CAT 2020 Slot 1 QA paper with solutions

Solution

Take $\log_2$ of both sides: $Y^2 \log_3 5 = \log_2 3 \cdot \log_2 5$.

$\dfrac{\log_2 5}{\log_3 5} = \log_2 3$, so $Y^2 = (\log_2 3)^2$.

$Y$ is negative: $Y = -\log_2 3 = \log_2\dfrac{1}{3}$.

Solution

Work backwards. The fifth child gets half plus one and nothing is left: $\dfrac{R}{2} - 1 = 0 \Rightarrow R = 2$.

Each earlier stock $= 2(\text{next} + 1)$: $2 \to 6 \to 14 \to 30 \to 62$.

Initial stock $= 62$.

Solution

Products 3, 4, 5 or 6 with non-zero digits:

3: $\{1, 1, 3\}$ (3 numbers). 4: $\{1, 1, 4\}$ (3), $\{1, 2, 2\}$ (3). 5: $\{1, 1, 5\}$ (3). 6: $\{1, 1, 6\}$ (3), $\{1, 2, 3\}$ (6).

Total $= 21$.

Solution

Left side: $2^x + 2^{-x} \geq 2$, equal to 2 only at $x = 0$.

Right side: $2 - (x - 2)^2 \leq 2$, equal to 2 only at $x = 2$.

They can only be equal if both are 2, which needs $x = 0$ and $x = 2$ at once. No real solutions: $0$.

Solution

$A^B = 1$: base 1, exponent 0, or base $-1$ with even exponent.

Base 1: $x^2 - 7x + 10 = 0 \Rightarrow x = 2, 5$.

Exponent 0: $x^2 - 13x + 42 = 0 \Rightarrow x = 6, 7$ (bases 5 and 11, non-zero) ✓.

Base $-1$: $x^2 - 7x + 12 = 0 \Rightarrow x = 3, 4$; exponents 12 and 6 are even ✓.

Total $= 6$.

Solution

The top piece is a cone of height 9, similar with ratio $\dfrac{1}{3}$: volume $= \dfrac{V}{27}$. The bottom piece is $\dfrac{26V}{27}$.

Difference $= \dfrac{25V}{27} = 225 \Rightarrow V = 243$ cc.

Solution

The region is $|x| \leq 1 + y$ for $0 \leq y \leq 1$: a trapezium.

Width at $y = 0$ is 2 and at $y = 1$ is 4.

Area $= \dfrac{1}{2}(2 + 4) \times 1 = 3$.

Solution

Painted (circle) $= A$, unpainted $= \dfrac{2}{3}A$, total $= \dfrac{5}{3}A = 135 \Rightarrow A = 81$, so $r = \dfrac{9}{\sqrt{\pi}}$.

The circle touches two opposite sides, so one side $= 2r = \dfrac{18}{\sqrt{\pi}}$; the other $= \dfrac{135\sqrt{\pi}}{18} = 7.5\sqrt{\pi}$.

Perimeter $= \dfrac{36}{\sqrt{\pi}} + 15\sqrt{\pi} = 3\sqrt{\pi}\left(5 + \dfrac{12}{\pi}\right)$.

Solution

Side $= \sqrt{6^2 + 8^2} = 10$; area $= \dfrac{12 \times 16}{2} = 96$.

Inradius $= \dfrac{\text{height}}{2} = \dfrac{96/10}{2} = 4.8$.

Ratio $= \dfrac{\pi(4.8)^2}{96} = \dfrac{23.04\pi}{96} = \dfrac{6\pi}{25}$.

Solution

100 students in 12 months: if every month had at most 8, the total would be at most 96.

So some month has at least 9, and 9 is achievable (e.g. 4 months of 9 and 8 months of 8).

Smallest $x_0 = 9$.

Solution

After meeting, the times are 45 min and 20 min. $\dfrac{v_2}{v_1} = \sqrt{\dfrac{45}{20}} = \dfrac{3}{2}$.

Car 2 speed $= 60 \times 1.5 = 90$ km/h.

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

Dye $= 16$ L, water $= 24$ L. To get $2 : 5$, water becomes 40 L (total 56 L).

Remove one-fourth: dye $= 12$, water $= 30$.

Add $d$ L dye: $\dfrac{12 + d}{30} = \dfrac{2}{3} \Rightarrow d = 8$ L.

Solution

$4096 = 64^2$, so $x = 64^{2(7 + 4\sqrt{3})}$. We need $x^p = 64$, i.e. $p = \dfrac{1}{2(7 + 4\sqrt{3})}$.

Since $(7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 1$: $p = \dfrac{7 - 4\sqrt{3}}{2} = \dfrac{7}{2} - 2\sqrt{3}$.

So $64 = \dfrac{x^{7/2}}{x^{2\sqrt{3}}}$.

Solution

Weight $\propto$ volume $\times$ density: A : B : C $= 3 \times 5 : 4 \times 2 : 7 \times 6 = 15 : 8 : 42$ (total 65).

Metal C in 130 kg $= \dfrac{42}{65} \times 130 = 84$ kg.

Solution

Let $t = x + \dfrac{1}{x}$: $t^2 - 3t + 2 = 0 \Rightarrow t = 1$ or $2$.

$x + \dfrac{1}{x} = 1$ has no real solution (for real $x$, $|t| \geq 2$).

$x + \dfrac{1}{x} = 2 \Rightarrow x = 1$.

Number of distinct real roots $= 1$.

Solution

$\log_6\sqrt{5} = \dfrac{1}{2}\log_6 5$. Using natural logs:

$\dfrac{\ln 5}{\ln 4} = \dfrac{\ln y}{\ln 4} \cdot \dfrac{\ln 5}{2\ln 6} \Rightarrow \ln y = 2\ln 6$.

$y = 36$.

Solution

The difference in time is 35 min: $\dfrac{d}{8} - \dfrac{d}{15} = \dfrac{35}{60} \Rightarrow \dfrac{7d}{120} = \dfrac{7}{12} \Rightarrow d = 10$ km.

From 9:10 to 10:00 is 50 min: speed $= \dfrac{10}{50/60} = 12$ km/h.

Solution

At one-third speed the trip takes $3T$, so it is late by $2T = 30$ min: $T = 15$ min.

Return: 5 min at usual speed covers one-third of the route, then a 4-min stop. Time left $= 15 - 9 = 6$ min.

The remaining two-thirds normally takes 10 min, so speed must be $\dfrac{10}{6}$ times usual.

Increase $\approx 66.7\% \approx 67\%$.

Solution

$\overline{aabb} = 1100a + 11b$ with $a = 1$–9 and $b \in \{0, 2, 4, 6, 8\}$ (even).

All combinations occur equally, so mean $= 1100 \times 5 + 11 \times 4 = 5544$.

Solution

Let the train speed be $v$ km/h and length $L$.

$L = (v - 2) \times \dfrac{90}{3600} = (v - 4) \times \dfrac{100}{3600} \Rightarrow 90v - 180 = 100v - 400 \Rightarrow v = 22$.

$L = 20 \times \dfrac{90}{3600} = 0.5$ km.

Time to pass a post $= \dfrac{0.5}{22}$ h $\approx 81.8$ s, nearest 82.

Solution

$\dfrac{ab}{bc} = \dfrac{a}{c} = \dfrac{432}{96} = \dfrac{9}{2}$, so $a = 4.5c$, meaning $c$ is even.

$c = 2$: $b = 48$, $a = 9$, sum 59. $c = 4$: $b = 24$, $a = 18$, sum 46. $c = 6$: $b = 16$, $a = 27$, sum 49. $c = 8$: $b = 12$, $a = 36$, sum 56.

Smallest $= 46$.

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

Let $t$ be years after Veeru invests.

Veeru: $10000 + 500t$. Joy: $8000 + 800(t - 2)$.

$10000 + 500t = 6400 + 800t \Rightarrow 300t = 3600 \Rightarrow t = 12$.

Solution

$f(5 + x) = f(5 - x)$ means the graph is symmetric about $x = 5$.

Roots come in pairs $5 + a$ and $5 - a$, each pair adding to 10.

Four roots: sum $= 2 \times 10 = 20$.

Solution

$A + \dfrac{B + C}{2} = 5 \Rightarrow 2A + B + C = 10$.

$B + \dfrac{A + C}{2} = 7 \Rightarrow A + 2B + C = 14$.

Subtracting: $B - A = 4$. Then $2A + (A + 4) + C = 10 \Rightarrow 3A + C = 6$.

Positive integers: $A = 1$, $C = 3$, $B = 5$.

$A + B = 6$.