Quantitative Aptitude PYQs

Shortcut Trick

In cyclic or symmetric algebraic expressions, substitute specific values for variables.

Let a = 1, b = 0, and c = 2 (Satisfies a ≠ b, b ≠ c, c ≠ a).

Term 1 = 1² ÷ [(2 − 1)(2 + 1 + 0)] = 1 ÷ 3 = 1/3.

Term 2 = [(1 + 0) × 2] ÷ [2² − 0 − 1² − 0] = 2 ÷ 3 = 2/3.

Term 3 = (1 + 0 + 2) ÷ [2(2 − 1)] = 3 ÷ 2 = 1.5.

Expression = 1/3 + 2/3 − 1.5 = 1 − 1.5 = −0.5 = −1/2.

∴ The correct answer is −1/2.

Alternate Method

Given: a ≠ b, b ≠ c, c ≠ a

Formula Used: Rational Expression Simplification and Substitution.

Term 1 1/3 + Term 2 2/3 − Term 3 1.5 Result: 1/3 + 2/3 − 1.5 = −0.5

⇒ Let a = 1, b = 0, c = 2.

⇒ Term 1 = (1 + 0)² ÷ [(2 − 1)(2 + 1 + 0)] = 1 ÷ [1 × 3] = 1/3.

⇒ Term 2 = (1 + 0) × 2 ÷ [2² − (0 × 2) − 1² − (1 × 0)] = 2 ÷ [4 − 1] = 2/3.

⇒ Term 3 = (1 + 2 × 0 + 2) ÷ [2(2 − 1)] = 3 ÷ 2 = 1.5.

⇒ Expression = 1/3 + 2/3 − 3/2 = 1 − 1.5 = −0.5.

∴ The correct answer is −1/2.

Additional Information

Substitution Method in Algebra

Complex algebraic expressions that are independent of variables can be solved by substituting convenient integers that satisfy the given constraints.

Cyclic and Symmetric Expressions

Expressions involving terms like (a−b), (b−c), and (c−a) are cyclic. If the degree of the numerator and denominator match across terms, the result is often a constant.

Rationalization Property

Always check for common factors in denominators, such as (c² − a²) = (c − a)(c + a), to simplify sums of fractions.

Given:

a^b = b^a

Find: \( \dfrac{a \times \left(\frac{a}{b}\right)^{\frac{a}{b}}}{a^{\frac{a}{b}}} \)

Formula used:

Properties of exponents: (u^p)/(v^p) = (u/v)^p and from a^b = b^a ⇒ a^1/a = b^1/b

Calculations:

Let E = \( \dfrac{a \times \left(\frac{a}{b}\right)^{\frac{a}{b}}}{a^{\frac{a}{b}}} \)

⇒ E = a × \( \dfrac{\left(\frac{a}{b}\right)^{\frac{a}{b}}}{a^{\frac{a}{b}}} \)

⇒ E = a × \( \left(\dfrac{a/b}{a}\right)^{\frac{a}{b}} \) = a × \( \left(\dfrac{1}{b}\right)^{\frac{a}{b}} \)

⇒ E = a × b^-a/b

From a^b = b^a ⇒ a^1/a = b^1/b

⇒ raise both sides to power a: b^a/b = a

⇒ b^-a/b = 1/a

⇒ E = a × (1/a) = 1

∴ Correct answer = 1.

Given:

x = 2 + 2^1/2 + 2^3/2

2^1/2 = √2 and 2^3/2 = 2√2

⇒ x = 2 + √2 + 2√2 = 2 + 3√2

Formula used:

(a + b)² = a² + 2ab + b²

Calculation:

x = 2 + 3√2

x² = (2 + 3√2)²

⇒ x² = 4 + 12√2 + 18 = 22 + 12√2

Now compute x² − 4x − 10

⇒ (22 + 12√2) − 4(2 + 3√2) − 10

⇒ 22 + 12√2 − 8 − 12√2 − 10

⇒ 4

∴ Correct answer = 4.

Given:

\(\dfrac{\sqrt{p+x} + \sqrt{p-x}}{\sqrt{p+x} - \sqrt{p-x}} = p\)

Formula used:

Cross multiplication and algebraic manipulation.

Calculation:

\(\dfrac{\sqrt{p+x} + \sqrt{p-x}}{\sqrt{p+x} - \sqrt{p-x}} = p\)

⇒ \(\sqrt{p+x} + \sqrt{p-x} = p(\sqrt{p+x} - \sqrt{p-x})\)

⇒ \(\sqrt{p+x} + \sqrt{p-x} = p\sqrt{p+x} - p\sqrt{p-x}\)

⇒ \(\sqrt{p+x} - p\sqrt{p+x} = -\sqrt{p-x} - p\sqrt{p-x}\)

⇒ \((1-p)\sqrt{p+x} = -(1+p)\sqrt{p-x}\)

⇒ Squaring both sides:

\((1-p)^2(p+x) = (1+p)^2(p-x)\)

⇒ \((1-p)^2p + (1-p)^2x = (1+p)^2p - (1+p)^2x\)

⇒ \((1-p)^2x + (1+p)^2x = (1+p)^2p - (1-p)^2p\)

⇒ \(x[(1-p)^2 + (1+p)^2] = p[(1+p)^2 - (1-p)^2]\)

⇒ \(x = \dfrac{p[(1+p)^2 - (1-p)^2]}{(1-p)^2 + (1+p)^2}\)

⇒ \(x = \dfrac{p[1 + 2p + p^2 - (1 - 2p + p^2)]}{1 - 2p + p^2 + 1 + 2p + p^2}\)

⇒ \(x = \dfrac{p[4p]}{2(1 + p^2)} \)

⇒ \(x = \dfrac{2p^2}{p^2+1}\)

∴ The correct answer is option (4).

Given:

\(\left(\frac{a-b}{2}\right)x^2-\left(\frac{a+b}{2}\right)x+b=0\)

Formula used:

If x = r is a root, then substituting x = r makes LHS = 0

Calculation:

Multiply whole equation by 2 to simplify

⇒ (a − b)x² − (a + b)x + 2b = 0

Check x = 1

⇒ (a − b) − (a + b) + 2b

⇒ a − b − a − b + 2b

⇒ 0 

So one root is x = 1

Sum of roots = (a + b)/(a − b)

Let second root = r

⇒ 1 + r = (a + b)/(a − b)

⇒ r = (a + b − a + b)/(a − b)

⇒ r = 2b/(a − b)

∴ The roots are 1 and 2b/(a − b).

Given:

\(\rm x-\frac{1}{x}=2\)

Formula used:

(x - 1/x)² = x² + 1/x² - 2

Calculation:

⇒ (x − 1/x)² = 2² = 4

⇒ x² + 1/x² − 2 = 4

⇒ x² + 1/x² = 6

We need x² − 1/x²

⇒ (x − 1/x)(x + 1/x) = x² − 1/x²

⇒ x + 1/x = √(x² + 1/x² + 2)

⇒ x + 1/x = √(6 + 2) = √8 = 2√2

⇒ x² − 1/x² = 2 × 2√2 = 4√2

∴ x² − 1/x² = 4√2.

Given:

(a − b)² + (b − c)² + (c − a)² = 6

a² + b² + c² = 29

Formula used:

(a − b)² + (b − c)² + (c − a)² = 2(a² + b² + c² − ab − bc − ca)

Calculation:

6 = 2(a² + b² + c² − ab − bc − ca)

⇒ 3 = a² + b² + c² − (ab + bc + ca)

Given a² + b² + c² = 29

⇒ 3 = 29 − (ab + bc + ca)

⇒ ab + bc + ca = 26

Now,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ (a + b + c)² = 29 + 52 = 81

⇒ a + b + c = ±9

∴ The correct answer is Option (1).

Given:

\(\rm p=\frac{\sqrt5-2}{\sqrt5+2}\) and \(\rm q=\frac{\sqrt5+2}{\sqrt5-2}\)

Formula used:

\(\dfrac{p}{q}+\dfrac{q}{p} = \dfrac{p^2+q^2}{pq}\)

Calculation:

\(\dfrac{p}{q} = \left(\dfrac{\sqrt5-2}{\sqrt5+2}\right)^2\)

⇒ \(\dfrac{( \sqrt5-2 )^2}{( \sqrt5+2 )^2}\)

⇒ \(\dfrac{5+4-4\sqrt5}{5+4+4\sqrt5}\)

⇒ \(\dfrac{9-4\sqrt5}{9+4\sqrt5}\)

Similarly, \(\dfrac{q}{p} = \dfrac{9+4\sqrt5}{9-4\sqrt5}\)

⇒ Required value = \(\dfrac{9-4\sqrt5}{9+4\sqrt5}+\dfrac{9+4\sqrt5}{9-4\sqrt5}\)

⇒ \(\dfrac{(9-4\sqrt5)^2+(9+4\sqrt5)^2}{9^2-(4\sqrt5)^2}\)

⇒ \(\dfrac{(81+80)+(81+80)}{81-80}\)

⇒  \(\dfrac{322}{1}\)

⇒ 322

∴ The value of (p/q + q/p) = 322.

Given:

Find digit at hundreds place of 25¹⁰

Formula used:

Last 3 digits of N determine hundreds digit; 25ⁿ for n ≥ 2 ends with 625.

Calculations:

25² = 625

⇒ 25ⁿ ≡ 625 (mod 1000) for all n ≥ 2

⇒ 25¹⁰ ≡ 625 (mod 1000)

Last 3 digits = 625 ⇒ hundreds digit = 6

∴ Required digit = 6.

Given:

A number N leaves remainder 3 when divided by 4, 6, 7 and 9

Formula used:

If a number leaves same remainder r when divided by several numbers, then

N − r is divisible by LCM of those numbers

Calculation:

LCM of 4, 6, 7, 9

LCM = 2² × 3² × 7 = 252

N − 3 = multiple of 252

⇒ N = 252k + 3

Smallest 4-digit N:

252 × 3 + 3 = 759 (not 4-digit)

252 × 4 + 3 = 1008 + 3 = 1011 

∴ The smallest 4-digit number is 1011.

Given:

\(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+....}}}}=cosecθ\)

Formula used:

For infinite nested radicals:

\(\sqrt{k+\sqrt{k+\sqrt{k+....}}} = x \text{ where } x^2 = k+x\)

Also, \(cosec \theta = \frac{1}{\sin \theta}\)

Calculation:

\(\sqrt{2+\sqrt{2+\sqrt{2+....}}} = x\)

⇒ \(cosec \theta = x\)

⇒ \(cosec \theta = x \text{ and } x^2 = 2+x\)

⇒\(cosec \theta = x\), and solving \(x^2 - x - 2 = 0\)

⇒ \((x-2)(x+1) = 0 \Rightarrow x = 2 \text{ (taking positive root as \(cosec \theta > 0\))}\)

⇒ \(cosec \theta = 2\)

⇒ \(\sin \theta = \frac{1}{cosec \theta} = \frac{1}{2}\)

∴ The correct answer is option (4).

Given:

8 sinθ − cosθ = 4, 0 < θ < π/2

Formula used:

cosθ = 8 sinθ − 4; sin²θ + cos²θ = 1; solve quadratic in sinθ

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Calculations:

⇒ cosθ = 8 sinθ − 4

⇒ Since sin²θ + cos²θ = 1 

⇒ sin²θ + (8 sinθ − 4)² = 1

⇒ sin²θ + 64 sin²θ − 64 sinθ + 16 = 1

⇒ 65 sin²θ − 64 sinθ + 15 = 0

⇒ Discriminant = 64² − 4 × 65 × 15 = 196, √196 = 14

⇒ sinθ = (64 ± 14)/130

⇒ sinθ = 78/130 = 3/5 (valid)

⇒ sinθ = 50/130 = 5/13 (invalid, cosθ negative)

⇒ Valid sinθ = 3/5 ⇒ cosecθ = 1 / sinθ = 5/3

∴ cosecθ = 5/3.

Given:

2 tanθ = sec² θ - 2 , 0 < θ < π/2

Formula used:

sec² θ = 1 + tan² θ

cotθ = 1 / tanθ

Calculations:

⇒ 2 tanθ = sec² θ - 2

⇒ 2 tanθ = (1 + tan² θ) - 2

⇒ 2 tanθ = tan² θ - 1

⇒ tan² θ - 2 tanθ - 1 = 0

Let t = tanθ

⇒ t² - 2t - 1 = 0

⇒ t = (2 ± √(4 + 4)) / 2 = (2 ± √8)/2 = 1 ± √2

In 0 < θ < π/2, tanθ > 0 ⇒ t = 1 + √2

⇒ cotθ = 1 / (1 + √2)

⇒ cotθ = √2 - 1 (after rationalisation)

∴ Correct option: √2 - 1.

Given:

\((\sec \theta - \tan \theta) - \sqrt{\dfrac{1-\sin \theta}{1+\sin \theta}}\)

Formula used:

\(\sec^2 \theta - \tan^2 \theta = 1\)

 \(\sqrt{\dfrac{1-\sin \theta}{1+\sin \theta}} = \sec \theta - \tan \theta\)

Calculation:

\((\sec \theta - \tan \theta) - \sqrt{\dfrac{1-\sin \theta}{1+\sin \theta}}\)

⇒ \((\sec \theta - \tan \theta) - (\sec \theta - \tan \theta)\)

⇒ 0

∴ The correct answer is option (1).

Given:

cot θ = √7

Formula used:

cosec²θ = 1 + cot²θ

sec²θ = 1 + tan²θ

tan θ = 1 / cot θ

Calculation:

⇒ tan θ = 1 / √7

⇒ sec²θ = 1 + 1/7 = 8/7

⇒ cosec²θ = 1 + 7 = 8

⇒ Numerator = 8 − 8/7 = (56/7 − 8/7) = 48/7

⇒ Denominator = 8 + 8/7 = (56/7 + 8/7) = 64/7

⇒ Value = (48/7) ÷ (64/7)

⇒ 48/64 = 3/4

∴ The correct answer is 3/4.

Given:

The difference between the two acute angles in a right-angled triangle is \(\dfrac{\pi}{12}\) radian.

One of the acute angles of the triangle is to be determined.

Formula used:

In a right-angled triangle:

\(\text{Sum of acute angles} = \dfrac{\pi}{2}\)

Let one angle be \(\theta\) and the other be \(\theta - \dfrac{\pi}{12}\).

\(\theta + \left(\theta - \dfrac{\pi}{12}\right) = \dfrac{\pi}{2}\)

Calculations:

\(\theta + \theta - \dfrac{\pi}{12} = \dfrac{\pi}{2}\)

⇒ \(\theta = \dfrac{\pi}{4} + \dfrac{\pi}{24}\)

⇒ \(\theta = \dfrac{6\pi + \pi}{24}\)

⇒ \(\theta = \dfrac{7\pi}{24}\)

Converting \(\theta\) to degrees:

\(\theta = \dfrac{7 \times 180}{24} = 52.5^\circ\)

∴ The correct answer is option (3): 52.5º.

Given:

\(\log_{10}\big[998+\sqrt{x^2-18x+76}\big]=3 \)

Formula used:

If log₁₀(A) = 3 ⇒ A = 10³

Calculations:

⇒ 998 + √(x² - 18x + 76) = 10³ = 1000

⇒ √(x² - 18x + 76) = 1000 - 998 = 2

⇒ x² - 18x + 76 = 2² = 4

⇒ x² - 18x + 72 = 0

⇒ x2 - 12x - 6x + 72 = 0

⇒ (x - 12) (x - 6) = 0

⇒ x = 12 or x = 6

⇒ Let α = 12, β = 6

⇒ (α - β)² = (12 - 6)² = 6² = 36

∴ (α - β)² = 36.

Given:

x⁴ + y⁴ = 14x²y²

Formula used:

x⁴ + y⁴ = (x² + y²)² − 2x²y²

log a + log b = log(ab)

Calculation:

x⁴ + y⁴ = (x² + y²)² − 2x²y²

⇒ (x² + y²)² − 2x²y² = 14x²y²

⇒ (x² + y²)² = 16x²y²

⇒ x² + y² = 4xy

Check Statement I:

log(x² + y²) = log(4xy)

⇒ = log4 + logx + logy

⇒ = 2log2 + logx + logy 

Statement I is correct

Check Statement II:

x² − y² = (x − y)(x + y)

From x² + y² = 4xy

⇒ (x − y)² = 2xy

⇒ x − y = √(2xy)

x + y = √(x² + y² + 2xy)

⇒ x + y = √(6xy)

⇒ x² − y² = √(12x²y²)

⇒ = 2√3 xy

log(x² − y²) = log(2√3 xy)

⇒ logx + logy + log2 + 0.5log3 

Statement II is correct

∴ Both I and II are correct.

Given:

(3x + y)² + (3x + y)(x + 5y) − 20(x + 5y)²

Formula used:

a² + ab − 20b² = (a + 5b)(a − 4b)

Calculation:

Let a = (3x + y) and b = (x + 5y)

⇒ Given expression = a² + ab − 20b²

⇒ (a + 5b)(a − 4b)

⇒ [(3x + y) + 5(x + 5y)] × [(3x + y) − 4(x + 5y)]

⇒ (3x + y + 5x + 25y) × (3x + y − 4x − 20y)

⇒ (8x + 26y) × (−x − 19y)

⇒ 2(4x + 13y) × (−1)(x + 19y)

⇒ Factors are (4x + 13y) and (x + 19y)

∴ The correct answer is Both I and II.

Given:

\(\dfrac{\dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x}}{\dfrac{x+y}{x-y}+\dfrac{y+z}{y-z}+\dfrac{z+x}{z-x}+3}\) = ?

Formula used:

y = x - (x - y) ⇒ y/(x - y) = x/(x - y) - 1 

Calculations:

Let A = \(\dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x}\)

Then sum \(\dfrac{y}{x-y}+\dfrac{z}{y-z}+\dfrac{x}{z-x}\) =

⇒ using y/(x-y) = x/(x-y) - 1:

⇒ \(\dfrac{y}{x-y}+\dfrac{z}{y-z}+\dfrac{x}{z-x} = \dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x} - 3\)

So denominator D = \(\dfrac{x+y}{x-y}+\dfrac{y+z}{y-z}+\dfrac{z+x}{z-x}+3\) =

⇒ \(\Big(\dfrac{x}{x-y}+\dfrac{y}{y-z}+\dfrac{z}{z-x}\Big) + \Big(\dfrac{y}{x-y}+\dfrac{z}{y-z}+\dfrac{x}{z-x}\Big) + 3\)

⇒ substitute previous result:

⇒ D = A + (A - 3) + 3 = 2A

Therefore the whole expression =

⇒ \(\dfrac{A}{2A} = \dfrac{1}{2}\)

∴ The value of the given expression is 1/2.