CAT 2021 Slot 2 QA paper with solutions

Solution

$0.25 \leq 2^{n+2} \leq 200 \Rightarrow -2 \leq n + 2 \leq 7 \Rightarrow -4 \leq n \leq 5$.

$3 + 3^{n+1}$ is an integer only when $n + 1 \geq 0$, i.e. $n = -1, 0, \dots, 5$.

Number of integer values $= 7$.

Solution

Let the numbers be $y - d, y, y + d$. Then $xyz = y(y^2 - d^2) = 5 \times 3y \Rightarrow y^2 - d^2 = 15$.

$(y - d)(y + d) = 15 = 1 \times 15$ or $3 \times 5$. Since $d > 2$: $y - d = 1$, $y + d = 15$, so $d = 7$.

$z - x = 2d = 14$.

Solution

Let the number be $ABCD$. $A + B + C = 14$ and $B + C + D = 15 \Rightarrow D = A + 1$.

$C = D + 4 = A + 5 \leq 9 \Rightarrow A \leq 4$, and $B = 14 - A - C = 9 - 2A$.

The largest number takes $A = 4$: $C = 9$, $D = 5$, $B = 1$.

Answer: $4195$.

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

Balloons: give each child 4 first, then share 3 freely: $\binom{5}{2} = 10$.

Pencils: give each child 1 first, then share 3 freely: $\binom{5}{2} = 10$.

Erasers: share 3 freely: $\binom{5}{2} = 10$.

Total $= 10 \times 10 \times 10 = 1000$.

Solution

Relative speed $= 8v$ (speeds $5v$ and $3v$).

From front-front to front of A passing rear of B, relative distance $=$ length of B: $L_B = 8v \times 46$.

From front-front to rear-rear $= L_A + L_B = 8v \times 115$, so $L_A = 8v \times 69$.

$L_A : L_B = 69 : 46 = 3 : 2$.

Solution

Coefficients are rational, so $2 - \sqrt{3}$ is the other root: sum $= 4$, product $= 1$.

$\dfrac{b}{a} = 4$ and $\dfrac{c}{a} = 1$, so $b = 4a$, $c = a$.

$b = c^3 \Rightarrow 4a = a^3 \Rightarrow a^2 = 4 \Rightarrow |a| = 2$.

Solution

Let the capacity be $C$. After two replacements, milk fraction $= \left(1 - \dfrac{9}{C}\right)^2 = \dfrac{16}{25}$.

$1 - \dfrac{9}{C} = \dfrac{4}{5} \Rightarrow C = 45$ litres.

Solution

$d_1 d_2 = 24$ and $d_1^2 + d_2^2 = 4 \times 25 = 100$.

$(d_1 + d_2)^2 = 148 \Rightarrow d_1 + d_2 = 2\sqrt{37}$; $(d_1 - d_2)^2 = 52 \Rightarrow d_1 - d_2 = 2\sqrt{13}$.

Longer diagonal $= \sqrt{37} + \sqrt{13}$.

Solution

$\log_2[\dots] = 2 \Rightarrow 3 + \log_3\{\dots\} = 4 \Rightarrow \log_3\{4 + \log_4(x - 1)\} = 1$.

$4 + \log_4(x - 1) = 3 \Rightarrow \log_4(x - 1) = -1 \Rightarrow x - 1 = \dfrac{1}{4}$.

$x = \dfrac{5}{4}$, so $4x = 5$.

Solution

ABPD is a parallelogram, so $DP = AB = a$, and P is the midpoint, so $PC = a$, $CD = 2a$. Let the height be $h$.

Parallelogram $= ah$; triangle BPC $= \dfrac{1}{2}ah$. Difference $= \dfrac{1}{2}ah = 10 \Rightarrow ah = 20$.

Trapezium $= \dfrac{1}{2}(a + 2a)h = 1.5ah = 30$.

Solution

Let $t = x^2 + 2x + 4 = (x + 1)^2 + 3 \geq 3$. The denominator is $2t + 1$.

$f = \dfrac{t}{2t + 1} = \dfrac{1}{2} - \dfrac{1}{2(2t + 1)}$, which increases with $t$.

At $t = 3$: $\dfrac{3}{7}$. As $t \to \infty$: approaches $\dfrac{1}{2}$ but never reaches it.

Range $= \left[\dfrac{3}{7}, \dfrac{1}{2}\right)$.

Solution

$(-1)^{n+1}x_n = (n^2 + 2n) - ((n - 1)^2 + 2(n - 1)) = 2n + 1$.

$x_{49} = 99$ (odd index, plus sign). $-x_{50} = 101 \Rightarrow x_{50} = -101$.

$x_{49} + x_{50} = -2$.

Solution

$|3x - 20| + |3x - 40| \geq |(3x - 20) - (3x - 40)| = 20$, with equality when $20 \leq 3x \leq 40$.

So the equation holds exactly for $\dfrac{20}{3} \leq x \leq \dfrac{40}{3}$ (about 6.67 to 13.33).

The interval in the options that lies fully inside is $7 < x < 12$.

Solution

Anil works all 24 days: $\dfrac{24}{60} = 0.4$ of the job. Bimal works 14 days: $\dfrac{14}{84} = \dfrac{1}{6}$.

Charu $= 1 - 0.4 - \dfrac{1}{6} = \dfrac{13}{30}$.

Charu's share $= \dfrac{13}{30} \times 21000 = 9100$.

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

Let him have played $n$ matches with $g$ goals.

$\dfrac{g + 1}{n + 10} = 0.15$ and $\dfrac{g + 2}{n + 10} = 0.2$.

Subtracting: $\dfrac{1}{n + 10} = 0.05 \Rightarrow n + 10 = 20 \Rightarrow n = 10$.

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

Write $d = x - y$. First: $x(d - 1) = 22$. Second: $y(1 - d) = 34$, i.e. $-y(d - 1) = 34$.

Adding: $(x - y)(d - 1) = 56 \Rightarrow d(d - 1) = 56$.

$d = 8$ (as $x > y$).

Solution

$\dfrac{[ADE]}{[ABC]} = \dfrac{AD}{AB} \times \dfrac{AE}{AC} = \dfrac{2}{3} \times \dfrac{2}{5} = \dfrac{4}{15}$.

$[ABC] = 8 \times \dfrac{15}{4} = 30$.

Solution

Let the total investment be $T$.

Anil's share falls by $70\% \times 3\%$ of $T$: $0.021T = 420 \Rightarrow T = 20000$, so Anil invested 14000.

Chintu's share rises by $2\%$ of his investment $= 80 \Rightarrow$ Chintu invested 4000.

Bobby $= 20000 - 14000 - 4000 = 2000$.

Solution

Let A fill $a$ and B drain $b$ per hour.

Case 1 (A 8 h, B 7 h): $8a - 7b = 1$. Case 2 (A 4 h, B 2 h): $4a - 2b = 1 \Rightarrow b = 2a - 0.5$.

$8a - 14a + 3.5 = 1 \Rightarrow a = \dfrac{5}{12}$.

A alone takes $\dfrac{12}{5}$ hours $= 144$ minutes.