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 Halfyearly: 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 R_{1}%, R_{2}%, R_{3}% for 1^{st}, 2^{nd}and 3^{rd} year respectively. Then the amount = P (1+R_{1}/100) (1+R_{2}/100) (1+R_{3}/100)
 Present worth of Rs.x due n years hence is given by: Present worth = [X / (1+ R/100)].
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Test  Simple and Compound Interest
Simple and compound interest  Question and Answers
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Question 1

A

3.5

B

3.75

C

4

D

4.5

= 350 => t =3.5
Question 2

A

2%

B

4%

C

8%

D

2%

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

Rs. 2500

B

Rs. 7000

C

Rs. 4000

D

Rs. 3000

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

A

Rs. 250

B

Rs. 400

C

Rs. 500

D

Rs.100

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

A

Rs. 1500

B

Rs. 2000

C

Rs. 1000

D

Rs. 5000

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

A

Rs. 7654

B

Rs. 1236

C

Rs. 1543

D

Rs. 2765

Therefore, CI = 11236 – 10000 = Rs. 1236
Question 7

A

Rs. 301.2

B

Rs. 307.5

C

Rs. 387.2

D

Rs. 309.3

Therefore, CI = 3307.5 – 3000 = Rs. 307.5
Question 8

A

8%

B

5%

C

3%

D

9%

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

A

Rs. 65000

B

Rs. 63000

C

Rs. 67000

D

Rs. 61000

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

9%

B

3%

C

7%

D

2%

Time = 3 years
SI = PTR / 100
R = SI * 100 / PT
R = 2250 * 100 / 25000 * 3
R = 3%
Question 11

A

Rs. 62

B

Rs. 62.4

C

Rs. 61.5

D

Rs. 60.5

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

A

Rs. 11231

B

Rs. 11781

C

Rs. 11658

D

Rs. 11244

= 10000 (102 / 100) (105 / 100) (110 / 100) = (102 × 105 × 11) / 10
= Rs. 11781
Question 13

A

Rs. 80000

B

Rs. 40000

C

Rs. 50000

D

Rs. 60000

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

Rs. 4320

B

Rs. 2220

C

Rs. 4400

D

Rs. 4420

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

A

Rs. 10000

B

Rs. 12000

C

Rs. 6000

D

Rs. 8000

Compound Interest on P at 10% for 2 years when interest is compounded halfyearly
=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

Rs. 1500

B

Rs. 1400

C

Rs. 1200

D

Rs. 1000

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

A

Rs. 1764 and Rs. 1600

B

Rs. 1756 and Rs. 1608

C

Rs. 1722 and Rs. 1642

D

None of these

(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))^{(75)}
=> (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

A

12 years

B

18 years

C

16 years

D

14 years

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

A

40%

B

30%

C

20%

D

10%

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) + (R^{2 }/ 10000) – x] =2400
x (2R / 100) + (R^{2} / 10000) = 2400
(2xR /100) + (xR^{2} / 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%

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) + (R^{2 }/ 10000) – x] =2400
x (2R / 100) + (R^{2} / 10000) = 2400
(2xR /100) + (xR^{2} / 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 21

A

Rs. 8000

B

Rs.7500

C

Rs. 8500

D

Rs. 8200

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

A

38484

B

38266

C

38416

D

38226

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

A

12%

B

13%

C

14%

D

15%

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

A

Rs. 365

B

Rs. 325

C

Rs. 395

D

Rs. 375

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

A

4120

B

3300

C

4200

D

4420

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

Rs.1820

B

Rs.1640

C

Rs.1260

D

Rs.1440

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

Rs. 12500

B

Rs. 11200

C

Rs. 8840

D

Rs. 12600

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 1^{1}⁄_{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

Rs. 800

B

Rs. 822

C

Rs. 840

D

Rs. 816

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

A

Rs. 66

B

Rs. 82

C

Rs. 86

D

Rs. 88

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

A

Rs.3794

B

Rs.3714

C

Rs.4612

D

Rs.4634

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 × 11^{3} = 18634
Compound Interest = Rs.18634  Rs.14000 = Rs.4634
Question 31

A

2

B

3

C

8

D

6

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

A

A.
Rs. 2.04 
B

Rs. 3.06

C

Rs. 4.80

D

Rs. 8.30

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

A

Rs. 2160

B

Rs. 3120

C

Rs. 3972

D

Rs. 6240

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)31]
= Rs (12000 x 331 / 1000) = 3972
Question 34

A

A.
625 
B

630

C

640

D

650

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

Rs. 120

B

Rs. 121

C

Rs. 122

D

Rs. 123

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

A

6.06%

B

6.07%

C

6.08%

D

6.09%

Effective rate = (106.09  100)% = 6.09%
Question 37

A

Rs. 8620

B

A.
Rs. 8600 
C

Rs. 8820

D

None of these

= Rs. (8000 x 21 / 20 x 21 / 20)
= Rs. 8820
Question 38

A

3

B

4

C

5

D

6

Now, (6 / 5 x 6 / 5 x 6 / 5 x 6 / 5) >2
So, n = 4 years
Question 39

A

6.5%

B

6%

C

7%

D

8%

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

A

Rs. 9000.30

B

Rs. 9720

C

Rs. 10123.20

D

Rs. 10483.20

= 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

A

8

B

10

C

12

D

Cannot be determined

15000[(1+R/100)2 – 1 – 2R / 100] = 96
15000[(100+R)^{2} – (10000 / 10000) (200 x R)] = 96
R^{2} = (96 x 2) / 3 = 64
R = 8
Rate = 8%
Question 42

A

Rs. 2.50

B

Rs. 3

C

Rs. 3.75

D

Rs 4

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