Quantitative Aptitude PYQs

Calculation:

The arithmetic mean of n numbers = M

Sum of n numbers = n × M = nM

Sum of (n - 1) numbers = k 

Then,

n^th number = Sum of n numbers - sum of (n -1) numbers                               

n^th number = nM - k 

∴ The correct answer is nM - k

Formula used:

GM = ⁿ√ x₁ × x₂ × x₃ × x_n =  (x1 × x2 × x3 × xn)^1/n

Calculation:

GM = (3 × 9 × 27 × 81 × 243 × 729 × 2187)^1/7

⇒  (3¹ × 3² × 3³ × 3⁴ × 3⁵ × 3⁶ × 3⁷)^1/7

⇒  (3^28/7) = 3⁴

∴ The answer is 81.

Given:

A person purchases one kg of tea powder

• A's place cost of sugar 1 kg = ₹ 1000
• B's place cost of sugar 2 kg = ₹ 1000
• C's place cost of sugar 4 kg = ₹ 1000
• D's place cost of sugar 5 kg = ₹ 1000

​

​Calculation:

Total purchase = 4 kg   (A + B + C + D)

cost of 1 kg tea powder in A's place = 1000

cost of 1 kg tea powder in B's place = 1000 / 2 = 500

cost of 1 kg tea powder in C' place = 1000 / 4 = 250

cost of 1 kg tea powder in D' place = 1000 / 5 = 200

Total cost = 1000 + 500 + 250 + 200 = ₹ 1950

According to the above calculation,

At ₹ 1950 = 4 kg

At ₹ 1000 = (4 / 1950) × 1000 = 2.05 kg

∴ On average, he purchases 2.05 kg of sugar for every ₹1000 spent.

Calculation:

All single-digit prime no = 2, 3, 5, 7

⇒ Largest no. = 7532

⇒ Smallest no = 2357

Sum:

⇒ 7532 + 2357 = 9889

∴ The sum of the largest and the smallest 4-digit numbers made by using single digit prime numbers is 9889.

Shortcut Trick

Convert the base 3 to a power close to the divisor 28: 3³ = 27.

Rewrite the expression: 3²⁵⁵ = (3³)⁸⁵ = 27⁸⁵.

Since 27 is 1 less than 28, the remainder is (−1)⁸⁵ = −1.

Final positive remainder = 28 − 1 = 27.

∴ The correct answer is 27.

Alternate Method

Given: Dividend = 3²⁵⁵, Divisor = 28

Formula Used: Remainder of (a − 1)ⁿ ÷ a is (−1)ⁿ

⇒ 3²⁵⁵ = (3³)⁸⁵

⇒ 3³ = 27

⇒ Remainder [3²⁵⁵ ÷ 28] = Remainder [27⁸⁵ ÷ 28]

⇒ 27 ≡ −1 (mod 28)

⇒ Remainder = (−1)⁸⁵ = −1

⇒ Positive Remainder = 28 + (−1) = 27

∴ The correct answer is 27.

Additional Information

Remainder Theorem for Powers

If a number (a + k)ⁿ is divided by a, the remainder is kⁿ. In this case, (28 − 1)⁸⁵ divided by 28 leaves (−1)⁸⁵.

Negative Remainder Property

A negative remainder −r is equivalent to a positive remainder (Divisor − r). It is a useful intermediate step to simplify calculations.

Odd and Even Powers of −1

(−1)ⁿ = 1 if n is even, and (−1)ⁿ = −1 if n is odd. This rule determines the sign of the intermediate remainder.

Given:

(10097 + 10054 + x + 1) is divisible by 9

x, (0 < x < 8)

Concept:

The divisibility rule for 9 states that a number is divisible by 9 if the sum of its digits is

either 0 or a multiple of 9.

Calculation:

If 10⁴ = 10000, the sum of all digits is 1

10⁶ = 1000000 also the sum of all digits is 1

sum of all digit of (10097 + 10054 + x + 1) is x + 3

According to the divisibility rule for 9,

If we add 6 to x + 3 so that their sum is 9, then x + 3 will be divisible by 9 with 0

remainder.

∴ The value of x is 6.

Given:

AB.DC = AC.BD → \(\frac{AB}{BD}=\frac{AC}{DC}\)

∠ BAD = α ,  ∠ CAD = β 

Concept Used:

By Angle Bisector Theorem:

A line that bisects an angle divides the opposite side into two segments that are proportional to the other two sides of the triangle.

i.e., \(\frac{AB}{BD}=\frac{AC}{DC}\)

Calculation:

According to the above theorem, 

⇒ If \(\frac{AB}{BD}=\frac{AC}{DC}\) then we can say that D bisects ∠ BAC.

⇒ Then, ∠ BAD = ∠ CAD which is α = β 

∴ The correct relation between α and β is α = β

Given:

N = 12345678AB which is divisible by 9

Concept Used:

Any number whose summation of the digits is divisible by 9, then the number itself can be divisible by 9.

Calculation:

Statement I:

All digits 1+2+3+4+5+6+7+8 = 36 which is divisible by 9.

Then,

⇒ (A + B) will be divisible by 9.

⇒ Statement I is correct.

Statement II:

For A & B to be divisible by 9, 

A & B must be one odd & one even. 

Such as: (1,8), (2,7), (3,6), (4,5), (9,0)

⇒ Statement II is false.

Hence, The correct answer is option 1.

Given:

x³ + \(1 \over x³\) = 65/8

y³ + \(1 \over y³\) = 730/27

Calculation:

⇒ x³ + \(1 \over x³\)³ = 65/8 = 8\(\frac{1}{8}\)

⇒ x³ + \(1 \over x³\) = 8 + 1/8

We get x³ = 8 

⇒ x = 2

Also,

⇒ y³ + \(1 \over y³\) = 730/27 = 27\(\frac{1}{27}\)

⇒ y³ + \(1 \over y³\) = 27 + \(1 \over 27\)

We get, y³ = 27

⇒ y = 3

Now, 

⇒ xy = 2 × 3 = 6

∴ The correct answer is 6.

Given:

11x + 5y is a prime number where x, y are natural numbers.

Formula Used:

Find the minimum value of (x + y) such that 11x + 5y is a prime number.

Calculation:

Let's test the values of x and y to find the minimum sum (x + y).

For x = 1:

11 × 1 + 5y ⇒ 11 + 5y

11 + 5y should be a prime number.

For y = 1:

11 + 5 × 1 ⇒ 11 + 5 = 16 (not a prime number)

For y = 2:

11 + 5 × 2 ⇒ 11 + 10 = 21 (not a prime number)

For y = 3:

11 + 5 × 3 ⇒ 11 + 15 = 26 (not a prime number)

For y = 4:

11 + 5 × 4 ⇒ 11 + 20 = 31 (prime number)

Thus, the minimum value of (x + y) when x = 1 and y = 4:

x + y = 1 + 4 = 5

The minimum value of (x + y) is 5.

Given:

A 4-digit number N which has e×actly 15 distinct divisors.

Concept:

Let, N = P² × Q³

⇒ Its number of factors = (2 + 1) (3 + 1) = 3 × 4 = 12

Calculation:

According to the question,

Let the 4-digit number N = (pa) (qb)

⇒ Its number of factors = (a + 1) (b + 1)

As per the question, (a + 1) (b + 1) = 15

⇒ 15 can be written as (3 ×  5) or (1 × 15)

 

But, N= q¹⁴ cannot satisfy the 4-digit Number condition.

∴  The correct number of factors is 45.

Given:

p, q, and r are the lengths of the sides of a right-angled triangle.

Calculation:

Let's consider a right-angle triangle of sides (3, 4, 5)

i.e. p = 3, q = 4, r = 5

So,

(p - q - r) (q - r - p) (r - p - q)

⇒ (3 - 4 - 5)(4 - 5 - 3)(5 - 3 - 4)

⇒ (3 - 9)(4 - 8)(5 - 7)

⇒ (-6)(-4)(-2) 

After multiplication, we get a negative term.

Thus, (p - q - r) (q - r - p) (r - p - q) is always negative only.

∴ The correct answer is option (2).

Given:

Expression: \(\frac{\left(a^8+a^4+1\right)\left(b^8+b^4+1\right)}{a^4 b^4}\), where a>0a>0

, b>0b>0

.

Formula used:

AM-GM inequality: For positive numbers \((x_1, x_2, \dots, x_n\), \((x_1 + x_2 + \dots + x_n)/n \geq \sqrt[n]{x_1x_2\dots x_n}\).

Calculations:

Let \(x = a^4 \) and \( y = b\), so the expression becomes:

⇒ \(\frac{(x^2+x+1)(y^2+y+1)}{xy}\)

Apply AM-GM on \((x^2, x, 1)\):

⇒ \((x^2 + x + 1 \geq 3\sqrt[3]{x^2 \cdot x \cdot 1} = 3\sqrt[3]{x^3} = 3x.\)

Equality holds when x = 1.

Similarly, \((y^2 + y + 1 \geq 3y)\).

Now substitute into the expression:

⇒ \(\frac{(x^2+x+1)(y^2+y+1)}{xy} \geq \frac{(3x)(3y)}{xy}.\)

⇒ \(\frac{(x^2+x+1)(y^2+y+1)}{xy} \geq 9.\)

The minimum value is achieved when x = y = 1, i.e., a = b = 1.

⇒ Minimum value = 9.

∴ The correct answer is option (3): 9.

Given:

• Total no of students = 200
• students prefer both tea and coffee = n
• students prefer coffee = 2n
• students prefer tea = 3n
• students prefer neither tea nor coffee = 4n

 

Calculation:

Students prefer only Tea = students prefer Tea - students prefer both tea and coffee 

⇒ 3n - n = 2n

Students prefer only coffee = students prefer coffee - students prefer both tea and coffee

⇒ 2n - n = n

Students prefer only Tea + only coffee + both + neither tea nor coffee = 200

⇒ ​2n + n + n + 4n = 200

⇒ 8n = 200

⇒ n =  \(\frac{200}{8}\)

⇒ n = 25

∴ The correct answer is 25.

Given: 

AB = 9 cm, BC = 12 cm

∠ABC = θ 

Area of triangle = 36 cm²

Formula used:

Area of triangle = \(\frac{1}{2}\) × Ist side × 2^nd side × sin θ

Calculation:

Area of triangle = \(\frac{1}{2}\) × Ist side × 2^nd side × sin θ

36 =  \(\frac{1}{2}\) × 9 × 12 × sin θ

⇒ sin θ = \(\frac{2}{3}\)

We know that

cos²θ = 1 - sin²θ 

cos2θ = 1 - (2/3)2

⇒ cos2θ = 1 - 4/9 = 5/9

∴ Cos θ = \(\frac{\sqrt 5}{3}\)

Given:

Let n be a natural number.

The HCF of n and n + 10 is 10.

The LCM of n and n + 10 is x (a 2-digit number).

Formula Used:

For any two numbers a and b,

HCF(a, b) × LCM(a, b) = a × b

Calculation:

Let n = 10k. Therefore, n + 10 = 10k + 10 = 10(k + 1).

Given HCF(n, n + 10) = 10, it implies k and k + 1 are co-prime (i.e., HCF(k, k+1) = 1).

We know,

HCF(n, n + 10) × LCM(n, n + 10) = n × (n + 10)

⇒ 10 × x = 10k × 10(k + 1)

⇒ 10 × x = 100k(k + 1)

⇒ x = 10k(k + 1)

Since x is a 2-digit number,

10 ≤ 10k(k + 1) ≤ 99

⇒ 1 ≤ k(k + 1) ≤ 9.9

⇒ 1 ≤ k(k + 1) ≤ 9

Checking values of k,

For k = 1, k(k + 1) = 1 × 2 = 2

For k = 2, k(k + 1) = 2 × 3 = 6

For k = 3, k(k + 1) = 3 × 4 = 12 (not possible as it exceeds 9)

Possible values of k are 1 and 2.

Corresponding values of x are:

For k = 1, x = 10 × 1 × 2 = 20

For k = 2, x = 10 × 2 × 3 = 60

Hence, there are only two values of x: 20 and 60.

The correct answer is option 2.

Calculation:

Value putting approach:

Let's put a = 2,

a4 + 2a3 + 3a2 + 2a + 1

⇒ 2⁴ + 2(2)³ + 3(2)² + 2(2) + 1 = 49

a6 - 2a3 + 1

⇒ (2)⁶ - 2(2)³ + 1 = 49

So, HCF of 49 and 49 is 49.

Now, check options:

Option (1):

a3 + 3a2 + 2a + 1

⇒ (2)³ + 3(2)² + 2(2) + 1 = 25 (false)

Option (2):

a3 + a2 + a + 1

⇒ (2)3 + (2)2 + 2 + 1 = 15 (false)

Option (3):

(a2 + a + 1)2

⇒ (2² + 2 + 1)² = (4 + 2 + 1)² = 7² = 49 (true)

∴ The correct answer is option (3).

Additional Information

Here, we can not put a = 0 and 1 because at a = 0, all values become 1 and at a = 1, the second expression becomes negative.

Given:

The quadratic equation is:

\(x² - (k - 2)x + (k + 1) = 0\)

Roots are equal.

Formula used:

For a quadratic equation ax² + bx + c = 0, the roots are equal if:

\(D = b² - 4ac = 0\)

Here, a = 1, b = -(k - 2), c = k + 1.

Calculations:

Discriminant (D) = b² - 4ac

⇒ \(D = [-(k - 2)]² - 4 × 1 × (k + 1)\)

⇒ \(D = (k - 2)² - 4(k + 1)\)

⇒ \(D = k² - 4k + 4 - 4k - 4\)

⇒ \(D = k² - 8k\)

⇒ Set D = 0:

⇒ \(k² - 8k = 0\)

⇒ \(k(k - 8) = 0\)

⇒ \(k = 0\) or \(k = 8\)

The valid values of k are 0 and 8.

∴ The correct answer is option 2.

Given:

\(\dfrac{\cos\theta - \sin\theta + 1}{\cos\theta + \sin\theta - 1} \times (\cot\theta - \csc\theta)\)

Formula used:

\(\cot\theta = \dfrac{\cos\theta}{\sin\theta}, cosec\theta = \dfrac{1}{\sin\theta}\)

Calculations:

⇒ Rewrite \(\cot\theta - cosec\theta\)

⇒ \(\cot\theta - cosec\theta = \dfrac{\cos\theta}{\sin\theta} - \dfrac{1}{\sin\theta} = \dfrac{\cos\theta - 1}{\sin\theta}\)

⇒ \(\dfrac{\cos\theta - \sin\theta + 1}{\cos\theta + \sin\theta - 1} \times \dfrac{\cos\theta - 1}{\sin\theta}\)

 

⇒ \(\dfrac{(\cos\theta - \sin\theta + 1)(\cos\theta - 1)}{(\cos\theta + \sin\theta - 1)(\sin\theta)}\)

⇒ Factorize terms:

⇒ \(\dfrac{-(\cos\theta + \sin\theta - 1)}{\cos\theta + \sin\theta - 1}\)

⇒ Cancel common terms:

⇒ -1

∴ The value is -1.

Given:

\(\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}\)

Formula used:

\(\sin^3 \theta = \sin \theta \cdot \sin^2 \theta \) and \(cos^3 \theta = \cos \theta \cdot \cos^2 \theta\)

\(sin^2 \theta + \cos^2 \theta = 1\)

Calculations:

Numerator: \(sin \theta - 2 \sin^3 \theta\)

⇒ \(sin \theta (1 - 2 \sin^2 \theta)\)

⇒ \(sin \theta (1 - 2(1 - \cos^2 \theta)\)

⇒ \(sin \theta (2 \cos^2 \theta - 1)\)

Denominator: \((2 \cos^3 \theta - \cos \theta)\)

⇒ \(cos \theta (2 \cos^2 \theta - 1))\)

Now simplify the expression:

⇒ \(\frac{\sin \theta (2 \cos^2 \theta - 1)}{\cos \theta (2 \cos^2 \theta - 1)}\)

⇒ \((\frac{\sin \theta}{\cos \theta})\)

⇒ \(tan \theta\)

∴ The correct answer is tanθ.