True Discount Questions and Answers

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.

Answer: C

Correct answer: 8 months

\[60 = \frac{1860 \times 5T}{100 + 5T} \quad\Longrightarrow\quad 60(100 + 5T) = 9300T\]

\[6000 + 300T = 9300T \quad\Longrightarrow\quad 9000T = 6000 \quad\Longrightarrow\quad T = \tfrac{2}{3} \text{ year}\]

Two-thirds of a year is 8 months. The present worth is Rs. 1800, and 8 months of interest on it at 5% is Rs. 60.

Answer: D

Writing \( k = \dfrac{RT}{100} \), the four quantities are related by

\[ \text{BD} = A\,k, \qquad \text{TD} = \frac{A\,k}{1+k} = \text{PW}\,k, \qquad \text{BG} = \text{TD}\,k \]

With \( T = \dfrac{6}{12} = 0.5 \) year at 6%:

\[ k = \frac{6 \times 0.5}{100} = 0.03 \]

Since \( \text{BD} = A\,k \) and \( \text{TD} = \dfrac{A\,k}{1+k} \), the two differ only by the factor \( 1+k \):

\[ \text{TD} = \frac{\text{BD}}{1+k} = \frac{18.54}{1.03} = \text{Rs. } 18 \]

The banker's gain is the 54 paise difference - small here because the period is short and the rate low.

Answer: A

Successive discounts multiply the retained fractions.

\[ 10 \times (1 - 0.20) \times (1 - 0.10) = 10 \times 0.80 \times 0.90 \]

\[ = 10 \times 0.72 = \text{Rs. } 7.20 \]

The single equivalent discount is 28%, not 30% - the 10% is charged on Rs. 8 rather than on Rs. 10, so it takes off 80 paise instead of one rupee.

Answer: B

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

Here \( T = 2\tfrac{1}{2} = 2.5 \) years at 15%:

\[ \frac{RT}{100} = \frac{15 \times 2.5}{100} = 0.375 \]

\[ \text{PW} = \frac{2310}{1.375} = \text{Rs. } 1680 \]

Checking: the true discount is \( 2310 - 1680 = 630 \), and 37.5% of 1680 is indeed 630. Note the interest is reckoned on the present worth, not on the Rs. 2310 face value.

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

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

Answer: C

Apply the known discount first, then see what fraction of the remainder she actually paid.

\[ 560 \times (1 - 0.20) = \text{Rs. } 448 \]

\[ \frac{336}{448} = 0.75 \quad\Longrightarrow\quad \text{second discount} = 1 - 0.75 = 25\% \]

Checking: \( 448 \times 0.75 = 336 \).

Taking \( \dfrac{560 - 336}{560} = 40\% \) gives the combined discount, not the second one - and indeed \( 1 - 0.80 \times 0.75 = 0.40 \).

Answer: C

Four quantities are related throughout this topic. Writing \( k = \dfrac{RT}{100} \):

\[ \text{BD} = A\,k, \qquad \text{TD} = \frac{A\,k}{1+k} = \text{PW}\,k, \qquad \text{BG} = \text{BD} - \text{TD} = \text{TD}\,k \]

The banker's discount is interest on the face value; the true discount is interest on the present worth; the banker's gain is the interest on the true discount.

Here \( k = \dfrac{10 \times 2}{100} = 0.2 \).

\[ \text{BG} = \text{TD} \times k \quad\Longrightarrow\quad \text{TD} = \frac{24}{0.2} = \text{Rs. } 120 \]

\[ \text{TD} = \text{PW} \times k \quad\Longrightarrow\quad \text{PW} = \frac{120}{0.2} = \text{Rs. } 600 \]

The chain runs BG → TD → PW, dividing by \( k \) at each step. Stopping one step early leaves Rs. 120, and the face value would be Rs. 720.

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.

Answer: B

Take the cost price as 100, so the marked price is 120.

\[ \text{SP} = 120 \times (1 - 0.30) = 120 \times 0.70 = 84 \]

\[ \text{loss} = 100 - 84 = 16 \quad\Longrightarrow\quad \text{loss}\% = \frac{16}{100} \times 100 = 16\% \]

Loss percentage is always measured against the cost price, not the marked price. Against the marked price the fall would read \( \tfrac{36}{120} = 30\% \), which is just the discount again.