Quantitative Aptitude PYQs

Calculation:

Given,

The frequency distribution of the height of students is given as follows:

We need to find the total number of students whose height is less than or equal to 165 cm. Therefore, we will sum the students in the following height ranges:

160-162 cm: 12 students

162-164 cm: 15 students

164-166 cm: 24 students

The total number of students is: 12 + 15 + 24 = 51 

Hence, the correcr answer is Option 3

Calculation:

Given,

The frequency distribution of the height of students in the class is as follows:

Height (in cm) | Number of Students

 

Calculate the cumulative frequency.

Cumulative frequency for the 160–162 cm range: 12

Cumulative frequency for the 162–164 cm range: 12 + 15 = 27

Cumulative frequency for the 164–166 cm range: 27 + 24 = 51

Cumulative frequency for the 166–168 cm range: 51 + 13 = 64

The total number of students is 64, so the median class contains the 32^nd student. The cumulative frequency before 32 is 27 (for the 162–164 cm class), and after 32 is 51 (for the 164–166 cm class). Therefore, the median class is 164–166 cm.

The formula for finding the median in a grouped frequency distribution is:

\( \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h \)

L = lower boundary of the median class = 164

N = total number of students = 64

F = cumulative frequency before the median class = 27

f = frequency of the median class = 24

h = class width = 2

Substituting the values:

\( \text{Median} = 164 + \left( \frac{32 - 27}{24} \right) \times 2 = 164 + \left( \frac{5}{24} \right) \times 2 = 164 + 0.42 = 164.42 \)

∴ The median height is approximately 164.41 cm.

Hence, the correct answer is Option 3.

Calculation:

Given the frequency distribution of height of students:

The class interval with the highest frequency is 164–166, with 24 students.

The formula for calculating the mode for grouped data is:

\( \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)

Where:

L: Lower boundary of the modal class = 164

f₁: Frequency of the modal class = 24

f₀: Frequency of the class before the modal class = 15

f₂: Frequency of the class after the modal class = 13

h: Class width = 2

Substituting the values into the formula:

\( \text{Mode} = 164 + \left( \frac{24 - 15}{2(24) - 15 - 13} \right) \times 2 \)

\( = 164 + \left( \frac{9}{48 - 28} \right) \times 2 \)

\( = 164 + \left( \frac{9}{20} \right) \times 2 \)

\( = 164 + 0.9 \)

\( = 164.9 \)

∴ The mode is approximately 164.9 cm.

Hence, the correct answer is Option 4.

Solution:

The appropriate graphical representation for the given frequency distribution is a Histogram. The bar chart above shows the frequency of students' heights within the specified ranges. The highest frequency is for the height range of 164-166 cm, which has 24 students.

Calculation:

Given,

The sum and sum of squares of the observations corresponding to length X in cm and weight Y in gm of 50 tropical tubers are given as:

\(\Sigma X\) = 200, \(\Sigma Y\) = 250, \(\Sigma X^2\)= 900, \(\Sigma Y^2\) = 1400 

The formula for variance is:

\( \text{Variance} = \frac{\Sigma X^2}{N} - \left( \frac{\Sigma X}{N} \right)^2 \)

Where N = 50 is the number of observations.

Variance of X:

\( \text{Variance of } X = \frac{\Sigma X^2}{N} - \left( \frac{\Sigma X}{N} \right)^2 \)

Substituting the given values:

\( \text{Variance of } X = \frac{900}{50} - \left( \frac{200}{50} \right)^2 \)

\( \text{Variance of } X = 18 - 16 = 2 \)

Variance of Y:

\( \text{Variance of } Y = \frac{\Sigma Y^2}{N} - \left( \frac{\Sigma Y}{N} \right)^2 \)

Substituting the given values:

\( \text{Variance of } Y = \frac{1400}{50} - \left( \frac{250}{50} \right)^2 \)

\( \text{Variance of } Y = 28 - 25 = 3 \)

Conclusion:

\( \text{Variance of } X < \text{Variance of } Y \)

Hence, the correct answer is Option 2.

Calculation:

Explanation:

The Coefficient of Variation (C.V.) formula is given as:

\( C.V.(X) = \frac{\sigma_X}{M_X} \times 100 \)

Substituting values:

\( C.V.(X) = \frac{\sqrt{2}}{50} \times 100 = 2\sqrt{2} \)

The Coefficient of Variation for Y is:

\( C.V.(Y) = \frac{\sqrt{3}}{50} \times 100 = 2\sqrt{3} \)

∴ C.V.(X) < C.V.(Y)

Hence, the correct answer is Option 2.

Calculation:

Given,

Let X be a random variable following a binomial distribution with parameters n = 6 and p = k.

Further, it is given that:

\( 9P(X = 4) = P(X = 2) \)

The probability mass function for a binomial distribution is:

\( P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x} \)

For P(X = 4) , we have:

\( P(X = 4) = \binom{6}{4} k^4 (1 - k)^2 = 15 k^4 (1 - k)^2 \)

For P(X = 2) , we have:

\( P(X = 2) = \binom{6}{2} k^2 (1 - k)^4 = 15 k^2 (1 - k)^4 \)

We are given that:

\( 9P(X = 4) = P(X = 2) \)

Substitute the expressions for P(X = 4)  and P(X = 2) :

\( 9 \times 15 k^4 (1 - k)^2 = 15 k^2 (1 - k)^4 \)

Cancel the common factor of 15:

\( 9 k^4 (1 - k)^2 = k^2 (1 - k)^4 \)

\( 9 k^2 = (1 - k)^2 \)

\( 9 k^2 = 1 - 2k + k^2 \)

 

\( 8 k^2 + 2k - 1 = 0 \)

Use the quadratic formula to solve for k :

\( k = \frac{-2 \pm \sqrt{2^2 - 4 \times 8 \times (-1)}}{2 \times 8} \)

\( k = \frac{-2 \pm \sqrt{4 + 32}}{16} \)

\( k = \frac{-2 \pm \sqrt{36}}{16} \)

\( k = \frac{-2 \pm 6}{16} \)

Thus, the two possible values for k are:

\( k = \frac{4}{16} = \frac{1}{4} \) or \( k = \frac{-8}{16} = -\frac{1}{2} \)

Since k  represents a probability, it must be between 0 and 1. Therefore, the valid solution is:

\( k = \frac{1}{4} \)

Hence, the correct answer is Option 3.

Calculation:

Given,

Let X  be a random variable following a binomial distribution with parameters n = 6 and p = \(\frac{1}{4} \)

The probability mass function for a binomial distribution is:

\( P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x} \)

For P(X = 3) , we have:

\( P(X = 3) = \binom{6}{3} \left( \frac{1}{4} \right)^3 \left( \frac{3}{4} \right)^3 \)

Substitute the values:

\( P(X = 3) = 20 \times \frac{1}{64} \times \frac{27}{64} \)

Now simplify the equation:

\( P(X = 3) = 20 \times \frac{27}{4096} = \frac{540}{4096} \)

Simplifying the fraction:

\( P(X = 3) = \frac{135}{1024} \)

Hence, the correct answer is Option 1

Calculation:

Given,

A committee of 6 members is to be formed from a group of 7 gentlemen and 4 ladies.

The total number of ways to form the committee of 6 members is:

\( \binom{11}{6} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 462 \)

The number of ways to form a committee with exactly 3 gentlemen and 3 ladies is:

\( \binom{7}{3} \times \binom{4}{3} = 35 \times 4 = 140 \)

The probability of forming a committee with exactly 3 gentlemen is:

\( \text{Probability} = \frac{140}{462} = \frac{10}{33} \)

Hence, the correct answer is Option 1.

Calculation:

Given,

A committee of 6 members is formed from a group of 7 gentlemen and 4 ladies.

The total number of ways to form the committee is:

\( \binom{11}{6} = 462 \)

The favorable outcomes (at least 2 ladies) are calculated by the sum of the cases:

\( \binom{4}{2} \times \binom{7}{4} + \binom{4}{3} \times \binom{7}{3} + \binom{4}{4} \times \binom{7}{2} \)

This simplifies to:

\( \binom{4}{2} \times \binom{7}{4} = 6 \times 35 = 210 \)

\( \binom{4}{3} \times \binom{7}{3} = 4 \times 35 = 140 \)

\( \binom{4}{4} \times \binom{7}{2} = 1 \times 21 = 21 \)

The total favorable outcomes are:

\( 210 + 140 + 21 = 371 \)

The probability is:

\( P(E) = \frac{371}{462} = \frac{53}{66} \)

Hence, the correct answer is Option 4.

Calculation:

Given,

The probability that A , B , and C  become managers are:

\( P(A) = \frac{3}{10}, , P(B) = \frac{1}{2}, , P(C) = \frac{4}{5} \)

The conditional probabilities that the bonus scheme is introduced are:

\( P(D|A) = \frac{4}{9}, , P(D|B) = \frac{2}{9}, , P(D|C) = \frac{1}{3} \)

The total probability that the bonus scheme will be introduced is given by:

\( P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) \)

Substituting the values:

\( P(D) = \left(\frac{4}{9} \times \frac{3}{10}\right) + \left(\frac{2}{9} \times \frac{1}{2}\right) + \left(\frac{1}{3} \times \frac{4}{5}\right) \)

Now, simplifying the terms:

\( P(D) = \frac{12}{90} + \frac{2}{18} + \frac{4}{15} \)

The least common denominator (LCD) is 90. So:

\( P(D) = \frac{12}{90} + \frac{10}{90} + \frac{24}{90} \)

Adding the fractions gives:

\( P(D) = \frac{46}{90} = \frac{23}{45} \)

Hence, the correct answer is Option 3.

Calculation:

Given,

The probability that A , B , and C  become managers are:

\( P(A) = \frac{3}{10}, , P(B) = \frac{1}{2}, , P(C) = \frac{4}{5} \)

The conditional probabilities that the bonus scheme is introduced are:

\( P(D|A) = \frac{4}{9}, , P(D|B) = \frac{2}{9}, , P(D|C) = \frac{1}{3} \)

We need to find the probability that B  is the manager given that the bonus scheme has been introduced using Bayes' Theorem:

\( P(B|D) = \frac{P(D|B)P(B)}{P(D)} \)

First, calculate the total probability P(D)  that the bonus scheme is introduced:

\( P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) \)

Substitute the values:

\( P(D) = \left( \frac{4}{9} \times \frac{3}{10} \right) + \left( \frac{2}{9} \times \frac{1}{2} \right) + \left( \frac{1}{3} \times \frac{4}{5} \right) \)

Simplifying:

\( P(D) = \frac{12}{90} + \frac{2}{18} + \frac{4}{15} \)

Find a common denominator (LCD = 90):

\( P(D) = \frac{12}{90} + \frac{10}{90} + \frac{24}{90} = \frac{46}{90} = \frac{23}{45} \)

Now, use Bayes' Theorem to find P(B|D) :

\( P(B|D) = \frac{P(D|B)P(B)}{P(D)} = \frac{\left( \frac{2}{9} \times \frac{1}{2} \right)}{\frac{23}{45}} \)

Simplifying:

\( P(B|D) = \frac{\frac{2}{18}}{\frac{23}{45}} = \frac{\frac{1}{9}}{\frac{23}{45}} = \frac{5}{23} \)

The probability that the manager appointed was B , given that the bonus scheme has been introduced, is 5/23.

Hence, the correct answer is Option 1.

Calculation:

Given,

Number of observations, n = 100

Original arithmetic mean, \( \bar{x} = 50 \)

Transformation applied to each observation:

New value, \( x' = \frac{x - 5}{20} \)

The new arithmetic mean is,

\( \bar{x}' = \frac{\bar{x} - 5}{20} \)

\( \bar{x}' = \frac{50 - 5}{20} = \frac{45}{20} = 2.25 \)

∴ The new arithmetic mean is 2.25.

Hence, the correct answer is Option 1.

Calculation:

Given,

Number of observations, n = 100

Original standard deviation, \( \sigma = 10 \)

Transformation applied to each observation:

New value, \( x' = \frac{x + 5}{20} \)

The additive constant (5) does not affect the standard deviation, and scaling by 20 divides the standard deviation by 20, so the new standard deviation is,

\( \sigma' = \frac{\sigma}{20} \)

\( \sigma' = \frac{10}{20} = 0.5 \)

∴ The new standard deviation is 0.5.

Hence, the correct answer is Option 2.

Calculation:

Given,

P(A) = 1/3

P(B) = 1/2

P(A ∩ B) = 1/4

We need to calculate the probability P(B | A^C), the probability that B occurs given that A does not occur.

Using the formula for conditional probability:

P(B | A^C) = P(B ∩ A^C) / P(A^C)

Now,

P(A^C) = 1 - P(A) = 1 - 1/3 = 2/3

P(B ∩ A^C) = P(B) - P(A ∩ B) = 1/2 - 1/4 = 1/4

Substitute the values:

P(B | A^C) = (1/4) / (2/3) = (1/4) × (3/2) = 3/8

Hence, the correct answer is Option 2.

Calculation:

Given,

\( P(A) = \frac{1}{3} \)

\( P(B) = \frac{1}{2} \)

\( P(A \cap B) = \frac{1}{4} \)

We need to calculate\( P(A^C \cap B^C) \), the probability that neither A  nor B  occurs.

Using the complement rule:

\( P(A^C \cap B^C) = 1 - P(A \cup B) \)

Now, apply the inclusion-exclusion formula to find \(( P(A \cup B) \):

\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)

Substitute the given values:

\( P(A \cup B) = \frac{1}{3} + \frac{1}{2} - \frac{1}{4} \)

To simplify:

\( P(A \cup B) = \frac{4}{12} + \frac{6}{12} - \frac{3}{12} = \frac{7}{12} \)

Now, calculate \(P(A^C \cap B^C) \):

\( P(A^C \cap B^C) = 1 - \frac{7}{12} = \frac{5}{12} \)

Hence, the correct answer is Option 2.

Calculation:

Given,

Two fair six-sided dice are rolled.

Each die has faces numbered 1 through 6.

Total number of outcomes:

Since each die has 6 faces, the sample space size is

\(N = 6 \times 6 = 36\).

Favourable outcomes (sum > 7):

We list all ordered pairs ((i, j)) with (i + j > 7):

Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes

Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes

Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes

Sum = 11: (5,6), (6,5) → 2 outcomes

Sum = 12: (6,6) → 1 outcome

Total favourable outcomes = (5 + 4 + 3 + 2 + 1 = 15).

Probability calculation:

The probability that the sum is strictly greater than 7 is

\(P(\text{sum} > 7) = \frac{\text{favourable outcomes}}{\text{total outcomes}} = \frac{15}{36} = \frac{5}{12}\).

Hence, the correct answer is Option 2.

Calculation:

Given,

Probability of hitting the target per shot, \(p = \frac{1}{5}\)

Number of shots fired, \(n = 7\)

The number of hits follows a binomial distribution:

\(X \sim \mathrm{Binomial}(n=7,;p=\tfrac15)\)

Probability of zero hits:

\(P(X=0)=\left(\frac45\right)^{7}\)

Probability of exactly one hit:

\(P(X=1)=\binom71!\left(\frac15\right)!\left(\frac45\right)^{6} =\frac{7}{5}\left(\frac45\right)^{6}\)

Probability of at least two hits:

\(P(X\ge2)=1-\bigl[P(X=0)+P(X=1)\bigr] =1-\frac{11}{5}\left(\frac45\right)^{6} \)

∴ The probability of hitting the target at least twice is  \(1-\dfrac{11}{5}\left(\dfrac45\right)^{6} \).

Hence, the correct answer is Option 3.

Calculation:

Given,

Mean of the binomial distribution, \( \mu = 200 \)

Variance of the binomial distribution, \( \sigma^{2} = 160 \)

For a binomial random variable,

\( \mu = np \quad\text{and}\quad \sigma^{2} = np(1-p). \)

From the mean,

\( np = 200 ;\Longrightarrow; p = \dfrac{200}{n}. \)

Using the variance,

\( np(1-p) = 160 ;\Longrightarrow; n\bigl(\tfrac{200}{n}\bigr)!\Bigl(1-\tfrac{200}{n}\Bigr)=160 ;\Longrightarrow; 200\Bigl(1-\tfrac{200}{n}\Bigr)=160. \)

Simplifying,

\( 1-\tfrac{200}{n} = 0.8 ;\Longrightarrow; \tfrac{200}{n} = 0.2 ;\Longrightarrow; n = \dfrac{200}{0.2} = 1000. \)

∴ The number of trials is \( n = 1000 \).

Calculation:

Given,

The sequence is \(8^{2},,9^{2},,10^{2},,\dots,,15^{2}\)

Number of terms, \(n = 15 - 8 + 1 = 8\)

The sum of the first \(m\) squares is  \( \displaystyle \sum_{k=1}^{m} k^{2} = \frac{m(m+1)(2m+1)}{6}\)

Sum up to \(15^{2}\):

\(S_{15} = \frac{15 \times 16 \times 31}{6} = 1240\)

Sum up to \(7^{2}\):

\(S_{7} = \frac{7 \times 8 \times 15}{6} = 140\)

Sum of the required terms:

\(S = S_{15} - S_{7} = 1240 - 140 = 1100\)

The arithmetic mean is,

\( \displaystyle \text{Mean} = \frac{S}{n} = \frac{1100}{8} = 137.5\)

∴ The arithmetic mean is \(137.5\).

Hence, the correct answer is option 3.