Profit & Loss: CAT previous year questions

Solution

Let the marked price be $M$. Discounted price per chair $= 0.78M$.

The customer pays for 12 chairs and gets 13: revenue $= 12 \times 0.78M = 9.36M$.

Cost of 13 chairs $= 13 \times 100 = 1300$; a $26\%$ profit means revenue $= 1.26 \times 1300 = 1638$.

$9.36M = 1638 \Rightarrow M = 175$.

Solution

Selling price at the original discount $d$: $M(1 - d) = 1.2 \times 1650 = 1980$.

With the discount doubled: $M(1 - 2d) = 1650 + 110 = 1760$.

Subtracting: $Md = 220$, so $M = 1980 + 220 = 2200$.

Now find a discount $r$ where the profit $\%$ is also $r$:

$2200(1 - r) = 1650(1 + r) \Rightarrow 550 = 3850r \Rightarrow r = \dfrac{1}{7} \approx 14.3\%$.

Nearest: $14$.

Solution

Let the marked price be $M$.

Revenue $= M(120 \times 0.8 + 135 \times 0.9 + 150 \times 0.95 + 165) = M(96 + 121.5 + 142.5 + 165) = 525M$.

Units sold $= 570$, so cost $= 570 \times 240 = 136800$.

$525M - 136800 = 138825 \Rightarrow 525M = 275625 \Rightarrow M = 525$.

Solution

Let the cost be $c$; selling price $= 1.4c$.

New cost $= 0.6c$ and new selling price $= 1.4c - 5$, which gives $50\%$ profit:

$1.4c - 5 = 1.5 \times 0.6c = 0.9c \Rightarrow 0.5c = 5 \Rightarrow c = 10$.

Original selling price $= 1.4 \times 10 = 14$.

Solution

Bina's cost price $= \dfrac{4860}{0.81} = 6000$.

Hari's purchase price would give Bina $17\%$ profit: $1.17 \times 6000 = 7020$.

Shyam bought at 4860 and sold at 7020.

Shyam's profit $= 7020 - 4860 = 2160$.

Solution

A $10\%$ discount saves Rs 15, so the marked price $= 150$.

Marked price $= 1.2 \times$ cost $\Rightarrow$ cost $= 125$.

Ravi pays $150 - 15 = 135$.

Gopi's profit $= 135 - 125 = 10$.

Solution

Let the common selling price be $S$. Cost of A $= \dfrac{S}{1.2}$, cost of B $= \dfrac{S}{0.9}$.

New price giving $10\%$ profit on B: $S' = 1.1 \times \dfrac{S}{0.9}$.

Profit factor on A $= \dfrac{S'}{S/1.2} = \dfrac{1.1 \times 1.2}{0.9} = 1.4667$.

Profit on A $\approx 46.67\% \approx 47\%$.

Solution

Minu buys at 1000 and sells to Kanu at $20\%$ profit: 1200 (profit 200).

Kanu sells it back at a $20\%$ loss: $0.8 \times 1200 = 960$.

Let Minu sell to Tanu at $x$. Total profit: $200 + (x - 960) = 500 \Rightarrow x = 1260$.

Profit on the last sale $= \dfrac{1260 - 960}{960} \times 100 = 31.25\%$.

Solution

Buying: he pays Rs 100 but gets 105 cm, so cost per cm $= \dfrac{100}{105}$.

Selling: price after $5\%$ discount $= 0.95 \times 110 = 104.5$ for what should be 100 cm, but he gives only 95 cm.

Revenue per cm actually given $= \dfrac{104.5}{95} = 1.1$.

Profit $= \dfrac{1.1}{100/105} - 1 = 1.1 \times 1.05 - 1 = 0.155 = 15.5\%$.

Solution

$7c = 30p = 9a$. Take this common value as $630u$: cashew $c = 90u$, peanut $p = 21u$, almond $a = 70u$ per kg.

Cost $C = 4(90u) + 14(21u) + 6(70u) = 1074u$ for 24 kg of mixture.

Let the marked price be $m$ per kg. Selling everything at $m$: $24m - C = 1752$.

Actual: $4m + 20(0.8m) = 20m$, so $20m - C = 744$.

Subtracting: $4m = 1008 \Rightarrow m = 252$, so $C = 24 \times 252 - 1752 = 4296$.

$1074u = 4296 \Rightarrow u = 4$. Almonds cost $6 \times 70 \times 4 = 1680$.

Solution

Let juice cost $j$ per kg; syrup costs $0.8j$.

Total cost $= 110(0.8j) + 120j = 208j$. Required revenue $= 1.64 \times 208j = 341.12j$.

Revenue: syrup $10 \times 0.88j = 8.8j$; juice $20 \times 1.2j = 24j$; mixture $200 \times 308.32 = 61664$.

$32.8j + 61664 = 341.12j \Rightarrow 308.32j = 61664 \Rightarrow j = 200$.

Syrup cost $= 0.8 \times 200 = 160$ per kg.

Solution

Let the number of pens be $n$ and the wage $W$.

Case 1: $1200 + 11(n - 100) - 8n - W = 300$.

Case 2: $1200 + 9(n - 100) - 8n - W = -300$.

Subtracting: $2(n - 100) = 600 \Rightarrow n = 400$.

Case 1: $1200 + 3300 - 3200 - W = 300 \Rightarrow W = 1000$.

Solution

Take 2, 3 and 5 kg. Cost $= 2(800) + 3(500) + 5(300) = 4600$ for 10 kg.

For $50\%$ profit, revenue $= 6900$.

One-sixth of 10 kg $= \dfrac{5}{3}$ kg sold at 700: $\dfrac{3500}{3} \approx 1166.67$.

Remaining $\dfrac{25}{3}$ kg must bring $6900 - 1166.67 = 5733.33$.

Price $= 5733.33 \div \dfrac{25}{3} = 688$ per kg.

Solution

Let fixed cost $= F$ and cost per boarder $= v$. Revenue $= 1600n$.

$n = 50$: profit $= 200 \times 50 = 10000 \Rightarrow 80000 - F - 50v = 10000 \Rightarrow F + 50v = 70000$.

$n = 75$: profit $= 250 \times 75 = 18750 \Rightarrow 120000 - F - 75v = 18750 \Rightarrow F + 75v = 101250$.

So $25v = 31250 \Rightarrow v = 1250$, $F = 7500$.

$n = 80$: profit $= 128000 - 7500 - 100000 = 20500$.

Solution

Let the desktop cost $D$; the laptop costs $50000 - D$.

Overall profit $= 2\%$ of $50000 = 1000$.

$0.2D - 0.1(50000 - D) = 1000 \Rightarrow 0.3D = 6000 \Rightarrow D = 20000$.

Solution

Let the labelled price be $L$.

8 toys at $0.8L$: $6.4L$. 4 toys at $0.8 \times 0.75L = 0.6L$: $2.4L$.

$8.8L = 2112 \Rightarrow L = 240$.

Cost $= \dfrac{2112}{1.1} = 1920$. Without discounts, revenue $= 12 \times 240 = 2880$.

Profit $= \dfrac{2880 - 1920}{1920} \times 100 = 50\%$.

Solution

Take cost $= 1$ per kg, so total cost $= 35$ and marked price $= 1.2$.

5 kg at 1.2: $6$. 15 kg at $1.08$: $16.2$. 3 kg wasted, so 12 kg remain.

Target revenue for $15\%$ profit $= 1.15 \times 35 = 40.25$.

$12 \times 1.2(1 + p) = 40.25 - 22.2 = 18.05 \Rightarrow 1 + p = \dfrac{18.05}{14.4} \approx 1.2535$.

$p \approx 25\%$.