1. What sum invested for 2 years at 14% compounded annually will grow to Rs. 5458.32?
1. 4120
2. 3300
3. 4200
4. 4420
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Answer:- 3
Explanation :-
P(1+R/100)^T = 5458.32
P(1+14/100)2 = 5458.32
P(114/100)2 = 5458.32
P = (5458.32×100×100)/(114×114)
P = (47.88×100×100) / 114 = 0.42×100×100 = 4200
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2. If the difference between the simple interest and compound interests on some principal amount at 20% for 3 years is Rs. 48, then the principal amount is
1. Rs. 365
2. Rs. 325
3. Rs. 395
4. Rs. 375
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Answer:- 4
Explanation :-
Amount (for compound interest compounded annually) = P(1+R/100)^T
Amount (compounded annually) = x(1+20/100)^3
Amount = x(120/100)3 = x(6/5)^3
Compound Interest = [x(6/5)^3]−x
Compound Interest = x [(6/5)^3−1] = x [(216/125) −1] = (91x/125)
Simple Interest = (P*R*T)/100 = (x × 20 × 3) / 100=(3x)/5
Difference Between CI & SI = Rs. 48
i.e. [(91x)/125] – [(3x)/5] = 48
(91x/125) − (3x/5) = 48
(91x−75x)/125 = 48
16x/125=48
x=(48×125)/16 = 3×125 = Rs. 375
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3. Andrews earns an interest of Rs. 1596 for the third year and Rs. 1400 for the second year on the same sum. Find the rate of interest if it is lent at compound interest.
1. 12%
2. 13%
3. 14%
4. 15%
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Answer:- 3
Explanation :-
SI for 1 Year = 1596 – 1400 =Rs. 196
R=(SI×100)/(PT) = (100×196)/(1400×1) = 196/14 = 14%
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4. The population of a town is 40,000. It decreases by 20 per thousand per year. Find out the population after 2 years.
1. 38484
2. 38266
3. 38416
4. 38226
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Answer:- 3
Explanation :-
R ( per thousand per year ) = (20×100)/1000 = 2%
Population after 2 years = P(1−R/100)^T
=40000(1−2/100)^2
=40000(1−1/50)^2
=40000 (49/50)^2
=40000×(49×49)/(50×50) = 16×49×49 = 38416
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