Home Aptitude Simple and Compound Interest

Simple and Compound Interest

August 20, 2017

Explanation / Important formulas:

SIMPLE INTEREST

Principal: The money borrowed or lent out for a certain period is called the principal or the sum.

Interest: Extra money paid for using other’s money is called interest.

Simple Interest(S.I.): If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest.

Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then

Simple Interest = (P x R x T / 100)
P = (100 x S.I / R x T); R = (100 x S.l / P x T) and T= (100 x S.I / P x R)

COMPOUND INTEREST

To find compound interest Let Principal = P, Rate = R% per annum, Time = n years.

When interest is compounded Quarterly: Amount = P [1+(R/4) /100]⁴ⁿ
When interest is compounded Half-yearly: Amount = P [1+(R/2) / 100]²ⁿ
When interest is compound Annually: Amount = P (1 + R/100)ⁿ
When interest is compounded annually but time is in fraction, say 3(2/5) years: Amount = P (1+R/100)³ x (1+ (2/5R) / 100)
When Rates are different for different years, say R₁%, R₂%, R₃% for 1^st, 2^ndand 3^rd year respectively. Then the amount = P (1+R₁/100) (1+R₂/100) (1+R₃/100)
Present worth of Rs.x due n years hence is given by: Present worth = [X / (1+ R/100)].

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Question 1

A money lender lent Rs. 1000 at 3% per year and Rs. 1400 at 5% per year. The amount should be returned to him when the total interest comes to Rs. 350. Find the number of years?

A	3.5
B	3.75
C	4
D	4.5

Question 1 Explanation:

(1000 * t * 3 / 100) + (1400 * t * 5 / 100)
= 350 => t =3.5

Question 2

How long will it take a certain amount to increase by 30% at the rate of 15% simple interest?

A	2%
B	4%
C	8%
D	2%

Question 2 Explanation:

Let the principal be Rs. x
Simple interest = x * 30 / 100 = 3x / 10
T = 100 * SI / PR = 100 * (3x / 10) / x * 15 = 2%
Alternatively, this can be solved by considering
principal amount to be Rs. 100
Then simple interest becomes Rs. 30
Then, T = 100 * 30 / 100 * 15 = 2%

Question 3

A certain amount becomes Rs. 5760 in 2 years and Rs. 6912 in 3 years. What is the principal amount and the rate of interest?

A	Rs. 2500
B	Rs. 7000
C	Rs. 4000
D	Rs. 3000

Question 3 Explanation:

S.I on Rs. 5760 for 1 year = 6912 – 5760 = Rs. 1152
Therefore, Rate of interest for 1 year = 100 * 1152/5760 * 1 = 20%
Let the principal be p
Then, Principal = p[1 + 20 / 100]²= 5760

Solving which gives Principal = Rs. 4000

Question 4

Find the simple interest on Rs. 5000 at a certain rate if the compound interest on the same amount for 2 years is Rs. 253.125?

A	Rs. 250
B	Rs. 400
C	Rs. 500
D	Rs.100

Question 4 Explanation:

Let the rate of interest be r
5000[1+ r /100]² = 5000 + 253.125
[1+ r /100]² = 5253.125 / 5000
Solving which gives
[1+ r/100]² = 1681 / 1600
1+ r/100 = 41 / 40
r = 2.5
Therefore, SI = 5000 * 2 * 2.5 / 100 = Rs. 250

Question 5

The difference between SI and CI compounded annually on a certain sum of money for 2 years at 8% per annum is Rs. 12.80. Find the principal?

A	Rs. 1500
B	Rs. 2000
C	Rs. 1000
D	Rs. 5000

Question 5 Explanation:

Let the principal amount be x
SI = x * 2 * 8 / 100 = 4x / 25
CI = x[1+ 8 / 100]² – x = 104x / 625
Therefore, 104x / 625 – 4x / 25 = 12.80
Solving which gives x, Principal = Rs. 2000

Question 6

Find the compound interest on Rs. 10000 at 12% rate of interest for 1 year, compounded half-yearly?

A	Rs. 7654
B	Rs. 1236
C	Rs. 1543
D	Rs. 2765

Question 6 Explanation:

Amount with CI = 10000 [1+ (12 / 2 * 100)]²= Rs. 11236
Therefore, CI = 11236 – 10000 = Rs. 1236

Question 7

Find the compound interest on Rs. 3000 at 5% for 2 years, compounded annually?

A	Rs. 301.2
B	Rs. 307.5
C	Rs. 387.2
D	Rs. 309.3

Question 7 Explanation:

Amount with CI = 3000 (1+ 5 / 100)² = Rs. 3307.5
Therefore, CI = 3307.5 – 3000 = Rs. 307.5

Question 8

A certain principal amounts to Rs. 15000 in 2.5 years and to Rs. 16500 in 4 years at the same rate of interest. Find the rate of interest?

A	8%
B	5%
C	3%
D	9%

Question 8 Explanation:

Amount becomes 15000 in 2.5 years and 16500 in 4 years
Simple interest for (4 - 2.5) years = 16500 – 15000
Therefore, SI for 1.5 years = Rs. 1500
SI for 2.5 years = 1500 / 1.5 * 2.5 = 2500
Principal amount = 15000 – 2500 = Rs. 12500
Rate of Interest = 2500 * 100 / 12500 * 2.5
R = 8%

Question 9

Find the present worth of Rs. 78000 due in 4 years at 5% interest per year?

A	Rs. 65000
B	Rs. 63000
C	Rs. 67000
D	Rs. 61000

Question 9 Explanation:

Amount with interest after 4 years = Rs. 78000
Therefore, simple interest = 78000 – Principal
Let the principal amount be p
78000 – p = p * 4 * 5 / 100
p = 13000
Principal = 78000 – 13000 = Rs. 65000

Question 10

A sum of Rs. 25000 becomes Rs. 27250 at the end of 3 years when calculated at simple interest. Find the rate of interest?

A	9%
B	3%
C	7%
D	2%

Question 10 Explanation:

Simple interest = 27250 – 25000 = 2250
Time = 3 years
SI = PTR / 100
R = SI * 100 / PT
R = 2250 * 100 / 25000 * 3
R = 3%

Question 11

Simple interest on a sum at 5% per annum for 2 years is Rs. 60. The compound interest on the same sum?

A	Rs. 62
B	Rs. 62.4
C	Rs. 61.5
D	Rs. 60.5

Question 11 Explanation:

The difference between compound interest and simple interest on Rs. P for 2 years at R% per annum = (R × SI) / (2 × 100)
Difference between compound interest and simple interest = (R × SI) / (2 × 100) = (5 × 60) / (2 × 100) = 1.5
Hence, compound interest = Simple Interest + 1.5 = 60 + 1.5 = Rs. 61.5

Question 12

What will be the amount if a sum of Rs. 10000 is placed at compound interest for 3 years while rate of interest for the first, second and third years is 2, 5 and 10 percent, respectively?

A	Rs. 11231
B	Rs. 11781
C	Rs. 11658
D	Rs. 11244

Question 12 Explanation:

Amount after 3 years = 10000 (1 + 2 / 100) (1 + 5 / 100) (1 + 10 / 100)
= 10000 (102 / 100) (105 / 100) (110 / 100) = (102 × 105 × 11) / 10
= Rs. 11781

Question 13

If in a certain number of years Rs. 10000 amount to Rs. 160000 at compound interest, in half that time Rs. 10000 will amount to?

A	Rs. 80000
B	Rs. 40000
C	Rs. 50000
D	Rs. 60000

Question 13 Explanation:

Let the rate of interest be R% per annum
Assume that Rs. 10000 amount to Rs. 160000 in T years
10000(1+R/100)^T= 160000
=>(1+R/100)^T= 160000 / 10000 = 16
=>(1+R / 100)^T/2= 16 = 4 --- (1)
In T/2 years, Rs.10000 amounts to 10000(1+R/100)^T/2
= 10000×4 [∵ from (1)]
= 40000

Question 14

A sum of Rs. 6600 was taken as a loan. This is to be repaid in two equal annual instalments. If the rate of interest be 20% compounded annually then the value of each instalment is?

A	Rs. 4320
B	Rs. 2220
C	Rs. 4400
D	Rs. 4420

Question 14 Explanation:

Present worth of Rs. x due T years hence is given by
Present Worth (PW) = x / (1+R/100)^T
Let x be the annual payment
Then, present worth of x due 1 year hence + present worth of x due 2 year hence = 6600
x / (1 + (20/100))¹ + x(1+(20/100))² = 6600
X / (6 / 5)+ x (6 / 5)
² = 6600
5x / 6 + 25x/ 36 = 6600
55x /36 = 6600
x = (6600×36) / 55 = 4320

Question 15

The difference between the simple interest on a certain sum at the rate off 10% per annum for 2 years and compound interest which is compounded every 6 months is Rs. 124.05. What is the principal sum?

A	Rs. 10000
B	Rs. 12000
C	Rs. 6000
D	Rs. 8000

Question 15 Explanation:

Let the sum be P
Compound Interest on P at 10% for 2 years when interest is compounded half-yearly
=P(1+(R/2) /100)^2T– P = P(1+(10/2) /100)^{2 x 2} –P = P(1+1/20)⁴ – P =P(21/20)⁴ –P
Simple Interest on P at 10% for 2 years = PRT/100=(P×10×2) /100=P / 5
Given that difference between compound interest and simple interest = 124.05
=>P(21 / 20)⁴ – P –P/5 = 124.05
=>P[(21/20)⁴−1 −1/5] = 124.05
=>P[194481 / 160000 – 1 −1/5] = 124.05

=>P[(194481−160000−32000) / 160000]=124.05
=>P[2481 / 160000] = 124.05
=>P = (124.05 × 160000) / 2481 = 160000 / 20 = 8000

Question 16

A sum is invested for 3 years compounded at 5%, 10% and 20 % respectively. In three years, if the sum amounts to Rs. 1386, then find the sum?

A	Rs. 1500
B	Rs. 1400
C	Rs. 1200
D	Rs. 1000

Question 16 Explanation:

1386 = P(1 + (5 / 100)) (1 + (10 / 100)) (1+(20 / 100))
1386 = P(21 / 20) (11 / 10) (6 / 5)
P = (1386 × 20 × 10 × 5) / (21 × 11 × 6)
= (66 × 20 × 10 × 5) / (11 × 6) = 20 × 10 × 5 = Rs. 1000
i.e., the sum is Rs.1000

Question 17

Divide Rs. 3364 between A and B, so that A's Share at the end of 5 years may equal to B's share at the end of 7 years, compound interest being at 5 percent?

A	Rs. 1764 and Rs. 1600
B	Rs. 1756 and Rs. 1608
C	Rs. 1722 and Rs. 1642
D	None of these

Question 17 Explanation:

A's share after 5 years = B's share after 7 years
(A's present share)(1 + (5 / 100))⁵ = (B's present share)(1 + (5 / 100))⁷
=>(A's present share) / (B's present share)
=> (1 + (5 / 100))⁷/ (1 + (5 / 100))⁵ = (1 + (5 / 100))^(7-5)
=> (1 + (5 / 100))² = (21 /20)² = 441 / 400
i.e, A's present share : B's present share = 441 : 400
Since the total present amount is Rs.3364, A's share = 3364 × [441 (441 + 400)]
= 3364 × (441 / 841) = 4 × 441 = Rs. 1764
B's present share = 3364 - 1764 = Rs.1600

Question 18

If a sum on compound interest becomes three times in 4 years, then with the same interest rate, the sum will become 81 times in?

A	12 years
B	18 years
C	16 years
D	14 years

Question 18 Explanation:

Let the sum be P
The sum P becomes 3P in 4 years on compound interest
3P = P(1 + R / 100)⁴
⇒ 3 = (1 + R / 100)⁴
Let the sum P becomes 81P in n years
=> 81P = P(1 + R / 100)ⁿ
=> 81=(1 + R / 100)ⁿ
=> (3)⁴= (1 + R / 100)ⁿ
=> ((1 + R / 100)⁴)⁴ = (1 + R / 100)ⁿ
=> (1 + R / 100)¹⁶
=> (1 + R / 100)ⁿ = 16
i.e, the sum will become 81 times in 16 years

Question 19

Arun borrowed a certain sum from Manish at a certain rate of simple interest for 2 years. He lent this sum to Sunil at the same rate of interest compounded annually for the same period. At the end of two years, he received Rs. 2400 as compound interest but paid Rs. 2000 only as simple interest. Find the rate of interest?

A	40%
B	30%
C	20%
D	10%

Question 19 Explanation:

Let the sum be x
Simple interest on x for 2 years = Rs.2000
Simple interest = PRT / 100
2000 = (x * R * 2) / 100
⇒ xR = 100000--- (1)
Compound Interest on x for 2 years = 2400
P(1 + R / 100)^T– P = 2400
x(1 + R / 100)² – x = 2400
x[1 + (2R / 100) + (R²/ 10000) – x] =2400
x (2R / 100) + (R² / 10000) = 2400
(2xR /100) + (xR² / 10000 = 2400--- (2)
Substituting the value of xR from (1) in (2) ,we get
(2 × 100000) / 100 + (100000 × R) / 10000 = 2400
2000 + 10R = 2400
10R = 400
R = 40%

Question 20

A	40%
B	30%
C	20%
D	10%

Question 20 Explanation:

Question 21

A sum is invested at compounded interest payable annually. The interest in the first two successive years was Rs. 400 and Rs. 420. The sum is?

A	Rs. 8000
B	Rs.7500
C	Rs. 8500
D	Rs. 8200

Question 21 Explanation:

This means that, simple Interest on Rs.400 for 1 year = 420 - 400 = 20
Rate = 100×SI / PT = (100×20) / 400×1 = 5%
Rs.400 is the interest on the sum for 1st year
Hence, sum = (100×SI) / RT = (100×400) / (5×1)
= Rs. 8000

Question 22

The population of a town is 40,000. It decreases by 20 per thousand per year. Find out the population after 2 years?

A	38484
B	38266
C	38416
D	38226

Question 22 Explanation:

This problem is similar to the problems we saw in compound interest
We can use the formulas of compound interest here as well
In compound interest, interest (a certain percentage of the principal) will be added to the principal after every year
Similarly, in this problem, a certain count(a certain percentage of the population) will be decreased from
the total population after every year
i.e., the formula becomes, A = P(1 − R / 100)T
where Initial population = P, Rate = R% per annum,
Time = T years and A = the population after T years
Please note that we have to use the -ve sign here instead of the + sign as the population gets decreased
R = (20 × 100) / 1000 = 2%(∵ percentage is calculated for twenty per thousand)
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

Question 23

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?

A	12%
B	13%
C	14%
D	15%

Question 23 Explanation:

Interest earned in 3rd year = Rs. 1596

Interest earned in 2nd year = Rs. 1400
i.e, in 3rd year, Andrews gets additional interest of (Rs. 1596 - Rs. 1400) = Rs.196
This means, Rs.196 is the interest obtained for Rs.1400 for 1 year
R = (100×SI) / PT = (100×196) / (1400×1) = 196 / 14 = 14%

Question 24

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?

A	Rs. 365
B	Rs. 325
C	Rs. 395
D	Rs. 375

Question 24 Explanation:

Let the sum be Rs. x
Amount after 3 years on Rs.x at 20% per annum when interest is compounded annually
=P(1 + R / 100)^T = x(1 + 20 / 100)³ = x(120 / 100)³= x(6 / 5)3
Compound Interest = x(6 / 5)³ – x = x[(6 / 5)³ − 1]
= x [216 / 125 − 1] = 91x / 125
Simple Interest = PRT / 100 = (x * 20 * 3) / 100 = 3x / 5
Given that difference between compound interest and simple interest is Rs. 48
91x / 125 − 3x / 5 = 48
(91x − 75x) / 125 = 48
16x / 125 = 48
x = (48 × 125) / 16 = 3 × 125 = Rs. 375
i.e, the sum is Rs.375

Question 25

What sum invested for 2 years at 14% compounded annually will grow to Rs. 5458.32?

A	4120
B	3300
C	4200
D	4420

Question 25 Explanation:

Let sum be P
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)
= (47.88 × 100 × 100) / 114 = 0.42 × 100 × 100 = 4200

Question 26

A sum of money is borrowed and paid back in two annual instalments of Rs. 882 each allowing 5% compound interest. The sum borrowed was?

A	Rs.1820
B	Rs.1640
C	Rs.1260
D	Rs.1440

Question 26 Explanation:

Present worth of Rs. x due T years hence is given by
Present Worth (PW) = x / (1+R / 100)^T
The sum borrowed = Present Worth of Rs.882 due 1 year hence + Present Worth of Rs.882 due 2 year hence
=> 882 / (1 + (5 / 100))¹+ 882(1 + (5 / 100))²
=> 882 / (105 /100) + 882 / (105 / 100)²
=> [882(21 / 20) + 882 / (21 / 20)²
=> (882 × 20) / 21+(882 × 20 × 20) / (21 × 21)
=> 42 × 20 + (42 × 20 × 20) / 21 = 840 + 2 × 20 × 20
=> 840 + 800 = 1640
i.e., The sum borrowed = Rs.1640

Question 27

A sum put out at 4% compound interest payable half-yearly amounts to Rs. 13265.10 in 1¹⁄₂ years. The sum is?

A	Rs. 12500
B	Rs. 11200
C	Rs. 8840
D	Rs. 12600

Question 27 Explanation:

Let the sum be P
Time, T = 1 1/2 year = 3 / 2 year
Amount after 1 1/2 years = P(1 + (R / 2) / 100)^2T = P(1 + (4 / 2) / 100)^{2 x 3/2}
= P(1 + 2 / 100)³ = P(102 / 100) ³ = P(51 / 50)³
Given that amount after 1¹⁄₂ years = 13265.10
=> P(51 / 50)³= 13265.10
=> P = 13265.10(50 / 51)³ = (13265.10 × 50 × 50 × 50) / (51 × 51 × 51) = (260.1 × 50 × 50 × 50) / (51 × 51)
=> (5.1 × 50 × 50 × 50) / 51 = 0.1 × 50 × 50 × 50 = Rs. 12500

Question 28

A sum amounts to Rs. 882 in 2 years at 5% compound interest. The sum is?

A	Rs. 800
B	Rs. 822
C	Rs. 840
D	Rs. 816

Question 28 Explanation:

Let the sum be P
Amount After 2 years = P(1 + R / 100)^T= P(1 + 5 / 100)²=P(105 / 100)²= P(21 / 20)²
Given that amount After 2 years = 882
=> P(21 / 20)²= 882
=> P=(882 × 20 × 20) / (21 × 21) = 2 × 20 × 20 = Rs. 800

Question 29

The Simple interest on a certain sum for 2 years at 20% per annum is Rs. 80. The corresponding compound interest is?

A	Rs. 66
B	Rs. 82
C	Rs. 86
D	Rs. 88

Question 29 Explanation:

Principal, P = (100 × SI) / RT = (100 × 80) / (20 × 2) = Rs. 200
Amount after 2 year on Rs.200 at 20% per annum when interest is compounded annually
=> P(1 + R100)^T = 200(1 + 20 / 100)² = 200(120 / 100)²
=> (200 × 120 × 120) / (100 × 100) = (2 × 120 × 120) / 100
=> 2 × 12 × 12 = Rs. 288
Compound Interest = 288 - 200 = Rs.88

Question 30

There is 80% increase in an amount in 8 years at simple interest. What will be the compound interest of Rs. 14,000 after 3 years at the same rate?

A	Rs.3794
B	Rs.3714
C	Rs.4612
D	Rs.4634

Question 30 Explanation:

Let P = Rs.100
Simple Interest = Rs. 80 ( ∵ 80% increase is due to the simple interest)
Rate of interest = (100 × SI) / PT = (100 × 80) / (100 × 8) = 10% per annum
Now let's find out the compound interest of Rs. 14,000 after 3 years at 10%
P = Rs. 14000
T = 3 years
R = 10%
Amount after 3 years = P(1 + R / 100)^T = 14000(1 + 10 / 100)³ = 14000 (110 / 100)³
= 14000(11 / 10)³ = 14 × 11³ = 18634
Compound Interest = Rs.18634 - Rs.14000 = Rs.4634

Question 31

The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is?

A	2
B	3
C	8
D	6

Question 31 Explanation:

Amount = Rs. (30000 + 4347) = Rs. 34347
Let the time be n years
Then, 30000(1 + 7 / 100)ⁿ= 34347
=> (107 / 100) ⁿ= (34347 / 30000) = 11449 / 10000 = (107 / 100)
n = 2 years

Question 32

What is the difference between the compound interests on Rs. 5000 for 1 1/2 years at 4% per annum compounded yearly and half-yearly?

A	A. Rs. 2.04
B	Rs. 3.06
C	Rs. 4.80
D	Rs. 8.30

Question 32 Explanation:

Rs.(5000 x (26 / 25) x (51 / 50)) = Rs 5304
C.I when interest is compounded half- yearly = Rs [5000 x (1 + 2 / 100)³]
= Rs.(5000 x (51 / 50) x (51 / 50) x (51 / 50)) = Rs 5306.04
Difference = Rs. (5306.04 - 5304) = Rs. 2.04

Question 33

There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?

A	Rs. 2160
B	Rs. 3120
C	Rs. 3972
D	Rs. 6240

Question 33 Explanation:

Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years
R =(100 x 60 / 100 x 6) = 10% p.a
Now, P = Rs. 12000. T = 3 years and R = 10% p.a
C.I = Rs [12000 x (1+10/100)3-1]
= Rs (12000 x 331 / 1000) = 3972

Question 34

The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is?

A	A. 625
B	630
C	640
D	650

Question 34 Explanation:

Let the sum be Rs. x. Then,
C.I = [x (1 + (4 / 100) 2 - x)] = (676 / 625 x -x) = 51 / 625 x
S.I = (X * 4 * 2) / 100
= 2x / 25
(51x / 625) – (2x / 25) = 1
x = 625

Question 35

A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1^st January and 1^st July of a year. At the end of the year, the amount he would have gained by way of interest is?

A	Rs. 120
B	Rs. 121
C	Rs. 122
D	Rs. 123

Question 35 Explanation:

Amount = Rs.[1600 x (1 + (5 / 2x 100)²) + 1600 x (1+ (5 / 2x 100))]
Rs.[1600 x 41/40 x 41/40 + 1600 x 41/40]
Rs [1600 x 41/40(41/40 + 1)]
Rs [(1600 x 41 x 81)/ (40 x 40)]
= Rs. 3321
C.I. = Rs. (3321 - 3200) = Rs. 121

Question 36

The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is?

A	6.06%
B	6.07%
C	6.08%
D	6.09%

Question 36 Explanation:

Amount of Rs.100 for 1 year when compounded half-yearly = Rs [ 100 x (1 + 3 / 100)2] Rs.106.09
Effective rate = (106.09 - 100)% = 6.09%

Question 37

Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?

A	Rs. 8620
B	A. Rs. 8600
C	Rs. 8820
D	None of these

Question 37 Explanation:

Amount = Rs. [8000 x (1 + 5 / 100)²]
= Rs. (8000 x 21 / 20 x 21 / 20)
= Rs. 8820

Question 38

The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is?

A	3
B	4
C	5
D	6

Question 38 Explanation:

P(1 + 20 / 100)n > 2P = (6 / 5)n > 2

Now, (6 / 5 x 6 / 5 x 6 / 5 x 6 / 5) >2
So, n = 4 years

Question 39

At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?

A	6.5%
B	6%
C	7%
D	8%

Question 39 Explanation:

Let the rate be R% p.a
Then, 1200 x (1 + R / 100)² = 1348.32
(1 + R/100)² = 134832 / 120000 = 11236 / 10000
(1+ R / 100)² = (106 / 100)²

1 + (R /100) =106 / 100 R = 6%

Question 40

What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.?

A	Rs. 9000.30
B	Rs. 9720
C	Rs. 10123.20
D	Rs. 10483.20

Question 40 Explanation:

Amount = Rs. [25000 x (1+ 12/100)³]
= Rs. (25000 x 28/25 x 28/25 x 28/25)
= Rs. 35123.20
C.I. = Rs. (35123.20 - 25000) = Rs. 10123.20

Question 41

The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum?

A	8
B	10
C	12
D	Cannot be determined

Question 41 Explanation:

[15000 x (1+R/100)2 - 15000] – (15000 x R x 2) / 100 = 96

15000[(1+R/100)2 – 1 – 2R / 100] = 96
15000[(100+R)² – (10000 / 10000)- (200 x R)] = 96
R² = (96 x 2) / 3 = 64
R = 8
Rate = 8%

Question 42

The difference between simple interest and compound on Rs. 1200 for one year at 10% per annum reckoned half-yearly is?

A	Rs. 2.50
B	Rs. 3
C	Rs. 3.75
D	Rs 4

Question 42 Explanation:

S.I = Rs (1200 x 10 x 1) / 100 = Rs. 120
C.I = Rs (1200 x (1 + (5/100)2 - 1200)) = Rs 123
Difference = Rs. (123 - 120) = Rs. 3

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