Quantitative Aptitude PYQs

Given:

Expression: 4x² - kx + 1 > 0.

Statements:

Statement I: \(k < -2\).

Statement II: \(k > -4\).

Formula used:

For a quadratic equation ax² + bx + c, it is positive when:

1. Discriminant (D = b² - 4ac) is less than 0 (no real roots).
2. Coefficient of x² is positive (upward parabola).

Calculations:

Discriminant: D = (-k)² - 4(4)(1) = k² - 16.

For 4x² - kx + 1 > 0 to be positive:

⇒ k² - 16 < 0.

⇒ \(-4 < k < 4\).

Analyzing the

Statements:

Statement I: \(k < -2\).

This alone is insufficient, as k must also satisfy \(k > -4\).

Statement II: \(k > -4\).

This alone is insufficient, as k must also satisfy \(k < 4\).

Combining both

Statements:

From Statement I: \(k < -2\).

From Statement II: \(k > -4\).

⇒ \(-4 < k < -2\) satisfies the condition k² - 16 < 0.

∴ The question can be answered using both statements together but not with either statement alone. Correct option: 3.

Given:

Statement-I: A and B together can finish the work in 24 days

Statement-II: B and C together can finish the work in 36 days

Formula Used:

Work done by A and B together in 1 day = 1/24

Work done by B and C together in 1 day = 1/36

Work done by A, B, and C together in 1 day = Work done by A and B together + Work done by B and C together - Work done by B alone

Calculation:

Let the work done by A, B, and C together in 1 day be 1/x.

From Statement-I:

Work done by A and B together in 1 day = 1/24

From Statement-II:

Work done by B and C together in 1 day = 1/36

We need to find the work done by A, B, and C together in 1 day:

Let the work done by B alone in 1 day be y.

So, we have:

Work done by A, B, and C together in 1 day = Work done by A and B together + Work done by B and C together - Work done by B alone

⇒ \(\frac{1}{x} = \frac{1}{24} + \frac{1}{36} - y\)

Without knowing the value of y (work done by B alone in 1 day), we cannot determine the exact value of 1/x (work done by A, B, and C together in 1 day).

The Question cannot be answered even by using both the Statements together.

∴ The correct answer is option 4.

Given:

Statement-I : x2 - 26x + 133 = 0

Statement-II : x2 - 44x + 475 = 0

Formula Used:

For a quadratic equation ax² + bx + c = 0 , the roots can be found using the quadratic formula:

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

Calculation:

For Statement-I:

x2 - 26x + 133 = 0

a = 1, b = -26, c = 133

Discriminant D = b² - 4ac

⇒ D = (-26)² - 4(1)(133)

⇒ D = 676 - 532

⇒ D = 144

Roots: x = \(\frac{26 \pm \sqrt{144}}{2}\)

⇒ x = 38/2 = 19 or x = 14/2 = 7

For Statement-II:

x2 - 44x + 475 = 0

a = 1, b = -44, c = 475

Discriminant D = b² - 4ac

⇒ D = (-44)² - 4(1)(475)

⇒ D = 1936 - 1900

⇒ D = 36

Roots: x = \(\frac{44 \pm \sqrt{36}}{2}\)

⇒ x = 50/2 = 25 or x = 38/2 = 19

Both statements have a common prime solution x = 19.

We have used both statements to come to this conclusion.

∴ The correct answer is option 3.

Given:

Question: Is 327ⁿ + 173ⁿ divisible by 500?

Statement-I: n is odd natural number

Statement-II: n is a positive integer

Formula Used:

For a number to be divisible by 500, it must be divisible by both 5 and 100.

We will use modulo arithmetic to solve this problem.

Calculation:

Statement-I: n is odd natural number

Consider modulo 5:

327 ≡ 2 (mod 5)

173 ≡ 3 (mod 5)

So, 327ⁿ + 173ⁿ ≡ 2ⁿ + 3ⁿ (mod 5)

For n being an odd natural number:

2ⁿ + 3ⁿ ≡ 2 + 3 ≡ 0 (mod 5)

Hence, 327ⁿ + 173ⁿ is divisible by 5.

Consider modulo 4:

327 ≡ 3 (mod 4)

173 ≡ 1 (mod 4)

So, 327ⁿ + 173ⁿ ≡ 3ⁿ + 1ⁿ (mod 4)

Since 327n + 173n is divisible by both 4 and 5, it must be divisible by 20.

Consider modulo 25:

327 ≡ 2 (mod 25)

173 ≡ 23 (mod 25)

So, 327n + 173n ≡ 2n + 23n (mod 25)

Statement-II: n is a positive integer

Using this statement we can not say for certain that 327n + 173n is divisible by 500.

 

The correct answer is option 1.

Given:

Question: If the price of petrol goes up by 20%, by what percentage should the consumption be reduced so that the expenditure remains the same?

Statement-I: Price of petrol per litre was Rs. 90

Statement-II: Consumption was 24 litres before the price hike

Formula Used:

To keep the expenditure the same, the formula is:

New Consumption = (Old Price × Old Consumption) / New Price

Calculation:

Let the old price per litre be P and the old consumption be C.

New price per litre = P + 0.20P = 1.20P

New Consumption = (P × C) / (1.20P)

New Consumption = C / 1.20

Percentage reduction in consumption = [(C - (C / 1.20)) / C] × 100

Percentage reduction in consumption = [1 - (1 / 1.20)] × 100

Percentage reduction in consumption = [(1.20 - 1) / 1.20] × 100

Percentage reduction in consumption = (0.20 / 1.20) × 100

Percentage reduction in consumption = 0.1667 × 100

Percentage reduction in consumption ≈ 16.67%

We can answer is question without using any statements.

The correct answer is option 4.

Given:

The ratio of P's salary to Q's salary is 6 : 5.

Formula Used:

Ratio of P's saving to Q's saving = 3 : 2

Ratio of P's expenditure to Q's expenditure = 1 : 1

Calculation:

Let P's salary = 6x and Q's salary = 5x.

Statement I:

Let P's saving = 3y and Q's saving = 2y.

Therefore, P's expenditure = 6x - 3y and Q's expenditure = 5x - 2y.

Statement II:

Given the ratio of P's expenditure to Q's expenditure is 1 : 1, we have:

Expenditure of P and Q is equal.

Using both statements together:

P's expenditure = 6x - 3y and Q's expenditure = 5x - 2y.

P's expenditure to Q's expenditure is 1 : 1

6x - 3y = 5x - 2y

⇒ 6x - 5x = 3y - 2y

⇒ x = y

Using Statement I:

P's expenditure = 6x - 3y = 6x - 3x = 3x.

Therefore, P's expenditure = 3x.

Without the value of salary or savings we can not find the exact value of expenditure, although we can find the expenditure value in terms of salary or savings.

∴ The correct answer is option 4.

Given:

The largest of five different integers = 8

The least of five different integers = 2

Formula Used:

Average of integers = Sum of integers / Number of integers

Calculation:

Let the five different integers be a, b, c, d, and e where 2 ≤ a < b < c < d < 8.

Statement-I: The sum of all the 5 integers is a multiple of 5.

⇒ The sum of the five integers must end in 0 or 5 to be a multiple of 5.

Let’s consider possible sets of integers:

{2, 3, 4, 5, 6}

{2, 3, 5, 7, 8}

{2, 4, 5, 6, 8}

{3, 4, 5, 6, 7}

{4, 5, 6, 7, 8}

Different sets will have different averages.

Statement II: The number of odd integers is odd.

If the number of odd integers is odd, we can have 1, 3, or 5 odd numbers,

Possible sets satisfying both conditions could be:

Set 1: {2, 3, 5, 7, 8} (Odd count = 3)

Set 2: {2, 4, 5, 6, 8} (Odd count = 1)

Set 3: {2, 3, 4, 6, 8} (Odd count = 1) and so on...

In this case, we will have a lot of sets with different averages. 

Both statements together:

If the number of odd integers is odd, and The sum of all the 5 integers is a multiple of 5.

we can have 1, 3, or 5 odd numbers,

Possible sets satisfying both conditions could be:

Set 1: {2, 3, 5, 7, 8} (Odd count = 3)

Set 2: {2, 4, 5, 6, 8} (Odd count = 1)

Checking sum divisibility by 5:

Set 1: Sum = 2 + 3 + 5 + 7 + 8 = 25 (divisible by 5)

Set 2: Sum = 2 + 4 + 5 + 6 + 8 = 25 (divisible by 5)

Average for both sets:

Average = Sum of integers / 5

⇒ Average = 25 / 5

⇒ Average = 5

The correct answer is option 3.

Given:

There are three different weights such that:

1. All weights are integers.
2. Their sum is a prime number.

Statement-I: One of the weights is twice the another weight

Let the weights be x, 2x, and y

Sum of weights = 3x + y = prime number

We can not find the value of weights using this statement.

tatement-II : One of the weights is thrice the another weight

Let the weights be x, 3x, and y

Sum of weights = 4x + y = prime number

We can not find the value of weights using this statement.

Using both

Statements:

If one weight is twice another and one is thrice another:

There can be 2 cases:

1. Let the weights be x, 2x, and 3x.

Sum of weights: x + 2x + 3x = 6x.

For the sum to be a prime number, x must satisfy 6x = Prime.

⇒ If x = 1, 6 × 1 = 6 (not a prime).

⇒ If x = 2, 6 × 2 = 12 (not a prime).

⇒ If x = 3, 6 × 3 = 18 (not a prime).

⇒ For any x where 6x = Prime, no integer x satisfies the condition.

2. Let the weights be x, 2x, and 6x.

Sum of weights: x + 2x + 6x = 9x.

For the sum to be a prime number, x must satisfy 9x = Prime.

⇒ If x = 1, 9 × 1 = 9 (not a prime).

⇒ If x = 2, 9 × 2 = 18 (not a prime).

⇒ If x = 3, 9 × 3 = 27 (not a prime).

⇒ For any x where 9x = Prime, no integer x satisfies the condition.

The given conditions are inconsistent, and no valid solution exists based on the provided information.

∴ The correct answer is option 4.

Given:

Question: What is the amount at the end of 10 years?

Statement-I: The principal amount is ₹1,00,000.

Statement-II: The rate of interest is 10% per annum.

Calculations:

To calculate the amount, we need:

1. The principal amount (P)
2. The rate of interest (r)
3. The time period (t)

From Statement-I: Principal amount is given, but rate of interest is missing.

From Statement-II: Rate of interest is given, but principal amount is missing.

Both statements are required to calculate the amount using the formula.

∴ The correct answer is option (3): The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.

Additional Information

When the question doesn’t specify whether the interest is compound or simple, it’s generally safe to assume it’s simple interest unless otherwise stated. Simple interest is more straightforward and commonly used in basic financial problems.

Given:

p, q are integers.

Calculation:

Statement I: p + q is even.

If (p + q) is even, it means that both p and q must either be both even or both odd. This is because the sum of two even numbers is even, and the sum of two odd numbers is also even.

if p and q are even, then p2 + pq + q2 is even.

if p and q are odd, then p2 + pq + q2 is odd.

So, we can not conclude that p2 + pq + q2 is odd.

Statement II: p q is odd.

For pq to be odd, both p and q must be odd integers.

i.e. If p and q are odd, then p2 + pq + q2 is odd.

Thus, we can conclude that p2 + pq + q2 is odd.

Conclusion:

Since we can answer using statement (II), but can not answer using statement (I), So the question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone

∴ The correct answer is option (1).

Given:

• Total population = 10,000 where males = females = 5,000
• 30% of the total are newspaper readers
• 1/3 rd of total newspaper readers are English newspaper readers
• 20% of total English newspaper readers are females

 

Calculation:

According to the given data,

⇒ Total newspaper reader = 30% × 10,000 = 3,000

⇒ Total English newspaper readers = 1/3 × 3,000 = 1,000

⇒ Female who read English newspapers = 20% × 1000 = 200

Hence, Males who read English newspapers = 1000 - 200 = 800

∵ Total Males = 5,000

∴ Total Males who do not read English newspapers = 5000 - 800 = 4200.

Given: 

The perimeter of the rectangle is 400 cm.

Formula Used: 

Perimeter of a rectangle: P = 2 (l + b)

Area of a rectangle: A = l × b

For a fixed perimeter, area of a rectangle is maximum when it is a square, i.e. l = b

Calculation:

According to the formula,

P = 2 (l + b) = 400

⇒ 400 = 2 (l + b) 

⇒ l + b = 200

To Maximize the area, set l = b

⇒ 2l = 200 

⇒ l = 100

The area is A = l × b = 100 × 100 = 10,000 cm².

∴ The maximum area of the rectangle is 10000 cm².

Given:

Base = 3, Exponent = 10, Modulus = 7

Formula used:

The remainder when a number is divided by 7 is equivalent to the result of raising the base to the exponent, then finding the remainder when divided by 7, i.e., \((a^b) \mod m\)

Calculations:

3¹⁰ mod 7 = ?

We can first reduce the powers of 3 modulo 7:

3¹ mod 7 = 3

3² mod 7 = 3 × 3 = 9 mod 7 = 2

3³ mod 7 = 3 × 2 = 6 mod 7 = 6

3⁴ mod 7 = 3 × 6 = 18 mod 7 = 4

3⁵ mod 7 = 3 × 4 = 12 mod 7 = 5

3⁶ mod 7 = 3 × 5 = 15 mod 7 = 1

So, 3⁶ ≡ 1 (mod 7)

3¹⁰ = 3⁶ × 3⁴ ≡ 1 × 4 = 4 (mod 7)

∴ The remainder when 3¹⁰ is divided by 7 is 4.

Calculation:

First, express 64% as a fraction = \(\frac{64}{100}\)

Taking the square root of, we get

⇒ √\(\frac{64}{100}\) = \(\frac{8}{10}\)

⇒ \(\frac{8}{10}\)× 100 = 80%

∴ The square root of 64% is 80%.

Concept:

A number is divisible by 3 if and only if the sum of its digits is divisible by 3 or a multiple of 3.

For example,

123 is divisible by 3 because 1 + 2 + 3 = 6, which is divisible by 3 and also the multiple of 3

Calculation:

According to the question,

Difference of (1031 - 5) and (1030 + p) can be written as,

⇒ (1031 - 5) - (1030 + p)

⇒ 1031 - 10³⁰ - 5 - p

⇒ 10 × 10³⁰ - 10³⁰ - (5 + p)

⇒ 10³⁰(10 - 1) - (5 + p)

⇒ 9 × 10³⁰ - (5 + p)

To make the above value divisible by 3, both these values should be multiple of 3.

Where 9 × 10³⁰ is a multiple of 3, it should be divisible.

(5 + p):

⇒ p gets only 3 possible values.

∴ The correct answer is 3.

Concept used:

aⁿ - bⁿ is divisible by (a + b) and (a - b) if n = even number

aⁿ + bⁿ is divisible by (a+ b) if n = odd number

Calculation:

Statement I:

107100 - 76100 = aⁿ - bⁿ

It is divisible by (a + b)

So, (a + b) = (107 + 76) = 183 and 183 is divisible by 61

Thus, 107100 - 76100 is also divisible by 61. (true)

Statement II:

675 + 335 = aⁿ + bⁿ

It is divisible by (a + b)

So, (a + b) = (67 + 33) = 100

Thus, 675 + 335 is divisible by 100.  (true)

∴ The correct answer is option (3).

Given:

The average temperature from Monday to Sunday (7 days) is 31°C.

The lowest temperature recorded is 30°C.

Calculation:

According to the question,

Total temperature of seven days = 31° × 7 

⇒ D₁ + D₂ + D₃ + D₄ + D₅ + D₆ + D₇  = 217°

The lowest temperature recorded is 30°C

Let all si× days temperature be 30° C          

⇒ 30° + 30° + 30° + 30° + 30° + 30° + D7 = 217°

⇒ D₇ = 217° - 180°

⇒ D₇ = 37° 

∴ The ma×imum temperature is 37°

Given:

x + z ≠ 2y

Calculation:

\(\left(x+\frac{1}{y z}\right)-\left(y+\frac{1}{z x}\right) = \) \(\left(y+\frac{1}{z x}\right)-\left(z+\frac{1}{x y}\right) \)

⇒ \(\left(x+\frac{1}{y z}\right)+\left(z+\frac{1}{xy}\right) = \) \(\left(y+\frac{1}{z x}\right)+\left(y+\frac{1}{z x}\right)\)

⇒ \(\frac{(xyz + 1)}{yz} + \frac{(xyz + 1)}{xy}\) = 2(\(\frac{xyz+1}{zx}\))

⇒ \(\frac{(x^2yz + x +xyz^2+z)}{xyz}\) = 2(\(\frac{xyz+1}{zx}\))

⇒ x²yz + x + xyz² + z = 2 × y (xyz+ 1)

⇒ x2yz + x + xyz2 + z = 2xy²z + 2y

⇒ x2yz + xyz2 - 2xy²z = - x - z + 2y

⇒ xyz (x + z - 2y) = -1(x + z - 2y)

⇒ xyz = -1

∴ The correct answer is option (2).

Given:

p = n(n + 1) (n + 2) (n + 3) + 1

Calculation:

Statement I:

Since there are four consecutive numbers, at least two of them are even.

Thus, (n(n + 1)(n + 2)(n + 3)) is always an even number because the product of at least one or more even numbers is even.

Adding 1 to an even number results in an odd number.

Hence statement I is true.

Statement II:

p = n(n + 1) (n + 2) (n + 3) + 1

For n = 1, p = 1.2.3.4 + 1 = 24 + 1 = 25 

Which is equal to (5²)

For n = 2, p = 2.3.4.5 + 1 = 120 + 1 = 121 

Which is equal to (112)

For n = 3, p = 3.4.5.6 + 1 = 360 + 1 = 361

Which is equal to (192)

From the calculations above, we observe that pp p

 appears to be a perfect square for several values of n.n.

Hence statement II is true.

Option 3 is the correct answer.

Formula Used:

From Arithmetic Progression,

S_n = \(n \over 2\) [a + l]

Calculation:

Average of the first 50 even natural numbers,

2 + 4 + 6 +....... + 100

where,

a = 2, l = 100 and n = 50

S_n = \(50 \over 2\) × [2 + 100]

S_n = 25 × 102 = 2550 

It's average = 2550/50 = 51.

Average of the first 50 odd natural numbers,

1 + 3 + 5 + ..... + 99

where,

a = 1, l = 99 and n = 50

S_n = \(50 \over 2\) × [1 + 99]

S_n = 25 × 100 = 2500

It's average = 2500/50 = 50.

Now, the difference of both average values = 51 - 50 = 1.

∴ The correct answer is 1.