165 + Interview Questions in COMPOUND INTEREST IN GENERAL AWARENESS Page 3 InterviewSolution

101.	Compute the amount and the compound interest by using the formulae when : Principal = Rs 160000, Rate = 10 paise per rupee per annum compounded half yearly, Time = 2 years.
Answer» Given, P = Rs 160000, rate = 10% = 10/2% = 5% (half yearly), time = 2years = 2×2 = 4 quarters A = P (1 + R/100)ⁿ = 160000 (1 + 5/100)⁴ = 160000 (105/100)⁴ = Rs 194481 Compound interest (CI) = A-P = Rs 194481 – 160000 = Rs 34481

101.

Compute the amount and the compound interest by using the formulae when : Principal = Rs 160000, Rate = 10 paise per rupee per annum compounded half yearly, Time = 2 years.

Answer»

Given, P = Rs 160000, rate = 10% = 10/2% = 5% (half yearly), time = 2years = 2×2 = 4 quarters

A = P (1 + R/100)ⁿ

= 160000 (1 + 5/100)⁴

= 160000 (105/100)⁴

= Rs 194481

Compound interest (CI) = A-P = Rs 194481 – 160000 = Rs 34481

Discussion

102.	At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs. 410 in 2 years?
Answer» Given details are, Principal = Rs 4000 Time = 2years CI = Rs 410 Rate be = R% per annum By using the formula, CI = P [(1 + R/100)ⁿ – 1] 410 = 4000 [(1 + R/100)² – 1] 410 = 4000 (1 + R/100)² – 4000 410 + 4000 = 4000 (1 + R/100)² (1 + R/100)² = 4410/4000 (1 + R/100)² = 441/400 (1 + R/100)² = (21/20)² By cancelling the powers on both the sides, 1 + R/100 = 21/20 R/100 = 21/20 – 1 = (21-20)/20 = 1/20 R = 100/20 = 5 ∴ Required Rate is 5% per annum.

102.

At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs. 410 in 2 years?

Answer»

Given details are,

Principal = Rs 4000

Time = 2years

CI = Rs 410

Rate be = R% per annum

By using the formula,

CI = P [(1 + R/100)ⁿ – 1]

410 = 4000 [(1 + R/100)² – 1]

410 = 4000 (1 + R/100)² – 4000

410 + 4000 = 4000 (1 + R/100)²

(1 + R/100)² = 4410/4000

(1 + R/100)² = 441/400

(1 + R/100)² = (21/20)²

By cancelling the powers on both the sides,

1 + R/100 = 21/20

R/100 = 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

∴ Required Rate is 5% per annum.

Discussion

103.	Compute the amount and the compound interest by using the formulae when : Principal = Rs 2000, Rate = 4 paise per rupee per annum, Time = 3 years.
Answer» Given, P = Rs 2000, rate = 4%, time = 3years A = P (1 + R/100)ⁿ = 2000 (1 + 4/100)³ = 2000 (104/100)³ = Rs 2249.72 Compound interest (CI) = A-P = Rs 2249.72 – 2000 = Rs 249.72

103.

Compute the amount and the compound interest by using the formulae when : Principal = Rs 2000, Rate = 4 paise per rupee per annum, Time = 3 years.

Answer»

Given, P = Rs 2000, rate = 4%, time = 3years

A = P (1 + R/100)ⁿ

= 2000 (1 + 4/100)³

= 2000 (104/100)³

= Rs 2249.72

Compound interest (CI) = A-P = Rs 2249.72 – 2000 = Rs 249.72

Discussion

104.	Compute the amount and the compound interest by using the formulae when : Principal = Rs 12800, Rate = 7 ½ %, Time = 3 years.
Answer» Given, P = Rs 12800, rate = 7 ½ % = 15/2% = 7.5%, time = 3years A = P (1 + R/100)ⁿ = 12800 (1 + 7.5/100)³ = 12800 (107.5/100)³ = Rs 15901.4 Compound interest (CI) = A-P = Rs 15901.4 – 12800 = Rs 3101.4

104.

Compute the amount and the compound interest by using the formulae when : Principal = Rs 12800, Rate = 7 ½ %, Time = 3 years.

Answer»

Given, P = Rs 12800, rate = 7 ½ % = 15/2% = 7.5%, time = 3years

A = P (1 + R/100)ⁿ

= 12800 (1 + 7.5/100)³

= 12800 (107.5/100)³

= Rs 15901.4

Compound interest (CI) = A-P = Rs 15901.4 – 12800 = Rs 3101.4

Discussion

105.	The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.
Answer» Given details are, Present population is = 25000 First year growth R₁ = 4% Second year growth R₂ = 5% Third year growth R₃ = 8% Number of years = 3 By using the formula, A = P (1 + R/100)ⁿ So, population after three years = P (1 + R₁/100) (1 + R₂/100) (1 + R₃/100) = 25000(1 + 4/100) (1 + 5/100) (1 + 8/100) = 25000 (1.04) (1.05) (1.08) = 29484 ∴ Population after 3 years will be 29484.

105.

The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.

Answer»

Given details are,

Present population is = 25000

First year growth R₁ = 4%

Second year growth R₂ = 5%

Third year growth R₃ = 8%

Number of years = 3

By using the formula,

A = P (1 + R/100)ⁿ

So, population after three years = P (1 + R₁/100) (1 + R₂/100) (1 + R₃/100)

= 25000(1 + 4/100) (1 + 5/100) (1 + 8/100)

= 25000 (1.04) (1.05) (1.08)

= 29484

∴ Population after 3 years will be 29484.

Discussion

106.	Nirmala deposited Rs. 20,000 in a bank where compound interest of rate 7% is computed annually Rs. 5000 was with drawn after one year. How much amount she will get after 2 years.
Answer» Amount deposited in the bank = Rs. 2000 Rate of interest = 7% principal after one year = 20000 + [20000 × \(\frac{7}{100}\)] = 20000 + 1400 = 21400 Amount withdrawn after one year = Rs. 5000 . Principal for the 2^nd year = 21400 – 5000 = Rs. 16400 Amount she gets after 2 years = 16400 + [16400 × \(\frac{7}{100}\)] = 16400 + 1148 = Rs. 17548

106.

Nirmala deposited Rs. 20,000 in a bank where compound interest of rate 7% is computed annually Rs. 5000 was with drawn after one year. How much amount she will get after 2 years.

Answer»

Amount deposited in the bank = Rs. 2000

Rate of interest = 7%

principal after one year

= 20000 + [20000 × \(\frac{7}{100}\)] = 20000 + 1400 = 21400

Amount withdrawn after one year = Rs. 5000 .

Principal for the 2^nd year

= 21400 – 5000 = Rs. 16400

Amount she gets after 2 years

= 16400 + [16400 × \(\frac{7}{100}\)] = 16400 + 1148

= Rs. 17548

Discussion

107.	Rs 8000 invested at compound interest gives Rs 1261 as interest after 3 years. The rate of interest per annum is (a) 25% (b) 17.5% (c) 10% (d) 5%
Answer» (d) 5% P = Rs 8000, C.I. = Rs 1261 \(\Rightarrow\) Amount = Rs 9261, n = 3, r = ? \(\therefore\) 9261 = 8000\(\big(1+\frac{r}{100}\big)^3\) \(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^3\) = \(\frac{9261}{8000}\) = \(\big(\frac{21}{20}\big)^3\) \(\Rightarrow\) 1 + \(\frac{r}{100}\) = \(\frac{21}{20}\) \(\Rightarrow\) \(\frac{r}{100}\) = \(\frac{21}{20}\) - 1 =\(\frac{1}{20}\) \(\Rightarrow\) r = \(\frac{100}{20}\)% = 5% p.a.

107.

Rs 8000 invested at compound interest gives Rs 1261 as interest after 3 years. The rate of interest per annum is (a) 25% (b) 17.5% (c) 10% (d) 5%

Answer»

(d) 5%

P = Rs 8000, C.I. = Rs 1261

\(\Rightarrow\) Amount = Rs 9261, n = 3, r = ?

\(\therefore\) 9261 = 8000\(\big(1+\frac{r}{100}\big)^3\)

\(\Rightarrow\) \(\big(1+\frac{r}{100}\big)^3\) = \(\frac{9261}{8000}\) = \(\big(\frac{21}{20}\big)^3\)

\(\Rightarrow\) 1 + \(\frac{r}{100}\) = \(\frac{21}{20}\)

\(\Rightarrow\) \(\frac{r}{100}\) = \(\frac{21}{20}\) - 1 =\(\frac{1}{20}\)

\(\Rightarrow\) r = \(\frac{100}{20}\)% = 5% p.a.

Discussion

108.	In how many years will a sum of Rs 800 at 10% per annum compound interest, compounded semiannually become Rs 926.10 ? (a) 1 year (b) 3 years(c) 2 years(d) 1\(\frac{1}{2}\) years
Answer» (d) \(1\frac{1}{2}\) years P = Rs 800, r = 10% p.a. = 5% per half year, A = Rs 926.10, Time = 2n \(\therefore\) 929.10 = 800\(\big(1+\frac{5}{100}\big)^{2n}\) \(\Rightarrow\) \(\frac{9261}{8000}\) = \(\big(1+\frac{1}{20}\big)^{2n}\) \(\Rightarrow\) \(\big(\frac{21}{20}\big)^3\) = \(\big(\frac{21}{20}\big)^{2n}\) \(\Rightarrow\) 2n = 3 \(\Rightarrow\) n = \(\frac{3}{2}\)= \(1\frac{1}{2}\) years.

108.

In how many years will a sum of Rs 800 at 10% per annum compound interest, compounded semiannually become Rs 926.10 ? (a) 1 year (b) 3 years(c) 2 years(d) 1\(\frac{1}{2}\) years

Answer»

(d) \(1\frac{1}{2}\) years

P = Rs 800, r = 10% p.a. = 5% per half year,

A = Rs 926.10, Time = 2n

\(\therefore\) 929.10 = 800\(\big(1+\frac{5}{100}\big)^{2n}\)

\(\Rightarrow\) \(\frac{9261}{8000}\) = \(\big(1+\frac{1}{20}\big)^{2n}\) \(\Rightarrow\) \(\big(\frac{21}{20}\big)^3\) = \(\big(\frac{21}{20}\big)^{2n}\)

\(\Rightarrow\) 2n = 3 \(\Rightarrow\) n = \(\frac{3}{2}\)= \(1\frac{1}{2}\) years.

Discussion

109.	The half life of Uranium - 233 is 160000 years, i.e., Uranium 233 decays at a constant rate in such a way that it reduces to 50% in 160000 years. In how many years will it reduce to 25% ? (a) 80000 years (b) 240000 years (c) 320000 years (d) 40000 years
Answer» (c) 320000 years Let the rate of decay of Uranium be R per cent per year. Also, let the initial amount of Uranium be 1 unit. Since, the half life of Uranium - 233 is 160000 years, therefore \(\big(1-\frac{R}{100}\big)^{160000}\) = \(\frac{1}{2}\) ..........(i) Suppose Uranium - 233 reduces to 25% in t years. Then, \(\big(1-\frac{R}{100}\big)^t\) = \(\frac{25}{100}\) = \(\frac{1}{4}\) = \(\big(\frac{1}{2}\big)^2\) = \(\Big(\big(1-\frac{R}{100}\big)^{160000}\Big)^2\) = \(\big(1-\frac{R}{100}\big)^{320000}\) \(\Rightarrow\) t = 320000 years.

109.

The half life of Uranium - 233 is 160000 years, i.e., Uranium 233 decays at a constant rate in such a way that it reduces to 50% in 160000 years. In how many years will it reduce to 25% ? (a) 80000 years (b) 240000 years (c) 320000 years (d) 40000 years

Answer»

(c) 320000 years

Let the rate of decay of Uranium be R per cent per year. Also, let the initial amount of Uranium be 1 unit. Since, the half life of Uranium - 233 is 160000 years, therefore

\(\big(1-\frac{R}{100}\big)^{160000}\) = \(\frac{1}{2}\) ..........(i)

Suppose Uranium - 233 reduces to 25% in t years. Then,

\(\big(1-\frac{R}{100}\big)^t\) = \(\frac{25}{100}\) = \(\frac{1}{4}\) = \(\big(\frac{1}{2}\big)^2\)

= \(\Big(\big(1-\frac{R}{100}\big)^{160000}\Big)^2\) = \(\big(1-\frac{R}{100}\big)^{320000}\)

\(\Rightarrow\) t = 320000 years.

Discussion

110.	What will be the compound interest on Rs. 4000 in two years when rate of interest is 5% per annum?
Answer» Given, Principal = Rs.4000 Time = 2 years Rate = 5 % per annum Compound interes \(={p}[({1}+\frac{R}{100})^T-{1}]\) \(={4000}[({1}+\frac{5}{100})^2-{1}]\) \(={4000}[(\frac{21}{20})^2-{1}]\) \(={4000}\times\frac{41}{400}\) Rs. 410

110.

What will be the compound interest on Rs. 4000 in two years when rate of interest is 5% per annum?

Answer»

Given,

Principal = Rs.4000

Time = 2 years

Rate = 5 % per annum

Compound interes

\(={p}[({1}+\frac{R}{100})^T-{1}]\)

\(={4000}[({1}+\frac{5}{100})^2-{1}]\)

\(={4000}[(\frac{21}{20})^2-{1}]\)

\(={4000}\times\frac{41}{400}\)

Rs. 410

Discussion

111.	Find the compound interest when principal = Rs 3000, rate = 5% per annum and time = 2 years.
Answer» Given details are, Principal (p) = Rs 3000 Rate (r) = 5% Time = 2years Interest for the first year = (3000×5×1)/100 = 150 Amount at the end of first year = Rs 3000 + 300 = Rs 3150 Principal interest for the second year = (3150×5×1)/100 = 157.5 Amount at the end of second year = Rs 3150 + 157.5 = Rs 3307.5 ∴ Compound Interest = Rs 3307.5 – Rs 3000 = Rs 307.5

111.

Find the compound interest when principal = Rs 3000, rate = 5% per annum and time = 2 years.

Answer»

Given details are,

Principal (p) = Rs 3000

Rate (r) = 5%

Time = 2years

Interest for the first year = (3000×5×1)/100 = 150

Amount at the end of first year = Rs 3000 + 300 = Rs 3150

Principal interest for the second year = (3150×5×1)/100 = 157.5

Amount at the end of second year = Rs 3150 + 157.5 = Rs 3307.5

∴ Compound Interest = Rs 3307.5 – Rs 3000 = Rs 307.5

Discussion

112.	Compute the amount and the compound interest by using the formulae when : Principal = Rs 3000, Rate = 5%, Time = 2 years.
Answer» Given, P = Rs 3000, rate = 5%, time = 2years A = P (1 + R/100)ⁿ = 3000 (1 + 5/100)² = 3000 (105/100)² = Rs 3307.5 Compound interest (CI) = A-P = Rs 3307.5 – 3000 = Rs 307.5

112.

Compute the amount and the compound interest by using the formulae when : Principal = Rs 3000, Rate = 5%, Time = 2 years.

Answer»

Given, P = Rs 3000, rate = 5%, time = 2years

A = P (1 + R/100)ⁿ

= 3000 (1 + 5/100)²

= 3000 (105/100)²

= Rs 3307.5

Compound interest (CI) = A-P = Rs 3307.5 – 3000 = Rs 307.5

Discussion

113.	What will be the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs. 1200 as simple interest.
Answer» Given details are, Rate = 5 % per annum Simple Interest (SI) = Rs 1200 Time (t) = 3 years By using the formula, SI = (PTR)/100 P = (SI×100) / (T×R) = (1200×100) / (3×5) = 120000/15 = Rs 8000 Now, P = Rs 8000 R = 5% T = 3years By using the formula, A = P (1 + R/100)ⁿ = 8000 (1 + 5/100)³ = 8000 (105/100)³ = Rs 9261 ∴ CI = Rs 9261 – 8000 = Rs 1261

113.

What will be the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs. 1200 as simple interest.

Answer»

Given details are,

Rate = 5 % per annum

Simple Interest (SI) = Rs 1200

Time (t) = 3 years

By using the formula,

SI = (PTR)/100

P = (SI×100) / (T×R)

= (1200×100) / (3×5)

= 120000/15

= Rs 8000

Now,

P = Rs 8000

R = 5%

T = 3years

By using the formula,

A = P (1 + R/100)ⁿ

= 8000 (1 + 5/100)³

= 8000 (105/100)³

= Rs 9261

∴ CI = Rs 9261 – 8000 = Rs 1261

Discussion

114.	If the rate of interest be 4% per annum for first year, 5% per annum for second year and 6% per annum for third year, then what is the compound interest on Rs 10000 for 3 years ?
Answer» A=P\(\big(1+\frac{r_1}{100}\big)\big(1+\frac{r_2}{100}\big)\big(1+\frac{r_3}{100}\big)\) \(\Rightarrow\) A = 10000\(\big(1+\frac{4}{100}\big)\big(1+\frac{5}{100}\big)\big(1+\frac{6}{100}\big)\) =10000 x \(\frac{104}{100}\) x \(\frac{105}{100}\) x \(\frac{106}{100}\) = Rs 11575.20 \(\therefore \) C.I. = Rs 11575.20 – Rs 10000 = Rs 1575.20.

114.

If the rate of interest be 4% per annum for first year, 5% per annum for second year and 6% per annum for third year, then what is the compound interest on Rs 10000 for 3 years ?

Answer»

A=P\(\big(1+\frac{r_1}{100}\big)\big(1+\frac{r_2}{100}\big)\big(1+\frac{r_3}{100}\big)\)

\(\Rightarrow\) A = 10000\(\big(1+\frac{4}{100}\big)\big(1+\frac{5}{100}\big)\big(1+\frac{6}{100}\big)\)

=10000 x \(\frac{104}{100}\) x \(\frac{105}{100}\) x \(\frac{106}{100}\) = Rs 11575.20

\(\therefore \) C.I. = Rs 11575.20 – Rs 10000 = Rs 1575.20.

Discussion

115.	Find the compound interest on Rs. 64000 for 1 year at the rate of 10% per annum compounded quarterly.
Answer» Given details are, Principal (p) = Rs 64000 Rate (r) = 10 % = 10/4 % (for quarterly) Time = 1year = 1× 4 = 4 (for quarter in a year) By using the formula, A = P (1 + R/100)ⁿ = 64000 (1 + 10/4×100)⁴ = 64000 (410/400)⁴ = Rs 70644.03 ∴ Compound Interest = A – P = Rs 70644.03 – Rs 64000 = Rs 6644.03

115.

Find the compound interest on Rs. 64000 for 1 year at the rate of 10% per annum compounded quarterly.

Answer»

Given details are,

Principal (p) = Rs 64000

Rate (r) = 10 % = 10/4 % (for quarterly)

Time = 1year = 1× 4 = 4 (for quarter in a year)

By using the formula,

A = P (1 + R/100)ⁿ

= 64000 (1 + 10/4×100)⁴

= 64000 (410/400)⁴

= Rs 70644.03

∴ Compound Interest = A – P = Rs 70644.03 – Rs 64000 = Rs 6644.03

Discussion

116.	The compound interest on Rs. 30,000 at 7% per annum is Rs 4347. The period (in years) is (a) 2 (b) \(2\frac{1}{2}\) (c) 3 (d) 4
Answer» (a) 2 P = Rs 30000, r = 7% p.a., C.I. = Rs 4347, n = ? \(\Rightarrow\) Amount = Rs 30000 + Rs 4347 = Rs 34347 \(\therefore\) 34347 = 30000\(\big(1+\frac{7}{100}\big)^n\) \(\Rightarrow\) \(\big(\frac{107}{100}\big)^n\) = \(\frac{34347}{30000}\) = \(\frac{11449}{10000}\) \(\Rightarrow\) \(\big(\frac{107}{100}\big)^n\) = \(\big(\frac{107}{100}\big)^2\) \(\Rightarrow\) n=2

116.

The compound interest on Rs. 30,000 at 7% per annum is Rs 4347. The period (in years) is (a) 2 (b) \(2\frac{1}{2}\) (c) 3 (d) 4

Answer»

(a) 2

P = Rs 30000, r = 7% p.a., C.I. = Rs 4347, n = ?

\(\Rightarrow\) Amount = Rs 30000 + Rs 4347 = Rs 34347

\(\therefore\) 34347 = 30000\(\big(1+\frac{7}{100}\big)^n\)

\(\Rightarrow\) \(\big(\frac{107}{100}\big)^n\) = \(\frac{34347}{30000}\) = \(\frac{11449}{10000}\)

\(\Rightarrow\) \(\big(\frac{107}{100}\big)^n\) = \(\big(\frac{107}{100}\big)^2\) \(\Rightarrow\) n=2

Discussion

117.	The difference between simple interest and compound interest on Rs 1200 for one year at 10% per annum reckoned half yearly is (a) Rs 2.50 (b) Rs 3 (c) Rs 3.75 (d) Rs 4
Answer» (b) Rs 3 P = Rs 1200, R = 10% p.a., T = 1 year \(\therefore\) S.I. = \(\frac{P\times R\times T}{100}\) = \(\frac{1200\times 10\times 1}{100}\) = Rs 120 For C.I. reckoned half yearly, P = Rs 1200, r = 5% per half year, n = 2 half year \(\therefore\) C.I = Rs \(\Big(1200\big(1+\frac{5}{100}\big)^2-1200\Big)\) = Rs\(\big(1200\times\frac{21}{20}\times\frac{21}{20}-1200\big)\) = Rs 1323- Rs 1200 = Rs 123. \(\therefore\) Required difference = Rs 123 – Rs 120 = Rs 3.

117.

The difference between simple interest and compound interest on Rs 1200 for one year at 10% per annum reckoned half yearly is (a) Rs 2.50 (b) Rs 3 (c) Rs 3.75 (d) Rs 4

Answer»

(b) Rs 3

P = Rs 1200, R = 10% p.a., T = 1 year

\(\therefore\) S.I. = \(\frac{P\times R\times T}{100}\) = \(\frac{1200\times 10\times 1}{100}\) = Rs 120

For C.I. reckoned half yearly,

P = Rs 1200, r = 5% per half year, n = 2 half year

\(\therefore\) C.I = Rs \(\Big(1200\big(1+\frac{5}{100}\big)^2-1200\Big)\)

= Rs\(\big(1200\times\frac{21}{20}\times\frac{21}{20}-1200\big)\)

= Rs 1323- Rs 1200 = Rs 123.

\(\therefore\) Required difference = Rs 123 – Rs 120 = Rs 3.

Discussion

118.	Find the compound interest at the rate of 5% for three years on that principal which in three years at the rate of 5% per annum gives Rs. 12000 as simple interest.
Answer» Given, Simple interest = Rs.12000 Rate = 5% per annum Time = 3 years So, = simple interest \(=\frac{P \times R \times \times T}{100}\) \(=\frac{ P \times 5 \times 3}{100}\) = 12000 \(={P}=\frac{12000\times100}{15}\) = Rs. 80000 We get , Principal = Rs.80000 Rate = 5% per annum Time = 3 years Compound interest \({P}[({1}+\frac{R}{100})^T-{1}]\) \(={80000}[({1}+\frac{5}{100})^3-{1}]\) \(={80000}[(\frac{21}{20})^3-{1}]\) \(={80000}\times\frac{1261}{8000}\) = Rs. 12610

118.

Find the compound interest at the rate of 5% for three years on that principal which in three years at the rate of 5% per annum gives Rs. 12000 as simple interest.

Answer»

Given,

Simple interest = Rs.12000

Rate = 5% per annum

Time = 3 years

So,

= simple interest \(=\frac{P \times R \times \times T}{100}\)

\(=\frac{ P \times 5 \times 3}{100}\) = 12000

\(={P}=\frac{12000\times100}{15}\) = Rs. 80000

We get ,

Principal = Rs.80000

Rate = 5% per annum

Time = 3 years

Compound interest \({P}[({1}+\frac{R}{100})^T-{1}]\)

\(={80000}[({1}+\frac{5}{100})^3-{1}]\)

\(={80000}[(\frac{21}{20})^3-{1}]\)

\(={80000}\times\frac{1261}{8000}\) = Rs. 12610

Discussion

119.	Find the compound interest when principal = Rs. 3000, rate = 5% per annum and time = 2 years.
Answer» Given: Principal =Rs.3000 Rate = 5% Time = 2 years Hence, Compound interest \(={p}[({1}+\frac{R}{100})^r-{1}]\) \(={3000}[({1}+\frac{5}{100})^2-{1}]\) \(={3000}((\frac{21}{20})^2-{1})\) \(={3000}(\frac{441-400}{400})\) \(={15}\times\frac{41}{2}\) =307.5

119.

Find the compound interest when principal = Rs. 3000, rate = 5% per annum and time = 2 years.

Answer»

Given:

Principal =Rs.3000

Rate = 5% Time = 2 years

Hence,

Compound interest

\(={p}[({1}+\frac{R}{100})^r-{1}]\)

\(={3000}[({1}+\frac{5}{100})^2-{1}]\)

\(={3000}((\frac{21}{20})^2-{1})\)

\(={3000}(\frac{441-400}{400})\)

\(={15}\times\frac{41}{2}\)

=307.5

Discussion

120.	The difference between the compound interest and simple interest on a certain sum for 2 years at 6% per annum is ₹ 90. Find the sum.
Answer» Given: Interest rate= 6% per annum Time=2 years Simple interest (SI) = PTR/100 Where P is principle amount, T is time taken, R is rate per annum Let sum is P SI = (P × 2× 6)/ 100 ⇒ SI = (12P)/ 100 ⇒ SI = (3P)/ 25 —————- equation 1 To find the amount we have the formula, Amount (A) = P (1+(R/100))ⁿ Where P is present value, r is rate of interest, n is time in years. Now substituting the values in above formula we get, ∴ CI= P (1 + 6/100)²-P ⇒ CI = P (1+3/50)²-P ⇒ CI = P (53/50)²-P ⇒ CI = P (2809) / (2500) – P ⇒ CI = 309P/2500——– equation 2 Now the difference is (CI – SI) = 309P/2500 – (3P)/ 25 ⇒ 90 = 309P/2500 – (3P)/ 25 ⇒ 90 = 309P – (300P)/2500 ⇒ 90 = 9P/2500 ⇒ P= 90 × 2500/9 ⇒ P= 10 × 2500 ⇒ P=25000 ∴ Sum = 25000

120.

The difference between the compound interest and simple interest on a certain sum for 2 years at 6% per annum is ₹ 90. Find the sum.

Answer»

Given:

Interest rate= 6% per annum

Time=2 years

Simple interest (SI) = PTR/100

Where P is principle amount, T is time taken, R is rate per annum

Let sum is P

SI = (P × 2× 6)/ 100

⇒ SI = (12P)/ 100

⇒ SI = (3P)/ 25 —————- equation 1

To find the amount we have the formula,

Amount (A) = P (1+(R/100))ⁿ

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

∴ CI= P (1 + 6/100)²-P

⇒ CI = P (1+3/50)²-P

⇒ CI = P (53/50)²-P

⇒ CI = P (2809) / (2500) – P

⇒ CI = 309P/2500——– equation 2

Now the difference is

(CI – SI) = 309P/2500 – (3P)/ 25

⇒ 90 = 309P/2500 – (3P)/ 25

⇒ 90 = 309P – (300P)/2500

⇒ 90 = 9P/2500

⇒ P= 90 × 2500/9

⇒ P= 10 × 2500

⇒ P=25000

∴ Sum = 25000

Discussion

121.	Mewa lal borrowed Rs. 20000 from his friend Rooplal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.
Answer» Given details are, Principal (p) = Rs 20000 Rate (r) = 18 % Time = 2 years By using the formula, Interest amount Mewa lal has to pay, By using the formula, Simple interest = P×T×R/100 = (20000×18×2)/100 = 7200 Interest amount Rampal has to pay to Mewa lal, By using the formula, A = P (1 + R/100)ⁿ = 20000 (1 + 18/100)² = 20000 (118/100)² = Rs 27848 – 20000 (principal amount) = Rs 7848 ∴ Mewa lal gain = Rs (7848 – 7200) = Rs 648

121.

Mewa lal borrowed Rs. 20000 from his friend Rooplal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.

Answer»

Given details are,

Principal (p) = Rs 20000

Rate (r) = 18 %

Time = 2 years

By using the formula,

Interest amount Mewa lal has to pay,

By using the formula,

Simple interest = P×T×R/100

= (20000×18×2)/100 = 7200

Interest amount Rampal has to pay to Mewa lal,

By using the formula,

A = P (1 + R/100)ⁿ

= 20000 (1 + 18/100)²

= 20000 (118/100)²

= Rs 27848 – 20000 (principal amount)

= Rs 7848

∴ Mewa lal gain = Rs (7848 – 7200) = Rs 648

Discussion

122.	How much should a sum of Rs 16000 approximately amount to in 2 years at 10% p.a. compounded half yearly ? (a) Rs 17423 (b) Rs 18973 (c) Rs 19448 (d) Rs 19880
Answer» (c) Rs 19448 P = Rs 16000, r = 10% p.a., = 5% per half year n = 2 years = 4 half years \(\therefore\) Amount = 16000\(\big(1+\frac{5}{100}\big)^4\) = 16000 x \(\big(\frac{21}{20}\big)^4\) = \(\frac{16000\times194481}{160000}\) = 19448.10 = Rs 19448 (approx)

122.

How much should a sum of Rs 16000 approximately amount to in 2 years at 10% p.a. compounded half yearly ? (a) Rs 17423 (b) Rs 18973 (c) Rs 19448 (d) Rs 19880

Answer»

(c) Rs 19448

P = Rs 16000, r = 10% p.a., = 5% per half year n = 2 years = 4 half years

\(\therefore\) Amount = 16000\(\big(1+\frac{5}{100}\big)^4\) = 16000 x \(\big(\frac{21}{20}\big)^4\)

= \(\frac{16000\times194481}{160000}\)

= 19448.10 = Rs 19448 (approx)

Discussion

123.	Rachna borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs. 1290 as interest compounded annually, find the sum she borrowed.
Answer» Given, Rate = 15% p.a Time = 2 years C.I = Rs. 1290 Let principal = P So, Compound interest \(={P}[({1}+\frac{R}{100})^T-{1}]\) \(={P}[({1}+\frac{15}{100})^2-{1}]\) = 1290 \(={P}[(\frac{23}{20})^2-{1}]\) = 1290 \(={P}\times\frac{129}{400}\) = 1290 \(={P}=\frac{1290\times400}{129}\) = Rs. 4000 Hence, Principal = Rs. 4000

123.

Rachna borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs. 1290 as interest compounded annually, find the sum she borrowed.

Answer»

Given,

Rate = 15% p.a

Time = 2 years

C.I = Rs. 1290

Let principal = P

So,

Compound interest \(={P}[({1}+\frac{R}{100})^T-{1}]\)

\(={P}[({1}+\frac{15}{100})^2-{1}]\) = 1290

\(={P}[(\frac{23}{20})^2-{1}]\) = 1290

\(={P}\times\frac{129}{400}\) = 1290

\(={P}=\frac{1290\times400}{129}\) = Rs. 4000

Hence,

Principal = Rs. 4000

Discussion

124.	The difference between the compound interest and simple interest on a certain sum for 3 years at 10% per annum is ₹ 93. Find the sum.
Answer» Given: Interest rate= 10% per annum Time=3 years Simple interest (SI) = PTR/100 Where P is principle amount, T is time taken, R is rate per annum Let sum is P SI = (P × 3× 10)/ 100 ⇒ SI = (30P)/ 100 ⇒ SI = (3P)/ 10 —————- equation 1 To find the amount we have the formula, Amount (A) = P (1+(R/100))ⁿ Where P is present value, r is rate of interest, n is time in years. Now substituting the values in above formula we get, ∴ CI= P (1 + 10/100)³-P ⇒ CI = P (1+1/10)³-P ⇒ CI = P (11/10)³-P ⇒ CI = P (1331) / (1000) – P ⇒ CI = 331P/1000——– equation 2 Now the difference is (CI – SI) = 331P/1000 – (3P)/ 10 ⇒ 93 = 331P- 300P/1000 ⇒ 93 = 31P/1000 ⇒ P= 93 × 1000/31 ⇒ P= 3 × 1000 ⇒ P= 3000 ∴ Sum = ₹3000

124.

The difference between the compound interest and simple interest on a certain sum for 3 years at 10% per annum is ₹ 93. Find the sum.

Answer»

Given:

Interest rate= 10% per annum

Time=3 years

Simple interest (SI) = PTR/100

Where P is principle amount, T is time taken, R is rate per annum

Let sum is P

SI = (P × 3× 10)/ 100

⇒ SI = (30P)/ 100

⇒ SI = (3P)/ 10 —————- equation 1

To find the amount we have the formula,

Amount (A) = P (1+(R/100))ⁿ

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

∴ CI= P (1 + 10/100)³-P

⇒ CI = P (1+1/10)³-P

⇒ CI = P (11/10)³-P

⇒ CI = P (1331) / (1000) – P

⇒ CI = 331P/1000——– equation 2

Now the difference is

(CI – SI) = 331P/1000 – (3P)/ 10

⇒ 93 = 331P- 300P/1000

⇒ 93 = 31P/1000

⇒ P= 93 × 1000/31

⇒ P= 3 × 1000

⇒ P= 3000

∴ Sum = ₹3000

Discussion

125.	Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum given Rs. 200 as simple interest.
Answer» Given, Rate of simple interest = 10% Time = 2 years Simple interest = RS.200 So, = simple interest \(=\frac{P\times R \times T}{100}\) \(={200}=\frac{P \times 2 \times 10}{100}\) = P = Rs.1000 Rate of compound interest = 10% Time = 2 years = Compound interest \(={p}[({1}+\frac{R}{100})^T-{1}]\) \(={1000}[({1}+\frac{10}{100})^2-{1}]\) \(={1000}[(\frac{11}{10})^T-{1}]\) \(={1000}\times\frac{21}{100}\) = RS. 210

125.

Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum given Rs. 200 as simple interest.

Answer»

Given,

Rate of simple interest = 10%

Time = 2 years

Simple interest = RS.200

So,

= simple interest \(=\frac{P\times R \times T}{100}\)

\(={200}=\frac{P \times 2 \times 10}{100}\)

= P = Rs.1000

Rate of compound interest = 10%

Time = 2 years

= Compound interest \(={p}[({1}+\frac{R}{100})^T-{1}]\)

\(={1000}[({1}+\frac{10}{100})^2-{1}]\)

\(={1000}[(\frac{11}{10})^T-{1}]\)

\(={1000}\times\frac{21}{100}\) = RS. 210

Discussion

126.	Find the compound interest on Rs. 8000 for 9 months at 20% per annum compounded quarterly.
Answer» Given details are, Principal (p) = Rs 8000 Rate (r) = 20 % = 20/4 = 5% (for quarterly) Time = 9 months = 9/3 = 3 (for quarter year) By using the formula, A = P (1 + R/100)ⁿ = 8000 (1 + 5/100)³ = 8000 (105/100)³ = Rs 9261 ∴ Compound Interest = A – P = Rs 9261 – Rs 8000 = Rs 1261

126.

Find the compound interest on Rs. 8000 for 9 months at 20% per annum compounded quarterly.

Answer»

Given details are,

Principal (p) = Rs 8000

Rate (r) = 20 % = 20/4 = 5% (for quarterly)

Time = 9 months = 9/3 = 3 (for quarter year)

By using the formula,

A = P (1 + R/100)ⁿ

= 8000 (1 + 5/100)³

= 8000 (105/100)³

= Rs 9261

∴ Compound Interest = A – P = Rs 9261 – Rs 8000 = Rs 1261

Discussion

127.	At what rate per cent per annum will Rs 32000 yield a compound interest of Rs 5044 in 9 months interest being compounded quarterly ?
Answer» Principal = Rs 32000 Amount = Rs (32000 + 5044) = Rs 37044 Rate = r% p.a. or \(\frac{r}{4}\)% per quarter Time = 9 months = 3 quarters, i.e., n = 3 \(\therefore\) Applying A= P\(\big(1+\frac{r}{100}\big)^n\) , we have 37044 = 32000\(\big(1+\frac{r}{400}\big)^3\) \(\Rightarrow\) \(\frac{37044}{32000}\) = \(\big(1+\frac{r}{400}\big)^3\) \(\Rightarrow\) \(\frac{9261}{8000}\) = \(\big(1+\frac{r}{400}\big)^3\) \(\Rightarrow\) \(\big(\frac{21}{20}\big)^3\) = \(\big(1+\frac{r}{400}\big)^3\) \(\Rightarrow\) 1 + \(\frac{r}{400}\) = \(\frac{21}{20}\) \(\Rightarrow\) \(\frac{r}{100}\) = \(\frac{21}{20}\)-1 = \(\frac{1}{20}\) \(\Rightarrow\) r = \(\frac{400}{20}\) = 20% p.a

127.

At what rate per cent per annum will Rs 32000 yield a compound interest of Rs 5044 in 9 months interest being compounded quarterly ?

Answer»

Principal = Rs 32000

Amount = Rs (32000 + 5044) = Rs 37044

Rate = r% p.a. or \(\frac{r}{4}\)% per quarter

Time = 9 months = 3 quarters, i.e., n = 3

\(\therefore\) Applying A= P\(\big(1+\frac{r}{100}\big)^n\) , we have

37044 = 32000\(\big(1+\frac{r}{400}\big)^3\) \(\Rightarrow\) \(\frac{37044}{32000}\) = \(\big(1+\frac{r}{400}\big)^3\)

\(\Rightarrow\) \(\frac{9261}{8000}\) = \(\big(1+\frac{r}{400}\big)^3\) \(\Rightarrow\) \(\big(\frac{21}{20}\big)^3\) = \(\big(1+\frac{r}{400}\big)^3\)

\(\Rightarrow\) 1 + \(\frac{r}{400}\) = \(\frac{21}{20}\) \(\Rightarrow\) \(\frac{r}{100}\) = \(\frac{21}{20}\)-1 = \(\frac{1}{20}\) \(\Rightarrow\) r = \(\frac{400}{20}\) = 20% p.a

Discussion

128.	Michael borrowed Rs.16000 from a finance company at 10% per annum, compounded half-yearly. What amount of money will discharge his debt after 1 (1/2) years?
Answer» Present value = Rs.16000 Interest rate = 10% per annum Time = (3/2) years ∵ Interest is compounded half-yearly. ∴ Amount (A) = P [1 + (R/2)/100]²ⁿ [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 16000 [1 + (10/2)/100]³ ⇒ A = 16000 [1 + 5/100]³ ⇒ A = 16000 [1 + 1/20]³ ⇒ A = 16000 [21/20]³ ⇒ A = 16000 × 21/20 × 21/20 × 21/20 ⇒ A = 2 × 21 × 21 × 21 ⇒ A = 18522 ∴ Amount = Rs.18522

128.

Michael borrowed Rs.16000 from a finance company at 10% per annum, compounded half-yearly. What amount of money will discharge his debt after 1 (1/2) years?

Answer»

Present value = Rs.16000

Interest rate = 10% per annum

Time = (3/2) years

∵ Interest is compounded half-yearly.

∴ Amount (A) = P [1 + (R/2)/100]²ⁿ [Where, P = Present value

R = Annual interest rate

n = Time in years]

∴ A = 16000 [1 + (10/2)/100]³

⇒ A = 16000 [1 + 5/100]³

⇒ A = 16000 [1 + 1/20]³

⇒ A = 16000 [21/20]³

⇒ A = 16000 × 21/20 × 21/20 × 21/20

⇒ A = 2 × 21 × 21 × 21

⇒ A = 18522

∴ Amount = Rs.18522

Discussion

129.	The interest on a sum of Rs. 2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs. 163.20.
Answer» Given details are, Rate = 4 % per annum CI = Rs 163.20 Principal (P) = Rs 2000 By using the formula, CI = P [(1 + R/100)ⁿ – 1] 163.20 = 2000[(1 + 4/100)ⁿ – 1] 163.20 = 2000 [(1.04)ⁿ -1] 163.20 = 2000 × (1.04)ⁿ – 2000 163.20 + 2000 = 2000 × (1.04)ⁿ 2163.2 = 2000 × (1.04)ⁿ (1.04)ⁿ = 2163.2/2000 (1.04)ⁿ= 1.0816 (1.04)ⁿ = (1.04)² So on comparing both the sides, n = 2 ∴ Time required is 2years.

129.

The interest on a sum of Rs. 2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs. 163.20.

Answer»

Given details are,

Rate = 4 % per annum

CI = Rs 163.20

Principal (P) = Rs 2000

By using the formula,

CI = P [(1 + R/100)ⁿ – 1]

163.20 = 2000[(1 + 4/100)ⁿ – 1]

163.20 = 2000 [(1.04)ⁿ -1]

163.20 = 2000 × (1.04)ⁿ – 2000

163.20 + 2000 = 2000 × (1.04)ⁿ

2163.2 = 2000 × (1.04)ⁿ

(1.04)ⁿ = 2163.2/2000

(1.04)ⁿ= 1.0816

(1.04)ⁿ = (1.04)²

So on comparing both the sides, n = 2

∴ Time required is 2years.

Discussion

130.	A sum of Rs 24000 is borrowed for \(1\frac{1}{2}\) years at the rate of interest 10% per annum compounded semi anually. What is the compound interest (x) ?(a) x < Rs 3000(b) Rs 3000 < x < Rs 4000(c) Rs 4000 < x < Rs 5000(d) x > Rs 5000
Answer» (b) Rs 3000 < x < Rs 4000 P = Rs 24000, Time =\(1\frac{1}{2}\) years = 3 half years, r = 10% p.a. = 5% per half year \(\therefore\) A = 24000\(\big(1+\frac{5}{100}\big)^3\) = 24000 x \(\big(\frac{21}{20}\big)^3\) = 24000 x \(\frac{9261}{8000}\) = Rs 27783. \(\therefore\) C.I. = Rs 27783 – Rs 24000 = Rs 3783 \(\therefore\) C.I. (x) lies between Rs 3000 and Rs 4000.

130.

A sum of Rs 24000 is borrowed for \(1\frac{1}{2}\) years at the rate of interest 10% per annum compounded semi anually. What is the compound interest (x) ?(a) x < Rs 3000(b) Rs 3000 < x < Rs 4000(c) Rs 4000 < x < Rs 5000(d) x > Rs 5000

Answer»

(b) Rs 3000 < x < Rs 4000

P = Rs 24000, Time =\(1\frac{1}{2}\) years = 3 half years,

r = 10% p.a. = 5% per half year

\(\therefore\) A = 24000\(\big(1+\frac{5}{100}\big)^3\) = 24000 x \(\big(\frac{21}{20}\big)^3\)

= 24000 x \(\frac{9261}{8000}\) = Rs 27783.

\(\therefore\) C.I. = Rs 27783 – Rs 24000 = Rs 3783

\(\therefore\) C.I. (x) lies between Rs 3000 and Rs 4000.

Discussion

131.	If the simple interest on a sum of money at 10% per annum for 3 years is Rs. 1500, then the compound interest on the same sum at the same rate for the same period is A. Rs. 1655 B. Rs. 1155 C. Rs. 1555 D. Rs. 1855
Answer» Simple interest = Rs.1500 Interest rate = 10% per annum Time = 3 years Simple interest (SI) = PRT/100 [where, P = Present value R = Interest rate ∴ 1500 = (P × 10 × 3)/100 T = Time] ⇒ 1500 = P × 30/100 ⇒ 1500 = P × 3/10 ⇒ P = 1500 × 10/3 ⇒ P = 500 × 10 ⇒ P = 5000 ∴ Sum = Rs.5000 Now, Amount (A) = P (1 + R/100)ⁿ [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 5000 [1 + 10/100]³ ⇒ A = 5000 [1 + 1/10]³ ⇒ A = 5000 [11/10]³ ⇒ A = 5000 × 11/10 × 11/10 × 11/10 ⇒ A = 5000 × 1331/1000 ⇒ A = 5 × 1331 ⇒ A = 6655 ∴ Amount = Rs.6655 ∴ Compound interest = Rs.(6655 – 5000) = Rs.1655

131.

If the simple interest on a sum of money at 10% per annum for 3 years is Rs. 1500, then the compound interest on the same sum at the same rate for the same period is A. Rs. 1655 B. Rs. 1155 C. Rs. 1555 D. Rs. 1855

Answer»

Simple interest = Rs.1500

Interest rate = 10% per annum

Time = 3 years

Simple interest (SI) = PRT/100 [where, P = Present value

R = Interest rate

∴ 1500 = (P × 10 × 3)/100 T = Time]

⇒ 1500 = P × 30/100

⇒ 1500 = P × 3/10

⇒ P = 1500 × 10/3

⇒ P = 500 × 10

⇒ P = 5000

∴ Sum = Rs.5000

Now,

Amount (A) = P (1 + R/100)ⁿ [Where, P = Present value

R = Annual interest rate

n = Time in years]

∴ A = 5000 [1 + 10/100]³

⇒ A = 5000 [1 + 1/10]³

⇒ A = 5000 [11/10]³

⇒ A = 5000 × 11/10 × 11/10 × 11/10

⇒ A = 5000 × 1331/1000

⇒ A = 5 × 1331

⇒ A = 6655

∴ Amount = Rs.6655

∴ Compound interest = Rs.(6655 – 5000)

= Rs.1655

Discussion

132.	The compound interest on Rs. 40000 at 6% per annum for 6 months, compounded quarterly, is A. Rs. 1209 B. Rs. 1902 C. Rs. 1200 D. Rs. 1306
Answer» Present value, P = Rs.40000 Interest rate, R = 6% per annum Time, n = 6 months = 1/2 years ∵ Compounded quarterly. ∴ Amount (A) = P [1 + (R/4)/100]⁴ⁿ [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 40000 [1 +(6/4) /100]² [4n = 4 × 1/2] ⇒ A = 40000 [1 + 3/200]² ⇒ A = 40000 [1 + 3/200]² ⇒ A = 40000 [203/200]² ⇒ A = 40000 × 203/200 × 203/200 ⇒ A = 40000 × 203/200 × 203/200 ⇒ A = 200 × 203 × 203/200 ⇒ A = 1 × 203 × 203 ⇒ A = 41209 ∴ Amount = Rs.41209 ∴ Compound interest = Rs.(41209 – 40000) [∵CI = A – P] = Rs.1209

132.

The compound interest on Rs. 40000 at 6% per annum for 6 months, compounded quarterly, is A. Rs. 1209 B. Rs. 1902 C. Rs. 1200 D. Rs. 1306

Answer»

Present value, P = Rs.40000

Interest rate, R = 6% per annum

Time, n = 6 months = 1/2 years

∵ Compounded quarterly.

∴ Amount (A) = P [1 + (R/4)/100]⁴ⁿ [Where, P = Present value

R = Annual interest rate

n = Time in years]

∴ A = 40000 [1 +(6/4) /100]² [4n = 4 × 1/2]

⇒ A = 40000 [1 + 3/200]²

⇒ A = 40000 [203/200]²

⇒ A = 40000 × 203/200 × 203/200

⇒ A = 200 × 203 × 203/200

⇒ A = 1 × 203 × 203

⇒ A = 41209

∴ Amount = Rs.41209

∴ Compound interest = Rs.(41209 – 40000) [∵CI = A – P]

= Rs.1209

Discussion

133.	Find the compound interest on Rs. 8000 for 9 months at 20% per annum compounded quarterly
Answer» Principal = Rs.8000 Time =9 months Rate = 20% per annum Interest is compounded quarterly, So Rate of interest will be counted as 20/4 = 5% and time will be 9/3 = 3 Quarter We know that, \({A}={P}\times(1+\frac{R}{100})^t\) \(\Rightarrow={8000}\times({1}+\frac{5}{100})^3\) \(\Rightarrow{A}={8000}\times({1}+\frac{105}{100})^3\) =Rs. 9261 Hence, Compound Interest = Rs. 9261 – Rs 8000 = Rs. 1261

133.

Find the compound interest on Rs. 8000 for 9 months at 20% per annum compounded quarterly

Answer»

Principal = Rs.8000

Time =9 months

Rate = 20% per annum

Interest is compounded quarterly, So Rate of interest will be counted as 20/4 = 5% and time will be 9/3 = 3 Quarter

We know that, \({A}={P}\times(1+\frac{R}{100})^t\)

\(\Rightarrow={8000}\times({1}+\frac{5}{100})^3\)

\(\Rightarrow{A}={8000}\times({1}+\frac{105}{100})^3\) =Rs. 9261

Hence, Compound Interest = Rs. 9261 – Rs 8000 = Rs. 1261

Discussion

134.	Swati borrowed Rs. 40960 from a bank to buy a piece of land. If the bank charges 12 (1/2)% per annum, compounded half-yearly, what amount will she have to pay after 1 (1/2)% years? Also, find the interest paid by her.
Answer» Present value, P = Rs.40960 Interest rate, R = (25/2)% per annum Time, n = 3/2 years ∵ Compounded half-yearly. ∴ Amount (A) = P [1 + (R/2)/100]²ⁿ [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 40960 [1 + (25/4)/100]³ [R = 25/2 and n = 3/2 years] ⇒ A = 40960 [1 + 1/16]³ ⇒ A = 40960 [17/16]³ ⇒ A = 40960 × 4913/4096 ⇒ A = 10 × 4913 ⇒ A = 49130 ∴ Amount = Rs.49130 ∴ Compound interest = Rs.(49130 – 40960) [∵CI = A – P] = Rs.8170

134.

Swati borrowed Rs. 40960 from a bank to buy a piece of land. If the bank charges 12 (1/2)% per annum, compounded half-yearly, what amount will she have to pay after 1 (1/2)% years? Also, find the interest paid by her.

Answer»

Present value, P = Rs.40960

Interest rate, R = (25/2)% per annum

Time, n = 3/2 years

∵ Compounded half-yearly.

∴ Amount (A) = P [1 + (R/2)/100]²ⁿ [Where, P = Present value

R = Annual interest rate

n = Time in years]

∴ A = 40960 [1 + (25/4)/100]³ [R = 25/2 and n = 3/2 years]

⇒ A = 40960 [1 + 1/16]³

⇒ A = 40960 [17/16]³

⇒ A = 40960 × 4913/4096

⇒ A = 10 × 4913

⇒ A = 49130

∴ Amount = Rs.49130

∴ Compound interest = Rs.(49130 – 40960) [∵CI = A – P]

= Rs.8170

Discussion

135.	Find the rate percent per annum, if Rs. 2000 amount to Rs. 2315.25 in an year and a half, interest being compounded six monthly.
Answer» Given, Principal = Rs.2000 Amount = Rs.2315.25 Time \(={1}\frac{1}{2}\) years \(=\frac{3}{2}\) years Let rate = R % per annum A \(={P}[({1}+\frac{R}{100})^T]\) \(={2000}[({1}+\frac{R}{100})^{\frac{3}{2}}]\) = 2315.25 \(=({1}+\frac{R}{100})^{\frac{3}{2}}\) = 1.1576 \(={1}+\frac{R}{100}\) = 1.1025 \(=\frac{R}{100}\) = 0.1025 = R = 10.25% Hence , Rate = 10.25%

135.

Find the rate percent per annum, if Rs. 2000 amount to Rs. 2315.25 in an year and a half, interest being compounded six monthly.

Answer»

Given,

Principal = Rs.2000

Amount = Rs.2315.25

Time \(={1}\frac{1}{2}\) years \(=\frac{3}{2}\) years

Let rate = R % per annum

A \(={P}[({1}+\frac{R}{100})^T]\)

\(={2000}[({1}+\frac{R}{100})^{\frac{3}{2}}]\) = 2315.25

\(=({1}+\frac{R}{100})^{\frac{3}{2}}\) = 1.1576

\(={1}+\frac{R}{100}\) = 1.1025

\(=\frac{R}{100}\) = 0.1025

= R = 10.25%

Hence ,

Rate = 10.25%

Discussion

136.	Find the rate percent per annum, if Rs. 2000 amount to Rs. 2315.25 in a year and a half, interest being compounded six monthly.
Answer» Given details are, Principal = Rs 2000 Amount = Rs 2315.25 Time = 1 ½ years = 3/2 years Let rate be = R % per annum By using the formula, A = P (1 + R/100)ⁿ 2315.25 = 2000 (1 + R/100)^3/2 (1 + R/100)^3/2 = 2315.25/2000 (1 + R/100)^3/2 = (1.1576) (1 + R/100) = 1.1025 R/100 = 1.1025 – 1 = 0.1025 × 100 = 10.25 ∴ Required Rate is 10.25% per annum.

136.

Find the rate percent per annum, if Rs. 2000 amount to Rs. 2315.25 in a year and a half, interest being compounded six monthly.

Answer»

Given details are,

Principal = Rs 2000

Amount = Rs 2315.25

Time = 1 ½ years = 3/2 years

Let rate be = R % per annum

By using the formula,

A = P (1 + R/100)ⁿ

2315.25 = 2000 (1 + R/100)^3/2

(1 + R/100)^3/2 = 2315.25/2000

(1 + R/100)^3/2 = (1.1576)

(1 + R/100) = 1.1025

R/100 = 1.1025 – 1

= 0.1025 × 100

= 10.25

∴ Required Rate is 10.25% per annum.

Discussion

137.	Neha borrowed Rs. 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months?
Answer» Present value = Rs.24000 Interest rate = 10 % per annum Time = 2 years 3 month = (2 + 1/4) years = \(2\frac{1}{4}\) years. Amount (A) = P (1 + R/100)ⁿ× [1 + (1/4 × R)/100] [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 24000 (1 + 10/100)² × [1 + (1/4 × 10)/100] ⇒ A = 24000 (1 + 1/10)² × [1 + 1/40] ⇒ A = 24000 (11/10)² × [41/40] ⇒ A = 24000 × 121/100 × 41/40 ⇒ A = 24 × 121 × 41/4 ⇒ A = 6 × 121 × 41 ⇒ A = 29766 ∴ Amount = Rs.29766 ∴ Neha should pay Rs. 29766 to the bank after 2 years 3 months.

137.

Neha borrowed Rs. 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months?

Answer»

Present value = Rs.24000

Interest rate = 10 % per annum

Time = 2 years 3 month = (2 + 1/4) years = \(2\frac{1}{4}\) years.

Amount (A) = P (1 + R/100)ⁿ× [1 + (1/4 × R)/100]

[Where, P = Present value

R = Annual interest rate

n = Time in years]

∴ A = 24000 (1 + 10/100)² × [1 + (1/4 × 10)/100]

⇒ A = 24000 (1 + 1/10)² × [1 + 1/40]

⇒ A = 24000 (11/10)² × [41/40]

⇒ A = 24000 × 121/100 × 41/40

⇒ A = 24 × 121 × 41/4

⇒ A = 6 × 121 × 41

⇒ A = 29766

∴ Amount = Rs.29766

∴ Neha should pay Rs. 29766 to the bank after 2 years 3 months.

Discussion

138.	Find the amount and the compound interest on Rs. 31250 for 1 (1/2)% years at 8% per annum, compounded half-yearly.
Answer» Present value, P = Rs.31250 Interest rate, R = 8% per annum Time, n = (3/2) years ∵ Compounded half-yearly. ∴ Amount (A) = P [1 + (R/2)/100]²ⁿ [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 31250 [1 + (8/2)/100]³ [2n = 2 × 3/2] ⇒ A = 31250 [1 + 4/100]³ ⇒ A = 31250 [1 + 1/25]³ ⇒ A = 31250 [26/25]³ ⇒ A = 31250 × 17576/15625 ⇒ A = 2 × 17576 ⇒ A = 35152 ∴ Amount = Rs.35152 ∴ Compound interest = Rs.(35152 – 31250) [∵CI = A – P] = Rs.3902

138.

Find the amount and the compound interest on Rs. 31250 for 1 (1/2)% years at 8% per annum, compounded half-yearly.

Answer»

Present value, P = Rs.31250

Interest rate, R = 8% per annum

Time, n = (3/2) years

∵ Compounded half-yearly.

∴ Amount (A) = P [1 + (R/2)/100]²ⁿ [Where, P = Present value

R = Annual interest rate

n = Time in years]

∴ A = 31250 [1 + (8/2)/100]³ [2n = 2 × 3/2]

⇒ A = 31250 [1 + 4/100]³

⇒ A = 31250 [1 + 1/25]³

⇒ A = 31250 [26/25]³

⇒ A = 31250 × 17576/15625

⇒ A = 2 × 17576

⇒ A = 35152

∴ Amount = Rs.35152

∴ Compound interest = Rs.(35152 – 31250) [∵CI = A – P]

= Rs.3902

Discussion

139.	At what rate per cent per annum will 5000 amount to Rs. 5832 in 2 years, compounded annually? A. 11% B. 10% C. 9% D. 8%
Answer» Present value, P = Rs.5000 Amount, A = Rs.5832 Time, n = 2 years Now, Amount (A) = P (1 + R/100)ⁿ [Where, A = Amount with compound interest P = Present value R = Annual interest rate n = Time in years] ∴ Amount (A) = P (1 + R/100)ⁿ ⇒ 5832 = 5000 (1 + R/100)² ⇒ (1 + R/100)² = 5832/5000 ⇒ (1 + R/100)² = 2916/2500 ⇒ (1 + R/100)²= (54/50)2 ⇒ 1 + R/100 = 54/50 ⇒ R/100 = (54/50) - 1 ⇒ R/100 = (54 – 50)/50 ⇒ R/100 = 4/50 ⇒ R = 400/50 ⇒ R = 8 ∴ Rate = 8 %.

139.

At what rate per cent per annum will 5000 amount to Rs. 5832 in 2 years, compounded annually? A. 11% B. 10% C. 9% D. 8%

Answer»

Present value, P = Rs.5000

Amount, A = Rs.5832

Time, n = 2 years

Now,

Amount (A) = P (1 + R/100)ⁿ [Where, A = Amount with compound interest

P = Present value

R = Annual interest rate

n = Time in years]

∴ Amount (A) = P (1 + R/100)ⁿ

⇒ 5832 = 5000 (1 + R/100)²

⇒ (1 + R/100)² = 5832/5000

⇒ (1 + R/100)² = 2916/2500

⇒ (1 + R/100)²= (54/50)2

⇒ 1 + R/100 = 54/50

⇒ R/100 = (54/50) - 1

⇒ R/100 = (54 – 50)/50

⇒ R/100 = 4/50

⇒ R = 400/50

⇒ R = 8

∴ Rate = 8 %.

Discussion

140.	Find the compound interest on Rs. 15625 for 9 months, at 16% per annum, compounded quarterly.
Answer» Given, Principal = Rs. 15625 Rate = 16% per annum \(=\frac{16}{4}\) = 4% quarterly Time = 9 months \(=\frac{9}{12}\) years \(=\frac{9}{12}\times{4}\) = 3 quarters Hence, Compound interest \(={P}[({1}+\frac{R}{100})^T-{1}]\) \(={15625}[({1}+\frac{4}{100})^3-{1}]\) \(={15625}[(\frac{26}{25})^3-{1}]\) = \({15625}\times\frac{1951}{15625}\) = Rs. 1951

140.

Find the compound interest on Rs. 15625 for 9 months, at 16% per annum, compounded quarterly.

Answer»

Given,

Principal = Rs. 15625

Rate = 16% per annum \(=\frac{16}{4}\) = 4% quarterly

Time = 9 months \(=\frac{9}{12}\) years

\(=\frac{9}{12}\times{4}\) = 3 quarters

Hence,

Compound interest \(={P}[({1}+\frac{R}{100})^T-{1}]\)

\(={15625}[({1}+\frac{4}{100})^3-{1}]\)

\(={15625}[(\frac{26}{25})^3-{1}]\)

= \({15625}\times\frac{1951}{15625}\)

= Rs. 1951

Discussion

141.	Swati borrowed ₹ 40960 from a bank to buy a piece of land. If the bank charges 12 ½ % per annum, compounded half yearly. What amount will she have to pay after 1 ½ years? Also, find the interest paid by her.
Answer» Given: Present value= ₹ 40960 Interest rate= 12 ½ % per annum Time=1 ½ = 3/2 years and compounded half yearly To find the amount we have the formula, Amount (A) = P (1+(R/100)²ⁿ Where P is present value, r is rate of interest, n is time in years. Now substituting the values in above formula we get, ∴ A = 40960 (1 + (25/4)/100)³ ⇒ A = 40960 (1+1/16)³ ⇒ A = 40960 (17/16)³ ⇒ A = 40960 × 4913/4096 ⇒ A = ₹ 49130 ∴ Compound interest = A – P = 49130 – 40960 = ₹ 8170

141.

Swati borrowed ₹ 40960 from a bank to buy a piece of land. If the bank charges 12 ½ % per annum, compounded half yearly. What amount will she have to pay after 1 ½ years? Also, find the interest paid by her.

Answer»

Given: Present value= ₹ 40960

Interest rate= 12 ½ % per annum

Time=1 ½ = 3/2 years and compounded half yearly

To find the amount we have the formula,

Amount (A) = P (1+(R/100)²ⁿ

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

∴ A = 40960 (1 + (25/4)/100)³

⇒ A = 40960 (1+1/16)³

⇒ A = 40960 (17/16)³

⇒ A = 40960 × 4913/4096

⇒ A = ₹ 49130

∴ Compound interest = A – P

= 49130 – 40960 = ₹ 8170

Discussion

142.	Find the rate percent per annum if Rs. 2000 amount to Rs. 2662 in 1 1/2 years, interest being compounded half-yearly?
Answer» Given, Principal = Rs.2000 Amount = Rs.2662 Time \(={1}\frac{1}{2}\) years \(=\frac{3}{2}\times2\) = 3 half years Let rate = R% per annum, \(\frac{R}{2}\text%\) half yearly So, A = \({P}[({1}+\frac{R}{100})^T]\) \(={2000}[({1}+\frac{R}{2\times100})^3]\) = 2662 \(=({1}+\frac{R}{100})^3\) \(=\frac{2662}{2000}\) \(=\frac{1331}{1000}\) \(=(\frac{11}{10})^3\) \(={1}+\frac{R}{200}\) \(=\frac{11}{10}\) \(=\frac{R}{200}\) \(=\frac{11}{10}-{1}\) \(=\frac{1}{10}\) = R \(=\frac{1\times200}{10}\) = 20% per annum Hence , Rate = 20% per annum

142.

Find the rate percent per annum if Rs. 2000 amount to Rs. 2662 in 1 1/2 years, interest being compounded half-yearly?

Answer»

Given,

Principal = Rs.2000

Amount = Rs.2662

Time \(={1}\frac{1}{2}\) years \(=\frac{3}{2}\times2\) = 3 half years

Let rate = R% per annum, \(\frac{R}{2}\text%\) half yearly

So,

A = \({P}[({1}+\frac{R}{100})^T]\)

\(={2000}[({1}+\frac{R}{2\times100})^3]\) = 2662

\(=({1}+\frac{R}{100})^3\) \(=\frac{2662}{2000}\) \(=\frac{1331}{1000}\) \(=(\frac{11}{10})^3\)

\(={1}+\frac{R}{200}\) \(=\frac{11}{10}\)

\(=\frac{R}{200}\) \(=\frac{11}{10}-{1}\) \(=\frac{1}{10}\)

= R \(=\frac{1\times200}{10}\) = 20% per annum

Hence ,

Rate = 20% per annum

Discussion

143.	Ramu borrowed Rs. 15625 from a finance company to buy scooter. If the rate of interest be 16% per annum compounded annually, what payment will he have to make after 2 ¼ years?
Answer» Given details are, Principal (p) = Rs 15625 Rate (r) = 16% Time (t) = 2 ¼ years By using the formula, A = P (1 + R/100 × 1 + R/100) = 15625 (1 + 16/100)² × (1 + (16/4)/100) = 15625 (1 + 16/100)² × (1 + 4/100) = 15625 (1.16)² × (1.04) = Rs 21866 ∴ Amount after 2 ¼ years is Rs 21866.

143.

Ramu borrowed Rs. 15625 from a finance company to buy scooter. If the rate of interest be 16% per annum compounded annually, what payment will he have to make after 2 ¼ years?

Answer»

Given details are,

Principal (p) = Rs 15625

Rate (r) = 16%

Time (t) = 2 ¼ years

By using the formula,

A = P (1 + R/100 × 1 + R/100)

= 15625 (1 + 16/100)² × (1 + (16/4)/100)

= 15625 (1 + 16/100)² × (1 + 4/100)

= 15625 (1.16)² × (1.04)

= Rs 21866

∴ Amount after 2 ¼ years is Rs 21866.

Discussion

144.	The compound interest on ₹ 6250 at 8% per annum for 1 year, compounded half yearly, is(a) ₹ 500 (b) ₹ 510 (c) ₹ 550 (d) ₹ 512.50
Answer» (b) ₹ 510 Explanation: Present value= ₹ 6250 Interest rate= 8% per annum Time=1 year To find the amount we have the formula, Amount (A) = P (1+(R/100))²ⁿ Where P is present value, r is rate of interest, n is time in years. Now substituting the values in above formula we get, ∴ A = 6250 (1 + (8/2)/100)² ⇒ A = 6250 (1+4/100)² ⇒ A = 6250 (1+1/25)² ⇒ A = 6250 (26/25)² ⇒ A = 10 × 26 × 26 ⇒ A = ₹ 6760 ∴ Compound interest = A – P = 6760 – 6250 = ₹ 510

144.

The compound interest on ₹ 6250 at 8% per annum for 1 year, compounded half yearly, is(a) ₹ 500 (b) ₹ 510 (c) ₹ 550 (d) ₹ 512.50

Answer»

(b) ₹ 510

Explanation:

Present value= ₹ 6250

Interest rate= 8% per annum

Time=1 year

To find the amount we have the formula,

Amount (A) = P (1+(R/100))²ⁿ

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

∴ A = 6250 (1 + (8/2)/100)²

⇒ A = 6250 (1+4/100)²

⇒ A = 6250 (1+1/25)²

⇒ A = 6250 (26/25)²

⇒ A = 10 × 26 × 26

⇒ A = ₹ 6760

∴ Compound interest = A – P

= 6760 – 6250 = ₹ 510

Discussion

145.	Sandeep deposited Rs 25000 in a bank which pays 8% interest compounded annually. How much would he get back after two years?
Answer» Interest in the first year = 25000 × \(\frac{8}{100}\) = Rs 2000 Principal in the second year = 25000 + 2000 = 27000 Interest in the second year = 27000 × \(\frac{8}{100}\) = 2160 Total amount gets back at the end of 2 year = 27000 + 2160 = 29160

145.

Sandeep deposited Rs 25000 in a bank which pays 8% interest compounded annually. How much would he get back after two years?

Answer»

Interest in the first year

= 25000 × \(\frac{8}{100}\)

= Rs 2000

Principal in the second year

= 25000 + 2000

= 27000

Interest in the second year

= 27000 × \(\frac{8}{100}\)

= 2160

Total amount gets back at the end of 2 year

= 27000 + 2160

= 29160

Discussion

146.	Find the amount and the compound interest on Rs. 2500 for 2 years at 10% per annum, compounded annually.
Answer» Present value = Rs.2500 Interest rate = 10% per annum Time = 2 years Amount (A) = P (1 + R/100)ⁿ [Where, P = Present value R = Annual interest rate n = Time in years] ∴ A = 2500 (1 + 10/100)² ⇒ A = 2500 (11/10)² ⇒ A = 2500 × 121/100 ⇒ A = 25 × 121 ⇒ A = 3025 ∴ Amount = Rs.3025 ∴ Compound interest = Rs.(3025 – 2500) = Rs.525

146.

Find the amount and the compound interest on Rs. 2500 for 2 years at 10% per annum, compounded annually.

Answer»

Present value = Rs.2500

Interest rate = 10% per annum

Time = 2 years

Amount (A) = P (1 + R/100)ⁿ

[Where, P = Present value

R = Annual interest rate

n = Time in years]

∴ A = 2500 (1 + 10/100)²

⇒ A = 2500 (11/10)²

⇒ A = 2500 × 121/100

⇒ A = 25 × 121

⇒ A = 3025

∴ Amount = Rs.3025

∴ Compound interest = Rs.(3025 – 2500)

= Rs.525

Discussion

147.	At what rate per cent per annum will Rs. 4000 amount to Rs. 4410 in 2 years when compounded annually?
Answer» Let rate = R % per annum P = Rs.4000 A = Rs.4410 Time = 2 years Now, Amount (A) = P (1 + R/100)ⁿ [Where, A = Amount with compound interest P = Present value R = Annual interest rate n = Time in years] ∴ A = P (1 + R/100)² ⇒ 4410 = 4000 (1 + R/100)² ⇒ (1 + R/100)² = 4410/4000 ⇒ (1 + R/100)² = 441/400 ⇒ (1 + R/100) = √(441/400) ⇒ R/100 = (21/20) - 1 ⇒ R/100 = (21 – 20)/20 ⇒ R/100 = 1/20 ⇒ R = 100/20 ⇒ R = 5 ∴ Rate = 5% per annuam.

147.

At what rate per cent per annum will Rs. 4000 amount to Rs. 4410 in 2 years when compounded annually?

Answer»

Let rate = R % per annum

P = Rs.4000

A = Rs.4410

Time = 2 years

Now,

Amount (A) = P (1 + R/100)ⁿ [Where, A = Amount with compound interest

P = Present value

R = Annual interest rate

n = Time in years]

∴ A = P (1 + R/100)²

⇒ 4410 = 4000 (1 + R/100)²

⇒ (1 + R/100)² = 4410/4000

⇒ (1 + R/100)² = 441/400

⇒ (1 + R/100) = √(441/400)

⇒ R/100 = (21/20) - 1

⇒ R/100 = (21 – 20)/20

⇒ R/100 = 1/20

⇒ R = 100/20

⇒ R = 5

∴ Rate = 5% per annuam.

Discussion

148.	Find the amount that David would receive if he invests Rs. 8192 for 18 months at 12% per annum, the interest being compounded half-yearly.
Answer» Given, Principal = Rs.8192 Rate = 12% p.a =\(\frac{12}{2}\)= 6% half yearly Time = 18 months \(=\frac{18}{12}\)\(={1}\frac{1}{2}\) years = 3 half years Hence, Amount \(={P}[({1}+\frac{R}{100})^{time}]\) \(={8192}[({1}+\frac{6}{100})^3]\) \(={8192}\times(\frac{53}{50})^3\) \(={8192}\times\frac{53}{50}\times\frac{53}{50}\times\frac{53}{50}\)= Rs.9826 So, David receives Rs.9826 after 18 months

148.

Find the amount that David would receive if he invests Rs. 8192 for 18 months at 12% per annum, the interest being compounded half-yearly.

Answer»

Given,

Principal = Rs.8192

Rate = 12% p.a =\(\frac{12}{2}\)= 6% half yearly

Time = 18 months \(=\frac{18}{12}\)\(={1}\frac{1}{2}\) years = 3 half years

Hence,

Amount \(={P}[({1}+\frac{R}{100})^{time}]\)

\(={8192}[({1}+\frac{6}{100})^3]\)

\(={8192}\times(\frac{53}{50})^3\)

\(={8192}\times\frac{53}{50}\times\frac{53}{50}\times\frac{53}{50}\)= Rs.9826

So,

David receives Rs.9826 after 18 months

Discussion

149.	Find the amount and the compound interest on ₹ 12800 for 1 year at 7 ½ % per annum, compounded half-yearly.
Answer» Given: Present value= ₹ 12800 Interest rate= 7 ½ % per annum Time=1 year and compounded half yearly To find the amount we have the formula, Amount (A) = P (1+(R/100)²ⁿ Where P is present value, r is rate of interest, n is time in years. Now substituting the values in above formula we get, ∴ A = 12800 (1 + (15/4)/100)² ⇒ A = 12800 (1+3/80)² ⇒ A = 12800 (83/80)² ⇒ A = 128 × 6889/64 ⇒ A = ₹ 13778 ∴ Compound interest = A – P = 13778 – 12800 = ₹ 978

149.

Find the amount and the compound interest on ₹ 12800 for 1 year at 7 ½ % per annum, compounded half-yearly.

Answer»

Given:

Present value= ₹ 12800

Interest rate= 7 ½ % per annum

Time=1 year and compounded half yearly

To find the amount we have the formula,

Amount (A) = P (1+(R/100)²ⁿ

Where P is present value, r is rate of interest, n is time in years.

Now substituting the values in above formula we get,

∴ A = 12800 (1 + (15/4)/100)²

⇒ A = 12800 (1+3/80)²

⇒ A = 12800 (83/80)²

⇒ A = 128 × 6889/64

⇒ A = ₹ 13778

∴ Compound interest = A – P

= 13778 – 12800 = ₹ 978

Discussion

150.	Find the rate percent per annum if Rs. 2000 amount to Rs. 2662 in 1 ½ years, interest being compounded half-yearly?
Answer» Given details are, Principal = Rs 2000 Amount = Rs 2662 Time = 1 ½ years = 3/2 × 2 = 3 half years Let rate be = R% per annum = R/2 % half yearly By using the formula, A = P (1 + R/100)ⁿ 2662 = 2000 (1 + R/2×100)³ (1 + R/200)³ = 1331/1000 (1 + R/100)³ = (11/10)³ By cancelling the powers on both sides, (1 + R/200) = (11/10) R/200 = 11/10 – 1 = (11-10)/10 = 1/10 R = 200/10 = 20% ∴ Required Rate is 20% per annum.

150.

Find the rate percent per annum if Rs. 2000 amount to Rs. 2662 in 1 ½ years, interest being compounded half-yearly?

Answer»

Given details are,

Principal = Rs 2000

Amount = Rs 2662

Time = 1 ½ years = 3/2 × 2 = 3 half years

Let rate be = R% per annum = R/2 % half yearly

By using the formula,

A = P (1 + R/100)ⁿ

2662 = 2000 (1 + R/2×100)³

(1 + R/200)³ = 1331/1000

(1 + R/100)³ = (11/10)³

By cancelling the powers on both sides,

(1 + R/200) = (11/10)

R/200 = 11/10 – 1

= (11-10)/10

= 1/10

R = 200/10

= 20%

∴ Required Rate is 20% per annum.

Discussion

Explore topic-wise InterviewSolutions in .

Compute the amount and the compound interest by using the formulae when : Principal = Rs 160000, Rate = 10 paise per rupee per annum compounded half yearly, Time = 2 years.

At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs. 410 in 2 years?

Compute the amount and the compound interest by using the formulae when : Principal = Rs 2000, Rate = 4 paise per rupee per annum, Time = 3 years.

Compute the amount and the compound interest by using the formulae when : Principal = Rs 12800, Rate = 7 ½ %, Time = 3 years.

The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.

Nirmala deposited Rs. 20,000 in a bank where compound interest of rate 7% is computed annually Rs. 5000 was with drawn after one year. How much amount she will get after 2 years.

Rs 8000 invested at compound interest gives Rs 1261 as interest after 3 years. The rate of interest per annum is (a) 25% (b) 17.5% (c) 10% (d) 5%

In how many years will a sum of Rs 800 at 10% per annum compound interest, compounded semiannually become Rs 926.10 ? (a) 1 year (b) 3 years(c) 2 years(d) 1\(\frac{1}{2}\) years

The half life of Uranium - 233 is 160000 years, i.e., Uranium 233 decays at a constant rate in such a way that it reduces to 50% in 160000 years. In how many years will it reduce to 25% ? (a) 80000 years (b) 240000 years (c) 320000 years (d) 40000 years

What will be the compound interest on Rs. 4000 in two years when rate of interest is 5% per annum?

Find the compound interest when principal = Rs 3000, rate = 5% per annum and time = 2 years.

Compute the amount and the compound interest by using the formulae when : Principal = Rs 3000, Rate = 5%, Time = 2 years.

What will be the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs. 1200 as simple interest.

If the rate of interest be 4% per annum for first year, 5% per annum for second year and 6% per annum for third year, then what is the compound interest on Rs 10000 for 3 years ?

Find the compound interest on Rs. 64000 for 1 year at the rate of 10% per annum compounded quarterly.

The compound interest on Rs. 30,000 at 7% per annum is Rs 4347. The period (in years) is (a) 2 (b) \(2\frac{1}{2}\) (c) 3 (d) 4

The difference between simple interest and compound interest on Rs 1200 for one year at 10% per annum reckoned half yearly is (a) Rs 2.50 (b) Rs 3 (c) Rs 3.75 (d) Rs 4

Find the compound interest at the rate of 5% for three years on that principal which in three years at the rate of 5% per annum gives Rs. 12000 as simple interest.

Find the compound interest when principal = Rs. 3000, rate = 5% per annum and time = 2 years.

The difference between the compound interest and simple interest on a certain sum for 2 years at 6% per annum is ₹ 90. Find the sum.

Mewa lal borrowed Rs. 20000 from his friend Rooplal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.

How much should a sum of Rs 16000 approximately amount to in 2 years at 10% p.a. compounded half yearly ? (a) Rs 17423 (b) Rs 18973 (c) Rs 19448 (d) Rs 19880

Rachna borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs. 1290 as interest compounded annually, find the sum she borrowed.

The difference between the compound interest and simple interest on a certain sum for 3 years at 10% per annum is ₹ 93. Find the sum.

Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum given Rs. 200 as simple interest.

Find the compound interest on Rs. 8000 for 9 months at 20% per annum compounded quarterly.

At what rate per cent per annum will Rs 32000 yield a compound interest of Rs 5044 in 9 months interest being compounded quarterly ?

Michael borrowed Rs.16000 from a finance company at 10% per annum, compounded half-yearly. What amount of money will discharge his debt after 1 (1/2) years?

The interest on a sum of Rs. 2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs. 163.20.

A sum of Rs 24000 is borrowed for \(1\frac{1}{2}\) years at the rate of interest 10% per annum compounded semi anually. What is the compound interest (x) ?(a) x &lt; Rs 3000(b) Rs 3000 &lt; x &lt; Rs 4000(c) Rs 4000 &lt; x &lt; Rs 5000(d) x &gt; Rs 5000

If the simple interest on a sum of money at 10% per annum for 3 years is Rs. 1500, then the compound interest on the same sum at the same rate for the same period is A. Rs. 1655 B. Rs. 1155 C. Rs. 1555 D. Rs. 1855

The compound interest on Rs. 40000 at 6% per annum for 6 months, compounded quarterly, is A. Rs. 1209 B. Rs. 1902 C. Rs. 1200 D. Rs. 1306

Find the compound interest on Rs. 8000 for 9 months at 20% per annum compounded quarterly

Swati borrowed Rs. 40960 from a bank to buy a piece of land. If the bank charges 12 (1/2)% per annum, compounded half-yearly, what amount will she have to pay after 1 (1/2)% years? Also, find the interest paid by her.

Find the rate percent per annum, if Rs. 2000 amount to Rs. 2315.25 in an year and a half, interest being compounded six monthly.

Find the rate percent per annum, if Rs. 2000 amount to Rs. 2315.25 in a year and a half, interest being compounded six monthly.

Neha borrowed Rs. 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months?

Find the amount and the compound interest on Rs. 31250 for 1 (1/2)% years at 8% per annum, compounded half-yearly.

At what rate per cent per annum will 5000 amount to Rs. 5832 in 2 years, compounded annually? A. 11% B. 10% C. 9% D. 8%

Find the compound interest on Rs. 15625 for 9 months, at 16% per annum, compounded quarterly.

Swati borrowed ₹ 40960 from a bank to buy a piece of land. If the bank charges 12 ½ % per annum, compounded half yearly. What amount will she have to pay after 1 ½ years? Also, find the interest paid by her.

Find the rate percent per annum if Rs. 2000 amount to Rs. 2662 in 1 1/2 years, interest being compounded half-yearly?

Ramu borrowed Rs. 15625 from a finance company to buy scooter. If the rate of interest be 16% per annum compounded annually, what payment will he have to make after 2 ¼ years?

The compound interest on ₹ 6250 at 8% per annum for 1 year, compounded half yearly, is(a) ₹ 500 (b) ₹ 510 (c) ₹ 550 (d) ₹ 512.50

Sandeep deposited Rs 25000 in a bank which pays 8% interest compounded annually. How much would he get back after two years?

Find the amount and the compound interest on Rs. 2500 for 2 years at 10% per annum, compounded annually.

At what rate per cent per annum will Rs. 4000 amount to Rs. 4410 in 2 years when compounded annually?

Find the amount that David would receive if he invests Rs. 8192 for 18 months at 12% per annum, the interest being compounded half-yearly.

Find the amount and the compound interest on ₹ 12800 for 1 year at 7 ½ % per annum, compounded half-yearly.

Find the rate percent per annum if Rs. 2000 amount to Rs. 2662 in 1 ½ years, interest being compounded half-yearly?

A sum of Rs 24000 is borrowed for \(1\frac{1}{2}\) years at the rate of interest 10% per annum compounded semi anually. What is the compound interest (x) ?(a) x < Rs 3000(b) Rs 3000 < x < Rs 4000(c) Rs 4000 < x < Rs 5000(d) x > Rs 5000