Practice Basic problems using formula - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) 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-2) 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-3) 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-4) 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-5) At what rate per cent per annum will a sum of Rs.1,000 amounts to Rs.1,102.50 in 2 years at compound interest ?
(a)
(b)
(c)
(d)
Using Rule 1,
A = P$(1 + R/100)^T$
Let rate be 'r'
${1102.50}/1000 = (1 + r/100)^2$
$11025/10000 = (1 + r/100)^2$
$(105/100)^2 = (1 + r/100)^2$
1 + $r/100 = 105/100$
$r/100 = 5/100$ = 5%
Q-6) The compound interest on Rs.6,000 at 10% per annum for 1$1/2$ years, when the interest being compounded annually, is
(a)
(b)
(c)
(d)
Amount = 6000$(1 + 10/100) × (1 + {{1/2} × 10}/100)$
= $6000 × 11/10 × 21/20$ = Rs.6930
Using Rule 4,If the time is in fractional form i.e.,t = nF, thenA = P$(1 + r/100)^n(1 + {rF}/100)$e.g. t =3$5/7$ yrs, thenA = P$(1 + r/100)^3(1 + r/100 × 5/7)$
Here, t = nF
A = P$(1 + r/100)^n(1 + {rF}/100)$
CI = Rs.(6930 - 6000) = Rs.930
Q-7) The principal, which will amount to Rs.270.40 in 2 years at the rate of 4% per annum compound interest, is
(a)
(b)
(c)
(d)
Using Rule 1,
Let the principal be Rs.P.
270.40 = P $(1 + 4/100)^2$
270.40 = P $(1 + 0.04)^2$
P = ${270.40}/{1.04 × 1.04}$ = Rs.250
Q-8) 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-9) 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-10) 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