CAT 2025 Lesson : Profit & Loss - Other Types

Leena sold a chair at a loss of $17 \%$. If she had sold it at a profit of $3 \%$, she would have received Rs. $25$ more than what she did. At what price should she have sold to make a profit of $20 \%$?

Solution

Let the Cost Price be $c$.

Amount realised when sold
at a loss of $17 \% = (1 - 0.17) \times c = 0.83 c$
at a profit of $3 \% = (1 + 0.03) \times c = 1.03 c$

$1.03c - 0.83c = 25$
⇒ $c = \dfrac{25}{0.2} = 125$

$\text{@}20 \%$ profit, SP $= 125 \times (1 + 0.2) = 150$

Answer: Rs. $150$

Ramlal purchased a certain number of mangoes at the rate of Rs. $25$ for $8$ mangoes. He sold half of these mangoes at the rate of Rs. $35$ for $12$ mangoes and the rest at Rs. $55$ for $16$ mangoes. Then overall he made a

(1) loss of $1.67 \%$            (2) profit of $1.67 \%$            (3) loss of $2.33 \%$            (4) profit of $2.33 \%$           

Solution

We can assume a value for the number of mangoes purchased such that our calculations will be easy. For the different rates, the costs per mango are $\dfrac{25}{8}, \dfrac{35}{12}$ and $\dfrac{55}{16}$.

LCM($8$, $12$, $16$) $= 48$. As the numbers are halved, let $96$ (twice the LCM) be the number of mangoes purchased.

Total CP $= 96 \times \dfrac{25}{8} = \bm{300}$

Half of these mangoes, i.e. $48$ are sold at Rs. $35$ for $12$ and the other $48$ at Rs. $55$ for $16$.

Total SP of mangoes $= 48 \times \dfrac{35}{12} + 48 \times \dfrac{55}{16} = 140 + 165 = \bm{305}$

Profit $= 305 - 300 =$ Rs. $5$

Profit $\% = \dfrac{5}{300} \times 100 \% = 1.67 \%$

Alternatively

You can assume the number of mangoes to be a variable, say $2n$.

Total CP of mangoes $= \dfrac{2n \times 25}{8} = \bm{\dfrac{25n}{4}}$

Total SP of mangoes $= n \times \dfrac{35}{12} + n \times \dfrac{55}{16} = \dfrac{35n}{12} + \dfrac{55n}{16} = \bm{\dfrac{305 n}{48}}$

Profit $= \dfrac{305n}{48} - \dfrac{25n}{4} = \dfrac{5n}{48}$

Profit $\% = \dfrac{\dfrac{5n}{48}}{\dfrac{25n}{4}} \times 100 \% = \dfrac{5}{48} \times \dfrac{4}{25} \times 100 \% = \bm{1.67 \%}$

We get the same answer with variables, however, as noted above, this is a little cumbersome. It is always easier to assume values in these kinds of problems where we are asked to find the profit % and not the absolute profit in Rupees.

Answer: (2) profit of $1.67 \%$

If Jamshed had increased the selling price on his cycle by $20 \%$, he would have made a profit of $38 \%$. What was his actual profit $\%$?

Solution

Let the actual CP and SP be $c$ and $s$ respectively.

$(1 + 0.2) \times s = (1 + 0.38) \times c$

⇒ $\dfrac{s}{c} = \dfrac{1.38}{1.2} = \dfrac{138}{120} = \dfrac{23}{20} = 1.15$

⇒ $s = 1.15 \times c$

∴ Profit $= 15 \%$

Answer: $15 \%$

On a certain day, Baba stores had a one-day offer of a discount voucher of Rs. $1,000$ for every person who shopped for Rs. $5,000$ or more. This voucher could be used on any subsequent purchase (on a different day of that month) for Rs. 4,000 or more. This scheme could result in a maximum possible discount of

(1) $10 \%$            (2) $11.1 \%$            (3) $12.5 \%$            (4) $25 \%$           

Solution

Discount $\% = \dfrac{\text{Discount}}{\text{Marked Price}}$

Discount is fixed at Rs. $1,000$. For Discount $\%$ to be maximum, the denominator, i.e. Marked Price has to be minimum.

As a shopper needs to shop for Rs. $5,000$ to get the voucher and further shop for Rs. $4,000$ more to use it,

Minimum purchase required $= 5000 + 4000 = 9,000$

Discount $\% = \dfrac{1000}{9000} \times 100 \% = 11.11 \%$

Answer: (2) $11.1 \%$