Practice Questions for Arithmetic - Simple & Compound Interest

1. PE 1 - SICI | Arithmetic - Simple & Compound Interest

The cost of a computer depreciates by Rs 720 during the second year and Rs 648 during the third year. What is the rate of depreciation per annum and the original cost of computer?

(a)
100/9%, Rs. 800
(b)
100/9%, Rs. 888.8
(c)
10%, Rs. 8000
(d)
10%, Rs. 800

2. PE 1 - SICI | Arithmetic - Simple & Compound Interest

The value of a fridge which costs Rs 6,000 depreciates by 10% each year. After how many years will its depreciated value be Rs 4374?

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

3. PE 1 - SICI | Arithmetic - Simple & Compound Interest

A sum of Rs 6,600 was taken as a loan. This is to be repaid in two equal annual installments. If the rate of interest is 20% compounded annually then find the value of each installment.

(a)
Rs. 4400
(b)
Rs. 2220
(c)
Rs. 4320
(d)
Rs. 4420

Answer: Option C

Join our Telegram Channel for CAT/MBA Preparation.

Explanation :

P = Rs. 6600
r = 20%
t = 2 years

Let the equal annual installment be x

∴ $P {(1 + \frac{r}{100})}^{2}$ = $x {(1 + \frac{r}{100})}^{1}$ + x

⇒ $6600 {(1 + \frac{20}{100})}^{2}$ = $x {(1 + \frac{20}{100})}^{1}$ + x

⇒ 6600 × 1.44 = 2.2x

Solving this we get

x = 4320

Hence, option (c).

4. PE 1 - SICI | Arithmetic - Simple & Compound Interest

A man borrowed a certain sum of money and paid back in two equal installments. If the rate of compound interest is 4% and if he paid back Rs 1014 annually. What sum did he borrow?

(a)
Rs. 1921.5
(b)
Rs. 1912.5
(c)
Rs. 1192.5
(d)
None of these

Answer: Option B

Join our Telegram Channel for CAT/MBA Preparation.

Explanation :

Let P be the sum.

Total amount due at the end of 1^st year = $x (1 + \frac{4}{100})$
Installment for 1^st year = 1014
Amount due after payment of installment = $x (1 + \frac{4}{100})$ - 1014.
⇒ $x (\frac{26}{25})$ - 1014

Similarly,
Amount due at the of 2^nd year after payment of installment = $[x (\frac{26}{25}) - 1014] (\frac{26}{25})$ - 1014 = 0

Solving the above equation, we get
x = Rs. 1912.5

Hence, option (b).

5. PE 1 - SICI | Arithmetic - Simple & Compound Interest

What equal installment will discharge a debt of Rs. 1007 due in 5 years, rate of interest being 3% p.a.?

(a)
RS. 200
(b)
Rs. 190
(c)
Rs. 210
(d)
None of these

Answer: Option B

Join our Telegram Channel for CAT/MBA Preparation.

Explanation :

A = 1007,
t = 5 years,
r = 3%

∴ A = $x (1 + \frac{R \times (T - 1)}{100})$ + $x (1 + \frac{R \times (T - 2)}{100})$ + ... + x

⇒ 1007 = $x (1 + \frac{3 \times 4}{100})$ + $x (1 + \frac{3 \times 3}{100})$ + $x (1 + \frac{3 \times 2}{100})$ + $x (1 + \frac{3 \times 1}{100})$ + x

⇒ 1007= $x (5 + \frac{3 \times 10}{100})$

⇒ x = $\frac{10070}{53}$ = 190

Hence, option (b).

6. PE 1 - SICI | Arithmetic - Simple & Compound Interest

Krishna borrows Rs. 45K from a bank at 10% compound interest. He repays it in three annual installments that are in arithmetic progression. He ends up paying 54K totally. How much did he pay in year 1?

(a)
Rs. 16,500
(b)
Rs. 19,500
(c)
Rs. 21,000
(d)
Rs. 18,000

Answer: Option B

Join our Telegram Channel for CAT/MBA Preparation.

Explanation :

Let the repayments be Rs "a – d", Rs "a" and Rs. "a + d"

∴ (a – d) + a + (a + d) = 54000
⇒ 3a = 54000
⇒ a = 18000

P = 45,000
r = 10% pa
t = 3 years

∴ 45000 ${(1 + \frac{10}{100})}^{3}$ = (18000 - d) ${(1 + \frac{10}{100})}^{2}$ + 18000 $(1 + \frac{10}{100})$ + (18000 + d)

⇒ 45000 × 1.331 = (18000 – d) × 1.21 + 18000 × 1.1 + (18000 + d)

⇒ 45000 × 1.331 = 18000 × (1.21 + 1.1 + 1) + d × (-1.21 + 1)

⇒ 0.21d = 18000 × 3.31 – 45000 × 1.331

⇒ 0.21d = 9000(2 × 3.31 – 5 × 1.331)

⇒ 21d = 900000(6.62 – 6.655)

⇒ d = -1500

∴ The payments are Rs. 19500, Rs. 18000 and Rs. 16,500.

Hence, option (b).

7. PE 1 - SICI | Arithmetic - Simple & Compound Interest

A bicycle is sold by a shopkeeper for Rs. 19,200 cash or for Rs. 4,800 cash down payment together with five equal monthly instalments. If the rate of interest charged by the shopkeeper is 12% per annum find each instalment. [Enter the nearest possible integer as your answer in Rs.]

Backspace

7 8 9 4 5 6 1 2 3 0 . -

Clear All Submit

Answer: 2965

Join our Telegram Channel for CAT/MBA Preparation.

Explanation :

Using the formula for installments for SI:

P $(1 + \frac{R \times T}{10})$ = x $(1 + \frac{R \times (T - 1)}{10})$ + x $(1 + \frac{R \times (T - 2)}{10})$ + ⋯ + x

Where, P = amount lent
x = equal installment,
R = interest rate and
T = total duration of loan.

⇒14,400 $(1 + \frac{1 \times 5}{10})$ = x $(1 + \frac{1 \times 4}{10})$ + x $(1 + \frac{1 \times 3}{10})$ + x $(1 + \frac{1 \times 2}{10})$ + $(1 + \frac{1 \times 1}{10})$ + x

⇒ 14,400 $(1 + \frac{5}{10})$ = x $(5 + \frac{1 \times (4 + 3 + 2 + 1)}{100})$

⇒ 14,400 $(\frac{21}{20})$ = x $(5 + \frac{1}{10})$

⇒ x ≈ 2965

Hence, 2965.

8. PE 1 - SICI | Arithmetic - Simple & Compound Interest

George borrows Rs 2,100 at the rate of 10% per annum compounded annually and has to repay it in two equal instalments in 2 years. What is the value of each instalment?

(a)
Rs. 1100
(b)
Rs. 1080
(c)
Rs. 1070
(d)
None of these

Answer: Option D

Join our Telegram Channel for CAT/MBA Preparation.

Explanation :

We know,
P ${(1 + \frac{R}{100})}^{2}$ = x ${(1 + \frac{R}{100})}^{1}$ + x

⇒ 2100 ${(1 + \frac{10}{100})}^{2}$ = x ${(1 + \frac{R}{100})}^{1}$ + x

⇒ 2100 × 1.21 = 1.1x + x

⇒ 2100 × 1.21 = 2.1x

⇒ x = 1210

Hence, option (d).

9. PE 1 - SICI | Arithmetic - Simple & Compound Interest

John purchased a new bike worth Rs 96,000 on instalment at 5% p.a. interest compounded annually. At the end of one year, he repaid a certain amount and he repaid the balance amount of Rs 63000 at the end of the second year. What amount did he repay at the end of the first year?

(a)
Rs. 40,000
(b)
Rs. 40,800
(c)
Rs. 41,400
(d)
Rs. 42,000

Answer: Option B

Join our Telegram Channel for CAT/MBA Preparation.

Explanation :

We know,
P ${(1 + \frac{R}{100})}^{2}$ = (1^st installment) ${(1 + \frac{R}{100})}^{1}$ + (2^nd installment)
⇒ 96,000 ${(1 + \frac{5}{100})}^{2}$ = x ${(1 + \frac{5}{100})}^{1}$ + 63,000

⇒ 96000 × 1.05 = x + 60000

⇒ x = 40,800

Hence, option (b).

10. PE 1 - SICI | Arithmetic - Simple & Compound Interest

Find the effective rate of interest if rate of interest is 40% p.a. and interest being compounded quarterly?

(a)
45%
(b)
46.41%
(c)
48%
(d)
47.5%

Answer: Option B

Join our Telegram Channel for CAT/MBA Preparation.

Explanation :

We know,
P ${(1 + \frac{R / n}{100})}^{(n \times t)}$ = P ${(1 + \frac{E_{r}}{100})}^{t}$

Since interest is compounded quarterly, n = 4 and t = 1, R = 40% and Er = effective rate of interest.

⇒ P ${(1 + \frac{40 / 4}{100})}^{4}$ = P ${(1 + \frac{E_{r}}{100})}^{1}$

⇒ (1.1)⁴ = (1 + E_r/100)

⇒ 1.4641 = 1 + E_r/100

⇒ E_r = 46.41%

Hence, option (b).

11. PE 1 - SICI | Arithmetic - Simple & Compound Interest

Find the CI realized in the year at a rate of 40% p.a., with compounding being done quarterly, on a sum of Rs 1000?

(a)
420
(b)
400
(c)
424.20
(d)
464.10

Answer: Option D

Join our Telegram Channel for CAT/MBA Preparation.

Explanation :

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

⇒ A = 1000 ${(1 + \frac{10}{100})}^{4}$ = 1464.1

⇒ Compound Interest = 1464.1 – 1000 = 464.1

Hence, option (d).

PE 1 - SICI | Arithmetic - Simple & Compound Interest

Feedback