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Arithmetic I

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Interest & Growth

Interest And Growth

MODULES

Basics & Simple Interest
Advanced Simple Interest
Basics of Compound Interest
Non-Annual Compounding
Present Value & EMI
Growth & CAGR
Common Types
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Interest and growth 1
-/10
Interest and growth 2
-/10

PRACTICE

Interest & Growth : Level 1
Interest & Growth : Level 2
Interest & Growth : Level 3
ALL MODULES

CAT 2025 Lesson : Interest & Growth - Past Questions

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6. Past Questions

Question 1

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:
[XAT 2015]

Rs. 16,00016,00016,000
Rs. 18,00018,00018,000
Rs. 20,00020,00020,000
Rs. 25,00025,00025,000
None of the above

Observation/Strategy
1) Compound interest also grows at the same rate as that of the interest rate.
2) For the next 333 years, the interest is higher by Rs. 5,0005,0005,000 because interest rate is being applied on the first 333 year's interest of Rs. 10,00010,00010,000.

Let
rrr be the interest rate.

Interest for the first
333 years =Rs.10,000= \text{Rs.} 10,000=Rs.10,000
Interest for the next
333 years =25,000−10,000=Rs.15,000.= 25,000 - 10,000 = \text{Rs.} 15,000.=25,000−10,000=Rs.15,000.

From Point
222, we form the following equation.

10000×(1+r100)3=1500010000 \times \left(1 + \dfrac{r}{100} \right)^{3} = 1500010000×(1+100r​)3=15000

⇒
(1+r100)3=1500010000=1.5 \left(1 + \dfrac{r}{100} \right)^{3} = \dfrac{15000}{10000} = 1.5(1+100r​)3=1000015000​=1.5

For a principal
ppp, the interest at the end of 333 years is Rs. 10,00010,00010,000.

A=p(1+r100)nA = p \left(1 +\dfrac{r}{100} \right)^{n}A=p(1+100r​)n

⇒
p+10000=p×(1+r100)3 p + 10000 = p \times \left(1 + \dfrac{r}{100} \right)^{3}p+10000=p×(1+100r​)3

⇒
p+10000=p×1.5 p + 10000 = p \times 1.5p+10000=p×1.5

⇒
0.5p=10000 0.5p = 100000.5p=10000

⇒
p=20000 p = 20000p=20000

Answer: (
333) Rs. 20,00020,000 20,000

Question 2

Three years ago, your close friend had won a lottery of Rs. 111 crore. He purchased a flat for Rs. 404040 lakhs, a car for Rs. 202020 lakhs and shares worth Rs. 101010 lakhs. He put the remaining money in a bank deposit that pays compound interest @ 121212 percent per annum. If today, he sells off the flat, the car and the shares at certain percentage of their original value and withdraws his entire money from the bank, the total gain in his assets is 5%5 \%5%. The closest approximate percentage of the original value at which he sold off the three items is
[IIFT 2013]

606060 percent
757575 percent
909090 percent
105105105 percent

Observation/Strategy
111) We are asked to find the combined percentage applicable for the 333 items - flat, car and shares. Therefore, these can be treated as one set of assets worth Rs. 70 lakhs. And, the bank deposit is Rs. 303030 lakhs.
222) The answers options are quite far apart (intervals of 15%15\%15%). Therefore, we can approximate numbers.
333) All figures can be expressed as lakhs to simplify. So, let the amount won from lottery be Rs. 100100100.

Initial investment
=100= 100=100
Final Return
=100×1.05= 100 \times 1.05=100×1.05 = 105

On simple interest at
12%12 \%12% p.a. for 333 years amounts to 36%36 \%36% of interest. Compound interest will be a little more and approximated to 40%.

Amount earned from Bank deposit
=30×(1.12)3=30×1.4= 30 \times (1.12)^{3} = 30 \times 1.4=30×(1.12)3=30×1.4 = 42

Revenue from other assets
=105−42= 105 - 42=105−42 = 63

Revenue from other assets as
% \%% of original value =6370×100%= \dfrac{63}{70} \times 100 \%=7063​×100% = 90

Answer: (
333) 909090 percent

Question 3

ICICI bank offers a 111-year loan to a company at an interest rate of 202020 percent payable at maturity, while Citibank offers on a discount basis at a 19%19 \%19% interest rate for the same period. How much should the ICICI Bank decrease/increase the interest rate to match up the effective interest rate of Citibank?
[FMS 2009]

Increase by 3.5%3.5 \%3.5%
Decrease by 1.8%1.8 \%1.8%
Increase by 1%1 \%1%
Decrease by 1.4%1.4 \%1.4%

Observation/Strategy
111) Interest is normally paid at maturity, as in the case of the ICICI loan.
222) In the Citibank loan, the interest is deducted/discounted at the time of issuance.

For instance, in a Rs.
100100100 loan issued on discount basis at 19%19 \%19%, the interest of Rs. 191919 is deducted from Rs. 100100100 and disbursed.

Amount disbursed on discount basis
=100−19=Rs.81= 100 - 19 = \text{Rs}. 81=100−19=Rs.81
Amount repaid on maturity
=Rs.100= \text{Rs}. 100=Rs.100

So, interest of Rs.
191919 is effectively on a principal of Rs. 818181.

Effective interest rate
=1981×100%=190081%= \dfrac{19}{81} \times 100 \% = \dfrac{1900}{81} \%=8119​×100%=811900​% ~ 23.5

ICICI which currently has interest of
20%20 \%20% should increase its rate by 3.5%.

Answer: (
111) Increase by 3.5%3.5 \%3.5%


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