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

You can practice all previous year OMET questions from the topic Arithmetic - Simple & Compound Interest. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

**XAT 2019 QADI | Arithmetic - Simple & Compound Interest**

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

**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$\left(1+\frac{1\times 5}{100}\right)$ = x$\left(1+\frac{1\times 4}{100}\right)$ + x$\left(1+\frac{1\times 3}{100}\right)$ + x$\left(1+\frac{1\times 2}{100}\right)$ + x$\left(1+\frac{1\times 1}{100}\right)$ + x

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

⇒ 15120 = 5.1x

⇒ x = 15120/5.1 ≈ 2965.

Hence, option (a).

Workspace:

**IIFT 2019 QA | Arithmetic - Simple & Compound Interest**

Ms. Debjani after her MBA graduation wants to have a start-up of her own. For this, she uses ₹ 8,00,000 of her own savings and borrows ₹ 12,00,000 from a public sector bank under MUDRA Scheme. As per the agreement with the bank, she is supposed to repay the principle of this loan equally over the period of the loan which is 25 years. Two years after taking the first loan, she borrowed an additional loan of ₹8,00,000 to finance expansion plan of her start-up. If Ms. Debjani clears all her loans in 25 years from the date of taking the first loan, how much total interest she has to pay on her initial borrowing ? Assume simple interest rate at 8 percent per annum.

- A.
Rs. 12,48,000

- B.
Rs. 12,84,000

- C.
Rs. 14,20,000

- D.
Rs. 12,96,000

Answer: Option A

**Explanation** :

Since the principal is to be paid equally over 25 years, principal amount paid per year = (12 lakhs)/25 = Rs. 48000.

For the first year, interest to be paid = (12 lakhs) × 0.08 ....(I)

For the second year, amount remaining = (12 lakhs) − 48000 = Rs. 11,52,000.

So interest for the second year = (Rs. 11,52,000) × 0.08 .... (II)

On similar lines, interest for the third year = (48000) × 0.08

Total interest = [(12 lakhs) × 0.08] + [(Rs. 11,52,000) × 0.08] + ...... [(48000) × 0.08]

This is an AP with first term as (48000) × 0.08 = 3840 with 25 terms and common difference of Rs. 3840. [{(12 lakhs) × 0.08} − {(Rs. 11,52,000) × 0.08} = Rs. 3840]

∴ Total interest = (25/2)[(2 × 3840) + {(25 − 1) × 3840}]

= Rs. 12,48,000.

Hence option (a).

Workspace:

**IIFT 2019 QA | Arithmetic - Simple & Compound Interest**

Monika buys a Samsung’s 360 litre refrigerator from M/s Coldrush Agencies for ₹42,000. She makes a down payment of ₹12,000 and the remaining amount in 4 equal half yearly instalments. If M/s Coldrush Agencies charge an interest of 10% per annum, approximately what amount Monika has to pay every six months?

- A.
Rs. 8230

- B.
Rs. 8600

- C.
Rs. 8460

- D.
Rs. 8620

Answer: Option C

**Explanation** :

Cost of refrigerator = Rs. 42000.

Down payment = Rs. 12000.

Loan amount for Monika = 42000 − 12000 = Rs. 30,000.

Interest charged every six months = 10/2 = 5%

∴ 30000 × 1.05^{4} = x × 1.05^{3} + x × 1.05^{2} + x × 1.05 + x.

∴ 36465 = 4.31x

∴ x ≈ 8460.

Hence option (c).

Workspace:

**IIFT 2017 QA | Arithmetic - Simple & Compound Interest**

Swarn, an SME enterprise, borrowed a sum of money from a nationalised bank at 10% simple interest per annum and the same amount at 8% simple interest per annum from a microfinance firm for the same period. It cleared the first loan 6 months before the scheduled date of repayment and repaid the second loan just at the end of scheduled period. If, in each case, it had to pay Rs. 62,100 as amount, then how much money and for what time period did it borrow?

- A.
Rs. 55,750, 2 years

- B.
Rs. 52,500, 2 years

- C.
Rs. 51,750, 2.5 years

- D.
Rs. 55,750, 2.5 years

Answer: Option C

**Explanation** :

Let the amount borrowed from each source be Rs. P for n months.

Since Swarn pays Rs. 62,100 to each firm, it pays the same simple interest to each firm.

Since it repays the bank 6 months before repayment, its effective loan tenure is (n − 6) months; while for the loan with the microfinance firm, its tenure is n months.

Considering same interest: [P × 10 × (n − 6)]/(12 × 100) = [P × 8 × n]/(12 × 100)

∴ 5(n − 6) = 4n i.e. n = 30

Hence, time period of borrowing = 30 months = 2.5 years. Hence, options 1 and 2 are eliminated.

Considering the loan with the microfinance firm:

62100 − P = (P × 8 × 2.5)/100

∴ 62100 = 1.2P i.e. P = Rs. 51,750

Hence, option 3.

Alternatively,

Consider the loan with the microfinance firm. Let the amount borrowed be Rs. P for n years.

∴ 62100 − P = (P × 8 × n)/100

∴ 62100 = P(1 + 0.08n)

Now, observe that there are only two values of n (2 and 2.5) in the options). Substitute each value in the above equation and check if the Principal value given in that option is obtained.

When n = 2; P = 62100/1.16 = Rs. 53,535 (approximately). SInce this value is not in the options, this case is invalid.

When n = 2.5; P = 62100/1.2 = Rs. 51,750

Hence, option 3.

Workspace:

**XAT 2015 QA | Arithmetic - Simple & Compound Interest**

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

**Explanation** :

Let P be the principal.

By the given conditions,

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

$\mathrm{}\phantom{\rule{0ex}{0ex}}P\times {\left(1+\frac{r}{100}\right)}^{6}-P=25000$

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

Thus,

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

P(X^{2 }- 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 3.

Workspace:

**XAT 2012 QA | Arithmetic - Simple & Compound Interest**

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

Answer: Option C

**Explanation** :

After one year, amount due = 6000 × 1.05 – 1200 = 5100

Hence, amount due after two years = 5100 × 1.05 – 1200 = 4155

Hence, option 3.

Workspace:

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