CAT 2022 Slot 3 QA paper with solutions

Solution

Let $t = \dfrac{x}{y}$ (any non-zero real). $c = 16t + \dfrac{49}{t}$.

$t > 0$: by AM–GM, $c \geq 2\sqrt{16 \times 49} = 56$. $t < 0$: $c \leq -56$.

So $c$ cannot lie in $(-56, 56)$; of the options, $-50$ is impossible.

Solution

$2x^2 + kx + 5$ has no real roots: $k^2 < 40 \Rightarrow -6 \leq k \leq 6$.

$x^2 + (k - 5)x + 1$ has two distinct real roots: $(k - 5)^2 > 4 \Rightarrow k < 3$ or $k > 7$.

Together: $k = -6, -5, \dots, 2$: $9$ values.

Solution

$\left(\dfrac{4a}{b}\right)^{6x - y} = \left(\dfrac{2a}{b}\right)^{y - 6x}$ for all $a, b$ forces $6x - y = 0$, so $y = 6x$.

$\dfrac{875}{2401} = \dfrac{5^3 \cdot 7}{7^4} = \left(\dfrac{7}{5}\right)^{-3}$, and $\left(\sqrt{\dfrac{7}{5}}\right)^{3x - y} = \left(\dfrac{7}{5}\right)^{\frac{3x - y}{2}}$.

$\dfrac{3x - y}{2} = -3 \Rightarrow 3x - 6x = -6 \Rightarrow x = 2$, $y = 12$.

$x + y = 14$.

Solution

Two smallest sum to 28 and two largest sum to 56. To make the middle two as large as possible:

Second smallest as large as possible: $13, 15$. Fifth number as large as possible: $27, 29$.

Middle two must lie between 15 and 27: take $25, 26$.

Sum $= 28 + 51 + 56 = 135$; average $= 22.5$.

Solution

O is the centroid, and E, D are midpoints of AB and AC, so $ED \parallel BC$ and $ED = \dfrac{BC}{2}$.

Triangle BOC has area $\dfrac{1}{3} \times 108 = 36$, and triangle EOD is similar to it with ratio $\dfrac{1}{2}$.

Area of EOD $= \dfrac{36}{4} = 9$.

Solution

With integer coefficients, $3 - 2\sqrt{2}$ is also a root: sum $= 6$, product $= 1$, so $b = -6a$, $c = a$.

Similarly $4 \pm 2\sqrt{3}$: sum $= 8$, product $= 4$, so $m = -8a$, $n = 4a$.

$\dfrac{b}{m} + \dfrac{c - 2b}{n} = \dfrac{6}{8} + \dfrac{13a}{4a} = \dfrac{3}{4} + \dfrac{13}{4} = 4$.

Solution

Work is proportional to person-hours.

$35\%$ took $70N$ person-hours; $65\%$ took $140(N - 10)$.

$\dfrac{70N}{35} = \dfrac{140(N - 10)}{65} \Rightarrow 130N = 140N - 1400 \Rightarrow N = 140$.

Solution

Cup after the first transfer: 500 water + 150 milk (650 cc).

150 cc of this goes to the glass: water $= 150 \times \dfrac{500}{650} = \dfrac{1500}{13}$, milk $= \dfrac{450}{13}$.

Water in glass $= \dfrac{1500}{13}$. Milk left in cup $= 150 - \dfrac{450}{13} = \dfrac{1500}{13}$.

Ratio $= 1 : 1$ (the volumes swapped are always equal).

Solution

Interest from A $= 8000 \times 5.5\% = 440$; from B $= 5000 \times 5.6\% = 280$.

Total required $= 5\%$ of 20000 $= 1000$, so C gives $280$ on 7000: $x = 4\%$.

All 20000 in C: $20000 \times 4\% = 800$.

Solution

Let the faster car's speed be $a$. Gap $D = 1.5(a + 60) = 10.5(a - 60)$.

$1.5a + 90 = 10.5a - 630 \Rightarrow a = 80$.

In 1.5 h the slower car covers $60 \times 1.5 = 90$ km.

Solution

Distinct arrangements of 1, 1, 2, 4: $\dfrac{4!}{2!} = 12$.

By symmetry each place has average digit $\dfrac{1 + 1 + 2 + 4}{4} = 2$.

Mean $= 2222$.

Solution

Each side must be shorter than the sum of the other three.

$x < 1 + 2 + 4 = 7$ and $4 < 1 + 2 + x \Rightarrow x > 1$.

$x = 2, 3, 4, 5, 6$: $5$ possible lengths.

Solution

Terms are $38 + 17k$. First 3-digit term: $k = 4$, giving 106. Last: $38 + 17k \leq 999 \Rightarrow k = 56$, giving 990.

These form an AP, so the average $= \dfrac{106 + 990}{2} = 548$.

Solution

Let section B have $n$ students and A have $n - 10$.

Combined average $= \dfrac{32(n - 10) + 60n}{2n - 10} = \dfrac{46n - 160}{n - 5} = 46 + \dfrac{70}{n - 5}$.

So $n - 5$ divides 70, and A $\geq 1$ means $n - 5 \geq 6$: $n - 5 \in \{7, 10, 14, 35, 70\}$.

A $= (n - 5) - 5 \in \{2, 5, 9, 30, 65\}$.

Difference $= 65 - 2 = 63$.

Solution

For $x < r$: $f(x) = r$ and $f(r) = 2r - r = r$, so $f(f(x)) = f(x)$ ✓.

For $x \geq r$: $f(x) = 2x - r \geq r$, so $f(f(x)) = 2(2x - r) - r = 4x - 3r$. This equals $2x - r$ only when $x = r$.

So the equation holds exactly for $x \leq r$.

Solution

A lies on the circle with diameter BC, so $\angle BAC = 90^\circ$ and $BC = 8\sqrt{2}$; circle 1 has radius $4\sqrt{2}$, circle 2 has radius 8.

On A's side of BC, the half of circle 1 lies entirely within circle 2: area $= \dfrac{1}{2}\pi(4\sqrt{2})^2 = 16\pi$.

On the other side, the overlap is circle 2's minor segment over chord BC: $\dfrac{90}{360}\pi(64) - \dfrac{1}{2}(8)(8) = 16\pi - 32$.

Total $= 32\pi - 32 = 32(\pi - 1)$.

Solution

LCM of 9, 10, 12, 25 $= 900$, so $N = 900k + 4 < 5000$.

$N$ divisible by 11: $900 \equiv 9 \pmod{11}$, so $9k + 4 \equiv 0 \Rightarrow k \equiv 2 \pmod{11}$, giving $k = 2$, $N = 1804$.

Teams of 12: $\left\lfloor \dfrac{1804}{12} \right\rfloor = 150$.

Solution

Let $t = 3 - x > 0$. Then $x^2 - 6x + 10 = (x - 3)^2 + 1 = t^2 + 1$.

Expression $= \dfrac{t^2 + 1}{t} = t + \dfrac{1}{t} \geq 2$, with equality at $t = 1$ ($x = 2$).

Minimum $= 2$.

Solution

Let $x, y, z$ be the numbers of 100, 250 and 500 cheques.

$x + y + z = 100$ and $100x + 250y + 500z = 15250$.

Subtract $100 \times$ the first: $150y + 400z = 5250 \Rightarrow 3y + 8z = 105$.

Largest $z$ with $y$ a whole number: $z = 12$, $y = 3$ (then $x = 85$). $z = 13$ gives $3y = 1$ ✗.

Maximum $= 12$.

Solution

Let the escalator speed be $e$ and walking speed $w$. Length $L = 30(e + w) = 20(e + 2w)$.

$10e = 10w \Rightarrow e = w$, so $L = 60e$.

Standing still: $\dfrac{L}{e} = 60$ seconds.

Solution

The routes meet at the port at $60^\circ$, each ship starting 24 km away.

The slower ship is now 16 km from the port. For a right angle at one ship, with $60^\circ$ at the port, the faster ship is either 32 km (impossible) or $16 \cos 60^\circ = 8$ km away.

So the faster ship covered 16 km while the slower covered 8: speeds are $2 : 1$.

When the faster ship covers all 24 km, the slower covers 12 km and is $24 - 12 = 12$ km from the port.

Solution

Take the job as 120 units. Rates per day: Bob $= 3$, Alex $= 6$, Cole $= 2$.

Days 1–3: $9 + 5 + 8 = 22$ units. Five cycles (15 days) $= 110$ units.

Day 16 (Alex + Bob): 119. Day 17 (Bob + Cole) finishes the last unit.

Alex worked 2 days per cycle $\times 5 + 1 = 11$ days.