CAT 2025 Lesson : Profit & Loss - Past Questions

Observation/Strategy
1) Two products and their loss$\%$ and gain$\%$ for two different scenarios are provided.
2) Loss$\%$ and Profit$\%$ are a function of the CP. So, the CP of A and B can be taken as the variables.
3) The two scenarios are in the form of linear equations.

Let the CP of products A and B be $a$ and $b$ respectively.

Scenario 1: $20 \%$ loss on A and $30 \%$ gain on B results in no profit or loss.
Overall Profit $= - 0.2a + 0.3b = 0 \longrightarrow (1)$

Scenario 2: $15 \%$ loss on A and $15 \%$ gain on B results in loss of Rs. $6$ million.
Overall Profit $= -0.5a + 0.15b = -6$
⇒ $b - a = -40$
⇒ $a = a + b = 40 \longrightarrow (2)$

Substituting ($2$) in ($1$),
$- 0.2b - 8 + 0.3b = 0$
⇒ $b = 80$

∴ CP of B = Rs. $80$ million

Answer: (4) Rs. $80$ million

Observation/Strategy
1) Number of units of apples and oranges are not provided and will form the variables
2) A linear equation with 2 variables exist, where both have to be positive integers, with apples more than oranges
3) We should be able to solve this equation and apply basic profit percentage formula.

Let the number of apples and oranges sold be $x$ and $y$ respectively. $x > y$ is given
$23x + 10y = 653$

$10y$ will always end in $0$. So, $23x$ has to end in $3$. Possible values of $x$ are $1$, $11$, $21$, $31$ etc.

$\bm{x = 21}$ results in $\bm{y = 17}$. These are the only values where $x > y$ and both are integers.

CP of A $= \dfrac{23}{1.15} = 20$ ; Profit of A $= 23 - 20 = 3$

CP of B $= \dfrac{10}{1.25} = 8$ ; Profit of B $= 10 - 8 = 2$

Total Profit $= 3 \times 21 + 2 \times 17 = 97$

Total CP $= 653 - 97 = 556$

Profit$\% = \dfrac{97}{556} \times 100 \% = \dfrac{2425}{139} \% = 17.45 \%$

Answer: (2) 17.4%

Observation/Strategy
1) Relationship CP of soaps and toothpastes are given at the end. So, $1$ variable for the unknown CP will be sufficient.
2) Scenario that results in profit of Rs. $385$ can be used to form a linear equation.
3) Solving this we can get the CPs, SPs and the overall loss.

Let CP of a soap $= s$
∴ CP of a toothpaste $= 0.6s$

Profit of a soap $= 0.15s$ and profit of a toothpaste $= 20$

$15$ soaps and $8$ toothpastes were sold at a profit of Rs. $38$5
⇒ $15 \times 0.15s + 8 \times 20 = 385$
⇒ $2.25s = 225$
⇒ $\bm{s = 100}$

CP of a soap $= 100$ and CP of a toothpaste $= 60$

As total of $20$ soaps and $12$ toothpastes were purchased,
Total CP $= 20 \times 100 + 12 \times 60 = 2720$

SP of a soap $= 115$ and SP of a toothpaste $= 80$

As total of $15$ soaps and $8$ toothpastes were sold,
Total SP $= 15 \times 115 + 8 \times 80 = 2365$

Overall Loss $= 2720 - 2365 = 355$

Answer: (1) Loss of Rs. $355$