Percentages: CAT previous year questions

Solution

Let the number of girls be $g$ and boys be $b$ ($b > 10$).

Girls remaining $= 0.6g$, boys remaining $= 0.4b$.

Given $0.6g - 0.4b = 8 \Rightarrow 3g - 2b = 40$.

Since $0.6g$ and $0.4b$ must be whole numbers, both $g$ and $b$ are multiples of 5.

Try $b = 15$: $3g = 70$ ✗. $b = 20$: $3g = 80$ ✗. $b = 25$: $3g = 90 \Rightarrow g = 30$ ✓.

Check: $18$ girls and $10$ boys remain, difference $8$ ✓.

Minimum total $= 30 + 25 = 55$.

Solution

Let investment in gold $= G$. Then bonds $= 0.25G$ and stocks $= 100000 - 1.25G$.

Total return: $0.10(100000 - 1.25G) + 0.06(0.25G) + 0.08G = 8200$

$10000 - 0.125G + 0.015G + 0.08G = 8200 \Rightarrow 0.03G = 1800 \Rightarrow G = 60000$.

Bonds $= 0.25 \times 60000 = 15000$.

Gain from bonds $= 6\%$ of $15000 = 900$.

Solution

Let the total be $100$.

Pinu $= 20$. Remaining $= 80$, so Meena $= 40\%$ of $80 = 32$.

Seema gets $20\%$ less than Pinu $= 16$.

Rinu $= 100 - 20 - 32 - 16 = 32$.

Pinu : Rinu $= 20 : 32 = 5 : 8$.

Solution

Take September incomes as $800, 600, 500$.

Rent $= 15\%(800) + 12\%(600) + 18\%(500) = 120 + 72 + 90 = 282$.

October incomes: $800 \times 1.10 = 880$, $600 \times 1.12 = 672$, $500 \times 1.15 = 575$. Total $= 2127$.

Rent is unchanged, so the share $= \dfrac{282}{2127} \times 100 \approx 13.26\%$.

Solution

Let the total stock be $N$. Mangoes $= 0.4N$; let apples $= A$ and bananas $= B$, with $A + B = 0.6N$.

Fruits sold $= 0.2N + 96 + 0.4A = 0.5N \Rightarrow 0.4A = 0.3N - 96 \Rightarrow A = 0.75N - 240$.

Then $B = 0.6N - A = 240 - 0.15N$.

Conditions: $A \geq 1 \Rightarrow N > 320$; $B \geq 96 \Rightarrow N \leq 960$.

Whole numbers: half the mangoes ($0.2N$) needs $N$ to be a multiple of 5, and $A = 0.75N - 240$ needs $N$ to be a multiple of 4. So $N$ is a multiple of 20 (then $0.4A$ is also whole).

Smallest multiple of 20 greater than 320 is $340$.

Check: $A = 15$, $B = 189$, mangoes $= 136$; sold $= 68 + 96 + 6 = 170 = 50\%$ ✓.

Solution

Let the first increment be $x$ (as a fraction); the second is $2x$.

$(1 + x)(1 + 2x) = 1.875 \Rightarrow 2x^2 + 3x - 0.875 = 0 \Rightarrow 16x^2 + 24x - 7 = 0$.

$x = \dfrac{-24 + \sqrt{576 + 448}}{32} = \dfrac{-24 + 32}{32} = 0.25$.

First increment $= 25\%$.

Solution

Initial salaries: $5, 6, 7$.

After year 1: Sita $= 6$, Gita $= 7.5$, Mita $= 8.4$.

After year 2: Sita $= 6 \times 1.4 = 8.4$, Mita $= 8.4 \times 1.25 = 10.5$.

Gita equals the mean: $G = \dfrac{8.4 + 10.5 + G}{3} \Rightarrow 2G = 18.9 \Rightarrow G = 9.45$.

Hike $= \dfrac{9.45 - 7.5}{7.5} \times 100 = 26\%$.

Solution

Take 100 employees and total salary $= 600$.

Manufacturing: 20 employees share $\dfrac{1}{6} \times 600 = 100$, so average $= 5$.

Others: 80 employees share $500$, so average $= 6.25$.

Ratio $= 5 : 6.25 = 4 : 5$.

Solution

Population in 2022 $= 100000\left(1 - \dfrac{y}{100}\right)\left(1 + \dfrac{x}{100}\right)$ with $x = y + 10$.

It must exceed 100000: $(100 - y)(110 + y) > 10000$

$\Rightarrow 11000 - 10y - y^2 > 10000 \Rightarrow y^2 + 10y - 1000 < 0$.

Root: $y = \dfrac{-10 + \sqrt{4100}}{2} \approx 27.02$, so the largest natural $y = 27$.

The 2021 population is lowest when $y$ is largest: $100000 \times 0.73 = 73000$.

Solution

Let the total stock be $N$. Mangoes $= 0.4N$; let apples $= A$ and bananas $= B$, with $A + B = 0.6N$.

Fruits sold $= 0.2N + 96 + 0.4A = 0.5N \Rightarrow A = 0.75N - 240$, so $B = 240 - 0.15N$.

Conditions: $A \geq 1 \Rightarrow N > 320$; $B \geq 96 \Rightarrow N \leq 960$.

For whole numbers, $N$ must be a multiple of 5 (half the mangoes) and of 4 (for $A$), i.e. a multiple of 20.

Smallest such $N > 320$ is $340$.

Check: $A = 15$, $B = 189$, mangoes $= 136$; sold $= 68 + 96 + 6 = 170 = 50\%$ ✓.

Solution

Let registered voters $= V$; votes cast $= 0.8V$.

One candidate gets $30\%$; the other $70\%$ is split $1 : 2 : 3$, i.e. $\dfrac{70}{6}\%, \dfrac{140}{6}\%, 35\%$.

Winner $= 35\%$, second $= 30\%$; the gap is $5\%$ of votes cast.

$0.05 \times 0.8V = 2512 \Rightarrow 0.04V = 2512 \Rightarrow V = 62800$.

Solution

Let the small box cost $p$ for $w$ grams and the large box cost $2p$ for $W$ grams.

Price per gram of large $= 0.88 \times$ price per gram of small:

$\dfrac{2p}{W} = 0.88 \cdot \dfrac{p}{w} \Rightarrow \dfrac{W}{w} = \dfrac{2}{0.88} \approx 2.2727$.

So the large box has about $127.27\%$ more chocolate. Nearest: $127$.

Solution

Let the loss in year 1 be $r\%$; growth in year 2 $= 5r\%$.

$\left(1 - \dfrac{r}{100}\right)\left(1 + \dfrac{5r}{100}\right) = 1.35$

$(100 - r)(100 + 5r) = 13500 \Rightarrow 5r^2 - 400r + 3500 = 0 \Rightarrow r^2 - 80r + 700 = 0$.

$r = 10$ or $r = 70$. The balance after year 1 exceeds ₹5000, so $r < 50$.

Loss in the first year $= 10\%$.

Solution

Let white balls $= W$ and black balls $= B$, with $W + B = 450$.

Metallic white $=$ metallic black: $0.4W = 0.5B \Rightarrow W = 1.25B$.

$2.25B = 450 \Rightarrow B = 200,\ W = 250$.

Metallic balls $= 100 + 100 = 200$.

Non-metallic $= 450 - 200 = 250$.

Solution

Matches won so far $= 30\%$ of $40 = 12$. Let the remaining matches be $R$.

$12 + 0.6R = 0.5(40 + R) \Rightarrow 0.1R = 8 \Rightarrow R = 80$.

If they win $90\%$ of 80 $= 72$, total wins $= 12 + 72 = 84$.

Solution

Let 1970 populations be $M$ (male) and $F$ (female).

$1.4M + 1.2F = 1.25(M + F) \Rightarrow 0.15M = 0.05F \Rightarrow F = 3M$.

Take $M = 100, F = 300$ (total 400).

1980: $M = 140$, $F = 360$. 1990: $F = 360 \times 1.25 = 450$ and $F = 2M \Rightarrow M = 225$.

1990 total $= 675$. Increase $= \dfrac{675 - 400}{400} \times 100 = 68.75\%$.

Solution

Take 100 people: young $= 28$, old $= 72$.

Literates $= 65$, of which young $= 25\%$ of $65 = 16.25$, so old literates $= 48.75$.

Illiterates $= 35$; old illiterates $= 72 - 48.75 = 23.25$.

Percentage $= \dfrac{23.25}{35} \times 100 \approx 66.4\% \approx 66$.

Solution

Rice in April = ₹450; a $20\%$ rise adds ₹90.

Total extra spend = ₹150, so wheat's extra = ₹60, which is $12\%$ of April's wheat spend.

April wheat $= \dfrac{60}{0.12} = 500$, i.e. ₹500.

May wheat $= 500 \times 1.12 = 560$, i.e. ₹560.

Solution

Let maximum marks $= M$ and Ramesh's marks $= R$.

$0.52M = R - 23$ and $0.64M = R + 34$.

Subtracting: $0.12M = 57 \Rightarrow M = 475$.

Geeta $= 84\%$ of $475 = 399$.