Practice Basic problems using formula - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A lent Rs.5000 to B for 2 years and Rs.3000 to C for 4 years on simple interest at the same rate of interest and received Rs.2200 in all from both as interest. The rate of interest per annum is
(a)
(b)
(c)
(d)
Using Rule 1,
Let the rate of interest per annum be r%
According to the question,
${5000 × 2 × r}/100 + {3000 × 4 × r}/100 = 2200$
100r + 120r = 2200
220 r = 2200
r = $2200/220$ = 10%
Q-2) A sum of Rs.1600 gives a simple interest of Rs.252 in 2 years and 3 months. The rate of interest per annum is:
(a)
(b)
(c)
(d)
Using Rule 1,
Principal (P) = Rs.1600
T = 2 years 3 months
= $(2 + 3/12) yrs. = (2 + 1/4) yrs. = 9/4 yrs.$
S.I = Rs.252
R = % rate of interest per annum
R = ${100 × S.I.}/{P × t}$
= ${100 × 252}/{1600 × 9/4}$
Rate of interest = 7% per annum.
Q-3) A sum of money amounts to Rs.5,200 in 5 years and to Rs.5,680 in 7 years at simple interest. The rate of interest per annum is
(a)
(b)
(c)
(d)
P + S.I. for 5 years = 5200 ..(i)
P + SI for 7 years = 5680 ...(ii)
On subtracting equation (i) from (ii),
SI for 2 years = 480
SI for 1 year = Rs.240
From equation (i),
P + 5 × 240 = 5200
P = 5200 - 1200 = Rs.4000
R = ${SI × 100}/{T × P}$
= ${240 × 100}/{1 × 4000}$ = 6%
Using Rule 12If certain sum P amounts to Rs. $A_1$ in $t_1$ years at rate of R% and the same sum amounts to Rs. $A_2$ in $t_2$ years at same rate of interest R%. Then,(i) R = $({A_1 - A_2}/{A_2T_1 - A_1T_2})$ × 100(ii) P = $({A_2T_1 - A_1T_2}/{T_1 - T_2})$
R = $({A_1 - A_2}/{A_2T_1 - A_1T_2})$ × 100
= $({5200 - 5680}/{5680 × 5 - 5200 × 7}) × 100$
= ${- 480}/{28400 - 36400}$ × 100
= ${- 480}/{- 8000} × 100$ = 6%
Q-4) A sum of money at simple interest amounts to Rs.1,012 in 2$1/2$ years and to Rs.1,067.20 in 4 years. The rate of interest per annum is :
(a)
(b)
(c)
(d)
Principal + S.I. for $5/2$ years = Rs.1012 ...(i)
Principal + S.I. for 4 years = Rs.1067.20 ...(ii)
Subtracting equation (i) from (ii)
S.I. for $3/2$ years = Rs.55.20
S.I. for $5/2$ years
= $55.20 × 2/3 × 5/2$ = Rs.92
Principal
= Rs.(1012 - 92) = Rs.920
Rate = ${92 × 100}/{920 × 5/2}$
= ${2 × 92 × 100}/{920 × 5}$ = 4%
Using Rule 12,
R = $({A_1 - A_2}/{A_2T_1 - A_1T_2})$ × 100
= $({1012 - 1067.20}/{1067.20 × 5/2 - 1012 × 4}) × 100$
= ${- 55.2}/({2668 - 4048}) × 100$
= ${- 55.2}/{- 1380} × 100$ = 4%
Q-5) A person deposited Rs.400 for 2 years, Rs.550 for 4 years and Rs.1,200 for 6 years. He received the total simple interest of Rs.1,020. The rate of interest per annum is
(a)
(b)
(c)
(d)
Using Rule 1,
Let the rate of interest be R per cent per annum.
${400 × 2 × R}/100 + {550 × 4 × R}/100$
+ ${1200 × 6 × R}/100 = 1020$
8R + 22 R +72 R = 1020
102 R= 1020
R = $1020/102$ = 10%
Q-6) At some rate of simple interest, A lent Rs.6,000 to B for 2 years and Rs.1,500 to C for 4 years and received Rs.9,00 as interest from both of them together. The rate of interest per annum was
(a)
(b)
(c)
(d)
Using Rule 1,
If rate of interest be R% p.a. then,
SI = ${\text"Principal × Rate × Time"/100$
${6000 × 2 × R}/100 + {1500 × 4 × R}/100$
= 900
120 R + 60R = 900
180R = 900
R = $900/180$ = 5%
Q-7) A certain sum of money amounts to Rs.756 in 2 years and to Rs.873 in 3$1/2$ years at a certain rate of simple interest. The rate of interest per annum is
(a)
(b)
(c)
(d)
S.I. for 1$1/2$ years
= Rs.(873 - 756) = Rs.117
S.I. for 2 years
= Rs.$(117 × 2/3 × 2)$ = Rs.156
Principal = 756 - 156 = Rs.600
Now, P = 600, T = 2, S.I. = 156
R = ${100 × S.I.}/{P × T}$
= ${100 × 156}/{600 × 2}$ = 13%
Using Rule 12,
Rate of interest
= $({A_1 - A_2}/{A_2T_1 - A_1T_2})$ × 100
= $({756 - 873}/{873 × 2 - 756 × 7/2}) × 100$
= $({- 117}/{1746 - 2646}) × 100$
= $({- 117}/{- 900})$ × 100 = 13%
Q-8) A sum of money lent at simple interest amounts to Rs.880 in 2 years and to Rs.920 in 3 years. The sum of money (in rupees) is
(a)
(b)
(c)
(d)
If the principal be x and rate of interest be r% per annum, then
SI after 1 year = 920 - 880 = Rs.40
SI after 2 years = Rs.80
880 = x + 80
x = Rs.(880 - 80) = Rs.800
Using Rule 12If certain sum P amounts to Rs. $A_1$ in $t_1$ years at rate of R% and the same sum amounts to Rs. $A_2$ in $t_2$ years at same rate of interest R%. Then,(i) R = $({A_1 - A_2}/{A_2T_1 - A_1T_2})$ × 100(ii) P = $({A_2T_1 - A_1T_2}/{T_1 - T_2})$
P = $({A_2T_1 - A_1T_2}/{T_1 - T_2})$
= $({920 × 2 - 880 × 3}/{2 - 3})$
= $({1840 - 2640}/{-1})$
= ${- 800}/{- 1}$ = Rs.800
Q-9) Rs.500 was invested at 12% per annum simple interest and a certain sum of money invested at 10% per annum simple interest. If the sum of the interest on both the sum after 4 years is Rs.480, the latter sum of money is :
(a)
(b)
(c)
(d)
Using Rule 1,
Simple interest gained from Rs.500
= ${500 × 12 × 4}/100$ = Rs.240
Let the other Principal be x.
S.I. gained = Rs.(480 - 240) = Rs.240
${x × 10 × 4}/100$ = 240
$x = {240 × 100}/40$ = Rs.600
Q-10) A sum of money lent out at simple interest amounts to Rs.720 after 2 years and to Rs.1020 after a further period of 5 years. The sum is :
(a)
(b)
(c)
(d)
Principal + SI for 2 years = Rs.720 .... (i)
Principal + SI for 7 years = Rs.1020 .....(ii)
Subtracting equation (i) from (ii) get,
SI for 5 years
= Rs.(1020 - 720) = Rs.300
SI for 2 years
= Rs.300 × $2/5$ = Rs.120
Principal = Rs.(720 - 120) = Rs.600
Using Rule 12,
P = $({A_2T_1 - A_1T_2}/{T_1 - T_2})$
= $({1020 × 2 - 720 × 7}/{2 - 7})$
= $({2040 - 5040}/{- 5})$
= ${- 3000}/{- 5}$ = Rs.600