Practice Difference in ci and si - 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) The difference between the compound interest and simple interest for the amount 5,000 in 2 years is 32. The rate of interest is
(a)
(b)
(c)
(d)
Using Rule 6,
Difference of 2 years
= ${p × r^2}/10000$
32 = ${5000 × r^2}/10000$
$r^2 = {32 × 10000}/5000$ = 64
r = $√{64}$ = 8%
Q-3) The difference between the compound interest and simple interest on 10,000 for 2 years is 25. The rate of interest per annum is
(a)
(b)
(c)
(d)
Using Rule 6,
Difference = ${PR}^2/10000$
25 = ${10000 × R^2}/10000$ ⇒ R = 5%
Q-4) The difference between the compound interest (compounded annually) and the simple interest on a sum of 1000 at a certain rate of interest for 2 years is 10. The rate of interest per annum is :
(a)
(b)
(c)
(d)
When difference between the compound interest and simple interest on a certain sum of money for 2 years at r% rate is x, then
x = Sum$(r/100)^2$
10 = 1000$(r/100)^2$
$(r/100)^2 = 10/1000$
$r/100 = √{1/100} = 1/10$
r = $100/10$ = 10%
Using Rule 6,
Here, C.I. - S.I. = Rs.10, R = ?, T= 2 years, P = Rs.1000
C.I. - S.I. = P$(R/100)^2$
10 = 1000$(R/100)^2$
10 = 1000$ × R/100 × R/100$
$R^2$ = 100
R = $√{100}$ = 10%
Q-5) The difference between the compound interest and the simple interest on a certain sum at 5% per annum for 2 years is 1.50. The sum is
(a)
(b)
(c)
(d)
Using Rule 6,
Difference = ${PR}^2/(100)^2$
1.50 = ${P × 5 × 5}/(100)^2$
P = 400 × 1.5 = Rs.600
Q-6) The difference between simple and compound interest compounded annually, on a certain sum of money for 2 years at 4% per annum is 1. The sum (in ) is :
(a)
(b)
(c)
(d)
Using Rule 6,
The difference between compound interest and simple interest for two years
= ${\text"Principal" × (Rate)^2}/{100 × 100}$
1 = ${\text"Principal" × (4)^2}/10000$
Principal = $10000/16$ = Rs.625
Q-7) The difference between simple and compound interest on a certain sum of money for 2 years at 4 per cent per annum is 1. The sum of money is :
(a)
(b)
(c)
(d)
Using Rule 6,
When difference between the CI and SI on a certain sum of money for 2 years at r % rate is x, then
Sum = x × $(100/r)^2$
= 1 × $(100/4)^2$ = Rs.625
Q-8) On what sum of money will the difference between S.I and C.I for 2 years at 5% per annum be equal to 25 ?
(a)
(b)
(c)
(d)
Using Rule 6,
Difference = ${PR^2}/10000$
25 = ${P × 5 × 5}/10000$
P = Rs.10000
Q-9) On what sum does the difference between the compound interest and the simple interest for 3 years at 10% is 31 ?
(a)
(b)
(c)
(d)
Let the sum be x
r = 10%, n = 3 years
S.I. = ${x × r × n}/100$
S.I.= ${x × 10 × 3}/100 = 3/10x$
C.I.= $[(1 + r/100)^n - 1]x$
= $[(1 + 10/100)^3 - 1]x$
= $[(11/10)^3 - 1]x$
$(1331/1000 - 1)x = 331/1000x$
$331/1000x - 3/10x$ = 31
or $({331 - 300})/1000x = 31$
or $31/1000x$ = 31
or x = 1000
Sum = Rs.1000
Using Rule 6,
Here, C.I. - S.I. = Rs.31, R = 10%, T = 3 years, P = ?
C.I. - S.I. = P × $(R/100)^2 × (3 + R/100)$
31 = P × $(10/100)^2(3 + 10/100)$
31 = P × $1/100 × 31/10$ ⇒ P = Rs.1000
Q-10) What is the difference between compound interest on 5,000 for 1 1 2 years at 4% per annum according as the interest is compounded yearly or halfyearly?
(a)
(b)
(c)
(d)
Compound Interest (when compounded yearly)
= $5000(1 + 4/100)^(1.5) - 5000$
= $5000(26/25)^(1.5) - 5000$
= 5302.9805 - 5000 = Rs.302.9805
C.I. (When compounded halfyearly).
= $5000(1 + 2/100)^3 - 50000$
= 5306.04 - 5000 = Rs.306.04
Required difference
= Rs.(306.04 - 302.9805)
= Rs.3.059 = Rs.3.06