## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 7 Percentage and Its applications Ex 7.4

Question 1.

Find the simple interest on:

(i) ₹ 350 for 2 years at 11% per annum

(ii) ₹ 20000 for 4\(\frac { 1 }{ 2 }\) years at 8\(\frac { 1 }{ 2 }\) % per annum

(iii) ₹ 648 for 8 months at 16\(\frac { 2 }{ 3 }\) % per annum. Also, find the amount in each case.

Solution:

(i) Principal (P) = ₹ 350

Rate (R) = 11% p.a.

Time (T) = 2 years

Amount = P + S.I. = ₹ 350 + ₹ 77 = ₹ 427

(ii) Principal (P) = ₹ 20000

Rate (R) = 8\(\frac { 1 }{ 2 }\) = \(\frac { 17 }{ 2 }\) % p.a.

Time (T) = 4\(\frac { 1 }{ 2 }\) years = \(\frac { 9 }{ 2 }\) years

and amount = P + S.I. = ₹ 20000 + ₹ 7650 = ₹ 27650

(iii) Principal (P) = ₹ 648

Amount = P + S.I. = ₹ 648 + ₹ 72 = ₹ 720

Question 2.

Find the time when:

(i) simple interest on ₹ 2500 at 4% per annum is ₹ 200

(ii) simple interest on ₹ 12000 at 6\(\frac { 1 }{ 2 }\) % per annum is ₹ 2730

Solution:

(i) S.I. = ₹ 200

Principal (P) = ₹ 2500

Rate (R) = 4% p.a.

Question 3.

Find the rate of interest when:

(i) simple interest on ₹ 1560 in 3 years is ₹ 585

(ii) simple interest on ₹ 1625 in 2\(\frac { 1 }{ 2 }\) years is ₹ 325.

Solution:

(i) S.I. = ₹ 585

Principal (P) = ₹ 1560

Time (T) = 3 years

Question 4.

Find the principal when:

(i) simple interest at 16% per annum for 2\(\frac { 1 }{ 2 }\) years is ₹ 3840

(ii) simple interest at 7\(\frac { 1 }{ 2 }\) % per annum for 2 years 4 months is ₹ 2730.

Solution:

(i) S.I. = ₹ 3840

Rate (R) = 16% p.a.

Question 5.

Find the rate of interest when:

(i) ₹ 1200 amounts to ₹ 1320 in 2 years

(ii) ₹ 300 amounts to ₹ 400 in 2 years.

Solution:

(i) Principal (P) = ₹ 1200

Amount (A) = ₹ 1320

S.I. = A – P = ₹ 1320 – ₹ 1200 = ₹ 120

Time (T) = 2 years

Question 6.

Find the time when:

(i) ₹ 1250 amounts to ₹ 1950 at 16% per annum

(ii) ₹ 6540 amounts to ₹ 8447.50 at 12\(\frac { 1 }{ 2 }\) per annum.

Solution:

(i) Principal (P) = ₹ 1250

Amount (A) = ₹ 1950

S.I. = A – P = ₹ 1950 – ₹ 1250 = ₹ 700

Rate = 16% p.a.

(ii) Principal (P) = ₹ 6540

Amount (A) = ₹ 8447.50

S.I. = A – P = ₹ 8447.50 – 6540.00 = ₹ 1907.50

Question 7.

₹ 14000 is invested at 4% per annum simple interest. How long will it take for the amount to reach ₹ 16240?

Solution:

Principal (P) = ₹ 14000

Amount (A) = ₹ 16240

S.I. = A – P = ₹ 16240 – ₹ 14000 = ₹ 2240

Rate (R) = 4%

Question 8.

An amount of money invested trebled in 6 years. Find the rate of interest earned.

Solution:

Time (T) = 6 years

Let principal = ₹ 100

Then amount = ₹ 100 × 3 = ₹ 300

S.I. = A – P = ₹ 300 – ₹ 100 = ₹ 200

Question 9.

Find the principal when:

(i) final amount is ₹ 4500 at 20% per annum for 5 years

(ii) final amount is ₹ 2420 at 4% per annum for 2\(\frac { 1 }{ 2 }\) years.

Solution:

(i) Amount (A) = ₹ 4500

Rate (R) = 20%

Time (T) = 5 years

Let principal (P) = ₹ 100

and amount = P + S.I = ₹ 100 + ₹ 100 = ₹ 200

If amount is ₹ 200 then principal = ₹ 100

and if amount is ₹ 4500 then principal

= \(\frac { 100 }{ 200 }\) × 4500

= ₹ 2250

(iii) Amount (A) = ₹ 2420

Rate (R) = 4% p.a.

Time (T) = 2\(\frac { 1 }{ 2 }\) = \(\frac { 5 }{ 2 }\) years

Let principal = ₹ 100

Amount = P + S.I. = ₹ 100 + ₹ 10 = ₹ 110

If amount is ₹ 110 then principal = ₹ 100

If amount is ₹ 2420, then principal

= \(\frac { 100\times 2420 }{ 110 }\)

= ₹ 2200