Practice Combination of si and ci - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) At a certain rate per annum, the simple interest on a sum of money for one year is Rs.260 and the compound interest on the same sum for two years is Rs.540.80. The rate of interest per annum is
(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$[(1 + r/100)^n - 1]$
Let the principal be x and rate of interest be r% per annum. Now,
S.I. = $\text"Principal × Time × Rate"/100$
260 = ${x × r}/100$ ....(i)
C.I.= P$[(1 + R/100)^T - 1]$
540.80 = $x[(1 + r/100)^2 - 1]$
540.80 = $x[1 + {2r}/100 + r^2/10000 - 1]$
540.80 = ${2xr}/100 + {xr^2}/10000$
540.80 = 2 × 260 + ${260 . r}/100$
260r = 54080 - 52000
260r = 2080
r = $2080/260$ = 8 %
Q-2) There is 40% increase in an amount in 8 years at simple interest. What will be the compound interest (in rupees) of Rs 30000 after 2 years at the same rate ?
(a)
(b)
(c)
(d)
According to the question,
If principal
= Rs.100 then interest = Rs.40.
Case I.
Rate = $\text"S.I. × 100"/ \text"Principal × Time"$
= ${40 × 100}/{100 × 8}$ = 5% per annum
Case II.
A = P$(1 + R/100)^T$
= 30000$(1 + 5/100)^2$
= 30000$(1 + 1/20)^2$
= 30000$({20 + 1}/20)^2$
= 30000$ × 21/20 × 21/20$
= Rs.33075
C. I. = Rs.(33075 - 30000) = Rs.3075
Q-3) A certain amount of money earns Rs.540 as Simple Interest in 3 years. If it earns a Compound Interest of Rs.376.20 at the same rate of interest in 2 years, find the amount (in Rupees).
(a)
(b)
(c)
(d)
S.I. for 2 years
= $2/3$ × 540 = Rs.360
C.I. - S.I.
= 376.20 - 360 = Rs.16.20
Rate of interest
= ${16.20}/180$ × 100 = 9% per annum
Principal = $\text"S.I. × 100"/\text"Time × Rate"$
= ${180 × 100}/{1 × 9}$ = Rs.2000
Q-4) 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-5) The compound interest on a certain sum of money for 2 years at 5% per annum is Rs.410. The simple interest on the same sum at the same rate and for the same time is
(a)
(b)
(c)
(d)
Compound interest = P $[(1 + R/100)^T - 1]$
410 = P$[(1 + 5/100)^2 - 1]$
410 = P$[(1 + 1/20)^2 - 1]$
410 = P$[(21/20)^2 - 1]$
410 = P$(441/400 - 1)$
410 = P$(41/400)$
P = ${410 × 400}/41$ = Rs.4000
S.I. = $\text"Principal × Time × Rate"/100$
= ${4000 × 2 × 5}/100$ = Rs.400
Using Rule 10,
Here, C.I. = Rs.410, R = 5%, S.I. = ?
C.I. = S.I.$(1 + R/200)$
410 = S.I.$(1 + 5/200)$
410 = S.I.$(205/200)$
S.I. = ${410 × 200}/205$ = Rs.400
Q-6) On a certain sum of money the compound interest for 2 years is Rs.282.15 and the simple interest for the same period of time is Rs.270. The rate of interest per annum is
(a)
(b)
(c)
(d)
Using Rule 10,
If SI on a certain sum for two years is x and CI is y, then
$y = x(r + /200)$
$282.15 = 270(1 + r/100)$
$1 + r/200 = 282.15/270$
$r/200 = 282.15/270$ - 1
$r/200 = {12.15}/270$
r = ${12.15 × 200}/270 = 9%$
Q-7) If the compound interest on a certain sum for 2 years at 4% p.a. is Rs.102, the simple interest at the same rate of interest for two years would be
(a)
(b)
(c)
(d)
If the sum be P, then
C.I. = P$[(1 + R/100)^T - 1]$
102 = $[(1 + 4/100)^2 - 1]$
102 = P$[(26/25)^2 - 1]$
102 = P$(676/625 - 1)$
102 = P$({676 - 625}/625)$
102 = P × $51/625$
P = ${102 × 625}/51$ = Rs.1250
S.I. = ${1250 × 2 × 4}/100$ = Rs.100
Q-8) A sum becomes Rs.2,916 in 2 years at 8% per annum compound interest. The simple interest at 9% per annum for 3 years on the same amount will be
(a)
(b)
(c)
(d)
Using Rule 1,
A = P$(1 + R/100)^T$
2916 = $x(1 + 8/100)^2$
2916 = $x(27/25)^2$
$x = {2916 × 25 × 25}/{27 × 27}$ = Rs.2500
S.I. = ${P × R × T}/100$
= ${2500 × 9 × 3}/100$ = Rs.675
Q-9) If the compound interest on a sum for 2 years at 12$1/2$% per annum is Rs.510, the simple interest on the same sum at the same rate for the same period of time is :
(a)
(b)
(c)
(d)
C.I. = P$[(1 + R/100)^T - 1]$
510 = P$[(1 + 25/200)^2 - 1]$
510 = P$(81/64 - 1)$
P = ${510 × 64}/17$ = 1920
S.I. = ${1920 × 2 × 25}/{100 × 2}$ = Rs.480
Using Rule 10,
Here, C.I. = Rs.510, R = 12$1/2$%, S.I. = ?
C.I. = S.I.$(1 + R/200)$
$510 = S.I.(1 + 25/400)$
S.I. = ${510 × 400}/425$ = Rs.480
Q-10) If the compound interest on a sum for 2 years at 12$1/2$ p.a is Rs.510, the simple interest on the same sum at the same rate for the same period of time is
(a)
(b)
(c)
(d)
Principal = Rs.P (let)
C.I. = P$[(1 + R/100)^T - 1]$
510 = P$[(1 + 25/200)^2 - 1]$
510 = P$[(1 + 1/8)^2 - 1]$
510 = P$[(9/8)^2 - 1]$
510 = P$(81/64 - 1)$
510 = P$({81 - 64}/64)$
510 = ${17P}/64$
P = ${510 × 64}/17$ = Rs.1920
S.I. = $\text"Principal × Time × Rate"/100$
= ${1920 × 2 × 25}/{100 × 2}$ = Rs.480
Using Rule 10,
Here, C.I. = Rs.510, R = 12$1/2$%, S.I. = ?
C.I. = S.I.$(1 + R/200)$
510 = S.I.$(1 + 25/400)$
510 = S.I.$(425/400)$
S.I. = ${510 × 400}/425$ = Rs.480