Practice Simple interest - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) 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-2) 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-3) A certain sum doubles in 7 years at simple interest. The same sum under the same interest rate will become 4 times in how many years.
(a)
(b)
(c)
(d)
Case I,
Interest = Principal
Rate = ${Interest × 100}/\text"Principal × Time"$
= $100/7%$ per annum
Case II,
Interest = 3 × Principal
Time = ${Interest × 100}/\text"Principal × Time"$
= ${3 × 100}/{100/7}$ = 3 × 7 = 21 years
Q-4) At what rate per cent per annum will the simple interest on a sum of money be $2/5$ of the amount in 10 years ?
(a)
(b)
(c)
(d)
Using Rule 1Simple Interest (S.I.)= ${\text"Principal × Rate × Time"/100$ orS.I. = ${\text"P × R × T"/100$P = ${\text"S.I." × 100}/\text"R × T"$, R = ${\text"S.I." × 100}/\text"P × T"$, T = ${\text"S.I." × 100}/\text"P × R"$ A = P + S.I. or S.I. = A - P
Let P be the principal and R% rate of interest.
S.I. = ${\text"PR" × 10}/100 = \text"PR"/10$
According to the question,
${PR}/10 = (P + {PR}/10) × 2/5$
$R/10 = (1 + R/10) × 2/5$
$R/10 = 2/5 + R/25$
$R/10 - R/25 = 2/5$
${5R - 2R}/50 = 2/5$
${3R}/50 = 2/5$
R = ${50 × 2}/{3 × 5} = 20/3 = 6{2}/3$%
Q-5) The rate of simple interest per annum at which a sum of money doubles itself in 16$2/3$ years is
(a)
(b)
(c)
(d)
According to the question,
Principal = Rs.x.
Interest = Rs.x.
Time = $50/3$ years
Rate = ${Interest × 100}/\text"Principal × Time"$
= ${x × 100}/{x × {50/3}}$
= ${100 × 3}/50$ = 6% per annum
Q-6) If ratio of principal and simple interest for 1 year is 25 : 1, then the rate of interest is
(a)
(b)
(c)
(d)
Using Rule 1,
Principal : Interest = 25 : 1
Interest : Principal = 1 : 25
Rate = $\text"S.I. × 100"/ \text"Principal × Time"$
= $1/25$ × 100 = 4% per annum
Q-7) A man loses Rs.55.50 yearly when the annual rate of interest falls from 11.5% to 10%. His capital (in rupees) is
(a)
(b)
(c)
(d)
Using Rule 1,
Let his capital be x.
According to the question,
${x × 11.5}/100 - {x × 10}/100$ = 55.50
or (11.5 - 10)x = 5550
or 1.5x = 5550
or $x = 5550/{1.5}$ = Rs.3700
Q-8) If a sum of money at simple interest doubles in 12 years, the rate of interest per annum is
(a)
(b)
(c)
(d)
Let the principal be x.
Amount = 2x
Interest = (2x - x) = x
Rate = ${S.I. × 100}/\text"Principal × Time"$
= ${x × 100}/{x × 12} = 25/3 = 8{1}/3%$
Using Rule 3,
R = ${(2 - 1)}/12 × 100%$
R = $25/3% = 8{1}/3$%
Q-9) The difference between the simple interest received from two different sources on Rs.1500 for 3 years is Rs.13.50. The difference between their rates of interest is:
(a)
(b)
(c)
(d)
Let $r_1$, and $r_2$ be the required rate of interest
Then, ${13.50} = {1500 × 3 × r_1}/100$
– ${1500 × 3 × r_2}/100 = 4500/100(r_1 - r_2)$
$r_1 - r_2 = 135/450 = 27/90$
= $3/10$ = 0.3%
Using Rule 13,
$P_1 = Rs.1500, R_1 , T_1$ = 3 years.
$P_2 = Rs.1500, R_2 , T_2$ = 3 years.
S.I. = Rs.13.50
13.50 = ${1500 × R_2 × 3 - 1500 × R_1 × 3}/100$
$1350/100 = {4500(R_2 - R_1)}/100$
$R_2 - R_1 = 1350/4500 = 27/90$
= $3/10$ = 0.3%
Q-10) A sum amounts to double in 8 years by simple interest. Then the rate of simple interest per annum is
(a)
(b)
(c)
(d)
Principal = Rs.x
Amount = Rs.2x
Interest = 2x - x = Rs.x
Rate = ${SI × 100}/\text"Principal × Time"$
= ${x × 100}/{x × 8} = 25/2$ = 12.5 % per annum
Using Rule 3,
R % = ${(n - 1)}/T × 100%$
= ${(2 - 1)}/8 × 100%$ = 12.5%
Q-11) Rs.800 becomes Rs.956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will 800 become in 3 years ?
(a)
(b)
(c)
(d)
Using Rule 1,
S.I. = 956 - 800 = Rs.156
Rate = ${S.I. × 100}/{Principal × Time}$
= ${156 × 100}/{800 × 3}$ = 6.5% per annum
New rate = 10.5%
S.I. = ${\text"Principal × Rate × Time"/100$
= ${800 × 3 × 10.5}/100$ = Rs.252
Amount = 800 + 252 = Rs.1052
Q-12) What sum of money must be given as simple interest for six months at 4% per annum in order to earn Rs.150 interest?
(a)
(b)
(c)
(d)
Using Rule 1Simple Interest (S.I.)= ${\text"Principal × Rate × Time"/100$ orS.I. = ${\text"P × R × T"/100$P = ${\text"S.I." × 100}/\text"R × T"$, R = ${\text"S.I." × 100}/\text"P × T"$, T = ${\text"S.I." × 100}/\text"P × R"$ A = P + S.I. or S.I. = A - P
P = ${150 × 100}/4 × 2/1$ = Rs.7500
Q-13) A sum of money at some rate of simple interest amounts to Rs.2,900 in 8 years and to Rs.3,000 in 10 years. The rate of interest per annum is
(a)
(b)
(c)
(d)
Principal + interest for 8 years= Rs.2900... (i)
Principal + interest for 10 years = Rs.3000 ... (ii)
Subtracting equation (i) from (ii)
Interest for 2 years = Rs.100
Interest for 8 years
= $100/2 × 8$ = Rs.400
From equation (i),
Principal = Rs.(2900 - 400) = Rs.2500
Rate = ${S.I × 100}/{\text"Time × Principal"}$
= ${400 × 100}/{8 × 2500} = 2%$
Using Rule 12,
R = $({A_1 - A_2}/{A_2T_1 - A_1T_2})$ × 100
= $({2900 - 3000}/{3000 × 8 - 2900 × 10})$ × 100
=$({- 100}/{24000 - 29000}) × 100$
= ${-100}/{-5000}$ × 100 = 2%
Q-14) A lends Rs.2500 to B and a certain sum to C at the same time at 7% annual simple interest. If after 4 years, A altogether receives Rs.1120 as interest from B and C, the sum lent to C is
(a)
(b)
(c)
(d)
Using Rule 1,
Let the sum lent to C be x
According to the question,
${2500 × 7 × 4}/100 + {x × 7 × 4}/100 = 1120$
or 2500 × 28 + 28x = 112000
or 2500 + x = 4000
or x = 4000 - 2500 = 1500
Q-15) What sum will amount to Rs.7000 in 5 years at 3$1/3$% simple interest ?
(a)
(b)
(c)
(d)
Using Rule 1,
P = ${A × 100}/{100 + r × t}$
=${7000 × 100}/{100 + 10/3 × 5}$
= ${7000 × 100 × 3}/350$ = Rs.6000
Q-16) A sum of Rs. 4000 is lent out in two parts, one at 8% simple interest and the other at 10% simple interest. If the annual interest is Rs. 352, the sum lent at 8% is
(a)
(b)
(c)
(d)
Principal lent at 8% S.I. = Rs.x.
Principal lent at 10% S.I. = Rs.(4000 - x)
S.I. = $\text"Principal × Time × Rate"/100$
${x × 8}/100 + {(4000 - x) × 10}/100$ = 352
8x + 40000 - 10x = 35200
2x = 40000 - 35200 = 4800
x = $4800/2$ = Rs.2400
Q-17) The simple interest on a sum at x% for x years is x. What is the sum?
(a)
(b)
(c)
(d)
P = $\text"100 × SI"/\text"R × T"$
= $\text"100 × x"/\text"x × x"$ = $100/x$
Q-18) A man had Rs.16,000, part of which he lent at 4% and the rest at 5% per annum simple interest. If the total interest received was Rs.700 in one year, the money lent at 4% per annum was
(a)
(b)
(c)
(d)
Using Rule 1,
Let the sum lent at 4% = Rs.x
Amount at 5%= (16000 - x )
According to the question,
${x × 4 × 1}/100 + {(16000 - X) × 5 × 1}/100$ = 700
4x + 80000 - 5x = 70000
x = 80000 - 70000 = Rs.10000
Q-19) The sum of money, that will give Rs.1 as interest per day at the rate of 5% per annum simple interest is
(a)
(b)
(c)
(d)
Using Rule 1,
The sum of money will give Rs.365 as simple interest in a year.
S.I. = $\text"PRT"/100$
365 = ${P × 5 × 1}/100$
P = ${365 × 100}/5$ = Rs.7300
Q-20) A money lender finds that due to fall in the annual rate of interest 8% to 7$3/4$% , his yearly income diminishes by Rs.61.50. His capital is
(a)
(b)
(c)
(d)
Difference in rate
= $(8 - 7{3}/4)% = 1/4$%
Let the capital be Rs.x.
$1/4$% of x = 61.50
x = 61.50 × 100 × 4 = Rs.24600