CAT 2022 Slot 2 QA paper with solutions

Solution

Let the capital be $C$. The remaining part is $1 - \dfrac{1}{5} - \dfrac{1}{3} = \dfrac{7}{15}$.

Annual interest $= C\left(\dfrac{6}{5} + \dfrac{10}{3} + \dfrac{7}{15}\right)\% = C \times \dfrac{18 + 50 + 7}{15}\% = 5\%$ of $C$.

To earn $100\%$ of $C$: $\dfrac{100}{5} = 20$ years.

Solution

Replacing $a_1$ by $6a_1$ adds $5a_1$ to the sum: $5a_1 = (400 - 300)N \Rightarrow a_1 = 20N$.

In a non-decreasing sequence the first term is at most the average: $20N \leq 300 \Rightarrow N \leq 15$.

$N = 1$ is impossible (then $a_1$ would be both 20 and 300).

So $N = 2, 3, \dots, 15$: $14$ possible values of $a_1$.

Solution

$A^B = 1$ when $A = 1$, or $B = 0$ (with $A \neq 0$), or $A = -1$ with $B$ even.

$x^2 - 10 = 1$: no integer $x$.

$x^2 - 3x - 10 = 0$: $x = 5$ or $-2$ (bases 15 and $-6$) ✓.

$x^2 - 10 = -1$: $x = \pm 3$; exponents $-10$ and $8$ are even ✓.

Total $= 4$ solutions.

Solution

Yearly savings target $= 12 \times 550 = 6600$.

First 9 months: $9 \times (4000 - 3500) = 4500$.

Needed in the last 3 months $= 2100$, i.e. 700 per month.

Earnings $= 3700 + 700 = 4400$.

Solution

$AD = AB\sin B = AB\sin\theta$ and $BE = AB\sin A = AB\sin 45^\circ$.

$\dfrac{AD}{BE} = \dfrac{\sin\theta}{1/\sqrt{2}} = \sqrt{2}\sin\theta$.

Solution

$f(x) \geq 0$ everywhere and $f(2) = 0$, so 2 is a double root: $f(x) = a(x - 2)^2$.

$f(4) = 4a = 6 \Rightarrow a = 1.5$.

$f(-2) = 1.5 \times 16 = 24$.

Solution

Let the roots be $r, -r, s$.

Sum of pairwise products: $-r^2 + rs - rs = -r^2 = -\dfrac{10}{5} \Rightarrow r^2 = 2$.

Product: $-r^2 s = -\dfrac{9}{5} \Rightarrow s = \dfrac{9}{10}$.

Sum of roots: $s = -\dfrac{c}{5} \Rightarrow c = -\dfrac{9}{2}$.

Solution

Let the speeds be $v$ and $v + 6$. After 2 h, the distances are $2v$ and $2v + 12$ at right angles.

$(2v)^2 + (2v + 12)^2 = 60^2 \Rightarrow v^2 + (v + 6)^2 = 900 \Rightarrow v^2 + 6v - 432 = 0$.

$v = 18$ km/h.

Solution

$f(x) + f(x - 1) = 1$ for all integers means $f$ alternates: $f(x + 2) = f(x)$.

$x^2 - x$ is always even, so $f(\text{even}) = 5$ and $f(\text{odd}) = 1 - 5 = -4$.

$f(g(5)) = f(25) = -4$. $g(f(5)) = g(-4) = 16$.

Sum $= -4 + 16 = 12$.

Solution

Let $c$ = correct, $w$ = wrong, $u$ = unattempted.

$c + w + u = 75$ and $3c - w + u = 97$. Subtracting: $2c - 2w = 22 \Rightarrow w = c - 11$.

Then $u = 75 - c - w = 86 - 2c$.

Unattempted exceeds attempted: $86 - 2c > 2c - 11 \Rightarrow c < 24.25$.

Maximum correct $= 24$.

Solution

Let A have $n$ sides and B have $2n$.

$\dfrac{180 - 360/n}{180 - 180/n} = \dfrac{3}{4} \Rightarrow 720 - \dfrac{1440}{n} = 540 - \dfrac{540}{n} \Rightarrow \dfrac{900}{n} = 180$.

$n = 5$, so B has $10$ sides.

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

On day $n$ the count is multiplied by $1 + \dfrac{1}{n} = \dfrac{n + 1}{n}$.

After day $m$: $100 \times \dfrac{3}{2} \times \dfrac{4}{3} \times \dots \times \dfrac{m + 1}{m} = 100 \times \dfrac{m + 1}{2}$.

$50(m + 1) = 1000 \Rightarrow m = 19$.

Solution

4-digit numbers above 2000: first digit 2–5: $4 \times 5 \times 4 \times 3 = 240$.

5-digit: $5 \times 5 \times 4 \times 3 \times 2 = 600$. 6-digit: $5 \times 5! = 600$.

Total $= 1440$.

Solution

$(n!)!$ divides $15000!$ exactly when $n! \leq 15000$.

$7! = 5040 \leq 15000$ but $8! = 40320 > 15000$.

Largest $n = 7$.

Solution

Times $5 : 8 : 10$ means rates $\dfrac{1}{5} : \dfrac{1}{8} : \dfrac{1}{10} = 8 : 5 : 4$ units per hour.

Together: 17 units/hour. Job $= 4 \times 8 \times 17 = 544$ units.

Anu and Tanu: $13 \times 6 \times \dfrac{20}{3} = 520$ units.

Remaining 24 units at 4 units/hour: Manu needs $6$ hours.

Solution

Take each container's capacity as 2: container 1 has 1 syrup, container 2 has 1 milk.

Half of container 1 to container 2: C2 $= 0.5$ syrup $+ 1$ milk.

Half of C2 back: C1 $= 0.75$ syrup $+ 0.5$ milk; C2 $= 0.25$ syrup $+ 0.5$ milk.

Half of C1 to C2: adds $0.375$ syrup $+ 0.25$ milk, so C2 $= 0.625$ syrup $+ 0.75$ milk.

Ratio $= 0.625 : 0.75 = 5 : 6$.

Solution

$A_n = \dfrac{n}{2}[6 + 4(n - 1)] = 2n^2 + n$.

$\sum_{n=1}^{25} A_n = 2 \times \dfrac{25 \times 26 \times 51}{6} + \dfrac{25 \times 26}{2} = 11050 + 325 = 11375$.

$\dfrac{11375}{25} = 455$.

Solution

Let $L = \log_2 n$; then $\log_4 n = \dfrac{L}{2}$.

$\dfrac{4 - L}{3 - L/2} = \dfrac{2(4 - L)}{6 - L} < 0 \Rightarrow 4 < L < 6 \Rightarrow 16 < n < 64$.

$n = 17, \dots, 63$: $47$ values.

Solution

$a = 6 - 2b \geq 0 \Rightarrow 0 \leq b \leq 3$.

$a + b = 6 - b$, which ranges from 3 (at $b = 3$) to 6 (at $b = 0$).

Average $= \dfrac{3 + 6}{2} = 4.5$.

Solution

Total marks $= 5 \times 38 = 190$. Three students score above 32; Amit and one other ($X$) score at most 32, with Amit $< X \leq 32$.

The top three are distinct and at most 50, so their sum $\leq 50 + 49 + 48 = 147$; hence Amit $+ X \geq 43$.

Also Amit $+ X \leq 31 + 32 = 63$.

Highest Amit: $31$ (with $X = 32$, top three summing to 127, e.g. 40, 43, 44).

Lowest Amit: $43 - 32 = 11$.

Difference $= 31 - 11 = 20$.

Solution

Triangles ADC and ABD share the height from A, so $DC = \dfrac{1}{2}BD$: $BD = 2$, $DC = 1$.

$AD^2 = 3^2 + 2^2 - 2 \times 3 \times 2\cos 60^\circ = 7$.

$AD = \sqrt{7}$.