CAT 2020 Slot 3 QA paper with solutions

Solution

Take 1 L of A and 3 L of B; doubling with A adds 4 L of A. Totals: A $= 5$ L, B $= 3$ L.

$5(0.6) + 3b = 8(0.72) \Rightarrow 3 + 3b = 5.76 \Rightarrow b = 0.92$.

B has $92\%$ alcohol.

Solution

Total over $n + 2$ innings $= 29(n + 2)$; the last two add $38 + 15 = 53$.

First $n$ innings: $29n + 58 - 53 = 30n \Rightarrow n = 5$, total $= 150$.

Each of these is less than 38 (at most 37). To make the lowest score as small as possible, let the other four be 37:

$x = 150 - 4 \times 37 = 2$.

Solution

Real roots: $m^2 \geq 8n$ and $4n^2 \geq 4m \Rightarrow n^2 \geq m$.

$n = 1$: $m \leq 1$ but $m^2 \geq 8$ ✗.

$n = 2$: $m \leq 4$ and $m^2 \geq 16 \Rightarrow m = 4$; $m + n = 6$.

Larger $n$ gives a larger sum (e.g. $n = 3$ needs $m \geq 5$). Smallest $m + n = 6$.

Solution

In 60 days, 140 people built 1.5 km: $8400$ person-days for 1.5 km.

The remaining 4.5 km needs $3 \times 8400 = 25200$ person-days in 140 days, i.e. 180 people.

Additional people $= 180 - 140 = 40$.

Solution

$x_{m+1} = x_m - (m + 1)$.

$x_{100} = x_1 - (2 + 3 + \dots + 100) = -1 - 5049 = -5050$.

Solution

$\log_2 a = \dfrac{1}{3} \Rightarrow \log_a 2 = 3$.

$A = \log_a 30 = 3 + \log_a 3 + \log_a 5$ and $-B = \log_a 5 - \log_a 3$.

Subtracting: $A + B - 3 = 2\log_a 3 \Rightarrow \log_a 3 = \dfrac{A + B - 3}{2}$.

$\log_3 a = \dfrac{2}{A + B - 3}$.

Solution

Let Tom $= t$. Dick $= 3t$, Harry $= 6t$; average $= \dfrac{10t}{3}$.

$3t = \dfrac{10t}{3} - 1 \Rightarrow 9t = 10t - 3 \Rightarrow t = 3$.

Harry $= 6 \times 3 = 18$.

Solution

$\dfrac{d}{35} - \dfrac{d}{40} = \dfrac{6}{60} \Rightarrow \dfrac{d}{280} = 0.1 \Rightarrow d = 28$ km; usual time $= 42$ min.

She covers two-thirds in 14 min and stops 8 min: 22 min used, 20 min left.

Remaining $\dfrac{28}{3}$ km in $\dfrac{1}{3}$ h: speed $= 28$ km/h.

Solution

$0.2 < \dfrac{m}{20} < 0.5 \Rightarrow 4 < m < 10$.

$0.2 < \dfrac{n}{11} < 0.5 \Rightarrow 2.2 < n < 5.5$, and $n$ even, so $n = 4$.

$0.2 < \dfrac{4}{m} < 0.5 \Rightarrow 8 < m < 20$. With $4 < m < 10$: $m = 9$.

$m - 2n = 9 - 8 = 1$.

Solution

3-digit numbers: 900. With all digits distinct: $9 \times 9 \times 8 = 648$.

At least one repeated digit $= 900 - 648 = 252$.

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 $475 = 399$.

Solution

Let $14^a = 36^b = 84^c = k$. Then $\dfrac{1}{a} = \dfrac{\ln 14}{\ln k}$, $\dfrac{1}{c} = \dfrac{\ln 84}{\ln k}$, $b = \dfrac{\ln k}{\ln 36}$.

$\dfrac{1}{c} - \dfrac{1}{a} = \dfrac{\ln 6}{\ln k}$ and $6b = \dfrac{6\ln k}{2\ln 6} = \dfrac{3\ln k}{\ln 6}$.

Product $= 3$.

Solution

Half-yearly rate $= 5\%$; $1.5$ years $= 3$ periods.

$P \times 1.05^3 = 18522 \Rightarrow P \times 1.157625 = 18522 \Rightarrow P = 16000$.

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 $= 1.15 \times 35 = 40.25$.

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

$p \approx 25\%$.

Solution

The diagonals of a parallelogram bisect each other, so the midpoint of $(2, 1)$ and $(-3, -4)$, which is $(-0.5, -1.5)$, lies on the other diagonal.

$-0.5 + 9(-1.5) + c = 0 \Rightarrow c = 14$.

Solution

By 10:30, train A has covered $1.5 \times 40 = 60$ km; 30 km remain between them.

Closing speed $= 40 + 20 = 60$ km/h, so they meet 30 min later, at $11{:}00$ am.

Solution

$x \in \{3, \dots, 9\}$ and $y \in \{15, \dots, 22\}$, so $N = x + y$ ranges over 18 to 31.

$N > 25$: $N = 26, 27, \dots, 31$: $6$ values.

Solution

A unique solution needs the determinant to be non-zero: $k \cdot k - 1 \cdot 4 \neq 0$.

$k^2 \neq 4 \Rightarrow |k| \neq 2$.

Solution

Count numbers divisible by 2, 5 or 7 using inclusion–exclusion:

$60 + 24 + 17 - 12 - 8 - 3 + 1 = 79$.

Numbers divisible by none $= 120 - 79 = 41$.

Solution

$4^{2017} = 2^{4034}$, which has 4035 divisors.

Each pair $a \leq b$ with $ab = 2^{4034}$ matches a divisor $a \leq 2^{2017}$: $\dfrac{4035 + 1}{2} = 2018$ pairs.

Solution

Anil's lap $= \dfrac{3}{15} = 0.2$ h; Sunil's lap $= \dfrac{3}{10} = 0.3$ h.

They are next together at the start after $\text{LCM}(0.2, 0.3) = 0.6$ h.

Ravi runs $8 \times 0.6 = 4.8$ km.

Solution

BC is perpendicular to DC (and AB), so the height is 4.

Drop a perpendicular from D to AB: with $\angle BAD = 45^\circ$, the horizontal part is also 4.

$AB = DC + 4 = 9$. Area $= \dfrac{1}{2}(9 + 5) \times 4 = 28$.

Solution

For $0 \leq x \leq 2$: $y = |x - 2| + 4 = 6 - x$.

Area $= \int_0^2 (6 - x)\,dx = 12 - 2 = 10$ (a trapezium with parallel sides 6 and 4, width 2).

Solution

$f(0) = f(0)^2 \Rightarrow f(0) = 1$ (since $f(5) \neq 0$).

$f(10) = f(5)^2 = 16$, and $f(10)f(-10) = f(0) = 1 \Rightarrow f(-10) = \dfrac{1}{16}$.

$f(10) - f(-10) = 16 - 0.0625 = 15.9375$.

Solution

Numerator $= 2 \times 4 \times 8 \times 16 = 2^{10} = 1024$.

$\log_2 4 = 2$, $\log_4 8 = \dfrac{3}{2}$, $\log_8 16 = \dfrac{4}{3}$.

Denominator $= 2^2 \times \left(\dfrac{3}{2}\right)^3 \times \left(\dfrac{4}{3}\right)^4 = 4 \times \dfrac{27}{8} \times \dfrac{256}{81} = \dfrac{128}{3}$.

Value $= 1024 \times \dfrac{3}{128} = 24$.

Solution

The centre lies on $x = 2$ (perpendicular bisector of $(0, 0)$ and $(4, 0)$). Let it be $(2, k)$.

$4 + k^2 = 1 + (k - 9)^2 \Rightarrow 18k = 78 \Rightarrow k = \dfrac{13}{3}$.

$R^2 = 4 + \dfrac{169}{9} = \dfrac{205}{9}$, so area $= \dfrac{205\pi}{9}$.