Quantitative Aptitude > Interest
SIMPLE INTEREST MCQs
Total Questions : 234
| Page 6 of 24 pages
Answer: Option D. -> 22 years
Answer: (d)Using Rule 1,Let the period of time be T years.800 + ${800 × 12 × T}/100$= 910 + ${910 × 10 × T}/100$800 + 96 T = 910 + 91T96 T - 91 T = 910 - 8005T = 110T = $110/5$ = 22 years.
Answer: (d)Using Rule 1,Let the period of time be T years.800 + ${800 × 12 × T}/100$= 910 + ${910 × 10 × T}/100$800 + 96 T = 910 + 91T96 T - 91 T = 910 - 8005T = 110T = $110/5$ = 22 years.
Answer: Option A. -> 5%
Answer: (a)Let the principal be x and rate be y% per annum.According to the question,SI = ${P × R × T}/100$$x/4 = {x × y × y}/100$$y^2 = 100/4$ = 25y = $√{25}$ = 5% per annumUsing Rule 5,n = $1/5$, R = TRT = n × 100$R^2 = 1/4 × 100$ = 25R = 5%
Answer: (a)Let the principal be x and rate be y% per annum.According to the question,SI = ${P × R × T}/100$$x/4 = {x × y × y}/100$$y^2 = 100/4$ = 25y = $√{25}$ = 5% per annumUsing Rule 5,n = $1/5$, R = TRT = n × 100$R^2 = 1/4 × 100$ = 25R = 5%
Answer: Option B. -> 60
Answer: (b)Using Rule 1,Let first part be x and second part be(1750 –x )According to the question,x × $8/100 = (1750 - x ) × 6/100$8x + 6x = 1750 × 614x = 1750 × 6$x = {1750 × 6}/14$ = Rs.750Interest = 8% of 750= 750 × $8/100 = Rs.60
Answer: (b)Using Rule 1,Let first part be x and second part be(1750 –x )According to the question,x × $8/100 = (1750 - x ) × 6/100$8x + 6x = 1750 × 614x = 1750 × 6$x = {1750 × 6}/14$ = Rs.750Interest = 8% of 750= 750 × $8/100 = Rs.60
Answer: Option D. -> 6 years
Answer: (d)Using Rule 1,Let the principal be x.Time = $\text"SI × 100"/\text"Principal × Rate"$= ${x × 100 × 3}/{x × 50}$ = 6 years
Answer: (d)Using Rule 1,Let the principal be x.Time = $\text"SI × 100"/\text"Principal × Rate"$= ${x × 100 × 3}/{x × 50}$ = 6 years
Answer: Option D. -> Rs.80
Answer: (d)Using Rule 1,True discount= $\text"Amount × Rate × Time"/ \text"100 +(Rate × Time)"$= ${2400 × 5 × 4}/{100 + (5 × 4)}$= ${2400 × 5 × 4}/120$ = Rs.400S.I. = ${2400 × 5 × 4}/100$ = Rs.480Required difference= Rs.(480 - 400) = Rs.80
Answer: (d)Using Rule 1,True discount= $\text"Amount × Rate × Time"/ \text"100 +(Rate × Time)"$= ${2400 × 5 × 4}/{100 + (5 × 4)}$= ${2400 × 5 × 4}/120$ = Rs.400S.I. = ${2400 × 5 × 4}/100$ = Rs.480Required difference= Rs.(480 - 400) = Rs.80
Answer: Option D. -> Rs.2000
Answer: (d)Let the sum lent be x.${x × 7.5 × 5}/100 - {x × 7.5 × 4}/100 = 150$${x × 7.5 × 1}/100$ = 150$x = {150 × 100}/{7.5}$ = Rs.2000Using Rule 13,Here, $P_1 = P, R_1 = 7.5%, T_1$ = 4 years.$P_2 = P, R_2 = 7.5%, T_2$ = 5 years.S.I. = Rs.150S.I. = ${P_2R_2T_2 - P_1R_1T_1}/100$150 = ${P × 7.5 × 5 - P × 7.5 × 4}/100$15000 = 7.5PP = $15000/{7.5}$P = $150000/75$ = Rs.2000
Answer: (d)Let the sum lent be x.${x × 7.5 × 5}/100 - {x × 7.5 × 4}/100 = 150$${x × 7.5 × 1}/100$ = 150$x = {150 × 100}/{7.5}$ = Rs.2000Using Rule 13,Here, $P_1 = P, R_1 = 7.5%, T_1$ = 4 years.$P_2 = P, R_2 = 7.5%, T_2$ = 5 years.S.I. = Rs.150S.I. = ${P_2R_2T_2 - P_1R_1T_1}/100$150 = ${P × 7.5 × 5 - P × 7.5 × 4}/100$15000 = 7.5PP = $15000/{7.5}$P = $150000/75$ = Rs.2000
Answer: Option D. -> Rs.750
Answer: (d)Using Rule 1,Let the sum lent out at 12.5% be xSum lent out at 10% = 1500 - xNow, ${(1500 - x) × 10 × 5}/100$= ${x × 12.5 × 4}/100$50 (1500 - x ) = 50x2x = 1500$x = 1500/2$ = Rs.750
Answer: (d)Using Rule 1,Let the sum lent out at 12.5% be xSum lent out at 10% = 1500 - xNow, ${(1500 - x) × 10 × 5}/100$= ${x × 12.5 × 4}/100$50 (1500 - x ) = 50x2x = 1500$x = 1500/2$ = Rs.750
Answer: Option D. -> 6$1/4$ years
Answer: (d)Using Rule 1,Let the period of time be T years. Then,${400 × 5 × T}/100 = {500 × 4 × 6.25}/100$T = ${500 × 4 × 6.25}/{400 × 5}$= $25/4 = 6{1}/4$ years
Answer: (d)Using Rule 1,Let the period of time be T years. Then,${400 × 5 × T}/100 = {500 × 4 × 6.25}/100$T = ${500 × 4 × 6.25}/{400 × 5}$= $25/4 = 6{1}/4$ years
Answer: Option A. -> 0.25%
Answer: (a)${500 × 2 × R_1}/100 - {500 × 2 × R_2}/100$ = 2.5where $R_1 & R_2$ are rate% of both banks10 $(R_1 - R_2 )$ = 2.5$R_1 - R_2 = {2.5}/10$= 0.25 % per annum Using Rule 7If the difference between two simple interests is 'x' calculated at different annual rates and times, then principal (P) isP = $ {x × 100}/{(\text"difference in rate") ×(\text"difference in time")}$
Answer: (a)${500 × 2 × R_1}/100 - {500 × 2 × R_2}/100$ = 2.5where $R_1 & R_2$ are rate% of both banks10 $(R_1 - R_2 )$ = 2.5$R_1 - R_2 = {2.5}/10$= 0.25 % per annum Using Rule 7If the difference between two simple interests is 'x' calculated at different annual rates and times, then principal (P) isP = $ {x × 100}/{(\text"difference in rate") ×(\text"difference in time")}$
Answer: Option B. -> 20 years
Answer: (b)Let the principal be P and rate of interest be r %According to the question,${30P}/100 = {P × R × 6}/100$30 = 6 RR = 5Now, let interest be equal to principal in T years.P = ${P × 5 × T}/100$T = $100/5$ = 20 years.Using Rule 5,Here, n = $30/100 = 3/10$, T = 6 years.RT = n × 100R × 6 = $3/10$ × 100R = 5%As,S.I. = PS.I. = ${P × R × T}/100$100 = RT100 = 5 × TThis is possible only when T = 20.
Answer: (b)Let the principal be P and rate of interest be r %According to the question,${30P}/100 = {P × R × 6}/100$30 = 6 RR = 5Now, let interest be equal to principal in T years.P = ${P × 5 × T}/100$T = $100/5$ = 20 years.Using Rule 5,Here, n = $30/100 = 3/10$, T = 6 years.RT = n × 100R × 6 = $3/10$ × 100R = 5%As,S.I. = PS.I. = ${P × R × T}/100$100 = RT100 = 5 × TThis is possible only when T = 20.