Practice Comparing sum in different years - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) 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-2) 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-3) 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-4) A sum of money invested at compound interest amounts in 3 years to Rs.2,400 and in 4 years to Rs.2,520. The interest rate per annum is :
(a)
(b)
(c)
(d)
S.I. on Rs.2400 for 1 year
= Rs.(2, 520 - 2, 400) = Rs.120
Rate = $\text"S.I. × 100"/ \text"Principal × Time"$%
= ${120 × 100}/{2400 × 1}$ = 5%
Using Rule 7(i),
Here, b - a = 4 - 3 = 1
B = Rs.2520, A = Rs.2400
R% = $(B/A - 1)$ × 100%
= $[2520/2400 - 1] × 100%$
= $[{2520 - 2400}/2400] × 100%$
= $120/2400 × 100%$ = 5%
Q-5) A certain sum of money amounts to Rs.2,420 in 2 years and Rs.2,662 in 3 years at some rate of compound interest, compounded annually. The rate of interest per annum is
(a)
(b)
(c)
(d)
Let the rate of interest = R% per annum.
We know that
A = P$(1 + R/100)^T$
2420 = P$(1 + R/100)^2$ ....(i)
2662 = P$(1 + R/100)^3$ ...(ii)
Dividing equation (ii) by (i),
$1 + R/100 = 2662/2420$
$R/100 = 2662/2420$ - 1
$R/100 = {2662 - 2420}/2420$
= $242/2420 = 1/10$
R = $1/10 × $100 = 10%
Using Rule 7(i),
Here, b - a = 3 - 2 = 1
B = Rs.2,662, A= Rs.2,420
R% = $(B/A - 1)$ × 100%
= $(2662/2420 -1)$ × 100%
= $[{2662 - 2420}/2420]$ × 100%
= $242/2420 × 100%$ = 10%
Q-6) On a certain sum of money, the simple interest for 2 years is Rs.350 at the rate of 4% per annum. 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
C.I. = P$[(1 + R/100)^T - 1]$
= 4375$[(1 + 4/100)^2 - 1]$
= 4375$[(1 + 1/25)^T - 1]$
= 4375$[(26/25)^2 - 1]$
= 4375$(676/625 - 1)$
= ${4375 × 51}/625$ = Rs.357
Required difference
= Rs.(357 - 350) = Rs.7
Q-7) An amount of money at compound interest grows up to Rs.3,840 in 4 years and up to Rs.3,936 in 5 years. Find the rate of interest.
(a)
(b)
(c)
(d)
A = P$(1 + R/100)^T$
3840 = P$(1 + R/100)^4$ ....(i)
3936 = P$(1 + R/100)^5$ ...(ii)
Dividing equation (ii) by equation (i),
$3936/3840 = 1 + R/100$
$R/100 = 3936/3840$ - 1
= ${3936 - 3840}/3840 = 96/3840$
R = $96/3840 × 100$ = 2.5%
Using Rule 7(i),
Here, b - a = 5 - 4 = 1
B = Rs.3,936, A = Rs.3,840
R%= $(B/A - 1)$ × 100%
= $(3936/3840 - 1)$ × 100%
= $({3936 - 3840}/3840) × 100%$
= $96/3840 × 100% = 10/4%$ = 2.5%
Q-8) 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-9) 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-10) 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 = $A_1(A_1/A_2)^n$