Selina Class 7 ICSE Solutions Mathematics : Chapter 10- Simple Interest

Class 7: Maths Chapter 10 solutions. Complete Class 7 Maths Chapter 10 Notes.

Selina Class 7 ICSE Solutions Mathematics : Chapter 10- Simple Interest

Selina 7th Maths Chapter 10, Class 7 Maths Chapter 10 solutions

Exercise 10 page: 116

1. Find the S.I. and the amount on:

(i) ₹ 150 for 4 years at 5% per year.

(ii) ₹ 350 for 3 ½ years at 8% p.a.

(iii) ₹ 620 for 4 months at 8 p per rupee per month.

(iv) ₹ 3,380 for 30 months at 4 ½ % p.a.

(v) ₹ 600 from July 12 to Dec. 5 at 10% p.a.

(vi) ₹ 850 from 10^th March to 3^rd August at 2 ½ % p.a.

(vii) ₹ 225 for 3 years 9 months at 16% p.a.

Solution:

(i) ₹ 150 for 4 years at 5% per year

We know that

P = ₹ 150

R = 5% per year

T = 4 years

Here

S.I = (P × R × T)/ 100

Substituting the values

= (150 × 5 × 4)/ 100

= ₹ 30

Amount = P + S.I

Substituting the values

= 150 + 30

= ₹ 180

(ii) ₹ 350 for 3 ½ years at 8% p.a.

We know that

P = ₹ 350

R = 8% p.a.

T = 3 ½ years = 7/2 years

Here

S.I = (P × R × T)/ 100

Substituting the values

= (350 × 8 × 7)/ (100 × 2)

= ₹ 98

Amount = P + S.I

Substituting the values

= 350 + 98

= ₹ 448

(iii) ₹ 620 for 4 months at 8 p per rupee per month

We know that

P = ₹ 620

R = 8 p per rupee per month = 8% p.m.

T = 4 months

Here

S.I = (P × R × T)/ 100

Substituting the values

= (620 × 8 × 4)/ 100

= ₹ 198.40

Amount = P + S.I

Substituting the values

= 620 + 198.40

= ₹ 818.40

(iv) ₹ 3,380 for 30 months at 4 ½ % p.a.

We know that

P = ₹ 3,380

R = 4 ½ % p.a. = 9/2 %

P = 30 months = 30/12 years

Here

S.I = (P × R × T)/ 100

Substituting the values

= (3380 × 9 × 30)/ (100 × 2 × 12)

= ₹ 380.25

Amount = P + S.I

Substituting the values

= 3380 + 380.25

= ₹ 3760.25

(v) ₹ 600 from July 12 to Dec. 5 at 10% p.a.

We know that

P = ₹ 600

R = 10% p.a.

T = July 12 to Dec 5

July = 19 days

Aug = 31 days

Sep = 30days

Oct = 31 days

Nov = 30 days

Dec = 05 days

Total = 146 days

T = 146/365 years = 2/5 years

Here

S.I = (P × R × T)/ 100

Substituting the values

= (600 × 10 × 2)/ (100 × 5)

= ₹ 24

Amount = P + S.I

Substituting the values

= 600 + 24

= ₹ 624

(vi) ₹ 850 from 10^th March to 3^rd August at 2 ½ % p.a.

We know that

P = ₹ 850

R = 2 ½% = 5/2% p.a.

T = 10^th March to 3^rd August

March = 21 days

April = 30 days

May = 31 days

June = 30 days

July = 31 days

Aug = 3 days

Total = 146 days

T = 146/365 = 2/5 years

Here

S.I = (P × R × T)/ 100

Substituting the values

= (850 × 5 × 2)/ (100 × 2 × 5)

= ₹ 8.50

Amount = P + S.I

Substituting the values

= 850 + 8.50

= ₹ 858.50

(vii) ₹ 225 for 3 years 9 months at 16% p.a.

We know that

P = ₹ 225

R = 16% p.a.

T = 3 years 9 months = 3 and 9/12 years = 3 ¾ years = 15/4 years

Here

S.I = (P × R × T)/ 100

Substituting the values

= (225 × 16 × 15)/ (100 × 4)

= ₹ 135

Amount = P + S.I

Substituting the values

= 225 + 135

= ₹ 360

2. On what sum of money does the S.I. for 10 years at 5% become ₹ 1,600?

Solution:

It is given that

S.I = ₹ 1,600

R = 5% p.a.

T = 10 years

We know that

P = (S.I × 100)/ (R × T)

Substituting the values

= (1600 × 100)/ (5 × 10)

So we get

= ₹ 3,200

3. Find the time in which ₹ 2,000 will amount to ₹ 2,330 at 11% p.a.

Solution:

It is given that

A = ₹ 2,330

P = ₹ 2,000

We know that

S.I = A – P

Substituting the values

= 2330 – 2000

= ₹ 330

Here

Time = (S.I × 100)/ (P × R)

Substituting the values

= (330 × 100)/ (2000 × 11)

So we get

= 3/2

= 1 ½ years

4. In what time will a sum of money double itself at 8% p.a.

Solution:

Consider the principal

P = ₹ 100

It is given that

A = 100 × 2 = ₹ 200

We know that

S.I = A – P

Substituting the values

= 200 – 100

= ₹ 100

R = 8% p.a.

Here

Time = (S.I × 100)/ (P × R)

Substituting the values

= (100 × 100)/ (100 × 8)

So we get

= 25/2

= 12 ½ years

5. In how many years will ₹ 870 amount to ₹ 1,044, the rate of interest being 2 ½% p.a.?

Solution:

It is given that

P = ₹ 870

A = ₹ 1044

We know that

S.I = A – P

Substituting the values

= 1044 – 870

= ₹ 174

R = 2 ½ = 5/2 % p.a.

We know that

Time = (S.I × 100)/ (P × R)

Substituting the values

= (174 × 100 × 2)/ (870 × 5)

So we get

= 8 years

6. Find the rate percent, if the S.I. on ₹ 275 in 2 years is ₹ 22.

Solution:

It is given that

P = ₹ 275

S.I = ₹ 22

T = 2 years

We know that

Rate = (S.I × 100)/ (P × T)

Substituting the values

= (22 × 100)/ (275 × 2)

So we get

= 4% p.a.

7. Find the sum which will amount to ₹ 700 in 5 years at 8% p.a.

Solution:

It is given that

Amount = ₹ 700

R = 8% p.a.

T = 5 years

Consider P = ₹ 100

We know that

S.I = (P × R × T)/ 100

Substituting the values

= (100 × 8 × 5)/ 100

= ₹ 40

Here

A = P + S.I

Substituting the values

= 100 + 40

= ₹ 140

If the amount is ₹ 140 then the principal is ₹ 100

If the amount is ₹ 700 then the principal = (100 × 700)/ 140 = ₹ 500

8. What is the rate of interest, if ₹ 3,750 amounts to ₹ 4,650 in 4 years?

Solution:

It is given that

P = ₹ 3,750

A = ₹ 4,650

We know that

S.I = A – P

Substituting the values

= 4650 – 3750

= ₹ 900

T = 4 years

Here

Rate = (S.I × 100)/ (P × T)

Substituting the values

= (900 × 100)/ (3750 × 4)

So we get

= 6% p.a.

9. In 4 years, ₹ 6,000 amounts to ₹ 8,000. In what time will ₹ 525 amount to ₹ 700 at the same rate?

Solution:

It is given that

P = ₹ 6,000

A = ₹ 8,000

We know that

S.I = A – P

Substituting the values

= 8000 – 6000

= ₹ 2000

T = 4 years

Here

Rate = (S.I × 100)/ (P × T)

Substituting the values

= (2000 × 100)/ (6000 × 4)

So we get

= 25/3%

= 8 1/3% p.a.

It is given that

P = ₹ 525

A = ₹ 700

We know that

S.I = A – P

Substituting the values

= 700 – 525

= ₹ 175

R = 25/3% p.a.

Here

Time = (S.I × 100)/ (P × R)

Substituting the values

= (175 × 100 × 3)/ (525 × 25)

So we get

= 4 years

10. The interest on a sum of money at the end of 2 ½ years is 4/5 of the sum. What is the rate percent?

Solution:

Consider the sum P = ₹ 100

S.I = 100 × 4/5 = ₹ 80

T = 2 ½ = 5/2 years

We know that

Rate = (S.I × 100)/ (P × T)

Substituting the values

= (80 × 100 × 2)/ (100 × 5)

So we get

= 32% p.a.

11. What sum of money lent out at 5% for 3 years will produce the same interest as ₹ 900 lent out at 4% for 5 years?

Solution:

It is given that

P = ₹ 900

R = 4%

T = 5 years

We know that

S.I = (P × R × T)/ 100

Substituting the values

= (900 × 4 × 5)/ 100

= ₹ 180

It is given that

S.I = ₹ 180

R = 5%

T = 3 years

We know that

Sum P = (S.I × 100)/ (R × T)

Substituting the values

= (180 × 100)/ (5 × 3)

So we get

= ₹ 1200

12. A sum of ₹ 1,780 becomes ₹ 2,136 in 4 years. Find:

(i) the rate of interest.

(ii) the sum that will become ₹ 810 in 7 years at the same rate of interest.

Solution:

(i) It is given that

P = ₹ 1780

A = ₹ 2136

We know that

S.I = A – P

Substituting the values

= 2136 – 1780

= ₹ 356

T = 4 years

Here

Rate = (S.I × 100)/ (P × T)

Substituting the values

= (356 × 100)/ (1780 × 4)

So we get

= 5% p.a.

(ii) Consider P = ₹ 100

R = 5% p.a.

T = 7 years

We know that

S.I = (P × R × T)/ 100

Substituting the values

= (100 × 5 × 7)/ 100

= ₹ 35

Here amount = P + S.I

Substituting the values

= 100 + 35

= ₹ 135

If the amount is ₹ 135 then the principal is ₹ 100

If the amount is ₹ 810 then principal = (100 × 810)/ 135 = ₹ 600

13. A sum amounts to ₹ 2,652 in 6 years at 5% p.a. simple interest. Find:

(i) the sum

(ii) the time in which the same sum will double itself at the same rate of interest.

Solution:

(i) Consider P = ₹ 100

R = 5% p.a.

T = 6 years

We know that

S.I = (P × R × T)/ 100

Substituting the values

= (100 × 5 × 6)/ 100

= ₹ 30

Here amount = 100 + 30 = ₹ 130

If the amount is ₹ 130 then principal is ₹ 100

If the amount is ₹ 2652 then principal = (100 × 2652)/ 130 = ₹ 2040

Consider sum P = ₹ 100

Amount = 100 × 2 = ₹ 200

We know that

S.I = A – P

Substituting the values

= 200 – 100

= ₹ 100

R = 5% p.a.

Here

T = (S.I × 100)/ (P × R)

Substituting the values

= (100 × 100)/ (100 × 5)

So we get

= 20 years

14. P and Q invest ₹ 36,000 and ₹ 25,000 respectively at the same rate of interest per year. If at the end of 4 years, P gets ₹ 3,080 more interest than Q, find the rate of interest.

Solution:

It is given that

P’s investment (P₁) = ₹ 36,000

Q’s investment (P₂) = ₹ 25,000

T = 4 years

Consider the rate of interest = x%

So we get

P’s interest (S.I) = (P × R × T)/ 100

Substituting the values

= (36000 × x × 4)/ 100

= ₹ 1440x

Q’s interest = (P × R × T)/ 100

Substituting the values

= (25000 × x × 4)/ 100

= ₹ 1000x

Here the difference in their interest = 1440x – 1000x = ₹ 440x

The difference given = ₹ 3080

So we get

440x = 3080

x = 3080/440

x = 7%

So the rate of interest = 7% p.a.

15. A sum of money is lent for 5 years at R% simple interest per annum. If the interest earned be one-fourth of the money lent, find the value of R.

Solution:

Consider the sum P = ₹ 100

We know that

S.I = 1/4 × 100 = ₹ 25

T = 5 years

Here

Rate = (S.I × 100)/ (P × T)

Substituting the values

= (25 × 100)/ (100 × 5)

So we get

= 5%

16. The simple interest earned on a certain sum in 5 years is 30% of the sum. Find the rate of interest.

Solution:

Consider sum P = ₹ 100

We know that

SI = 30/100 × 100 = ₹ 30

T = 5 years

Here

Rate = (S.I × 100)/ (P × T)

Substituting the values

= (30 × 100)/ (100 × 5)

So we get

= 6%

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Selina Class 7 ICSE Solutions Mathematics : Chapter 10- Simple Interest

Download PDF: Selina Class 7 ICSE Solutions Mathematics : Chapter 10- Simple Interest PDF

Chapterwise Selina Publishers ICSE Solutions for Class 7 Mathematics :

• Chapter 1- Integers
• Chapter 2- Rational Numbers
• Chapter 3- Fraction (Including Problems)
• Chapter 4- Decimal Fractions (Decimals)
• Chapter 5- Exponents (Including Laws of Exponents)
• Chapter 6- Ratio and Proportion (Including Sharing in a Ratio)
• Chapter 7- Unitary Method (Including Time and Work)
• Chapter 8- Percent and Percentage
• Chapter 9- Profit, Loss and Discount
• Chapter 10- Simple Interest
• Chapter 11- Fundamental Concepts (Including Fundamental Operations)
• Chapter 12- Simple Linear Equations (Including Word Problems)
• Chapter 13- Set Concepts
• Chapter 14- Lines and Angles (Including Construction of Angles)
• Chapter 15- Triangles
• Chapter 16- Pythagoras Theorem
• Chapter 17- Symmetry (Including Reflection and Rotation)
• Chapter 18- Recognition of Solids (Representing 3-D in 2-D)
• Chapter 19- Congruency: Congruent Triangles
• Chapter 20- Mensuration (Perimeter and Area of Plane Figures)
• Chapter 21- Data Handling
• Chapter 22- Probability

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