Algebra, Equations & Operations: previous-year questions

Answer: (C) 2:1

Let cows $= x$ and crows $= y$. Heads: $x + y = 30$. Legs: $4x + 2y = 100$. Subtracting twice the first equation from the second gives $2x = 40$, so $x = 20$ cows and $y = 10$ crows, a ratio of $20 : 10 = 2 : 1$. Shortcut: if all 30 were crows there would be 60 legs; the 40 extra legs come from cows at 2 extra legs each, so there are 20 cows. The trap is option A (1 : 2), which is the ratio of crows to cows; always check the order asked in the question.

Answer: (B) $x = 6$, $y = 8$

From $\frac{x}{y} = \frac{3}{4}$, $x = \frac{3y}{4}$. Substituting in $2x - 3y = -12$: $\frac{3y}{2} - 3y = -12$, so $-\frac{3y}{2} = -12$, giving $y = 8$ and $x = 6$. The quickest exam method is to test the options: only $x = 6$, $y = 8$ satisfies both conditions ($12 - 24 = -12$ and $\frac{6}{8} = \frac{3}{4}$). Option A fails the ratio ($\frac{6}{9} = \frac{2}{3}$), and options C and D give a ratio greater than 1.

Answer: (C) 6

Regroup the left side as $(x - 2)^2 + (x - z)^2 + (y - z)^2 = 0$. A sum of squares of real numbers is zero only when each square is zero, so $x = 2$, $z = x = 2$ and $y = z = 2$, giving $x + y + z = 6$. The values 2, 8 and 10 do not satisfy all three conditions.

Answer: (A) 0

$x + \frac{1}{x} = -1$ gives $x^2 + x + 1 = 0$. Since $x^3 - 1 = (x - 1)(x^2 + x + 1)$, it equals 0. The values 1, −1 and 2 ignore this factorisation.

Answer: (C) $-a, -b$

Move $\frac{1}{x}$ to the left: $\frac{1}{a + b + x} - \frac{1}{x} = \frac{a + b}{ab}$, which gives $\frac{-(a + b)}{x(a + b + x)} = \frac{a + b}{ab}$. So $x^2 + (a + b)x + ab = 0$, i.e. $(x + a)(x + b) = 0$ and $x = -a$ or $-b$. Positive $a$ or $b$ do not satisfy the equation.

Answer: (A) $16 - \sqrt{3}$

Rationalising, $\frac{2 + \sqrt{3}}{2 - \sqrt{3}} = 7 + 4\sqrt{3}$ and $\frac{2 - \sqrt{3}}{2 + \sqrt{3}} = 7 - 4\sqrt{3}$, which add to 14. Also $\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$. The total is $16 - \sqrt{3}$; option C is only the last term.

Answer: (C) 3

Writing everything as powers of 3, the left side is $\frac{3^{2n} \times 3^5 \times 3^9}{3^1 \times 3^{16}} = 3^{2n - 3}$. Setting $3^{2n - 3} = 3^3$ gives $2n = 6$, so $n = 3$. The other values give $3^{-3}$, $3^1$ or $3^5$.

Answer: (C) 8

Rationalising each term gives $\frac{1}{\sqrt{n} + \sqrt{n + 1}} = \sqrt{n + 1} - \sqrt{n}$, so the sum telescopes. Everything cancels except $\sqrt{81} - \sqrt{1} = 9 - 1 = 8$. Option A forgets to subtract $\sqrt{1}$.

Answer: (D) 8

By the remainder theorem the remainder is the value at $x = 1$: 1 + 3 + 3 + 1 = 8. Equivalently the polynomial is $(x + 1)^3$ and $2^3 = 8$. 0 would be the remainder on dividing by $(x + 1)$.

Answer: (B) 2

Pair the factors: $(2 - \sqrt{3})(4 + 2\sqrt{3}) = 2(2 - \sqrt{3})(2 + \sqrt{3}) = 2(4 - 3) = 2$. Also $(\sqrt{5} + 2)(\sqrt{5} - 2) = 5 - 4 = 1$. The product is 2 × 1 = 2.

Answer: (D) 27

Dividing by 4, $5^x = 125 = 5^3$, so $x = 3$. Then $x^x = 3^3 = 27$. 8 and 64 are $2^3$ and $4^3$, which use the wrong base.

Answer: (A) 5

If the roots are reciprocals their product is 1. The product of roots is $\frac{k}{5}$, so $\frac{k}{5} = 1$ and $k = 5$. $k = 0$ would make one root zero, which has no reciprocal.

Answer: (B) 3

The expression equals $\frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2]$. Here that is $\frac{1}{2}(1 + 1 + 4) = 3$. It would be 0 only if $a = b = c$.

Answer: (A) −1

$4^{x+2} = 2^{2x+4} = 2 \times 2^{2x+3}$, so the equation becomes $2^{2x+3} = 2$. Hence $2x + 3 = 1$ and $x = -1$. Check: $4^1 = 4 = 2^1 + 2$.

Answer: (D) ₹ 250

3H = 5C means a cow costs $\frac{3}{5}$ of a horse. Then 4H + 6 × $\frac{3}{5}$H = 7·6H = 1900, so H = ₹ 250 (and a cow ₹ 150). ₹ 150 is the cost of a cow.

Answer: (B) 6

The expression $\frac{x^2}{x}$ is not defined at $x = 0$, so 0 cannot be a root. For $x \ne 0$ it equals $x$, so $x = 6$. Option C wrongly keeps the root 0 that appears after multiplying through by $x$.

Answer: (B) 2800

With S sweets, $\frac{S}{140} - \frac{S}{175} = 4$, i.e. $\frac{35S}{24500} = 4$, so S = 2800. Check: 2800 ÷ 175 = 16 and 2800 ÷ 140 = 20, which is 4 more. 2400 does not divide evenly by 175.

Answer: (A) 1

Substituting gives 7 + 24 ÷ 8 − 6 ÷ 2 × 3. Doing division and multiplication first: 24 ÷ 8 = 3 and 6 ÷ 2 × 3 = 9, so the value is 7 + 3 − 9 = 1. Working strictly left to right without BODMAS gives a different, wrong value.

Answer: (C) $\frac{1001}{3}$

Each factor simplifies: $2 - \frac{1}{3} = \frac{5}{3}$, $2 - \frac{3}{5} = \frac{7}{5}$, …, $2 - \frac{997}{999} = \frac{1001}{999}$. In the product $\frac{5}{3} \times \frac{7}{5} \times \cdots \times \frac{1001}{999}$ every numerator cancels the next denominator, leaving $\frac{1001}{3}$. Option B is just the last factor.

Answer: (A) 20

Work from the innermost brackets. $7*2$: since $7>2$, it is $7-2=5$. Then $5*5$: the numbers are equal, so it is $2\times5=10$. Next $5*7$: since $5<7$, it is $5+7=12$. Then $12*2$: since $12>2$, it is $12-2=10$. Adding the two parts, $10+10=20$. The trap lies in the middle step: after the first bracket becomes 5, the operation $5*5$ must use the "equal" rule ($2x$), not addition or subtraction.