Practice Compound interest - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) At what rate per annum will Rs.32000 yield a compound interest of Rs.5044 in 9 months interest being compounded quarterly ?
(a)
(b)
(c)
(d)
Using Rule 1,
Let the rate of CI be R per cent per annum.
CI = P$[(1 + R/100)^T - 1]$
5044 = 32000$[(1 + R/400)^3 - 1]$
[Since, Interest is compounded quarterly]
$5044/32000 = (1 + R/400)^3 - 1$
$(1 + R/400)^3 - 1 = 1261/8000$
$(1 + R/400)^3 = 1 + 1261/8000$
$(1 + R/400)^3 = 9261/8000 = (21/20)^3$
1 + $R/400 = 21/20$
$R/400 = 21/20 - 1 = 1/20$
R = $400/20$ = 20
Q-2) A sum of money on compound interest amounts to Rs.10648 in 3 years and Rs.9680 in 2 years. The rate of interest per annum is :
(a)
(b)
(c)
(d)
Using Rule 1,
Let the sum be P and rate of interest be R% per annum. Then,
$P(1 + R/100)^2 = 9680$ ...(i)
$P(1 + R/100)^3 = 10648$ ...(ii)
On dividing equation (ii) by (i)
$1 + R/100 = 10648/9680$
$R/100 = 10648/9680$ -1
= ${10648 - 9680}/9680$
$R/100 = 968/9680 = 1/10$
R = $1/10 × 100$ = 10%
Q-3) At what percent per annum will Rs.3000/- amounts to Rs.3993/- in 3 years if the interest is compounded annually?
(a)
(b)
(c)
(d)
Using Rule 1,If A = Amount, P = Principal, r = Rate of Compound Interest (C.I.), n = no. of years then,A=P$(1 + r/100)^n$, C.I. = A - PC.I. = P$[(1 + r/100)^n - 1]$
P = Rs.3000, A = Rs.3993, n = 3 years
A = P$(1 + r/100)^n$
$(1 + r/100)^n = A/P$
$(1 + r/100)^3 = 3993/3000 = 1331/1000$
$(1 + r/100)^3 = (11/10)^3$
1 + $r/100 = 11/10$
$r/100 = 11/10$ - 1
$r/100 = 1/10 ⇒ r = 100/10$ = 10%
Q-4) If a sum of money compounded annually becomes 1.44 times of itself in 2 years, then the rate of interest per annum is
(a)
(b)
(c)
(d)
A = P$(1 + R/100)^T$
1.44P = P$(1 + R/100)^2$
$(1.2)^2 = (1 + R/100)^2$
$1 + R/100$ = 1.2
R = 0.2 × 100 = 20%
Using Rule 8,
Here, n = 1.44, t = 2 years
R% = $(n^{1/6} - 1) × 100%$
= $[(1.44)^{1/2} - 1] × 100%$
= [(1.2) - 1] × 100%
= 0.2 × 100% ⇒ R% = 20%
Q-5) If the amount is 2.25 times of the sum after 2 years at compound interest (compound annually), the rate of interest per annum is :
(a)
(b)
(c)
(d)
Suppose P = Rs.100
and amount A = Rs.225
A = P$(1 + r/100)^t$
or 225 = $100(1 + r/100)^2$
or $225/100 = [1 + r/100]^2$
or 1 + $r/100 = 15/10$
or ${100 + r}/100 = 15/10$
or 100 + r = 150
or r = 50%
Using Rule 8,If a sum becomes 'n' times of itself in 't' years on compound interest, then R% = $[n^{1/t} - 1] × 100%$
Here, n = 2.25, t = 2 years
R% = $(n{1/t} - 1) × 100%$
R% = $[(2.25)^{1/2} - 1]$ × 100%
= [1.5 - 1] × 100%
= 0.5 × 100% = 50%
Q-6) The difference between simple and compound interests on a sum of money at 4% per annum for 2 years is 8. The sum is
(a)
(b)
(c)
(d)
Using Rule 6,
Let the sum be x.
When difference between the compound interest and simple interest on a certain sum of money for 2 years at r% rate is x, then the sum is given by:
Sum = Difference × $(100/\text"Rate")^2$
= Rs.8 × $(100/4)^2$
= Rs.8 × 25 × 25 = Rs.5000
Q-7) If the difference between the compound interest, compounded every six months, and the simple interest on a certain sum of money at the rate of 12% per annum for one year is 36, the sum is :
(a)
(b)
(c)
(d)
Easy Trick
As the interest was compounded half-yearly,
we changed r to $r/2$ and t to 2t.
T = 1 year & R 6%
Sum = ${36 × 100 × 100}/{6 × 6}$ = Rs.10000
Q-8) The difference between the compound and the simple interest on a sum for 2 years at 10% per annum, when the interest is compounded annually, is 28. If the interest were compounded halfyearly, the difference in the two interests will be
(a)
(b)
(c)
(d)
Using Rule 6,The difference between C.I. and S.I. on a sum 'P' in 2 years at the rate of R% rate of compound interest will beC.I - S.I. = P$(R/100)^2 = {S.I. × R}/200$ For 3 years, C.I. - S.I. = P$(R/100)^2 × (3 + R/100)$
Using Rule 1,If A = Amount, P = Principal, r = Rate of Compound Interest (C.I.), n = no. of years then,A=P$(1 + r/100)^n$, C.I. = A - PC.I. = P$[(1 + r/100)^n - 1]$
If the difference between compound interest and simple interest at the rate of r% per annum for 2 years be x, then
Principal = $x(100/r)^2$
= 28$(100/10)^2$ = Rs.2800
If the interest is compounded half yearly, then
r = $10/2$ = 5%,
Time = 4 half years
Simple interest
= ${2800 × 5 × 4}/100$ = Rs.560
Compound interest
= $2800[(1 + 5/100)^4 - 1]$
= 2800 [1.2155 - 1]
= 2800 × 0.2155 = 603.41
Difference = Rs.(603.41–560) = Rs.43.41
Q-9) The difference between the simple and compound interest on a certain sum of money at 5% rate of interest per annum for 2 years is 15. Then the sum is :
(a)
(b)
(c)
(d)
Let the sum Rs.x. Then,
C.I. = $x(1 + 5/100)^2 - x$
= ${441x}/400 - x = {441x - 400x}/400$
= $41/400$x
Now, S.I. = ${x × 5 × 2}/100 = x/10$
(C.I.) - (S.I.)= ${41x}/400 - x/10$
= ${41x - 40x}/400 = x/400$
$x/400 = 15$
x = 15 × 400 = 6000
Hence, the sum is Rs.6000
Using Rule 6,The difference between C.I. and S.I. on a sum 'P' in 2 years at the rate of R% rate of compound interest will beC.I - S.I. = P$(R/100)^2 = {S.I. × R}/200$ For 3 years, C.I. - S.I. = P$(R/100)^2 × (3 + R/100)$
C.I. - S.I. = Rs.15, R = 5%, T = 2 years, P =?
C.I. - S.I. = P$(R/100)^2$
15 = P$(5/100)^2$
P = 15 × 400 = Rs.6000
Q-10) If the difference between the compound and simple interests on a certain sum of money for 3 years at 5% per annum is 15.25, then the sum is
(a)
(b)
(c)
(d)
Using Rule 6,
Difference between C.I. and S.I for 3 years
= $\text"PR"^2/(100)^2(R/100 + 3)$
15.25 = ${P × 25}/10000(5/100 + 3)$
15.25 = ${P × 305}/{400 × 100}$
P = ${15.25 × 400 × 100}/305$ = Rs.2000
Q-11) A sum of money invested at compound interest amounts to Rs.650 at the end of first year and Rs.676 at the end of second year. The sum of money is :
(a)
(b)
(c)
(d)
Interest on Rs.650 for 1 year
= 676 - 650 = Rs.26
So, r = $26/650 × 100$
r = 4% per annum
P = $A/[1 + r/100]^t = 650/[1 + 4/100]^1$
= $650/{26/25} = 650 × 25/26$ = Rs.625
Using Rule 7(i),
Here, b - a = 1
B = Rs.676, A = Rs.650
R% = $(B/A - 1)$ × 100%
= $[676/650 - 1] × 100%$
= $[{676 - 650}/650] × 100%$
= $26/650 × 100% = 100/25$ = 4%
Amount= P$(1 + R/100)^1$
650 = P$(1 + 4/100)$
P = ${650 × 100}/104$ = Rs.625
Note : A sum at a rate of interest compounded yearly becomes Rs.$A_1$, in n years and Rs. $A_2$ in (n + 1) years, then
P = $A_1(A_1/A_2)^n$
Q-12) The time in which Rs.80,000 amounts to Rs.92,610 at 10% p.a. compound interest, interest being compounded semi annually is :
(a)
(b)
(c)
(d)
Using Rule 1 and 2,
Time = t half year
and R = 5% per half year
A = P$(1 + R/100)^T$
$92610/80000 = (1 + 5/100)^T$
$9261/8000 = (21/20)^T$
T = 3 half years or 1$1/2$ years
$(21/20)^3 = (21/20)^T$
Q-13) The compound interest on Rs.8,000 at 15% per annum for 2 years 4 months, compounded annually is:
(a)
(b)
(c)
(d)
Using Rule 1,
Amount = P$(1 + R/100)^t$
= 8000$(1 + 15/100)^{2{1}/3}$
= 8000$(1 + 3/20)^2(1 + 3/{20 × 3})$
= $8000 × 23/20 × 23/20 × 21/20$ = Rs.11109
Compound Interest
= Rs.(11109 - 8000) = Rs.3109.
Q-14) In what time Rs.8,000 will amount to Rs.9,261 at 10% per annum compound interest, when the interest is compounded half yearly ?
(a)
(b)
(c)
(d)
Using Rule 1 and 2,
Interest is compounded half yearly.
Rate of interest = 5%
Time = $n/2$ years (let)
or n half-years
A = P$(1 + R/100)^T$
9261 = 8000$(1 + 5/100)^n$
$9261/8000 = (21/20)^n$
$(21/20)^3 = (21/20)^n$
n = 3 half years
= $3/2$ years = $1{1}/2$ years
Q-15) At what rate per cent per annum will Rs.2304 amount to Rs.2500 in 2 years at compound interest ?
(a)
(b)
(c)
(d)
Using Rule 1,
Let the rate per cent per annum be r. Then,
2500 = 2304$(1 + r/100)^2$
$(1 + r/100)^2 = 2500/2304 = (50/48)^2$
$1 + r/100 = 50/48 = 25/24$
$r/100 = 25/24 - 1 = 1/24$
r = $100/24 = 25/6 = 4{1}/6$%
Q-16) A sum of money doubles itself in 4 years at compound interest. It will amount to 8 times itself at the same rate of interest in :
(a)
(b)
(c)
(d)
A sum of Rs.x becomes Rs.2x in 4 years.
Similarly, Rs.2x will become 2 × 2x
= Rs.4x in next 4 years and Rs.4x will become
2 × 4x = Rs.8x in yet another 4 years.
So, the total time = 4 + 4 + 4 = 12 years
Using Rule 5,A certain sum becomes 'm' times of itself in 't' years on compound interest then the time it will take to become mn times of itself is t × n years.
Here, m = 2, t = 4
Time taken to become
$2^3$ = n × t years
= 3 × 4 = 12 years
Note : If a sum of money becomes n times in t years, it will become $t^1 = n^x$ times at the same rate of interest in $t^1$ years given by,$t^1$ = xt
Q-17) Compound interest on a sum of money for 2 years at 4 per cent per annum is Rs.2,448. Simple interest of the same sum of money at the same rate of interest for 2 years will be
(a)
(b)
(c)
(d)
C.I.= P$(1 + r/100)^t$ - P
2448 = P$[(1 + r/100)^t - 1]$
or 2448 = P$[(1 + 4/100)^2 - 1]$
2448 = P$[676/625 - 1]$
2448 = P$[51/625]$
P = ${2448 × 625}/51$ = Rs.30,000
S.I. = ${30000 × 4 × 2}/100$ = Rs.2400
Using Rule 10,
Here, C.I. = Rs.2448, R = 4%, S.I. = ?
C.I.= S.I.$(1 + R/200)$
2448 = S.I.$(1 + 4/200)$
2448 = S.I.$(1 + 1/50)$
2448 = S.I.$(51/50)$
S.I. = ${2448 × 50}/51$ = Rs.2400
Q-18) If the compound interest on a certain sum for two years at 12% per annum is Rs.2,544, the simple interest on it at the same rate for 2 years will be
(a)
(b)
(c)
(d)
C.I. = P$[(1 + R/100)^T - 1]$
2544 = P$[(1 + 12/100)^2 - 1]$
2544 = P$[(28/25)^2 - 1]$
2544 = P$(784/625 - 1)$
2544 = P$({784 - 625}/625)$
2544 = ${P × 159}/625$
P = ${2544 × 625}/159$ = Rs.10000
S.I. = ${P × R × T}/100$
= ${10000 × 2 × 12}/100$ = Rs.2400
Using Rule 10,
Here, C.I. = Rs.2544, R = 12%, S.I. = ?
C.I. = S.I.$(1 + R/200)$
2544 = S.I.$(1 + 12/200)$
2544 = S.I.$(212/200)$
S.I. = ${2544 × 200}/212$ = Rs.2400
Q-19) The simple interest on a sum of money at 4% per annum for 2 years is Rs.80. The compound interest in the same sum for the same period is
(a)
(b)
(c)
(d)
Principal = $\text"S.I. × 100"/ \text"Time × Rate"$
= ${80 × 100}/{2 × 4}$ = Rs.1000
C.I. = P$[(1 + R/100)^T - 1]$
= 1000$[(1 + 4/100)^2 - 1]$
= 1000$[(25/26)^2 - 1]$
= 1000$(676/625 - 1)$
= 1000$({676 - 625}/625)$
= ${1000 × 51}/625$= Rs.81.60
Using Rule 10,
Here, S.I. = Rs.80, R = 4%, C.I. = ?
C.I.= S.I.$(1 + R/200)$
C.I.= 80$(1 + 4/200)$
= $80(1 + 1/50)$
= 80 × $51/50$ = Rs.81.60
Q-20) If the compound interest on a certain sum for 2 years at 3% per annum is Rs.101.50, then the simple interest on the same sum at the same rate and for the same time will be
(a)
(b)
(c)
(d)
Let the sum be P.
101.50 = P$[(1 + 3/100)^2 - 1]$
[Since, C.I. = P$[(1 + r/100)^n - 1]$]
101.50 = P$[(103/100)^2 - 1]$
=P$({10609 - 10000}/10000)$
P = Rs.${101.50 × 10000}/609 = Rs.1015000/609$
S.I. = ${1015000 × 2 × 3}/{609 × 100}$ = Rs.100
Using Rule 10,The simple interest for a certain sum for 2 years at an annual rate interest R% is S.I., thenC.I. = S.I.$(1 + R/200)$
Here, C.I. = Rs.101.50, R = 3%, S.I. = ?
C.I. = S.I.$(1 + R/200)$
101.50 = S.I.$(1 + 3/200)$
S.I. = ${101.50 × 200}/203$ = Rs.100