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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
1). Rs. 77.212). Rs. 70.563). Rs. 76.324). Rs. 79.46 |
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Answer» Let the principal sum = RS. x We know the formula for compound interest- $( \Rightarrow {\rm{CI}} = \left[ {{\rm{P}}\left\{ {{{\left( {1 + \frac{{\rm{R}}}{{100}}} \right)}^t} - 1} \right\}} \right])$Where, CI = Compound interest P = Principal R = Rate of interest T = Time period Under SCHEME X, C.I. = 10 % p.a. $(\begin{array}{l} \Rightarrow {\rm{CI}} = \left[ {{\rm{P}}\left\{ {{{\left( {1 + \frac{{\rm{r}}}{{100}}} \right)}^t} - 1} \right\}} \right]\\ \Rightarrow 63 = \left[ {{\rm{x}}\left\{ {{{\left( {1 + \frac{{10}}{{100}}} \right)}^2} - 1} \right\}} \right]\\ \Rightarrow 63 = \left[ {{\rm{x}}\left\{ {{{\left( {\frac{{110}}{{100}}} \right)}^2} - 1} \right\}} \right]\\ \Rightarrow 63 = \left[ {\frac{{21}}{{100}}{\rm{x}}} \right] \end{array})$ ⇒ x = Rs. 300 Under scheme Y, C.I. = 12 % p.a. $(\begin{array}{l} \Rightarrow {\rm{CI}} = \left[ {{\rm{P}}\left\{ {{{\left( {1 + \frac{{\rm{r}}}{{100}}} \right)}^t} - 1} \right\}} \right]\\ \Rightarrow {\rm{CI}} = \left[ {300\left\{ {{{\left( {1 + \frac{{12}}{{100}}} \right)}^2} - 1} \right\}} \right]\\ \Rightarrow {\rm{CI}} = \left[ {300\left\{ {{{\left( {\frac{{112}}{{100}}} \right)}^2} - 1} \right\}} \right]\\ \Rightarrow {\rm{CI}} = \left[ {300 \times \frac{{2544}}{{10000}}} \right] \end{array})$ ⇒ CI = Rs. 76.32 |
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| 52. |
1). 6 percent2). 3 percent3). 12 percent4). 7 percent |
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Answer» We know the formula for compound interest - $(\Rightarrow {\rm{\;CI}} = \left[ {{\rm{P}}\left\{ {{{\left( {1{\rm{\;}} + {\rm{\;}}\frac{{\rm{r}}}{{100}}} \right)}^t} - 1} \right\}} \right])$ Where, CI = Compound interest P = Principal R = RATE of interest T = TIME period After 2nd year ⇒ P (1 + R/100)2 = 28090 After 3rd year ⇒ P (1 + R/100)3 = 29775.4 ⇒ 1 + R/100 = 29775.4/28090 ∴ R = 6% |
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| 53. |
The compound interest on a certain sum for first two successive years are Rs. 100 and Rs. 172 respectively. Find the rate of interest.1). 55%2). 10%3). 63%4). 72% |
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Answer» Let the SUM of money = Rs. x N = 1 year each Compound interest for first year = C1 = x [1 + r/100]1 – x = x [r/100]---(1) ∴ Amount obtained from first year = x [1 + r/100]1 Compound interest for second year C2 =x [1 + r/100]1 [1 + r/100]1 - x[1 + r/100]1 = x[1 + r/100]1 [r/100]---(2) From 1 and 2, C1 = Rs. 100 ⇒ x [r/100] = Rs. 100---(3) C2 = Rs. 172 x (r/100) × [1 + r/100] = Rs. 172 From 3, 100 × [1 + r/100] = 172 ⇒ 1 + (r/100) = 1.72 ⇒ r = 72% THUS, the rate is 72% p.a. |
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| 54. |
What is the difference (in Rs.) between compound interest (compounded annually) and simple interest for 3 years on a principal of Rs. 3000 at the annual rate of 20%?1). 4642). 3843). 3564). 424 |
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Answer» We know that, Difference between C.I and S.I for 3 years = P(R/100)2 (3 + R/100) We have given that, C.I – S.I = 3000 × (20/100)2 (3 + 20/100) = 3000 × (1/25) × (320/100) = Rs. 384 ∴ Difference for 3 years between C.I and S.I is Rs. 384 |
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| 55. |
An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 20%, the effective rate of interest becomes:1). 20%2). 20.25%3).4). 20.75% |
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Answer» Let the Principle VALUE be Rs. 100. Given, interest rate is 20% and is applied every six month. Formula used:Simple interest calculated annually$(,\;S.I = \FRAC{{P \times R \times T}}{{100}})$ For six MONTHS, T = ½ ∴Simple interest for first six month,$(\;S.{I_1} = \frac{{100 \times 20 \times \frac{1}{2}}}{{100}})$ = Rs. 10 New principle value = Rs. 110 ⇒Simple interst for next six month,$(\;S.{I_2} = \frac{{\left( {110 \times 20 \times \frac{1}{2}} \right)}}{{100}})$ = Rs. 11 Hence, TOTAL amount at the end of one year = Rs. 110 + 11 =Rs. 121 ⇒ Effective rate of interest$( = \frac{{121 - 100}}{{100}} \times 100)$ = 21% |
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| 56. |
1). 12 percent2). 6 percent3). 24 percent4). 18 percent |
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Answer» The formula for annual compound interest, including principal sum, is: $(CI\; = \;P{\left( {1\; + \;\FRAC{r}{{100\; \times \;N}}} \RIGHT)^{nt}} - P)$ Where: CI = Compound InterestP = the principal investment amount (the initial deposit or LOAN amount)r = the annual interest rate (decimal)n = the number of times that interest is compounded per yeart = the number of years the money is INVESTED or borrowed for Here n = 2 Let the principal be P SI = (P × r × 2)/100 = 2400 Pr = 120000 CI = P(1 + r/100)2 – P = 2544 P(1 + r2/10000 + 2r/100) – P = 2544 (Pr) × r/10000 + Pr/50 = 2544 120000r/10000 + 2400 = 2544 r = 12% |
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| 57. |
Calculate the simple interest rate per annum at which sum will get double in 4 years?1). 202). 253). 504). 35.5 |
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Answer» LET the principal be x, then amount will be 2x after 4 years Simple Interest (SI) = PRT/100, where P = Principal, R = Rate of Interest and T = Time Amount = P + SI ⇒ 2x = x + (xRT/100) ⇒ x = xRT/100 ⇒ R = 100/4 ∴ R = 25% |
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| 58. |
A bank offers 10% interest annually, compounded half yearly. A customer deposits Rs. 3600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is.1). Rs 5492). Rs 22323). Rs 5584). Rs 279 |
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Answer» We know that, AMOUNT received after compound interest = P(1 + r/100)n Where, P is the principal amount r is the rate of interest n is the number of intervals Amount received at the END of year is GIVEN by, A = 3600(1 + 10/200)2 + 3600 (1 + 10/200)1 A = 3600 (1 + 0.05)2 + 3600 (1 + 0.05)1 A = 3969 + 3780 = 7749 Principal Amount = 3600 + 3600 = Rs. 7200 ∴ Interest gained = 7749 - 7200 = Rs. 549 |
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| 59. |
1). Rs. 75002). Rs. 84003). Rs. 70004). Rs. 7400 |
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Answer» P = RS. 36000, r = 16.67% = 1/6 and t = 1 year 73 DAYS Interest GENERATED in 1 year = 36000(1 + 1/6) – 36000 = Rs. 6000 Interest generated in 2nd year = Interest on (SUM + interest of 1st year) ⇒ Interest on Rs. 42000 ⇒ 42000(1 + 1/6) – 42000 = Rs. 7000 But in 2nd year, we need to calculate the interest for 73 days: ∴ Interest for 73 days = 7000 × 73/365 = Rs. 1400 ∴ Total interest = 6000 + 1400 = Rs. 7400 |
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| 60. |
1). 1154.42). 136003). 9159.24). 9428.6 |
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Answer» P = 8000 R = 7% = 7/100 ∴ AMOUNT after 2 YEARS (at C.I.) = 8000 × 107/100 × 107/100 = 9159.2 |
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| 61. |
1). 102). 203). 254). 30 |
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Answer» <P>PRINCIPAL : INTEREST = 5 : 8 So, S.I. = 8a and P = 5a S.I. = (P × R × T)/100 ⇒ 8a = (5a × R × 8)/100 ∴ R = (100 × 8a)/(5a × 8) = 20% |
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| 62. |
How much simple interest will Rs. 7254 earn in 30 months at 10% per annum?1). Rs. 2165.502). Rs. 36003). Rs. 19254). Rs. 1813.50 |
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Answer» <P>Simple Interest = PRT/100 Where P = Principal AMOUNT = Rs. 7254 R = Rate = 10% p.a. Simple Interest = (7254 × 10 × 2.5)/100 = Rs. 1813.50 |
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| 63. |
1). 6%2). 5%3). 10%4). 16% |
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Answer» We KNOW the formula for compound interest - $(\Rightarrow {\rm{\;CI}} = \left[ {{\rm{P}}\left\{ {{{\left( {1{\rm{\;}} + {\rm{\;}}\frac{{\rm{R}}}{{100}}} \RIGHT)}^t} - 1} \right\}} \right])$ Where, CI = Compound interest P = Principal R = Rate of interest T = Time period Let the principal be P P(1 + r/100)2 = 2000 P(1 + r/100)3 = 2320 Dividing both the equations (1 + r/100) = 2320/2000 r = ((2320/2000) – 1) × 100 r = 16% |
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| 64. |
A sum of money is borrowed and paid back in two annual installments of Rs. 1764 each, allowing 5% compound interest. What was the sum borrowed?1). Rs. 40002). Rs. 33403). Rs. 32804). Rs. 3200 |
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Answer» Let’s assume that the amount borrowed is Rs. P. Also, let each installment be Rs. A. Now, the total amount to be repaid at the end of given time period $(= P\;{\left( {1 + \frac{R}{{100}}} \right)^T})$ Also, total amount $( = A + A\left( {1 + \frac{R}{{100}}} \right) + A{\left( {1 + \frac{R}{{100}}} \right)^2})$ ….. till T terms In this case, A = 1764, R = 5 and T = 2 $(\THEREFORE P\;{\left( {1 + \frac{R}{{100}}} \right)^2} = A + A\left( {1 + \frac{R}{{100}}} \right) = A\left( {2 + \frac{R}{{100}}} \right))$ SUBSTITUTING the given values: $(\begin{array}{l} \Rightarrow P{\left( {1 + \frac{5}{{100}}} \right)^2} = 1764\;\left( {2 + \frac{5}{{100}}} \right)\\ \Rightarrow P \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} = 1764 \times \frac{{41}}{{20}}\end{array})$ ⇒ P = 3280 |
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| 65. |
The compound interest on a certain sum for 2 years is Rs. 60.60 and simple interest is Rs. 60. Find the rate of interest.1). 1%2). 2%3). 3%4). 4% |
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Answer» Let the Principal be = P $({\RM{S}}.{\rm{I}} = \frac{{{\rm{P}} \times {\rm{R}} \times {\rm{T}}}}{{100}})$ $( \Rightarrow 60 = \frac{{{\rm{P}} \times {\rm{R}} \times 2}}{{100}})$ ⇒ 6000 = 2PR ⇒ P × R = 3000 C.I = A - P $(= {\rm{P}}{\left( {1 + \frac{{\rm{R}}}{{100}}} \right)^2} - {\rm{P}})$ $(\Rightarrow 60.60 = \left( {\frac{{3000}}{{\rm{R}}}} \right) \times {\left( {1 + \frac{{\rm{R}}}{{100}}} \right)^2} - \frac{{3000}}{R} = \left( {\frac{{3000}}{{\rm{R}}}} \right) \times \left( {1 + \frac{{{{\rm{R}}^2}}}{{10000}} + \frac{{2{\rm{R}}}}{{100}} - 1} \right))$ $(\Rightarrow 60.60 = \left( {\frac{{3000}}{{\rm{R}}}} \right) \times \left( {\frac{{{{\rm{R}}^2}}}{{10000}} + \frac{{2{\rm{R}}}}{{100}}} \right))$ ⇒ 606 = 600 + 3R ∴ R = 6/3 = 2% ∴ 2% is the rate of INTEREST. |
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| 66. |
1). Rs. 56802). Rs. 58603). Rs. 57804). Rs. 5800 |
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Answer» Let the SUM of money be Rs. x. We KNOW that, Amount compounded in ‘n’ years = P × [1 + (R/100)]n Where, P = Principle Amount, R = RATE of interest Rs. x AMOUNTS to Rs. 7719.8 the rate of 10% compounded annually for 3 years. ⇒ 7719.8 = x × [1 + (10/100)]3 ⇒ 7719.8 = x × 1.13 ⇒ x = 7719.8/1.13 = Rs. 5800 ∴ The sum of money = Rs. 5800 |
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| 67. |
1). 14.42). 12.43). 10.44). 16.4 |
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Answer» Compound interest if it was compounded ANNUALLY can be given as ⇒ CI = 4000{(1 + 12/100) - 1} ⇒ CI = 480 Compound interest if it was compounded half yearly can be given as ⇒ CI = 4000{(1 + 6/100)2 - 1} = 4000{(1.06)2 - 1} ⇒ CI = 4000 × 0.1236 = 494.4 So, DIFFERENCE = 494.4 - 480 = 14.4 |
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| 68. |
A certain bank offers 8% rate of compound interest on the 1st year and 9% on the 2nd year in a certain fixed deposit scheme. If Rs. 17,658 is received after investing for 2 years in this scheme, then what was the amount (in Rs) invested?1). 160002). 150003). 155004). 16500 |
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Answer» According to question, 17658 = P(1 + 8/100)(1 + 9/100) ⇒ 17658 = P × 108/100 × 109/100 ∴ The AMOUNT INVESTED = Rs. 15000 |
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| 69. |
The difference between compound interest and simple interest of a certain sum of money at 20% per annum for 2 years is Rs. 58. Then the sum is1). Rs. 1,0002). Rs. 1,4503). Rs. 1,5504). Rs. 2,000 |
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Answer» ⇒ Let, Principal be Rs. x simple interest = S.I, compound interest = C.I Rate = 20% and TIME (N) = 2 years. ⇒ Principal + C.I = AMOUNT ⇒ Amount = P × [1+ (r/100)] n ⇒ Amount = x × [1 + (20/100)]2 ⇒ Amount = (36x/25) ⇒ C.I = (36x/25) - x ⇒ C.I = (11X/25) ------ 1 ⇒ S.I = (P × R × T)/100 ⇒ S.I = (x × 20 × 2)/100 ⇒ S.I = 2x/5 ------ 2 ⇒ Given C.I - S.I = 58 ⇒ PUTTING the value of C.I and S.I from 1 and 2 we have ⇒ (11x/25) - (2x/5) = 58 ⇒ 11x - 10x = 58 × 25 ∴ x = Rs. 1450 |
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| 70. |
1). Rs. 26,4842). Rs. 37,5853). Rs. 63,5474). Rs. 49,664 |
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Answer» GIVEN that value of each installment = Rs. 25600 Rate R = 2% Number of investments = 2 ⇒ Value of each installment = $(\FRAC{P}{{\left( {\frac{{100}}{{100 + R}}} \right) + {{(\frac{{100}}{{100 + R}})}^2}}})$ = $(\frac{P}{{\left( {\frac{{100}}{{100 + 2}}} \right) + {{(\frac{{100}}{{100 + 2}})}^2}}})$ = P/1.94 ⇒ 25,600 × 1.94 = PRINCIPAL ∴ Principal = Rs. 49,664 |
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| 71. |
On a certain sum of money, a man paid an interest of Rs. 5044 after 3 years at a rate of 5% per annum compounded annually. Find the sum.1). Rs. 420002). Rs. 580003). Rs. 311004). Rs. 32000 |
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Answer» Compound interest = AMOUNT – Sum ⇒ 5044 = P(1 + 5/100)3 – P ⇒ 5044 = P[(21/20)3 – 1] ⇒ 5044 = P(9261 – 8000]/8000 ⇒ P = 5044 × 8000/1261 ⇒ P = 32000 ∴ The sum is Rs. 32000 |
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| 72. |
1). 12 years2). 13 years3). 9 years4). 16 years |
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Answer» ⇒ Let, Principal be Rs. x then Amount = Rs. 2x, RATE = r% and Time (n) = 3 years. ⇒ A = P× [1 + (r/100)]n ⇒ 2x = x × [1 + (r/100)] 3 ⇒ 2 = [1 + (r/100)] 3 ------ 1 ⇒ If SUM become 8 times in the time n years then, ⇒ 8 = [1 + (r/100)] n ⇒ 23 = [1 + (r/100)] n ⇒ putting the value of EQUATION 1 into above we GET, ⇒ ([1 + (r/100)] 3)3 = [1 + (r/100)] n ∴ n = 9 |
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| 73. |
An amount of Rs 1,00,000 is invested in two types of shares. The first yields an interest of 9% p.a. and the second, 11% p.a. If the total interest at the end of one year is \(9\frac{3}{4}\%\), then the amount invested in each share was:1). Rs. 52,500; Rs. 47,500 2). Rs. 62,500; Rs. 37,5003). Rs. 72,500; Rs. 27,5004). Rs. 82,500; Rs. 17,500 |
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Answer» Let x be the amount INVESTED at 9% and (1,00,000 - x) be invested at 11% According to the given condition, $(\Rightarrow \FRAC{{x \times 1 \times 9}}{{100}} + \frac{{\left( {1,00,000 - x} \right) \times 1 \times 11}}{{100}} = \frac{{1,00,000 \times 1 \times 9.75}}{{100}})$ ∴ 9X + 11,00,000 - 11x = 9,75,000 ∴ 2x = 1,25,000 ∴ x = 62,500 ∴ The amount invested in each share is RS. 62,500 and Rs. 37,500 |
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| 74. |
The difference of compound interest and simple interest for 3 years and for 2 years are in ratio 23 : 7 respectively. What is rate of interest per annum (in %)?1). 200/72). 100/73). 300/74). 400/7 |
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Answer» <P>⇒ (CI – SI)3/(CI – SI)2 = [P(R/100)2 × (300 + R)/100]/ P(R/100)2 = 23/7 ⇒ (CI – SI)3/(CI – SI)2 = (300 + R)/100 = 23/7 ⇒ (CI – SI)3/(CI – SI)2 = 2100 + 7R = 2300 ⇒ 7R = 200 ∴ R = 200/7 |
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| 75. |
At a certain rate per annum, the simple interest on a sum of money for one year is Rs. 260 and the compound interest on the same sum for two years is Rs. 540.80. The rate of interest per annum is 1).2). 6%3). 8%4). 10% |
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Answer» <P>SI = (P × R × t)/100 $(A\; = \;P\;{\left( {\;1\; + \frac{R}{{100\;}}\;} \right)^t})$ ; CI = A - P Where, CI = Compound interest SI = Simple interest A = Amount on compound interest P = Principal R = rate % t = time in years Given, rate is same for calculating both SI and CI Let the rate be ‘R’. SI is calculated for 1 year SI = (P × R × 1)/100 = PR/100 Given, SI = Rs. 260 ⇒ PR/100 = 260 -------------eq (1) CI is calculated for 2 year $(\therefore A\; = \;P\;{\left( {\;1\; + \frac{R}{{100\;}}\;} \right)^2}\;)$ As, CI = A – P $(\Rightarrow C.I = \;P\;{\left( {\;1\; + \frac{R}{{100\;}}\;} \right)^2} - P)$ $(\Rightarrow C.I = P\left( {\;{{\left( {1 + \frac{R}{{100}}} \right)}^2} - 1} \right))$ Given, CI = Rs. 540.80 $(\therefore P\left( {\;{{\left( {1 + \frac{R}{{100}}} \right)}^2} - 1} \right) = 540.80)$ -----------eq (2) DIVIDING eq 2 by eq 1 $(\frac{{P\left( {\;{{\left( {1 + \frac{R}{{100}}} \right)}^2} - 1} \right)}}{{\frac{{PR}}{{100}}}} = \frac{{540.80}}{{260}})$ $(\Rightarrow \frac{{\left( {\;{{\left( {1 + \frac{R}{{100}}} \right)}^2} - 1} \right)}}{{\frac{R}{{100}}}} = 2.08)$ Using a2 – B2 = (a + b)(a – b) $(\Rightarrow \frac{{\left( {1 + \frac{R}{{100}} + 1} \right)\left( {1 + \frac{R}{{100}} - 1} \right)}}{{\frac{R}{{100}}}} = 2.08)$ ⇒ 2 + R/100 = 2.08 ⇒ R = 0.08 × 100 = 8% |
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| 76. |
1). 14522). 13203). 15524). 1420 |
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Answer» Compound interest for 2nd year can be given as ⇒ CI for 2nd year = y{(1 + 10/100)2 – (1 + 10/100)} = 1200 ⇒ 1200 = y(1.12 – 1.1) ⇒ 1200/0.11 = y ⇒ y = Rs. 10909.1 Compound interest for 4th year can be given as ⇒ CI for 4th year = y{(1 + 10/100)4 – (1 + 10/100)3} ⇒ CI for 4th year = 10909.1(1.14 – 1.13) = 10909.1 × 0.1331 ⇒ CI for 4th year = Rs. 1452 ∴ Interest for 4th year is Rs. 1452 |
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| 77. |
What annual payment will discharge a debt of Rs. 38100 due in 3 years at 16 2/3 percent per annum compound interest?1). Rs. 113002). Rs. 127003). Rs. 108004). Rs. 9981 |
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Answer» Let’s assume that the Annual payment is Rs. M. Now, the debt of Rs. 38100 is due in 3 years, it means that if all the debt is cleared at once, an amount of Rs. 38100 should be paid after 3 years. But obviously, while clearing the debt through annual INSTALLMENTS, the total amount payable will be less. Once the first installment is paid, it will be excluded from interest calculation, and interest will be CALCULATED for the REMAINING amount next year. After the 2nd installment, the amount worth 2 installments will be excluded from interest calculations and so on. Hence, in order to balance the installments and actual amount payable, we write: $( \Rightarrow M{\left( {1 + \frac{R}{{100}}} \right)^2} + M\left( {1 + \frac{R}{{100}}} \right) + M = A)$ Where, M = annual installment, R = % rate of interest, A = total debt Here, A = 38100, R = 50/3 $( \Rightarrow M{\left( {1 + \frac{{\frac{{50}}{3}}}{{100}}} \right)^2} + M\left( {1 + \frac{{\frac{{50}}{3}}}{{100}}} \right) + M = 38100)$ $(\begin{array}{l} \Rightarrow M{\left( {1 + \frac{1}{6}} \right)^2} + M\left( {1 + \frac{1}{6}} \right) + M = 38100\\ \Rightarrow \frac{{49M}}{{36}} + \frac{{7M}}{6} + M = 38100\\ \Rightarrow \frac{{127M}}{{36}} = 38100 \end{array})$ ⇒ M = 10800 |
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| 78. |
An amount fetched a total simple interest of Rs. 3200 at the rate of 6.25% per year in 4 years. What is the amount (in Rs.)?1). 138002). 118003). 128004). 14800 |
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Answer» Simple INTEREST can be given as ⇒ 3200 = (y × 6.25 × 4)/100 ⇒ y = 12800 ∴ the amount is Rs. 12800 |
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| 79. |
A bank offers 20% compound interest per half year. A customer deposits Rs. 6000 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is1). Rs. 76802). Rs. 38403). Rs. 19204). Rs. 960 |
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Answer» For first half year Principal = RS. 6, 000 As rate is calculated semi annually Interest = (6000 × 20 × 1)/100 = Rs. 1200 For the second half year Principal = 6000 + 1200 + 6000 = Rs. 13200 Interest = (13200 × 20 × 1)/100 = Rs. 2640 Total interest = 1200 + 2640 = Rs. 3840 |
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| 80. |
Rahul invested 50% of his money at 13% per annum and rest at 19% per annum. What would be the annual rate of interest, if the interest is calculated on the whole sum?1). 16.5%2). 15.6%3). 16%4). 15% |
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Answer» Let the TOTAL amount is Rs. 100 50% of 100 = 50 and Rest amount = 50 According to the question, R = (13% of 50 + 19% of 50)/100 × 100 ⇒ R = (6.5 + 9.5)/100 × 100 ⇒ R = 16% |
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| 81. |
The amount (in Rs) received at 10% per annum compound interest after 3 yrs is Rs. 119790. What was the principal?1). 900002). 1000003). 800004). 75000 |
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Answer» From the PROBLEM statement ⇒ 119790 = p(1 + 0.1)3 ⇒ p = 119790/1.331 ⇒ p = 90000 ∴ Principal is Rs 90000. |
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| 82. |
A bank offers 15% simple interest per annum. A customer deposits Rs. 7200 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by means of interest is1). Rs. 68032). Rs.1660.53). Rs. 17014). Rs. 850 |
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Answer» Principal = RS. 7,200 Rate of interest = 15% As rate is calculated semi annually Interest = (7200 × 15 × 1)/200 = Rs. 540 For the second half year Principal = 7200 + 7200 = Rs. 14400 Interest = (14440 × 15 × 1)/200 = Rs. 1080 Total interest = 540 + 1080 = Rs. 1620 |
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| 83. |
A sum is invested at 10% annual rate of simple interest for 4 years. Rs. 355 is the increase in interest if the rate of interest is increased by 5% of the initial rate. What is the sum?1). Rs. 170002). Rs. 174503). Rs. 177504). Rs. 17550 |
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Answer» $(SI = \FRAC{{P \TIMES R \times T}}{{100}})$ Where, SI – Simple Interest P – Principal R – rate T – time Given, sum is INVESTED at a certain annual rate of simple interest for 4 years. Rs. 355 is the increase in interest if the rate of interest is increased by 5% of initial rate. $(SI\; = \frac{{\;P\; \times \;R\; \times \;4}}{{100}}\; = \frac{{PR}}{{25}})$ Increased rate = R + 5% of R = 1.05R $(SI\; = \frac{{P\; \times \;1.05R\; \times \;4}}{{100}} = \frac{{1.05PR}}{{25}})$ Given, difference between the two INTERESTS is Rs. 355 $(\therefore \frac{{1.05PR}}{{25}} - \frac{{PR}}{{25}} = 355)$ R = 10% ⇒0.05 × P × 10 = 355 × 25 ⇒ P = Rs. 17750 |
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| 84. |
A person invests in the stock market at 10% per annum, compounded annually, for 1 years. If the interest was compounded half-yearly, he would have received Rs. 75 more. Find the sum that invest in the stock market.1). Rs. 40,0002). Rs. 35,0003). Rs. 30,0004). Rs. 38,000 |
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Answer» When INTEREST compounded annually, then PERSON get 10% of the TOTAL sum If interest compounded half-annually, Interest rate = 5% and time t = 2 As we KNOW, 5 + 5 + (5 × 5)/100 = 10.25% If interest compounded half-annually, then person get 10.25% of the total sum According to the question ⇒ 10.25% – 10% = 75 ⇒ 0.25% = 75 ⇒ 100% = 30,000 ∴ The sum that invest in the stock market = Rs. 30000 |
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| 85. |
If a certain sum of money doubles itself in 7 years 8 months at simple interest, then what will be the yearly rate of interest (in %)?1). \(18\frac{3}{{24}}\)2). \(13\frac{1}{{23}}\)3). \(26\frac{2}{{23}}\)4). 30 |
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Answer» <P>Let the principal = P Then, Amount = 2P We know that, A = P + [P × R × $(7\frac{2}{3})$] /100 ⇒ 2P = P + [P × R × $(7\frac{2}{3})$] /100 ⇒ P = [P × R × $(7\frac{2}{3})$] /100 ⇒ 100 = R × $(7\frac{2}{3})$ ⇒ 100/ (23/3) = R ⇒ R = $(13\frac{1}{{23}})$ ∴ the CORRECT option is 2) |
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| 86. |
The rate of simple interest per annum of bank being decreased from 5% to \(3\frac{1}{2}\%\), the annual income of a person from interest was less by Rs. 105. The sum deposited at the bank was1). Rs. 7,0002). Rs. 6,0003). Rs. 7,2004). Rs. 6,800 |
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Answer» We know that, SI = P × R × T/100 Where, P = PRINCIPAL, R = % RATE of interest, T = time in years Here, SI1 = P × R1 × T/100 and SI2 = P × R2 × T/100 SI1 – SI2 = P × (R1 – R2) × T/100 Change in the interest rate = 1.5% Change in the interest OBTAINED = 105 ⇒ 105 = P × 1.5/100 = 7000 |
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| 87. |
The C.I. on a certain sum of money for the 4th year at 8% p.a. is Rs. 486. What was the compound interest for the third year on the same sum at the same rate?1). Rs. 4502). Rs. 4753). Rs. 4564). None of these |
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Answer» LET the basic AMOUNT for the 4th year be ‘x’ ∴ According to the first given condition, ⇒ 486 = x × (1 + 0.08) - x ∴ 0.08x = 486 ∴ x = 6075 ∴ Amount after 3rd year = 6075 Let the BASE amount for the 3rd year be ‘y’ ∴ 6075 = y × (1 + 0.08) ∴ y = 5625 ∴ C.I for the 3rd year = 6075 - 5625 = 450 ∴ The Compound Interest for the 3rd year is RS. 450 |
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| 88. |
1). Rs. 22002). Rs. 240003). Rs. 20004). Rs. 1600 |
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Answer» FORMULAS to be used:- SI, I = ( P × r × t ) / 100 For CI: $(A\; = \;P\;{\LEFT( {\;1\; + \frac{r}{{100}}\;} \right)^t})$; CI = A – P Where SI is Simple interest, A is the amount at the end of time t, P is the principal, t is time, r is rate Given, CI on a certain sum for 2 years at 6% PER annum. $(A = P{\left( {\;1 + \frac{6}{{100}}} \right)^2})$ ⇒ A = P(1.06)2 ⇒ A = 1.1236P ∴ CI = A – P ⇒ CI = 1.1236P – P ⇒ CI = 0.1236P SI on Rs. 412 for 10 years at 6% per annum $(\Rightarrow SI = \frac{{412 \times 10 \times 6}}{{100}})$ ⇒ SI = Rs. 247.20 Given, C.I. of 2 yrs be the same as the S.I. on Rs. 412 for 10 yrs ∴ 0.1236P = 247.20 ⇒ P = Rs. 2000 |
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| 89. |
At what rate of simple interest will a sum of money double itself in 18 years?1). \(5\frac{5}{9}{\rm{\% }}\)2). \(5\frac{2}{3}{\rm{\% }}\)3). \(5\frac{5}{2}{\rm{\% }}\)4). 5% |
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Answer» LET the principal amount be RS. X. Simple interest will a sum of money double itself in 18 years. So, the interest will also be Rs. x Simple interest = P × t × r/100 [Where, P = principal amount, t = duration, r = interest rate] x = x × 18 × r/100 $(\RIGHTARROW r = \frac{{100}}{{18}} = 5\frac{5}{9})$ |
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| 90. |
A certain sum invested at 4% per annum compound interest, compounded half-yearly, amounts to Rs 7803 after the end of one year. The sum is?1). 70002). 72003). 75004). 7700 |
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Answer» We have R = 4 % per annum or 2 % per HALF-year T = 2 CYCLES of half YEARS. Hence we take t = 2. P = ? Amount = RS 7803 We have the formula for compound interest Amount $(= P{\left( {1 + \frac{R}{{100}}} \right)^T})$ Where P is the principal T is the time R is the rate of interest ⇒ 7803 = P ( 1 + 2/100)2 ⇒ 7803 = P × 1.02 × 1.02 ⇒ P = 7803/1.0404 ⇒ P = Rs 7500 |
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| 91. |
An amount increased by 36% in 3 year invested on simple interest, what will be the compound interest earned on 10000 in 2 years at same rate?1). 23002). 25003). 25444). 2100 |
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Answer» LET Principal be Rs X. So after 3 years amount would be = 136x/100 ∴ SI = Amount – Principal = 136x/100 – x = 36x/100 ⇒ SI = (P × R × t) /100 ⇒ 36x/100 = (x × r × 3) /100 ⇒ r = 12% $(A = 10000{\left( {1 + \frac{{12}}{{100}}} \right)^2})$ ⇒ A = 12544 ⇒ Compound Interest = 12, 544 – 10, 000 = 2544 |
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| 92. |
The difference between compound interest and simple interest on a sum for 2 year at 20% per annum is Rs. 200. If the interest is compounded half yearly, then what is the difference (in Rs) between compound and simple interest for 1st year?1). 502). 753). 1004). 150 |
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Answer» Given, ⇒ CI – SI = Rs. 200 Let the principal be P So, we have ⇒ P(1 + R/100n)nt – P - (P × R × T)/100 = 200 ⇒ P(1 + 20/100)2 – P - (P × 20 × 2)/100 = 200 ⇒ 1.44P – P – 0.4P = 200 ⇒ 0.04P = 200 ⇒ P = 200/0.04 = Rs. 5000 Now, the DIFFERENCE when compounded HALF yearly ⇒ [5000(1 + 20/200)2 – 5000] – (5000 × 20 × 1)/100 ⇒ [5000(1.1)2 – 5000] – 1000 ⇒ [5000(1.21) – 5000] – 1000 ⇒ 6050 – 6000 = Rs. 50 ∴ the correct option is 1) |
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| 93. |
Ravi took a loan of certain amount from a bank. If the difference of compound interest and simple interest on that amount for 3 years at the rate of 5% per annum is Rs. 11.40, then the amount is1). Rs. 15002). Rs. 14753). Rs. 13954). Rs. 1495 |
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Answer» We know that, S.I. = Simple interest C.I. = Compound interest R = interest rate T = time CI - SI = P(R/100)2(R/100 + 3) ∴ P = [(CI - SI) × (100)3/{(R)2 × (300 + R)}] ⇒ (11.4 × (100)3)/{(5)2 × (300 + 5)} ⇒ Rs. 1495 |
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| 94. |
A sum of Rs. 9000 brings simple interest of Rs. 2970 in three years. If the interest rate is increased by 2%, what will be the interest in three years on the same capital?1). Rs. 35202). Rs. 35103). Rs. 35404). Rs. 3550 |
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Answer» Given, P = Rs.9000, N = 3YEARS and SIMPLE INTEREST = Rs. 2970 Using simple interest formula, Let P = principle, N = time and R = rate percent per annum Simple interest = (P × N × R)/100 ⇒ 2970 = (9000 × 3 × R)/100 ⇒ 270 × R = 2970 ⇒ R = 11 Rate of interest = 11% Given, If the interest rate is INCREASED by 2%, then New interest rate = = 11 + 2 = 13% Using simple interest formula, Simple interest = (9000 × 3 × 13)/100 = 3510 ∴ the simple interest in three years on the same capital is Rs. 3510. |
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| 95. |
A person borrows Rs. 6000 from a co-operative bank at 8% compound interest. At the end of every year he pays Rs. 1500 as part repayment of loan and interest. How much does he still owe to the co-operative bank after four such installments?1). Rs. 1403.7652). Rs. 2688.6723). Rs. 3878.44). Rs. 4980 |
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Answer» AMOUNT left after 1st installment = 6000(1 + (8/100)) – 1500 = 6480 – 1500 = 4980 Amount left to pay after 2nd installment = 4980(1 + (8/100)) – 1500 = 5378.4 – 1500 = 3878.4 Amount left to pay after 3rd installment = 3878.4(1 + (8/100)) – 1500 = 4188.672 – 1500 = 2688.672 Amount left to pay after 4th installment = 2688.672(1 + (8/100)) – 1500 = 2903.765 – 1500 = 1403.765 ∴ Amount left to be paid = RS. 1403.765 |
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| 96. |
1). 142). 163). 284). 20 |
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Answer» <P>Let suppose that principal is P Difference between S.I and C.I is (D) = Rs. 18 The rate of interest (R) = 10% ⇒? $(\frac{R}{{100\;}} = \;\sqrt {\frac{D}{P}} )$ ⇒? $(\frac{10}{{100\;}} = \;\sqrt {\frac{18}{P}} )$ P = 1800 Amount compounded half yearly = Rs. [1800 × (1 + 5/100)4] = Rs. 2187.91 = Rs. 2188 ⇒ C.I. = Rs. (2188 – 1800) = Rs. 388 ⇒ S.I. = Rs. (1800 × 10 × 2)/100 = Rs. 360 ∴ C.I – S.I = Rs. (388 – 360) = Rs. 28 |
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| 97. |
1). 11002). 25003). 45104). 3600 |
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Answer» Let x be the equal installment at the END of one year as the RATE is compounded annually ⇒ Total installment after 2 years = 2x For 1st year, ⇒ P = 5500, R = 30% and T = 1 year ⇒ Interest = (5500 × 20 × 1)/100 = 1100 At the beginning of 2nd year, ⇒ P = 5500 + 1100 – x ⇒ P = 6600 – x Interest at the end of 2ND year, ⇒ Interest = [(6600 – x) × 20 × 1]/100 = 1320 – x/5 Hence, ⇒ Total installment = 2x = 5500 + 1100 + 1320 – x/5 ⇒ 2x + x/5 = 7920 ⇒ x = 3600 ∴ The value of each installment is Rs. 3600 |
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| 98. |
The compound interest on a sum of money is three times the simple interest on the same amount of money when it is invested at the rate of 10%interest for a period of 2 years. Find the sum of money being invested upon?1). Rs. 13312). Rs. 15343). Rs. 34314). Cannot be determined |
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Answer» We know the FORMULA for COMPOUND interest $(CI = \LEFT[ {\left( {P{{\left\{ {1 + \frac{r}{{100}}} \right\}}^t} - 1} \right)} \right])$ Where CI = compound interest; P = principal; R = rate of interest and t = time period We know the formula for simple interest ⇒ SI = (p × r × t)/100 Where SI = Simple interest; P = principal; R = rate of interest and t = time period Given that Compound interest = 3 × Simple Interest $(\Rightarrow P \times r \times t = 3 \times \left[ {\left( {P{{\left\{ {1 + \frac{r}{{100}}} \right\}}^t} - 1} \right)} \right])$ In the above equation, P gets cancelled on both the sides of the equation ∴ VALUE of P cannot be DETERMINED |
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| 99. |
At what rate of interest will Rs. 20000 become Rs. 28800 after 2 year when interest is compounded annually?1). 5%2). 6%3). 20%4). 15% |
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Answer» P = 20000 A = 28800 T = 2 years Formula for COMPOUND interest, ⇒ Putting all values in above formula, we get- ⇒ 28800 = 20000{ 1 + R/100}2 ⇒ 28800/20000 = {1 + R/100}2 ⇒144/100 = {1 + R/100}2 $(\Rightarrow 1 + \frac{{\rm{R}}}{{100}} = \frac{{12}}{{10}})$ ⇒ R = 20% |
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| 100. |
Find the compound interest on Rs. 12550 for 9 months at 12% per annum compounded quarterly.1). Rs. 11642). Rs. 1125.183). Rs. 1198.724). Rs. 1164.32 |
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Answer» <P>We know that, Compound INTEREST $(= P \times {\left( {1 + \frac{r}{n}} \right)^{nt}} - P)$ ∴ ANSWER $(= 12550 \times {\left( {1 + \frac{{0.12}}{4}} \right)^{4 \times \frac{9}{{12}}}} - 12550 \approx 1164)$ |
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