Quantitative Aptitude PYQs

Shortcut TrickWe have, 

 \(\rm\sqrt{20+\sqrt{20+\sqrt{20+ \dots \ \infty}}}\),

Find two factors of 20 such that their difference is 1.

The larger factor will be the solution for expression.

Factors of 20 = 4 × 5

⇒ x2 - 20 = 5

∴ x = 5

Alternate MethodConcept:

If y = \(\sqrt {a + \sqrt{a + \dots \infty}}\)

⇒ y =  \(\frac{\sqrt{4a+1}+1}{2}\)

Calculation:

We have

x² - 20 =\(\rm\sqrt{20+\sqrt{20+\sqrt{20+ \dots \ \infty}}}\)

Using the above concept, (a = 20)

⇒ x² - 20 = \(​​\frac{\sqrt{4(20)+1}+1}{2} \) 

⇒ x2 - 20 = \( \frac{\sqrt 81+1}{2} = \frac{10}{2}=5\)

⇒ x² = 25

∴  x = 5

Given:

\(\rm \frac{a+b}{b+c}=\frac{c+d}{d+a}\) a ≠ c

Concept used:

Componendo and Dividendo:

If a : b = c : d then

​(a + b ) : (a – b) = (c + d) : (c – d) 

**Calculation:**​

\(\rm \frac{a+b}{b+c}=\frac{c+d}{d+a}\)​

According to the above concept

⇒ \(\rm \frac{a+b + b + c}{a + b - b-c}=\frac{c+d+ d + a}{c +d-d-a}\)

⇒ \(\rm \frac{a+2b + c}{a -c}=\frac{c+2d + a}{c -a}\)

⇒ \(\rm \frac{a+2b + c}{a -c}=\frac{c+2d + a}{-{(a-c)}}\)

⇒ a + 2b + c = -c - 2d - a

⇒ 2a + 2b + 2c + 2d = 0

⇒ a + b + c + d = 0

∴ The required value is a + b + c + d = 0

Shortcut Trick

Given that (a³ + b³) ∝ (a² − b²).

Expand both terms using algebraic identities: (a + b)(a² − ab + b²) ∝ (a + b)(a − b).

Cancel the common factor (a + b) from both sides of the proportionality relation.

We get (a² − ab + b²) ∝ (a − b).

∴ The correct answer is (a − b).

Alternate Method

Given: (a³ + b³) is directly proportional to (a² − b²).

Formula Used: If x ∝ y, then x = k × y, where k is a constant of proportionality.

Calculations:

⇒ (a³ + b³) = k × (a² − b²)

⇒ (a + b)(a² − ab + b²) = k × (a + b)(a − b)

⇒ Divide both sides by the common factor (a + b):

⇒ (a² − ab + b²) = k × (a − b)

⇒ Since k is a constant, we can write this in variation form:

⇒ (a² − ab + b²) ∝ (a − b)

∴ The correct answer is (a − b).

Additional Information

Sum of Cubes Identity

The expression a³ + b³ is factored as (a + b)(a² − ab + b²).

Difference of Squares Identity

The expression a² − b² is factored as (a − b)(a + b).

Direct Proportionality

If two quantities x and y are related such that x = ky, they are said to be directly proportional, written as x ∝ y.

Difference of Cubes Identity

The expression a³ − b³ is factored as (a − b)(a² + ab + b²).

Given:

\(\rm\frac{√{x+20}+√{x-1}}{√{x+20}-√{x-1}}=\frac{7}{3}\)

Shortcut TrickAt x = 5,

\(\rm\frac{√{5+20}+√{5-1}}{√{5+20}-√{5-1}}=\frac{5+2}{5-2}=\frac{7}{3}\)

Hence, the required value

\(⇒ \rm √{(x + 20)(x-1)} = √{(25\times4)}\)

⇒ √100 =  10

Alternate Method

Concept used:

Componendo and Dividendo (reverse):

If (a + b ) : (a – b) = (c + d) : (c – d) then

a : b = c : d 

Calculation:

\(\rm\frac{√{x+20}+√{x-1}}{√{x+20}-√{x-1}}=\frac{7}{3}\)

⇒ \(\rm\frac{√{x+20}}{√{x-1}}=\frac{7+3}{7-3}= \frac{10}{4}\)

⇒ \(\rm\frac{√{x+20}}{√{x-1}}=\frac{5}{2}\)

Squaring both sides, we get

⇒ \(\rm\frac{{x+20}}{{x-1}}=\frac{25}{4}\)

⇒ 4x + 80 = 25x – 25

⇒ (25x – 4x) = (80 + 25)

⇒ 21x = 105

⇒ x = 5

The value of \(\rm √{(x + 20)(x-1)}\)

⇒ \(\rm √{(5 + 20)(5-1)}\)

⇒ \(\rm √{(25\times 4)}\) = √100

⇒ \(\rm √{(x + 20)(x-1)}\) = 10

∴ The required value is 10.

Given:

The HCF of (x8 – y8) and (x7 – y7 + x5y2 – x2y5) 

Concept:

1. HCF of two numbers:  Product of the smallest power of each common prime factor in the numbers.

2. (aⁿ - bⁿ) = (a - b) (a^n-1+ ba^n-2 +  ....+ b^n-1)

Where n is a natural number. Hence,

For every value of n, (a - b) will be a factor of (aⁿ - bⁿ)

3. (a + b)(a  - b) = a² - b²

Calculation:

Using the  above formula

(x8 – y8) = (x⁴ - y⁴)(x4 + y4)

⇒ (x8 – y8) = (x² - y²)(x² + y²) (x⁴ + y⁴) 

⇒ (x8 – y8) = (x + y)(x – y)(x² + y²)(x⁴ + y⁴)

Now, (x7 – y7 + x5y2 – x2y5) 

⇒ x⁷ + x⁵y² – y⁷ – x²y⁵

⇒ x⁵(x² + y²) – y⁵(y² + x²)

⇒ (x² + y²) (x5 – y5)

Using the general formula of (aⁿ - bⁿ)

⇒ (x² + y²)(x - y)(x⁴ + yx³ + y²x² + y³x + y⁴)

Now, using the concept of HCF

The required HCF = (x – y)(x² + y²)

Multiplying the factor

⇒ (x3 – y3 – x2y + xy2) 

∴The required HCF is (x3 - y3 - x2y + xy2)

Given:

The sum of the squares of the roots of the equation x2 - 14x + k = 0 = 100

Formula used:

For a quadratic equation ax² + bx + c = 0

Sum of roots = \(\frac{-b}{{a}}\)

Product of roots = \(\frac{c}{{a}}\)

Calculation:

x2 - 14x + k = 0

α + β = \(\frac{-b}{{a}}\) = 14

αβ = \(\frac{c}{{a}}\) = k

Now, the sum of the squares of the roots 

⇒ α² + β² = (α + β)² - 2αβ

⇒ α2 + β2 = (14)² - 2k = 100      (Given)

⇒ 196 - 2k = 100

⇒ 2k = 96

⇒ k = 48

∴ The required value of k is 48.

Given: 

(x + k) is the HCF of x2 + px + q and x2 + qx + p

Concept:

1. HCF: Product of the smallest power of each common prime factor in the numbers.

2. If (x - a) is factor of p(x) then p(a) = 0

Calculation:

We have,

P(x) = x2 + px + q

Q(x) = x2 + qx + p

Since, (x + k) is HCF (common prime factor) of P(x) & Q(x)

⇒ P(k) =  k2 - kp + q = 0

⇒ k2 = kp - q          ------(1)

Again,

Q(k) = x2 + qx + p

⇒ k² = kq - p           ------(2)

From equation (1) & (2)

⇒ kp - q = kq - p 

⇒ k(p - q) = (q - p)

⇒ k = -1

∴ The required value of k is -1.

Additional Information

LCM: Product of the greatest power of each prime factor, involved in the numbers.

Formula used:

Dividend = Divisor ×  quotient + Remainder

Calculation:

Let the greater number be G & smaller number be S.

⇒ 3G = S × 6 + 6

⇒ 3G = 6S + 6           

⇒ G = 2S + 2           -----(1) 

Also,

⇒ 5S = G × 2 + 3

⇒ 5S = 2G + 3          

⇒ 5S = 2(2S + 2) + 3        [From equation (1)]

⇒ 5S = 4S + 7  

⇒ S = 7

Put this value in equation (1)

⇒ G = 2 × 7 + 2 = 16

∴ Difference = 16 - 7 = 9

Concept:

Factor: A number that divides another number.

For example, 3 and 6 are factors of 12

Shortcut TrickWe have  a2 (b - c)3 + b2 (c - a)3 + c2 (a - b)3

Let, a = 1, b = 2, c = 3

⇒ 1² (2 - 3)³ + 2²(3 - 1)³ + 3²(1 - 2)³

⇒ (-1) + 32 - 9 = 22

Statement:1 

(ab + bc + ca) = 1 × 2 + 2 × 3 + 3 × 1

⇒ (ab + bc + ca) = 11

Which is a factor of 22

Statement:2

(a + b + c) = 1 + 2 + 3 = 6

Which is not a factor of 22

∴ Only statement 1 is correct.

Shortcut Trick

We have,

\(\rm \frac{1}{x(x-y)(x-z)}+\frac{1}{y(y-z)(y-x)}+\frac{1}{z(z-x)(z-y)}\)

Put x = 1, y = 2, z = 3

⇒ \(\rm \frac{1}{1(1-2)(1-3)}+\frac{1}{2(2-3)(2-1)}+\frac{1}{3(3-1)(3-2)}\) 

⇒ \(\frac{1}{(-1)(-2)}+\frac{1}{2(-1)}+\frac{1}{3(2)}\)

⇒ \(\frac{1}{2}-\frac{1}{2}+\frac{1}{6} = \frac{1}{6}\)

From option (3), 

\(\frac{1}{xyz} = \frac{1}{1\times 2\times 3} = \frac{1}{6}\)

∴ \(\rm\frac{1}{xyz}\) is the correct value.

Calculation:

Statement 1. 

x4(y – z) + y4(z - x) + z4(x - y) is positive if x > y > z.

We have to solve this by Value putting method

Here, x > y > z.

Then, x = 2, y = 1 or z = 0

Now,

⇒ 2⁴(1 − 0) + 1⁴(0 − 2) + 0⁴(2 − 1)

⇒ 16 + 1(-2) + 0

⇒ 16 − 2 = 14

So, Statement 1 is positive.

Statement 2.

x4(y - z) + y4(z - x) + z4(x - y) is negative if x < y < z.

Here, x < y < z

Then, x = 0, y = 1 or z = 2

Now,

⇒ 0⁴(1 − 2) + 1⁴(2 − 0) + 2⁴(0 − 1)

⇒ 0 + 2 − 16 = -14

So, Statement 2 is negative.

∴ Both 1 and 2.

Shortcut TrickPut x = 0 

\(\rm \frac{x^6-24x^4+144x^2}{(x^2+4√3x+12)(x-2√3)^2} = 0\)

From option (1)

x² = (0²) = 0

Except (1), all other option gives a value other than 0. 

∴ The required value is x2

Alternate Method

Formula used:

(a + b)(a − b) = a² − b²

Calculation:

\(\rm \frac{x^6-24x^4+144x^2}{(x^2+4√3x+12)(x-2√3)^2}\)

⇒ \(\rm \frac{x^2(x^4-24x^2+144)}{(x^2+4√3x+12)(x^2 +12- 4√3x)}\)

⇒ \(\rm \frac{x^2(x^4-24x^2+144)}{(x^2+12)^2-(4√3x)^2}\)  

⇒ \(\rm \frac{x^2(x^4-24x^2+144)}{(x^4+144 + 24x^2 - 48x^2)}\)

⇒ \(\rm \frac{x^2(x^4-24x^2+144)}{(x^4+144 - 24x^2)}\)

⇒ x²

∴ The required value is x²

Shortcut Trick

Total work = LCM of 45 and 40 = 360 units.

Efficiency of X = 360 ÷ 45 = 8 units/day | Efficiency of Y = 360 ÷ 40 = 9 units/day.

Work done by Y alone in the last 23 days = 23 × 9 = 207 units.

Remaining work done by X and Y together = 360 − 207 = 153 units.

Combined efficiency = 8 + 9 = 17 units/day ⇒ n = 153 ÷ 17 = 9.

∴ The correct answer is n = 9.

Alternate Method

Given:

Time taken by X to complete the work = 45 days.

Time taken by Y to complete the work = 40 days.

X and Y work together for n days, and Y completes the remaining work in 23 days.

Formula Used:

Total Work = LCM of given days.

Efficiency = Total Work ÷ Number of Days.

Calculations:

⇒ Combined efficiency of X and Y = 8 + 9 = 17 units/day.

⇒ Work completed by Y alone in 23 days = 23 × 9 = 207 units.

⇒ Remaining work done by X and Y together = Total Work − Work done by Y alone.

⇒ Remaining work = 360 − 207 = 153 units.

⇒ Time taken by X and Y to complete 153 units (n) = 153 ÷ 17 = 9 days.

⇒ Hence, n = 9.

∴ The correct answer is n = 9.

Additional Information

LCM Method in Time and Work

Total work is considered to be the Lowest Common Multiple (LCM) of the distinct days taken by the individuals to complete the job. Efficiency of a person is given by: Total Work ÷ Time Taken.

Work and Wages

Wages are strictly distributed in the ratio of the total work done by each individual. If they work for the same number of days, the wages are distributed in the ratio of their efficiencies.

Pipes and Cisterns

The concept remains similar to time and work, with the only variation being that an inlet pipe does positive work (filling) while an outlet or leakage pipe does negative work (emptying).

Given:

480 persons working 10 hours per day complete one-fourth of a work in 10 days.

Formula used:

M₁ × D₁ × H₁/W₁ = M₂ × D₂ × H₂/W₂

Where, M = Number of men,

D = Number of days, H = Number of hours,

W = Work done

Calculation:

Let, y additional workers are required;

M₁ = 480, D₁ = 10, H₁ = 10, W₁ = \(\frac{1}{{4}}\)

M₂ = (480 + y), D₂ = 20, H₂ = 8, W₂ =  (1 − \(\frac{1}{{4}}\)) = \(\frac{3}{{4}}\)

Using the above formula

⇒ \(\frac{480\times 10\times10}{\frac{1}{4}} = \frac{(480+y)\times 20\times 8 }{\frac{3}{4}}\)

⇒ \(\frac{480\times 10\times10}{1} = \frac{160(480+y) }{3}\)

⇒ 480 + y = 900

⇒ y = (900 − 480) = 420

∴ The required number of persons is 420.

Shortcut Trick

Equate total work: 5 × (12W + 16M) = 4 × (13W + 24M)

⇒ 60W + 80M = 52W + 96M ⇒ 8W = 16M ⇒ 1W = 2M

Convert Group 1 to men only: 12(2M) + 16M = 40M, taking 5 days.

Target group in men only: 25(2M) + 50M = 100M.

Using M₁D₁ = M₂D₂ ⇒ 40 × 5 = 100 × D ⇒ D = 2 days.

∴ The correct answer is 2 days.

Alternate Method

Given:

Group 1: 12 Women (W) and 16 Men (M) can do the work in 5 days.

Group 2: 13 Women (W) and 24 Men (M) can do the work in 4 days.

Method Used:

We use the LCM method to determine the efficiency of each group and then solve for the individual efficiencies of men and women.

Calculations:

⇒ 12W + 16M = 4    (Equation 1)

⇒ 13W + 24M = 5    (Equation 2)

⇒ Multiply Equation 1 by 3 and Equation 2 by 2 to balance M:

⇒ 36W + 48M = 12

⇒ 26W + 48M = 10

⇒ Subtracting the two equations: 10W = 2 ⇒ W = 0.2 units/day

⇒ Substituting W = 0.2 into Equation 1: 12(0.2) + 16M = 4

⇒ 2.4 + 16M = 4 ⇒ 16M = 1.6 ⇒ M = 0.1 units/day

⇒ Efficiency of target group (25W + 50M) = 25(0.2) + 50(0.1) = 5 + 5 = 10 units/day

⇒ Time required = Total Work ÷ Efficiency = 20 ÷ 10 = 2 days

∴ The correct answer is 2 days.

Additional Information

Total Work Concept

Total work is the product of efficiency and the time taken to complete it. For equivalent groups completing the same task, the work relation is W = M₁ × D₁ = M₂ × D₂.

Combined Efficiency

When individuals with different capabilities work together, their one-day work (efficiency) is added directly. The combined efficiency of a mixed group is (Number of men × Men's efficiency) + (Number of women × Women's efficiency).

Pipes & Cisterns Analogy

The efficiency concept is closely mirrored in Pipes & Cisterns problems, where the work done corresponds to the volume filled, and efficiency equates to the filling rate of different pipes.

Given:

Average speed of first train = 60 km/hr

Average speed of second train = 80 km/hr

Formula used:

Distance = speed × time

Average speed = Total Distance between two trains/ Total Time

Calculation:

According to the question;

The first train starts from station X at 6 a.m.

6 a.m. to 8 p.m. = 8 hours

Distance travelled by X in 8 hour

⇒ (60 × 8) = 480 km  

Remaining distance = (1320 - 480) = 840 km

Now, X & Y travel toward each other

⇒ Relative Speed = 60 + 80 = 140 km/hr

The time is the same at station X and station Y i.e. 2 p.m.

Time when both trains will meet

⇒ t = \((\frac{840}{{140}})\) = 6 hours

∴ They meet at (2 p.m. + 6 p.m.) = 8 p.m.

Alternate Method Given:

Distance between X and Y = 1320 km

Train from X starts at 6 a.m., Speed = 60 km/hr

Train from Y starts at 2 p.m., Speed = 80 km/hr

Formula Used:

Distance = Speed × Time

Calculation:

The train from X travels for 8 hours before the train from Y starts (from 6 a.m. to 2 p.m.):

⇒ Distance covered by train from X = 60 km/hr × 8 hours = 480 km

Remaining distance between the two trains when the train from Y starts:

⇒ Remaining distance = 1320 km - 480 km = 840 km

Relative speed of both trains (since they are moving towards each other):

⇒ Relative speed = 60 km/hr + 80 km/hr = 140 km/hr

Time taken to meet (using the remaining distance and relative speed):

⇒ Time to meet = 840 km / 140 km/hr = 6 hours

The train from Y starts at 2 p.m.:

⇒ Meeting time = 2 p.m. + 6 hours = 8 p.m.

∴ The trains meet at 8 p.m.

Given:

The ratio of alcohol and water in the first glass = 3 : 2

The ratio of alcohol and water in the second glass = 4 : 1

Shortcut Trick

∴ The required ratio is 7 : 3

Alternate Method

According to the question

Alcohol contained in first glass = 3/5

Water contained in first glass = 2/5

Alcohol contained in second glass = 4/5

Water contained in second glass = 1/5

Now, the ratio of alcohol and water in the third glass

⇒ (3/5 + 4/5) : (2/5 + 1/5)

⇒ (7/5 : 3/5)

⇒ 7 : 3

∴ The required ratio is 7 : 3

Shortcut Trick

For fractions, HCF = HCF of numerators ÷ LCM of denominators.

For fractions, LCM = LCM of numerators ÷ HCF of denominators.

Numerators: 3, 6, 9, 27 | Denominators: 5, 25, 20, 50.

x = HCF = 3 ÷ 100 | y = LCM = 54 ÷ 5.

Ratio y/x = (54/5) ÷ (3/100) = (54/5) × (100/3) = 18 × 20 = 360.

∴ The correct answer is y = 360x.

Alternate Method

Given: Fractions are 3/5, 6/25, 9/20, and 27/50.

Formula Used:

⇒ HCF of Fractions = HCF(Numerators) / LCM(Denominators)

⇒ LCM of Fractions = LCM(Numerators) / HCF(Denominators)

Calculations:

⇒ x = HCF(3, 6, 9, 27) / LCM(5, 25, 20, 50) = 3 / 100

⇒ y = LCM(3, 6, 9, 27) / HCF(5, 25, 20, 50) = 54 / 5

⇒ To find the relation, divide y by x:

⇒ y/x = (54/5) / (3/100)

⇒ y/x = (54/5) × (100/3)

⇒ y/x = 18 × 20 = 360

⇒ y = 360x

∴ The correct answer is y = 360x.

Additional Information

HCF and LCM of Fractions

Always ensure fractions are in their simplest form before applying the HCF/LCM formulas to avoid errors.

Product Rule

For any two numbers a and b, HCF(a, b) × LCM(a, b) = a × b.

Note:

This rule does not directly apply to a set of more than two fractions in the same way.

LCM of Denominators

To find the LCM of 5, 25, 20, 50: 50 is the largest, its multiples are 50, 100... 100 is divisible by 5, 25, 20, and 50.

Shortcut Trick
By Remainder Theorem, we can directly substitute the remainders in place of the variables.
Remainder of x = 2 and Remainder of y = 3
Remainder of (x − y) ÷ 6 = (2 − 3) = −1
Since the remainder cannot be negative, add the divisor (6) to the negative remainder.
Actual Remainder = −1 + 6 = 5
∴ The correct answer is 5.

Alternate Method
Given:
When x is divided by 6, the remainder is 2.
When y is divided by 6, the remainder is 3.
x > y
Formula Used:
Dividend = (Divisor × Quotient) + Remainder
Calculation:
⇒ Let the quotient when x is divided by 6 be a. Then, x = 6a + 2
⇒ Let the quotient when y is divided by 6 be b. Then, y = 6b + 3
⇒ (x − y) = (6a + 2) − (6b + 3)
⇒ (x − y) = 6(a − b) + 2 − 3
⇒ (x − y) = 6(a − b) − 1
⇒ To find the positive remainder when divided by 6, we rewrite the expression by adding and subtracting 6:
⇒ (x − y) = 6(a − b) − 6 + 6 − 1
⇒ (x − y) = 6(a − b − 1) + 5
⇒ Since x > y and x leaves remainder 2 while y leaves remainder 3, we know a > b, making (a − b − 1) a non-negative integer.
⇒ Here, 6(a − b − 1) is completely divisible by 6, so the remainder left is 5.
∴ The correct answer is 5.

Additional Information
Negative Remainders Concept
When a calculation yields a negative remainder −R with a divisor D, the true positive mathematical remainder is given by (D − R).
Remainder Addition & Subtraction Rule
The remainder of an expression (A ± B) ÷ D is equivalent to [Remainder(A ÷ D) ± Remainder(B ÷ D)] evaluated again with modulo D.
Remainder Multiplication Rule
The remainder of (A × B) ÷ D is identical to the remainder obtained from [Remainder(A ÷ D) × Remainder(B ÷ D)] ÷ D.

Given:

The value of a 2-digit number is 5 times the sum of the digits. 

Shortcut Trick

∴ Product of digits = 5 × 4 = 20

Alternate Method

Let the two digits be x and y respectively.

According to the question

⇒ 10x + y = 5(x + y)

⇒ 10x + y = 5x + 5y

⇒ 5x = 4y

⇒ \(\frac{x}{{y}}\) = \(\frac{4}{{5}}\)

Now, the product of the digits

xy = (4 × 5) = 20

∴ The required product of the digits is 20.