of two rays as C.
5. ∆ ABC is now completed.

Question 2.
Construct ∆ PQR if PQ = 5 cm, m ∠PQR = 105° and m ∠QRP = 40°.
(Hint: Recall angle-sum property of a triangle).
Solution:
By angle-sum property of a triangle
m ∠RPQ + m ∠PQR + m ∠QRP = 180°
⇒ m ∠RPQ + 105° + 40° = 180°
⇒ m ∠RPQ + 145° = 180°
⇒ m ∠RPQ = 35°
Steps of Construction

1. Draw PQ of length 5 cm.
2. At Q, draw a ray QX making an angle of 105° with QP.
   
3. At P draw a ray PY making an angle of 35° with PQ.
4. Mark the point of intersection of two rays as R.

∆ PQR is now completed.

Question 3.
Examine whether you can construct ∆DEF such that EF = 7.2 cm, m ∠E = 110° and m ∠F = 80°. Justify your answer.
Solution:
m ∠E + m ∠F = 110° + 80° = 190° > 180°
This is not possible since the sum of the measures of the three angles of a triangle is 180°. As such, the sum of two angles of a triangle cannot exceed 180°.
Hence, ∆ DEF cannot be constructed.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.2
• Practical Geometry Class 7 Ex 10.3
• Practical Geometry Class 7 Ex 10.4

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 10
Chapter Name	Practical Geometry
Exercise	Ex 10.5
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

Question 1.
Construct the right-angled ∆ PQR where m ∠Q = 90°, QR = 8 cm and PR = 10 cm.
Solution:
Steps of Construction:

1. Draw QR of length 8 cm.
   
2. At Q, draw QX ⊥ QR.
3. With R as centre, draw an arc of radius 10 cm.
4. Mark the meeting point of these two as P.

∆ PQR is now obtained.

Question 2.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
Solution:
Steps of Construction

1. Draw QR of length 4 cm.
2. At Q, draw QX ⊥ QR.
   
3. With R as centre, draw an arc of radius 6 cm.
4. Mark the meeting point of arc and QX as P.

∆ PQR is now obtained.

Question 3.
Construct an isosceles right-angled triangle ABC where m ∠ACB = 90° and AC = 6 cm.
Solution:
Steps of Construction

1. Draw AC of length 6 cm.
2. At C, draw CX ⊥ CA.
   
3. With C as centre, draw an arc of radius 6 cm to intersect CX at B.
4. Join AB. ∆ACB is now obtained.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2.

• Rational Numbers Class 7 Ex 9.1
• Rational Numbers Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 9
Chapter Name	Rational Numbers
Exercise	Ex 9.2
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2

Question 1.
Find the sum:

Solution:

Question 2.
Find :

Solution:

Question 3.
Find the product :

Solution:

Question 4.
Find the value of :

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Ex 9.2, 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 8.3 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.

• Comparing Quantities Class 7 Ex 8.1
• Comparing Quantities Class 7 Ex 8.2
• Comparing Quantities Class 7 MCQ

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 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.

• Comparing Quantities Class 7 Ex 8.1
• Comparing Quantities Class 7 Ex 8.3
• Comparing Quantities Class 7 MCQ

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.