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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.