Simple Interest(Quantitative Aptitude > Interest ) Questions and answers for exam Preparation

Quantitative Aptitude > Interest

SIMPLE INTEREST MCQs

Total Questions : 234 | Page 4 of 24 pages

Question 31. Manoj deposited Rs.29400 for 6 years at a simple interest. He got Rs.4200 as interest after 6 years. The annual rate of interest was

2$7/20$%
2$8/21$%
4$8/21$%
3$8/21$%

Discuss Question

Answer: Option B. -> 2$8/21$%
Answer: (b)Using Rule 1,4200 = ${29400 × 6 × R}/100$R = $4200/{294 × 6} = 50/21 = 2{8}/21$%

Question 32. The simple interest on Rs.7,300 from 11 May, 1987 to 10 September, 1987 (both days included) at 5% per annum is

Rs.103
Rs.123
Rs.223
Rs.200

Discuss Question

Answer: Option B. -> Rs.123
Answer: (b)Using Rule 1,Time from 11 May to 10 September, 1987= 21 + 30 + 31 + 31 + 10= 123 daysTime = 123 days = $123/365$ yearS.I. = ${7300 × 123 × 5}/{365 × 100}$ = Rs.123

Question 33. What annual instalment will discharge a debt of Rs.6450 due in 4 years at 5% simple interest ?

Rs.1835
Rs.1500
Rs.1950
Rs.1935

Discuss Question

Answer: Option B. -> Rs.1500
Answer: (b)Let each instalment be x Then,$(x + {x × 5 × 1}/100) + (x + {x × 5 × 2}/100)$+ $(x + {x × 5 × 3}/100) + x = 6450$$(x + x/20) + (x + x/10)$+ $(x + {3x}/20)$ + x= 6450${21x}/20 + {11x}/10 + {23x}/20 + x = 6450$${21x + 22x + 23x + 20x}/20$= 6450${86x}/20 = 6450$$x = {6450 × 20}/86$ = Rs.1500 Using Rule 10If a sum is to be deposited in equal instalments, then,Equal instalment = ${A × 200}/{T[200 + (T - 1)r]}$where T = no. of years, A = amount, r = Rate of Interest.

Question 34. 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

Discuss Question

Answer: Option C. -> 800
Answer: (c)If the principal be x and rate of interest be r% per annum, thenSI after 1 year = 920 - 880 = Rs.40SI after 2 years = Rs.80880 = x + 80x = Rs.(880 - 80) = Rs.800Using 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 = $({A_2T_1 - A_1T_2}/{T_1 - T_2})$

Question 35. 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

Discuss Question

Answer: Option B. -> 5%
Answer: (b)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$= 900120 R + 60R = 900180R = 900R = $900/180$ = 5%

Question 36. Rs.1,000 is invested at 5% per annum simple interest. If the interest is added to the principal after every 10 years, the amount will become Rs.2,000 after

18 years
15 years
16$2/3$ years
20 years

Discuss Question

Answer: Option C. -> 16$2/3$ years
Answer: (c)Using Rule 1,After 10 years,SI = ${1000 × 5 × 10}/100$ = Rs.500Principal for 11th year= 1000 + 500 = Rs.1500SI = Rs.(2000 - 1500) = Rs.500T = ${SI × 100}/{P × R} = {500 × 100}/{1500 × 5}$= $20/3$ years = 6$2/3$ yearsTotal time = 10 + 6$2/3$= 16$2/3$ years

Question 37. Ram deposited a certain sum of money in a company at 12% per annum simple interest for 4 years and deposited equal amount in fixed deposit in a bank for 5 years at 15% per annum simple interest. If the difference in the interest from two sources is Rs.1350, then the sum deposited in each case is :

Rs.4000
Rs.3000
Rs.6500
Rs.5000

Discuss Question

Answer: Option D. -> Rs.5000
Answer: (d)Let amount invested in each company be Rs. x.S.I. = ${\text"Principal × Rate × Time"/100$According to the question,${x × 15 × 5}/100 - {x × 12 × 4}/100$ = 1350${75x}/100 - {48x}/100$ = 1350${27x}/100$ = 1350$x = {1350 × 100}/27$ = Rs. 5000Using Rule 13Here, $P_1 = Rs. P, R_1 = 12%, T_1$ = 4 years$P_2 = Rs. P, R_2 = 15%, T_2$ = 5 yearsS.I. = Rs. 1350S.I.= ${P_2R_2T_2 - P_1R_1T_1}/100$1350 = ${P × 15 × 5 - P × 12 × 4}/100$135000 = 75 P - 48P135000 = 75 PP = Rs. 5000

Question 38. Simple interest on a certain sum at a certain annual rate of interest is $16/25$ of the sum. If the number representing rate per cent and time in years be equal, then the rate of interest is

11$1/2$%
8%
12$1/4$%
12$1/2$%

Discuss Question

Answer: Option B. -> 8%
Answer: (b)Let the rate of interest be r% and principal be P.According to the question.${16P}/25 = {P × r × r}/100$[Since, r = t numerically]$r^2 = 1600/25$r = $40/5$ = 8 % Using Rule 5If Simple Interest (S.I.) becomes 'n' times of principal i.e.S.I. = P × n then.RT = n × 100

Question 39. The rate of interest per annum at which the total simple interest of a certain capital for 1 year is equal to the total simple interest of the same capital at the rate of 5% per annum for 2 years, is

10%
$5/2$%
12.5%
25%

Discuss Question

Answer: Option A. -> 10%
Answer: (a)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

Question 40. Prakash lends a part of Rs.20,000 at 8% simple interest and remaining at $4/3$% simple interest. His total income after a year was Rs.800. Find the sum lent at 8%.

Rs.12,000
Rs.8,000
Rs.10,000
Rs.6,000

Discuss Question

Answer: Option B. -> Rs.8,000
Answer: (b)Using Rule 1,Amount lent at 8% rate of interest = Rs.xAmount lent at $4/3$% rate of interest = Rs.(20,000 - x)S.I. = ${\text"Principal × Rate × Time"/100$${x × 8 × 1}/100 + {(20,000 - x) × 4/3 × 1}/100$ = 800${2x}/25 + {20,000 - x}/75 = 800$${6x + 20,000 - x}/75 = 800$5x + 20,000 = 75 × 800 = 60,0005x = 60,000-20,000 = 40,000$x = {40,000}/5$ = Rs.8000