Find the simple interest on Rs.500 for 9 months at 6 paisa per month?
Answer: B
"6 paisa per month" means 6 paise on each rupee each month, or 6% per month.
\[ \text{SI} = 500 \times \frac{6}{100} \times 9 \]
\[ = 500 \times 0.06 \times 9 = \text{Rs. } 270 \]
The rate is already monthly, so the nine months are used directly - no conversion to years is needed.
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At what rate of interest is an amount doubled in two years, when compounded annually?
Answer: B
Compounded annually, doubling in two years means
\[ \left(1 + \frac{R}{100}\right)^{2} = 2 \]
\[ 1 + \frac{R}{100} = \sqrt{2} = 1.4142 \]
\[ R = 41.42 \approx 41.4\% \]
Under simple interest the answer would be 50%; compounding needs less, because the first year's interest itself earns in the second.
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A man took loan from a bank at the rate of 12% p.a. S.I. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was?
Answer: C
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ 5400 = \frac{P \times 12 \times 3}{100} = 0.36P \]
\[ P = \frac{5400}{0.36} = \text{Rs. } 15{,}000 \]
Three years at 12% comes to 36% of the principal, so the principal is a little under three times the interest.
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Nitin borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a. for the next five years and 13% p.a. for the period beyond eight years. If the total interest paid by him at the end of eleven years is Rs. 8160, how much money did he borrow?
Answer: A
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Beyond eight years leaves \( 11 - 8 = 3 \) years at 13%, so the three slabs are:
\[ P\left(\frac{6 \times 3 + 9 \times 5 + 13 \times 3}{100}\right) = 8160 \]
\[ P\left(\frac{18 + 45 + 39}{100}\right) = 1.02P = 8160 \]
\[ P = \text{Rs. } 8000 \]
The final slab covers only the last three years, not all eleven.
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Vijay lent out an amount Rs. 10000 into two parts, one at 8% p.a. and the remaining at 10% p.a. both on simple interest. At the end of the year he received Rs. 890 as total interest. What was the amount he lent out at 8% pa.a?
Answer: B
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Let \( p \) be the part at 8%; the rest, \( 10000 - p \), earns 10%.
\[ 0.08p + 0.10(10000 - p) = 890 \]
\[ 0.08p + 1000 - 0.10p = 890 \quad\Longrightarrow\quad -0.02p = -110 \]
\[ p = \text{Rs. } 5500 \]
Checking: Rs. 440 from the 8% part and Rs. 450 from the Rs. 4500 at 10%.
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In 4 years the simple interest on certain sum of money is 9/25 of the principal. The annual rate of interest is
Answer: C
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Write the interest as a fraction of the principal.
\[ \frac{9}{25}P = \frac{P \times R \times 4}{100} \]
The \( P \) cancels:
\[ \frac{9}{25} = \frac{4R}{100} \quad\Longrightarrow\quad R = \frac{9 \times 100}{25 \times 4} = 9\% \]
Because the interest is quoted as a fraction of the principal, no actual sum is needed.
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Find the principal which yields a simple interest of Rs. 20 and compound interest of Rs. 21 in two years, at the same percent rate per annum?
Answer: D
Simple interest of Rs. 20 over two years is Rs. 10 a year. Compound interest exceeds it by Re. 1, which is one year's interest on that first Rs. 10:
\[ 10 \times \frac{R}{100} = 1 \quad\Longrightarrow\quad R = 10\% \]
Now use the simple interest to find the principal:
\[ 20 = \frac{P \times 10 \times 2}{100} = 0.2P \quad\Longrightarrow\quad P = \text{Rs. } 100 \]
Checking: at 10%, Rs. 100 gives compound interest of \( 100(1.21 - 1) = 21 \).
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If Rs.450 amount to Rs.540 in 4 years, what will it amount to in 6 years at the same rate % per annum?
Answer: C
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Find the rate from the four-year figure.
\[ \text{SI} = 540 - 450 = 90 \quad\Longrightarrow\quad 90 = \frac{450 \times R \times 4}{100} = 18R \]
\[ R = 5\% \]
\[ \text{SI over 6 years} = \frac{450 \times 5 \times 6}{100} = 135 \quad\Longrightarrow\quad \text{amount} = \text{Rs. } 585 \]
Since simple interest is linear in time, the shortcut is \( 90 \times \tfrac{6}{4} = 135 \).
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A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is?
Answer: C
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Under simple interest each year adds the same amount, so the gap between the two figures is one year's interest.
\[ 854 - 815 = \text{Rs. } 39 \text{ per year} \]
\[ P = 815 - 3 \times 39 = 815 - 117 = \text{Rs. } 698 \]
Working back three years from the three-year amount recovers the principal.
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If x is the interest on y and y is the interest on z, the rate and time is the same on both the cases. What is the relation between x, y and z?
Answer: C
Write both statements with the same \( R \) and \( T \), and let \( k = \tfrac{RT}{100} \).
\[ x = yk, \qquad y = zk \]
Dividing one by the other:
\[ \frac{x}{y} = \frac{y}{z} \quad\Longrightarrow\quad y^{2} = xz \]
So \( y \) is the geometric mean of \( x \) and \( z \). Substituting directly gives the same thing: \( x = zk^{2} \) and \( y^{2} = z^{2}k^{2} = xz \).
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