Monday, 4 April 2016

Aptitude questions on Interests - Compound Interest Concept & Solved Problems

Introduction:
If the interest at the end of a year or fixed period is added to the sum lent, and the amount thus obtained becomes the principal for the next period, then sum of the money is said to be lent at compound interest.
If P is the principal, T is the number of years and R is rate of interest per annum.
Amount= P [ 1 +
R / 100
]T
Note 1 : When the interest is compounded half yearly
 Amount= P[1 +
R/2 / 100
]2T = p[1+
R / 200
]2T
Note 2: When the interest is compounded quarterly
  Amount= P[1 +
R/4 / 100
]4T = p[1+
R / 400
]4T
Compound interest is obtained by when principal is subtracted from Amount.

Solved Problems on Compound Interest
1.    Find the compound interest on Rs  80000 for 3 years at 5% per annum rate of interest?
a)Rs 12000    b)Rs 12610    c)Rs 14800    c)Rs 15300
Answer:B
Explanation: Here the given principal is Rs 80000. Time period is 3 years and rate of interest is 5%.
A =P [ 1 +
R / 100
]T => A= 80000 [1+
5 / 100
]3
ð  A = 80000 x
21 / 20
x
21 / 20
x
21 / 20
ð  A= Rs 92610
Compound Interest = Amount – Principal 
                           =92610 – 80000 = Rs 12610

2.    Find the compound interest on Rs  24000 at 10% per annum for 2 years 6 months
        a)    Rs 6492    b)Rs 6200      c)Rs 6000     d)Rs 5825
Answer :A
Explanation: Principal = Rs 24000   Rate of interest = 10% p.a
Time Period = 2 ½ years
For 2 years , the rate of interest is 10% and for next 6 months rate of interest will be 10%/2= 5%
Amount => A = Rs 24000 ( 1 +
10 / 100
)2 ( 1 +
5 / 100
)
                A= 24000 x
110 / 100
x
110 / 100
x
105 / 100
                A = 24000 x
11 / 10
x
11 / 10
x
21 / 20
=30492
 Compound interest = Amount – Principal 
                           = Rs 30492 – Rs 24000 = Rs 6492

3.    Find the Compound interest on Rs 5000 in 2 years, the rate of interest being 5% for the first year and 10% for the second year ?
a)Rs 775       b) Rs 3875     c)Rs 1650      d)Rs 2150
Answer: A
Explanation: Given Principal = Rs 5000 T= 2 years and Rate of interest for 1st year is 5% and 2nd year is 10% => R1= 5%  and R2= 10%
             A= [1+
R1 / 100
][1+
R2 / 100
]
            A = 5000 x [ 1+
5 / 100
[1+
10 / 100
]
        =>A=5000 x
21 / 20
x
11 / 10
=
           => A= Rs 5775
     ∴ Compound interest= A – P => 5775 – 5000 = Rs 775

4.    At what rate percent per annum compound interest will Rs 12500 amount to Rs 13520 in 2 years ?
a)13%          b)10%          c)8%            d)4%
Answer: D
Explanation: Amount A= Rs 13520 Principal = Rs 12500 Rate of Interest= R and Time period T = 2 years
A =P [ 1 +
R / 100
]T => 13520 = 12500 [ 1 +
R / 100
]2
=>
13520 / 12500
13 = [ 1 +
R / 100
]2
=> 
676 / 675
= [ 1 +
R / 100
]2
 => (
26 / 25
)2= [ 1 +
R / 100
]2
=> [ 1 +
R / 100
] =
26 / 25
=>
R / 100
=
1 / 25
      Therefore, rate of interest is 4%.

5.    What sum will amount to Rs 30000 in 3 years at 25% p.a compound interest?
       a)    Rs 15000  b) Rs 15360   c)Rs 20000  d)Rs 24000
Answer :B
Explanation:  Given Amount= Rs 30000 Time period T= 3 years and Rate of interest= 25%
We have to find the princpali P
A =P [ 1 +
R / 100
]T =>30000= P[ 1 +
25 / 100
]3
  =>30000= P [
5 / 4
]3 => P= 30000 x
4 / 5
x
4 / 5
x
4 / 5
= Rs 15,360
On a sum of Rs 15,360, the amount we get in 3 years 25% rate of compound interest is Rs 30000

6.    At what rate percent compound interest, will Rs  20000 amount to Rs 22050 in 2 years?
a)    12%           b)8%          c)5%         d)2%
Answer: C
Explanation:  Given the Amount is Rs 22050 And Principal = Rs 20000
Time period T= 2 years
A =P [ 1 +
R / 100
]T => 22050 = 20000[ 1 +
R / 100
]2
ð 
22050 / 20000
= [ 1 +$\frac{R}{100}$ ]2
ð  $\frac{441}{400}$ =[ 1 +$\frac{R}{100}$ ]2
ð  ($\frac{21}{20}$ )2 = [ 1 +$\frac{R}{100}$ ]2
ð  $\frac{21}{20}$ = 1 +$\frac{R}{100}$
ð  R= 5%
At 5% p.a compound interest, Rs 20000 becomes Rs  22050 in 2 years

7.    Find the compound interest on Rs 32000 at 10% p.a for one and half years, the interest being compounded half yearly?
a)Rs 5044        b)Rs 6000    c)Rs 3822    d)Rs 4000
Answer:A
 Explanation: Given Principal P= Rs 32000  Rate of Interest =10% and Time period= one and half years = 1
1 / 2
years =
3 / 2
years
 Amount= P[1 +
R/2 / 100
]2T => A = 32000 [ 1 +
10 / 200
]2 x 3/2
ð  A= 32000 x $\left[\frac{21}{20}\right] ^{3} $
ð  A= 32000 x $ \frac{21}{20}$ x $ \frac{21}{20}$ x $ \frac{21}{20}$
ð  A = 4 x21x21x21 =37044

Compound Interest = Rs 37044 – Rs 32000 = 5044

8.    What is the difference between Compound interest and simple interest on Rs 12800 for 2 years at 10% p.a rate of interest?
a)    Rs 120     b)Rs 128       c)Rs 150        d)Rs 172
Answer:B
Explanation : Simple Interest =
PTR / 100
= 12800 X 2 X
10 / 100
  = Rs 2560
To find compound interest, first we find Amount
Amount= P[1 +
R / 100
]T = > 12800 x [ 1 +
10 / 100
]2 => 12800x
11 / 10
x
11 / 10
=15488
Compound interest = A – P => Rs 15488 – 12800 = Rs 2688
The difference between CI and SI = Rs 2688 – Rsd 2560 = Rs 128
Short cut : The difference between CI and SI for 2 years at R% p.a  is  D =
PR2 / 1002
 Therefore, D =
12800 X 102 / 1002
=
12800 X 10 X 10 / 100 X 100 2
= Rs 128

9.    The difference between CI and SI on Rs  4000 for 2 years is Rs 10. What is the rate of interest per annum?
a)    2%                 b)7%            c)9%            d)5%
Answer: D
Explanation: The difference between CI and SI for 2 years at R% p.a  is  D =
PR2 / 1002
ð  R2 =
D x 1002 / P
ð  R2 = $\frac{10\times100\times100}{100}$
ð  R= 5%

10. The difference between CI and SI on a certain sum of money for 3 years at 10% p.a rate of interest is Rs 1550. Find the principal?
a)    Rs 50000  b)Rs 34000    c)Rs 40000    d)Rs 42000
Answer : A
Explanation:
Shortcut method :When the difference between the simple interest and compound interest on P for 3 years at R% rate of interest, then P =
1003 D / R2 (300+R)
                 P=
100 x 100 x 100 x 1550 / 102 x (300+10)
                   P=
100x100x100x1550 / 100
x 310 = Rs 50000

11. The value of a land increases by 15% annually. If its present value is 1058000. What was its value 2 years ago?
a)    Rs 400000 b)Rs 500000   c)Rs 800000   d)Rs 1000000
Answer:C
Explanation:  Its present value is Rs 1058000 means  that is the amount.
        A= P [ 1+
R / 100
]T => Rs 1058000 = P [1+
15 / 100
]2
ð  1058000= P [ $\frac{23}{20}$ ]2
ð  1058000= P X $\frac{529}{400}$
ð  P = 1058000 X $\frac{400}{529}$ = Rs 800000

12. At Compound interest, a sum of money becomes 2 times itself in 4 years, In how many years will it become 8 times?
         a)    10 years        b)12 years     c)14 years     d)15 years
Answer: B
Explanation: At compound interest, a principal is  always multiplied.
           A sum of money becomes 2 times in 4 years. Means in every 4 years, the principal becomes 2 times.
To become 8 times => 23 times (for each 2 times, it takes 4 years) , it takes 3 x 4 = 12 years

13. The compound interest on a certain sum of money for 2 years at 10% pa. Is Rs   2520. Find the simple interest on the same sum of money at the same rate for 2 years?
         a)    Rs 2000    b)Rs 2200   c)Rs 2400  d)Rs 2540
Answer:C
Explanation: Given CI= Rs 2520 and Rate of intest R = 10% and T= 2 years
 Let Principal P be Rs 100
 => Amount at compound interest
=> A = 100 x (
110 / 100
)2 => A= Rs 121
Theefore CI for 2 years is Rs 21
If P is Rs 100  ---------CI is Rs 21
   ?                ---------CI is Rs 2520
P =
2520 / 21
x 100 = Rs 12000
Now SI on Rs  12000 for 2 years at 10% rate of interest per annum => SI =
PTR / 100
             =>   12000 x 2 x
10 / 100
= Rs 2400

14. The SI on a certain sum of money for 3 years at 8% per annum is Rs  1200. What is the compound interest on the same sum of money at the same rate of interest for 2 years?
        a)    Rs 840     b)Rs 800       c)Rs 1000      d)Rs 832
Answer:D
Explanation: Given SI= Rs 1200  T= 3 Years   R= 8% T= 3 years
    SI =
PTR / 100
=> 1200 = P x 3 x
8 / 100
=> P = Rs 5000
To find CI, first we find Amount => A = P[ 1+
R / 100
]T
ð  A = Rs 5000 x $\frac{108}{100}$ x $\frac{108}{100}$ = Rs 5832
Therefore, CI = A –P= Rs 5832 – Rs 5000 = Rs 832
Shortcut Method :
Given SI for 3 years is Rs 1200 . SI for 1 years is
1200 / 3
= Rs 400
SI and CI are same for 1st year.
Now CI for 2nd year = 400 + 8% on 1st year interest 400 = 432
 CI for 2 years = Rs 400+ Rs 432 = Rs 832

15. A sum of money amounts to Rs 2400 in 2 years and Rs 2640 in 3 years at CI. Find the rate of percent per annum?
a)10 %              B)12$        C)12.5%        D)14%
Answer:A
Explanation: A1 = Rs 2400  and A2= 2640
P + CI for 3 years – PI + CI for 2 years = Rs 2640 – 2400 = Rs 240
This Rs 240 is the simple interest obtained on Rs 2400 in the 3rd year.
ð  SI= $\frac{PTR}{100}$
ð  240 = 2400 x 1 x $\frac{R}{100}$
ð  R= 10%