Arithmetic - Simple & Compound Interest - Previous Year CAT/MBA Questions

The best way to prepare for Arithmetic - Simple & Compound Interest is by going through the previous year Arithmetic - Simple & Compound Interest XAT questions. Here we bring you all previous year Arithmetic - Simple & Compound Interest XAT questions along with detailed solutions.

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1. XAT 2023 QADI | Arithmetic - Simple & Compound Interest XAT Question

Jose borrowed some money from his friend at simple interest rate of 10% and invested the entire amount in stocks. At the end of the first year, he repaid 1/5th of the principal amount. At the end of the second year, he repaid half of the remaining principal amount. At the end of third year, he repaid the entire remaining principal amount. At the end of the fourth year, he paid the last three years’ interest amount. As there was no principal amount left, his friend did not charge any interest in the fourth year. At the end of fourth year, he sold out all his stocks. Later, he calculated that he gained Rs. 97500 after paying principal and interest amounts to his friend. If his invested amount in the stocks became double at the end of the fourth year, how much money did he borrow from his friend?

(a)
250000
(b)
200000
(c)
150000
(d)
125000
(e)
None of the above

2. XAT 2023 QADI | Arithmetic - Simple & Compound Interest XAT Question

Separately, Jack and Sristi invested the same amount of money in a stock market. Jack’s invested amount kept getting reduced by 50% every month. Sristi’s investment also reduced every month, but in an arithmetic progression with a common difference of Rs. 15000. They both withdrew their respective amounts at the end of the sixth month. They observed that if they had withdrawn their respective amounts at the end of the fourth month, the ratio of their amounts would have been the same as the ratio after the sixth month. What amount of money was invested by Jack in the stock market?

(a)
Rs. 100000
(b)
Rs. 120000
(c)
Rs. 150000
(d)
Rs. 180000
(e)
None of the above

Answer: Option A

Video Explanation

Text Explanation :

Let their initial amounts be Rs. x

After 4 months:
Jack's investment after 4 months = x × ${(\frac{1}{2})}^{4}$ = $\frac{x}{16}$
Sristi's investment after 4 months = x - 15000 × 4 = x - 60000

After 6 months:
Jack's investment after 6 months = x × ${(\frac{1}{2})}^{6}$ = $\frac{x}{64}$
Sristi's investment after 6 months = x - 15000 × 6 = x - 90000

Ratio of their investments after 4 months = Ratio of their investments after 6 months.

⇒ $\frac{x / 16}{x - 60000}$ = $\frac{x / 64}{x - 90000}$

⇒ 4(x - 90,000) = x - 60,000
⇒ 3x = 3,00,000
⇒ x = 1,00,000

Hence, option (a).

3. XAT 2021 QADI | Arithmetic - Simple & Compound Interest XAT Question

Mohan has some money (₹M) that he divides in the ratio of 1:2. He then deposits the smaller amount in a savings scheme that offers a certain rate of interest, and the larger amount in another savings scheme that offers half of that rate of interest. Both interests compound yearly. At the end of two years, the total interest earned from the two savings schemes is ₹830. It is known that one of the interest rates is 10% and that Mohan deposited more than ₹1000 in each saving scheme at the start. What is the value of M?

(a)
7500
(b)
6000
(c)
To solve this, the other interest rate must also be given.
(d)
4500
(e)
12000

Answer: Option B

Text Explanation :

Let the total amount be 3x
Case 1:
Smaller amount = x, rate of interest = 10
Larger amount = 2x, rate of interest = 5

Total amount received at the end of two years (smaller amount) = ( $1 + \frac{10}{100}$ )² = 1.21 x. CI = 0.21x

Total amount received at the end of two years (larger amount) = 2x ( $1 + \frac{5}{100}$ )² = 2.205x CI = 0.205x

Given, 0.21x + 0.205x = 830
⇒ x = 2000
2x = 4000

Case 2:
Smaller amount = x, rate of interest = 20
Larger amount = 2x, rate of interest = 10
Total amount received at the end of two years (smaller amount) = x( $1 + \frac{20}{100}$ )² = 1.44x. CI = 0.44x

Total amount received at the end of two years (larger amount) = 2x ( $1 + \frac{10}{100}$ )² = 2.42x CI = 0.42x

Given, 0.44x+0.42x = 830
⇒ x = 965.11 which is not valid since it should be greater than 1000

4. XAT 2019 QADI | Arithmetic - Simple & Compound Interest XAT Question

A computer is sold either for Rs.19200 cash or for Rs.4800 cash down payment together with five equal monthly installments. If the rate of interest charged is 12% per annum, then the amount of each installment (nearest to a rupee) is

(a)
Rs. 2965
(b)
Rs. 2990
(c)
Rs.3016
(d)
Rs. 2896
(e)
Rs. 2880

Answer: Option A

Video Explanation

Text Explanation :

Price of computer = Rs 19,200

Down payment = Rs. 4,800

∴ Amount borrowed (i.e., loan taken) = 19200 – 4800 = Rs. 14,400.

This amount has to be repaid equally over 5 months at 12% p.a.

Let Rs. x is paid at the end of every month.

Interest charged for 1 month = 12%/12 = 1%

∴ 14400 $(1 + \frac{1 \times 5}{100})$ = x $(1 + \frac{1 \times 4}{100})$ + x $(1 + \frac{1 \times 3}{100})$ + x $(1 + \frac{1 \times 2}{100})$ + x $(1 + \frac{1 \times 1}{100})$ + x

⇒ 15120 = 1.04x + 1.03x + 1.02x + 1.01x + x

⇒ 15120 = 5.1x

⇒ x = 15120/5.1 ≈ 2965.

Hence, option (a).

5. XAT 2015 QA | Arithmetic - Simple & Compound Interest XAT Question

In the beginning of the year 2004, a person invests some amount in a bank. In the beginning of 2007, the accumulated interest is Rs. 10,000 and in the beginning of 2010, the accumulated interest becomes Rs. 25,000. The interest rate is compounded annually and the annual interest rate is fixed. The principal amount is:

(a)
Rs. 16,000
(b)
Rs. 18,000
(c)
Rs. 20,000
(d)
Rs. 25,000
(e)
None of the above

Answer: Option C

Video Explanation

Text Explanation :

Let P be the principal.

By the given conditions,

$P \times {(1 + \frac{r}{100})}^{3} - P = 10000$

$ P \times {(1 + \frac{r}{100})}^{6} - P = 25000$

Let ${(1 + \frac{r}{100})}^{3} = X$

Thus,

P(X - 1) = 10000 ...(i)

P(X²- 1) = 25000 ...(ii)

Dividing (ii) by (i),

X + 1 = 5/2

∴ X = 3/2

Substituting value of X in (i), we get

P = Rs. 20,000

Hence, option (c).

6. XAT 2012 QA | Arithmetic - Simple & Compound Interest XAT Question

A man borrows Rs. 6000 at 5% interest, on reducing balance, at the start of the year. If he repays Rs. 1200 at the end of each year, find the amount of loan outstanding, in Rs., at the beginning of the third year.

(a)
3162.75
(b)
4125.00
(c)
4155.00
(d)
5100.00
(e)
5355.00

Arithmetic - Simple & Compound Interest - Previous Year CAT/MBA Questions

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