Practice Compound interest - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) The difference between compound interest and simple interest on 2500 for 2 years at 4% per annum is
(a)
(b)
(c)
(d)
S.I. = Rs.${2500 × 2 × 4}/100$ = Rs.200
C.I. = Rs.2500$[(1 + 4/100)^2 - 1]$
= Rs.2500$[(26/25)^2 - 1]$
= Rs.${(676 - 625)}/625$ × 2500
= Rs.$51/625 × 2500$ = Rs.204
The required difference
= C.I. - S.I. = Rs.(204 - 200) = Rs.4
Using Rule 6,
Here, C.I. - S.I.= ?, P = Rs.2500, R = 4%, T = 2
C.I. - S.I.= P$(R/100)^2$
= 2500$(4/100)^2$
= 2500 × $1/25 × 1/25$
C.I.–S.I. = Rs.4
Q-2) A sum of Rs.8000 will amount to Rs.8820 in 2 years if the interest is calculated every year. The rate of compound interest is
(a)
(b)
(c)
(d)
Using Rule 1,
If the rate of C.I. be r% per annum, then
A = P$(1 + R/100)^T$
8820 = 8000$(1 + r/100)^2$
$8820/8000 = (1 + r/100)^2$
$441/400 = (21/20)^2 = (1 + r/100)^2$
$1 + r/100 = 21/20$
$r/100 = 21/20 - 1 = 1/20$
r = $1/20$ × 100
r = 5% per annum
Q-3) A sum of money amounts to Rs.4,840 in 2 years and to Rs.5,324 in 3 years at compound interest compounded annually. The rate of interest per annum is :
(a)
(b)
(c)
(d)
Let the rate of interest be r% per annum,
According to the question,
4840 = P$(1 + r/100)^2$ ..... (i)
and 5324 = P$(1 + r/100)^3$....(ii)
On dividing equation (ii) by equation (i), we have,
$1 + r/100 = 5324/4840 = 1 + 484/4840$
$r/100 = 484/4840$ ⇒ r = 10%
Using Rule 7,If on compound interest, a sum becomes Rs.A in 'a' years and Rs.B in 'b' years then,(i) If b - a = 1, then, R% = $(B/A - 1)$ × 100%(ii) If b - a = 2, then, R% = $(√{B/A} - 1)$ × 100%(iii) If b - a = n then, R% = $[(B/A)^{1/n} - 1]$ × 100%where n is a whole number.
Here, b - a = 3 - 2 = 1, B = Rs.5,324, A = Rs.4,840
R% = $(B/A - 1)$ × 100%
= $(5324/4840 -1)$ × 100%
= $({5324 - 4840}/4840) × 100%$
= $484/4840 × $100% = 10%
Q-4) The compound interest on a certain sum for two successive years are Rs.225 and Rs.238.50. The rate of interest per annum is :
(a)
(b)
(c)
(d)
Difference = 238.50 - 225 = Rs.13.50
= S.I. on Rs.225 for 1 year
Rate = $\text"S.I. × 100"/\text"Principal × Time"$
= ${13.50 × 100}/{225 × 1}$ = 6% per annum
Using Rule 7(i),
Here, b - a = 1
B = Rs.238.50, A = Rs.225
R% = $(B/A - 1)$ × 100%
= $({238.50}/225 - 1) × 100%$
= $({238.50 - 225}/225) × 100%$
= $({13.5}/225) × 100%$ = 6%
Q-5) A certain amount of money at r%, compounded annually after two and three years becomes Rs.1440 and Rs.1728 respectively. r is
(a)
(b)
(c)
(d)
If the principal be Rs.P, then
A = P$(1 + R/100)^T$
1440 = P$(1 + R/100)^2$ ...(i)
and 1728 = P$(1 + R/100)^3$ ...(ii)
On dividing equation (ii) by (i),
$1728/1440 = 1 + r/100$
$r/100 = 1728/1440$ - 1
= ${1728 - 1440}/1440 = 288/1440$
r = ${288 × 100}/1440$
r = 20% per annum
Using Rule 7(i),
Here, b - a = 3 - 2 = 1
B = Rs.1728, A = Rs.1440
R% = $(B/A - 1)$ × 100%
= $(1728/1440 - 1) × 100%$
= $({1728 - 1440}/1440) × 100%$
= $[288/1440] × 100%$ = 20%
Q-6) On a certain sum of money lent out at 16% p.a. the difference between the compound interest for 1 year, payable half yearly, and the simple interest for 1 year is 56. The sum is
(a)
(b)
(c)
(d)
Using Rule 6,
Rate of interest = 8% per halfyear
Time = 2 half years
Difference of interests = ${PR^2}/100$
56 = $P × (8)^2/(100)^2$
P = ${56 × 10000}/64$ = 8750
Q-7) If the difference between the simple and compound interests on a sum of money for 2 years at 4% per annum is 80, the sum is :
(a)
(b)
(c)
(d)
Using Rule 6,
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
$x(100/r)^2$ Here x = Rs.80, r = 40%
Required sum = 80$(100/4)^2$
= 80 × 25 × 25 = Rs.50000
Q-8) The difference between simple and compound interest (compounded annually) on a sum of money for 2 years at 10% per annum is 65. The sum is
(a)
(b)
(c)
(d)
Let the sum be x. Then,
C.I. = $x(1 + 10/100)^2 - x = {21x}/100$
S.I. = ${x × 10 × 2}/100 = x/5$
C.I. - S.I. = ${21x}/100 - x/5 = x/100$
Given that, $x/100$ = 65
x = 6500
Hence, the sum is Rs.6500.
Using Rule 6,
Here, C.I. - S.I. = Rs.65, R = 10%, T = 2 years, P = ?
C.I. - S.I. = P$(R/100)^2$
65 = P$(10/100)^2$ ⇒ P = Rs.6500
Q-9) If the difference between the compound interest and simple interest on a sum at 5% rate of interest per annum for three years is 36.60, then the sum is
(a)
(b)
(c)
(d)
Simple Approach
Difference of SI and CI for 3 years
= ${PR(300 + R)}/{100^3}$
Since, ${P × 25 × 305}/{100 × 100 × 100}$ = 36.60
P = ${36.60 × 100 × 100 × 100}/{25 × 305}$ = Rs.4800
Using Rule 6,
C.I.–S.I. = Rs.36.60, R = 5%, P =?, T = 3yrs.
C.I. - S.I.= P$(R/100)^2 × (3 + R/100)$
36.60 = P$(5/100)^2 × (3 + 5/100)$
36.60 = P × $25/{100^2} × 305/100$
P = ${36.60 × 100 × 100 × 100}/{25 × 305}$
P = $36600000/{25 × 305}$ = Rs.4800
Q-10) On a certain sum of money, the simple interest for 2 years is Rs.350 at the rate of 4% per annum. If it was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned ?
(a)
(b)
(c)
(d)
Principal = $\text"S.I. × 100"/\text"Time × Rate"$
= ${350 × 100}/{2 × 4}$ = Rs.4375
Difference = ${PR^2}/10000$
= ${4375 × 4 × 4}/10000$ = Rs.7
Q-11) What sum will give 244 as the difference between simple interest and compound interest at 10% in 1 1 2 years compounded half yearly ?
(a)
(b)
(c)
(d)
Using Rule 6,
Time = $3/2 × 2 = 3$ half years
Rate = $10/2$ = 5% per half year
[Since, when r → $r/2$, then t → 2t]
Difference = P$(r^3/1000000 + {3r^2}/10000)$
244 = P$(125/1000000 + 75/10000)$
244 = P$(7625/1000000)$
P = ${244 × 1000000}/7625$ = Rs.32000
Q-12) A sum of 6,000 is deposited for 3 years at 5% per annum compound interest (compounded annually). The difference of interests for 3 and 2 years will be
(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$[(1 + r/100)^n - 1]$
C.I. after 3 years
= 6000$[(1 + 5/100)^3 - 1]$
= 6000$({9261 - 8000}/8000)$
= 6000 × $1261/8000$ = Rs.945.75
CI after 2 years
= $6000[(1 + 5/100)^2 - 1]$
= 6000$({441 - 400}/400)$
= 6000 × $41/400$ = Rs.615
Required difference
= Rs.(945.75 - 615) = Rs.330.75
Q-13) A sum of Rs.210 was taken as a loan. This is to be paid back in two equal instalments. If the rate of interest be 10% compounded annually, then the value of each instalment is
(a)
(b)
(c)
(d)
Using Rule 9(i),
Let the value of each instalment be Rs.x
Principal = Present worth of Rs.x due 1 year hence, present worth of Rs. x due 2 years hence
210 = $x/(1 + R/100) + x/(1 + R/100)^2$
210 = $x/(1 + 10/100) + x/(1 + 10/100)^2$
210 = $x/{1 + 1/10} + x/(1 + 1/10)^2$
210 = $x/{11/10} + x/(11/10)^2$
210 = ${10x}/11 + {100x}/121$
210 = ${110x + 100x}/121$
210 × 121 = 210 x
$x = {210 × 121}/210$ = Rs.121
Q-14) The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525 . The simple interest on the same sum for double the time at half the rate per cent per annum is :
(a)
(b)
(c)
(d)
C.I. = P$[(1 + R/100)^T - 1]$
525 = P$[(1 + 10/100)^2 - 1]$
525 = P$(121/100 - 1)$
525 = ${P × 21}/100$
P = ${525 × 100}/21$ = Rs.2500
Again, new rate = 5% per annum
S.I. = $\text"Principal × Time × Rate"/100$
= ${2500 × 5 × 4}/100$ = Rs.500
Q-15) A sum of money at compound interest will amount to Rs.650 at the end of the first year and Rs.676 at the end of the second year. The amount of money is
(a)
(b)
(c)
(d)
Principal = Rs.P (let)
Rate = R% per annum
A = P$(1 + R/100)^T$
650 = P$(1 + R/100)$
$650/P = (1 + R/100)$ ...(i)
Again, 676 = P$(1 + R/100)^2$
676 = P$(650/P)^2$
= ${P × 650 × 650}/P^2$
P = ${650 × 650}/676$ = Rs.625
Q-16) A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to four times itself ?
(a)
(b)
(c)
(d)
A = P$(1 + R/100)^T$
Let P be Rs.1, then A = Rs.2
2 = 1$(1 + R/100)^4$
$2^2 = (1 + R/100)^8$
Time = 8 years
Using Rule 11,
Here, $x = 2, n_1 = 4, y = 4, n_2$ = ?
Using $x^{1/n_1} = y^{1/n_2}$
$(2)^{1/4} = (4)^{1/n_2}$
$(2)^{1/4} = (2^2)^{1/n_2}$
$2^{1/4} = 2^{1/n_2}$
$1/4 = 2/n_2$
$n_2$ = 8 years
Q-17) A sum of money invested at compound interest doubles itself in 6 years. At the same rate of interest it will amount to eight times of itself in :
(a)
(b)
(c)
(d)
Let the sum be x. Then,
$2x = x(1 + r/100)^6$
2 = $(1 + r/100)^6$
Cubing both sides,
8 =$((1 + r/100)^6)^3$
8 = $(1 + r/100)^18$
$8x = x(1 + r/100)^18$
The sum will be 8 times in 18 years.
i.e., Time = 18 years
Using Rule 5,
Here, m = 2, t = 6 years
It will becomes 8 times of itself
= $2^3$ times of it self
in t × n years = 6 × 3 = 18 years
Q-18) A sum borrowed under compound interest doubles itself in 10 years. When will it become fourfold of itself at the same rate of interest ?
(a)
(b)
(c)
(d)
Let the sum be x which becomes 2x in 10 years.
Hence, 4x in 20 years
Method 2 :
Unitary Method can also be used.
Using Rule 5,
Here, m = 2, t = 10
Time taken to become 4 times = $2^2$ times
= t × n = 10 × 2 = 20 years
Q-19) A sum of money at compound interest doubles itself in 15 years. It will become eight times of itself in
(a)
(b)
(c)
(d)
A = P$(1 + R/100)^T$
2 = 1$(1 + R/100)^15$
Cubing on both sides, we have
8 = 1$(1 + R/100)^45$
Required time = 45 years
Using Rule 5,
Here, m = 2, t = 15 years
It becomes 8 times = $2^3$ times
in t × n years= 15 × 3 = 45 years
Q-20) If the amount is 3$3/8$ times the sum after 3 years at compound interest compounded annually, then the rate of interest per annum is
(a)
(b)
(c)
(d)
A = P$(1 + R/100)^T$
$27/8x = x(1 + R/100)^3$
$(3/2)^3 = (1 + R/100)^3$
$1 + R/100 = 3/2$
$R/100 = 3/2 - 1 = 1/2$
R = $1/2$ × 100 ⇒ R = 50%
Using Rule 8,
n= $27/8$, t = 3 years
R% = $(n^{1/t} - 1) × 100%$
= $((27/8)^{1/3} - 1) × 100%$
= $[(3/2) - 1]$ × 100% ⇒ R% = 50%