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True Discount Questions and Answers

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1.

If a debt is cleared by paying in two instalments, ₹.17,640 in each instalment, wherein the interest is allowed at the rate of 5% p.a. compounded annually, then the amount borrowed is:

Exam Year(s)

DSSSB PGT - 2023,

Answer: C

The amount borrowed is Rs. 32,800. The principal equals the present value of the two instalments at 5 per cent compound interest. \[P = \frac{17640}{1.05} + \frac{17640}{(1.05)^2}\] \[P = 16800 + 16000 = 32800\] The first instalment, paid after one year, is discounted once and the second, paid after two years, twice. The option Rs. 35,280 is the plain sum of the two instalments, which ignores interest.

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2.

A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600, allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum, then the man has a gain of:

Answer: A

The present worth (PW) is what the sum is really worth today; the true discount (TD) is the interest that PW would earn over the remaining time:

\[ \text{Amount} = \text{PW}\left(1 + \frac{RT}{100}\right), \qquad \text{TD} = \text{Amount} - \text{PW} \]

The Rs. 3600 is not received today but in 2 years, so compare its present worth with the Rs. 3000 he spent.

\[ \text{PW} = \frac{3600}{1 + \dfrac{10 \times 2}{100}} = \frac{3600}{1.2} = 3000 \]

The present worth exactly equals the cost price, so

\[ \text{gain} = 3000 - 3000 = 0 \quad\Longrightarrow\quad 0\% \]

Comparing Rs. 3600 with Rs. 3000 directly suggests a 20% profit, but that ignores two years of waiting - which is precisely what true discount measures.

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3.

A trader owes a merchant Rs. 10,028 due 1 year hence. The trader wants to settle the account after 3 months. If the rate of interest 12% per annum, how much cash should he pay?

Answer: B

The present worth (PW) is what the sum is really worth today; the true discount (TD) is the interest that PW would earn over the remaining time:

\[ \text{Amount} = \text{PW}\left(1 + \frac{RT}{100}\right), \qquad \text{TD} = \text{Amount} - \text{PW} \]

The debt falls due in 1 year and is being settled after 3 months, so it is being paid 9 months early. Only that unexpired period is discounted.

\[ T = \frac{9}{12} = \frac{3}{4} \text{ year}, \qquad \frac{RT}{100} = \frac{12 \times 3/4}{100} = 0.09 \]

\[ \text{PW} = \frac{10028}{1.09} = 9200 \]

So he should pay Rs. 9200. The commonest error is discounting for the full year rather than for the nine months actually saved.

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4.

A man wants to sell his scooter. There are two offers, one at Rs. 12,000 cash and the other a credit of Rs. 12,880 to be paid after 8 months, money being at 18% per annum. Which is the better offer?

Answer: A

The present worth (PW) is what the sum is really worth today; the true discount (TD) is the interest that PW would earn over the remaining time:

\[ \text{Amount} = \text{PW}\left(1 + \frac{RT}{100}\right), \qquad \text{TD} = \text{Amount} - \text{PW} \]

Bring the credit offer back to today's money and then compare like with like.

\[ T = \frac{8}{12} = \frac{2}{3} \text{ year}, \qquad \frac{RT}{100} = \frac{18 \times 2/3}{100} = 0.12 \]

\[ \text{PW of the credit offer} = \frac{12880}{1.12} = 11500 \]

\[ 12000 > 11500 \]

The cash offer is worth Rs. 500 more in today's money, so Rs. 12,000 in cash is the better deal. The larger headline figure loses because eight months at 18% costs more than the extra Rs. 880 is worth.

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5.

If Rs. 10 be allowed as true discount on a bill of Rs. 110 due at the end of a certain time, then the discount allowed on the same sum due at the end of double the time is:

Answer: D

The present worth (PW) is what the sum is really worth today; the true discount (TD) is the interest that PW would earn over the remaining time:

\[ \text{Amount} = \text{PW}\left(1 + \frac{RT}{100}\right), \qquad \text{TD} = \text{Amount} - \text{PW} \]

First find the rate–time product from the original bill.

\[ \text{PW} = 110 - 10 = 100 \quad\Longrightarrow\quad 100 \times \frac{RT}{100} = 10 \quad\Longrightarrow\quad \frac{RT}{100} = 0.1 \]

Doubling the time doubles that product to 0.2, so for the same Rs. 110:

\[ \text{PW}' = \frac{110}{1.2} = 91.67 \]

\[ \text{TD}' = 110 - 91.67 = 18.33 \]

True discount does not double when the time doubles - the Rs. 20 option is exactly that trap. It grows more slowly because the present worth shrinks at the same time.

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6.

The true discount on a bill due 9 months hence at 16% per annum is Rs. 189. The amount of the bill is:

Answer: B

The present worth (PW) is what the sum is really worth today; the true discount (TD) is the interest that PW would earn over the remaining time:

\[ \text{Amount} = \text{PW}\left(1 + \frac{RT}{100}\right), \qquad \text{TD} = \text{Amount} - \text{PW} \]

Rearranging gives the standard relation

\[ \text{TD} = \frac{\text{Amount} \times \frac{RT}{100}}{1 + \frac{RT}{100}} \]

Here \( T = \dfrac{9}{12} = 0.75 \) year at 16%:

\[ \frac{RT}{100} = \frac{16 \times 0.75}{100} = 0.12 \]

\[ 189 = \frac{A \times 0.12}{1.12} \quad\Longrightarrow\quad A = \frac{189 \times 1.12}{0.12} = 1764 \]

As a check, PW = 1764 − 189 = 1575, and 12% of 1575 is indeed 189.

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7.

A man buys a watch for Rs. 1950 in cash and sells it for Rs. 2200 at a credit of 1 year. If the rate of interest is 10% per annum, the man:

Answer: B

The present worth (PW) is what the sum is really worth today; the true discount (TD) is the interest that PW would earn over the remaining time:

\[ \text{Amount} = \text{PW}\left(1 + \frac{RT}{100}\right), \qquad \text{TD} = \text{Amount} - \text{PW} \]

The Rs. 2200 arrives a year late, so discount it back to today.

\[ \text{PW} = \frac{2200}{1 + \dfrac{10 \times 1}{100}} = \frac{2200}{1.1} = 2000 \]

\[ \text{gain} = 2000 - 1950 = \text{Rs. } 50 \]

Comparing 2200 with 1950 suggests a Rs. 250 profit, but Rs. 200 of that is merely compensation for waiting a year - only the remaining Rs. 50 is real gain.

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8.

The true discount on Rs. 1760 due after a certain time at 12% per annum is Rs. 160. The time after which it is due is:

Answer: D

The present worth (PW) is what the sum is really worth today; the true discount (TD) is the interest that PW would earn over the remaining time:

\[ \text{Amount} = \text{PW}\left(1 + \frac{RT}{100}\right), \qquad \text{TD} = \text{Amount} - \text{PW} \]

\[ \text{PW} = 1760 - 160 = 1600 \]

The true discount is simple interest on the present worth, so

\[ 160 = \frac{1600 \times 12 \times T}{100} = 192\,T \]

\[ T = \frac{160}{192} = \frac{5}{6} \text{ year} = 10 \text{ months} \]

Note the interest is charged on the present worth of Rs. 1600, not on the Rs. 1760 face value - using the face value gives about 9 months, which is why that option is listed.

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9.

Rs. 20 is the true discount on Rs. 260 due after a certain time. What will be the true discount on the same sum due after half of the former time, the rate of interest being the same?

Answer: B

The present worth (PW) is what the sum is really worth today; the true discount (TD) is the interest that PW would earn over the remaining time:

\[ \text{Amount} = \text{PW}\left(1 + \frac{RT}{100}\right), \qquad \text{TD} = \text{Amount} - \text{PW} \]

\[ \text{PW} = 260 - 20 = 240 \quad\Longrightarrow\quad 240 \times \frac{RT}{100} = 20 \quad\Longrightarrow\quad \frac{RT}{100} = \frac{1}{12} \]

Halving the time halves that product to \( \dfrac{1}{24} \), so for the same Rs. 260:

\[ \text{PW}' = \frac{260}{1 + \frac{1}{24}} = \frac{260 \times 24}{25} = 249.60 \]

\[ \text{TD}' = 260 - 249.60 = \text{Rs. } 10.40 \]

Halving the time does not simply halve the discount to Rs. 10 - that is the first distractor, and it is slightly too small.

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10.

The simple interest and the true discount on a certain sum for a given time and at a given rate are Rs. 85 and Rs. 80 respectively. The sum is:

Answer: C

Simple interest is reckoned on the amount, true discount on the present worth. Both use the same \( \dfrac{RT}{100} \), so

\[ \text{SI} = A \times \frac{RT}{100}, \qquad \text{TD} = \text{PW} \times \frac{RT}{100} = (A - \text{TD}) \times \frac{RT}{100} \]

Dividing one by the other:

\[ \frac{\text{SI}}{\text{TD}} = \frac{A}{A - \text{TD}} \quad\Longrightarrow\quad \frac{85}{80} = \frac{A}{A - 80} \]

\[ 85A - 6800 = 80A \quad\Longrightarrow\quad 5A = 6800 \quad\Longrightarrow\quad A = 1360 \]

The shortcut worth remembering is

\[ \text{Sum} = \frac{\text{SI} \times \text{TD}}{\text{SI} - \text{TD}} = \frac{85 \times 80}{5} = 1360 \]

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