Linear Equations: CAT previous year questions

Solution

Let B and C be the numbers of shares, with $B + C = 20$.

$10 \times 120 + 90B + 150C = 3300 \Rightarrow 90B + 150(20 - B) = 2100$.

$3000 - 60B = 2100 \Rightarrow B = 15$.

Solution

Let $u = b - c$. The equations become $a - 6u = 4$ and $6a + 3u = 50$.

From the first, $a = 4 + 6u$. Then $24 + 39u = 50 \Rightarrow u = \dfrac{2}{3}$, $a = 8$.

$2a + 3b - 3c = 2a + 3u = 16 + 2 = 18$.

Solution

Let $s, a, c$ be the numbers of science, arts and commerce students.

$s + a + c = 1500$ and $1100s + 1000a + 800c = 1550000$.

Subtract $800 \times$ the first: $300s + 200a = 350000 \Rightarrow 3s + 2a = 3500$.

Since $s \leq a$: $3500 = 3s + 2a \geq 5s \Rightarrow s \leq 700$.

$s = 700$ works ($a = 700$, $c = 100$). Maximum $= 700$.

Solution

Work backwards. The third customer takes half the remainder $+ 3$ and nothing is left: $\dfrac{R_3}{2} = 3 \Rightarrow R_3 = 6$.

Before the second: $\dfrac{R_2}{2} - 3 = 6 \Rightarrow R_2 = 18$.

Initially: $\dfrac{Q}{2} - 3 = 18 \Rightarrow Q = 42$ kg.

Solution

No solution means the lines are parallel but distinct: $\dfrac{p}{3} = \dfrac{-4}{k} \neq \dfrac{2}{a}$.

Parallel: $pk = -12$. Distinct: $\dfrac{-4}{k} \neq \dfrac{2}{a} \Rightarrow -4a \neq 2k \Rightarrow 2a + k \neq 0$.

So $2a + k \neq 0$ is a necessary condition.

Solution

Selling price $= 0.9 \times 1.25 = 1.125 \times$ average cost, so profit $= 12.5\%$ of total cost.

Total cost $= \dfrac{51000}{0.125} = 408000$.

$1000w + 1125b = 408000 \Rightarrow 8w + 9b = 3264$.

For the most shirts, buy as many cheaper (white) shirts as possible. Since $3264$ is divisible by 8, $b$ must be a multiple of 8; take $b = 8$.

$w = \dfrac{3264 - 72}{8} = 399$. Total $= 399 + 8 = 407$.

Solution

Infinitely many solutions: $\dfrac{a + 5}{1} = \dfrac{b^2 - 15}{1} = \dfrac{8b}{4} = 2b$.

$b^2 - 15 = 2b \Rightarrow b = 5$ or $b = -3$.

$b = 5 \Rightarrow a = 5$, $ab = 25$. $b = -3 \Rightarrow a = -11$, $ab = 33$.

Maximum $ab = 33$.

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 $x, y, z$ 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 prices be $a$ (apple), $o$ (orange), $m$ (mango).

$2a + 4o + 6m = a + 4o + 8m \Rightarrow a = 2m$.

$2a + 4o + 6m = 8o + 7m \Rightarrow 4o + 10m = 8o + 7m \Rightarrow o = \dfrac{3m}{4}$.

Basket cost $= 8o + 7m = 6m + 7m = 13m$: $13$ mangoes.

Solution

Let a large shirt cost $L$; a small one costs $L - 50$.

$\dfrac{5000}{L} + \dfrac{1800}{L - 50} = 64$.

Try $L = 125$: $40 + 24 = 64$ ✓.

Large + small $= 125 + 75 = 200$.

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

$2x + 5y = 99$ needs $y$ odd, with $x = \dfrac{99 - 5y}{2}$.

$x \geq y \Rightarrow 99 - 5y \geq 2y \Rightarrow y \leq 14.1$, so odd $y \leq 13$.

$y \geq -20$, so odd $y$ runs from $-19$ to $13$.

Count $= \dfrac{13 - (-19)}{2} + 1 = 17$.

Solution

Let a pencil cost $c$ (sharpener $c + 2$); Aron buys $p$ pencils and $s$ sharpeners.

$pc + s(c + 2) = 2pc + (s - 10)(c + 2) \Rightarrow pc = 10(c + 2) \Rightarrow p = 10 + \dfrac{20}{c}$.

So $p > 10$; the smallest whole $p$ is 11 (with $c = 20$).

Together they buy $p + 2p = 3p = 33$ pencils.

Solution

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

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