Quantitative Aptitude PYQs

Calculation:

Combinations:

\({}^{10}C_3 = 120\)

\({}^{10}C_4 = 210\)

\({}^{10}C_5 = 252\)

\({}^{10}C_6 = 210\)

\({}^{10}C_7 = 120\)

Sum of reciprocals:

\( \frac{1}{120} + \frac{1}{210} + \frac{1}{252} + \frac{1}{210} + \frac{1}{120} = 2\left(\frac{1}{120} + \frac{1}{210}\right) + \frac{1}{252} \)

\( = 2\left(\frac{7 + 4}{840}\right) + \frac{1}{252} = \frac{11}{420} + \frac{1}{252} = \frac{33 + 5}{1260} = \frac{38}{1260} = \frac{19}{630} \)

Harmonic Mean:

\( \text{H.M.} = \frac{5}{\frac{19}{630}} = \frac{3150}{19} \)

Hence, the correct answer is Option 1.

Calculation:

The probability that one farmer is in debt is given as P = 0.60

Since the events are independent, the probability that three randomly selected farmers are all in debt is:

P³ = (0.60)³ = 0.216

Hence, the correct answer is Option 3.

Calculation:

Given:

\( P(L) = 0.68 \)

\( P(D) = 0.56 \)

\( P(L \cap D) = 0.48 \)

Using the formula for union of two events:

⇒ \( P(L \cup D) = P(L) + P(D) - P(L \cap D) \)

Substituting values:

⇒ \( P(L \cup D) = 0.68 + 0.56 - 0.48 = 0.76 \)

Hence, the correct answer is Option 2.

Calculation:

Given:

White balls = 10,Red balls= 5 ,Total balls= 15 

Total ways to draw 2 balls from 15:

\( \binom{15}{2}=\frac{15\times 14}{2}=105 \)

Favourable ways to draw 2 red balls from 5 red balls:

\( \binom{5}{2}=\frac{5\times 4}{2}=10 \)

Therefore, required probability:

\( P=\frac{\binom{5}{2}}{\binom{15}{2}}=\frac{10}{105}=\frac{2}{21} \)

Hence, the correct answert is Option 1.

Calculation:

Given:

White balls=5, Red balls = 6, Blue balls = 4

\( \text{Total balls}=5+6+4=15 \)

Total ways to draw 3 balls from 15:

\( \binom{15}{3}=\frac{15\times14\times13}{6}=455 \)

Favourable ways to draw one white, one red and one blue ball:

\( \binom{5}{1}\times\binom{6}{1}\times\binom{4}{1} =5\times6\times4=120 \)

Therefore, required probability:

\( P=\frac{120}{455}=\frac{24}{91} \)

Hence, the correct answer is Option 3.

Calculation:

For a binomial distribution, the required conditions are:

• The number of trials must be finite and fixed
• The trials must be independent
• Each trial must have exactly two possible outcomes (success/failure)
• Probability of success remains constant

Hence, the correct answer is Option 4.

Calculation:

We are asked to compute:

\( P(X \cap Y') + P(X' \cap Y) \)

Using the formula:

\( = P(X)[1 - P(Y)] + [1 - P(X)]P(Y) \)

Substituting values:

\( = \frac{4}{5} \left(1 - \frac{5}{6}\right) + \left(1 - \frac{4}{5}\right) \cdot \frac{5}{6} \)

\( = \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{5} \cdot \frac{5}{6} = \frac{4}{30} + \frac{5}{30} = \frac{9}{30} = \frac{3}{10} \)

Hence, the correct answer is Option 1.

Calculation:

Let:

\( P(P)=\frac{2}{3},\quad P(P\cap E)=\frac{11}{15},\quad P(P\cup E)=\frac{4}{5} \)

Using the formula:

⇒ \( P(P\cup E)=P(P)+P(E)-P(P\cap E) \)

Substitute the given values:

⇒ \( \frac{4}{5}=\frac{2}{3}+P(E)-\frac{11}{15} \)

Convert fractions to a common denominator:

⇒ \( \frac{4}{5}=\frac{12}{15},\quad \frac{2}{3}=\frac{10}{15} \)

Simplifying:

⇒ \( \frac{12}{15}=\frac{10}{15}+P(E)-\frac{11}{15} \)

⇒ \( \frac{12}{15}=P(E)-\frac{1}{15} \)

Therefore,

 \( P(E)=\frac{12}{15}+\frac{1}{15}=\frac{13}{15} \)

Hence, the correct answer is Option 2.

Calculation:

Let:

\( P(X) = p \)

\( P(Y) = q \)

X and Y are independent ⇒ \( P(X \cap Y) = pq \)

Probability of exactly one event occurring:

\( P(\text{exactly one}) = P(X) + P(Y) - 2P(X \cap Y) = p + q - 2pq \)

So, statement I is incorrect.

Probability of at least one event occurring:

\( P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = p + q - pq \)

So, statement II is also incorrect.

Hence, the Correct answer is Option 4.

Calculation:

Given:

Number of black faces = 3

Number of white faces = 2

Number of red faces = 1

Probability that black, white, and red appear in first, second, and third positions respectively:

\( P = \frac{3}{6} \cdot \frac{2}{6} \cdot \frac{1}{6} = \frac{6}{216} = \frac{1}{36} \)

Hence, the correct answer is Option 1.

Calculation:

Total number of outcomes for 4 coin tosses:

\( n(S) = 2^4 = 16 \)

Favorable outcomes with two consecutive heads:

⇒ \( E = { \text{HHTT, THHT, TTHH, HHHT, HTHH, HHTH, THHH, HHHH} } \)

⇒ \( n(E) = 8 \Rightarrow n(E') = 16 - 8 = 8 \)

Probability of not getting two consecutive heads:

⇒ \( P(E') = \frac{8}{16} = \frac{1}{2} \)

Hence, the correct answer is Option 4.

Calculation:

Given:

Prime numbers = {2, 3, 5} ⇒ 3 choices

Composite numbers = {4, 6} ⇒ 2 choices

Neither prime nor composite = {1} ⇒ 1 choice

Probability of selecting one prime, one composite, and one neither prime nor composite:

\( P = \frac{3}{6} \cdot \frac{2}{6} \cdot \frac{1}{6} \cdot 3! \)

Calculating:

\( = \frac{6}{216} \cdot 6 = \frac{36}{216} = \frac{1}{6} \)

Hence, the correct answer is Option 4.

Calculation:

Given:

Numbers divisible by 5 = 10

Numbers divisible by 9 = 5

Numbers divisible by both 5 and 9 = 1

Using the inclusion-exclusion principle:

Divisible by 5 or 9 \( = 10 + 5 - 1 = 14 \)

Total numbers = 50

Not divisible by 5 or 9 \( = 50 - 14 = 36 \)

Required probability:

\( P = \frac{36}{50} = \frac{18}{25} \)

Hence, the correct answer is Option 2.

Calculation:

Out of 50 consecutive natural numbers:

Even numbers = 25

Odd numbers = 25

Sum of one even and one odd number is always odd.

Number of ways to choose one even and one odd:

\( = 25 \cdot 25 = 625 \)

Total number of ways to choose any two numbers from 50:

\( = {}^{50}C_2 = \frac{50 \cdot 49}{2} = 1225 \)

Required probability:

\( P(\text{sum is odd}) = \frac{625}{1225} = \frac{25}{49} \)

Hence, the Correct answer is Option 4.

Calculation:

Given:

Standard deviation of 100 observations = 10

A constant value \(20\) is added to each observation.

Property of standard deviation:

Adding or subtracting a constant from each observation does not change the standard deviation

Hence,

New standard deviation = Old standard deviation = 10

Hence, the correct answer is Option 1.

Calculation:

Given:

\( n = 5 \)

\( p = k \), \( q = 1 - k \)

Binomial probabilities:

⇒ \( P(X = 1) = {}^5C_1 \cdot k \cdot (1 - k)^4 = 0.4096 \)

⇒ \( P(X = 2) = {}^5C_2 \cdot k^2 \cdot (1 - k)^3 = 0.2048 \)

Taking ratio:

⇒ \( \frac{{}^5C_2 \cdot k^2 \cdot (1 - k)^3}{{}^5C_1 \cdot k \cdot (1 - k)^4} = \frac{0.2048}{0.4096} \)

Simplifying:

⇒ \( \frac{10 \cdot k}{5 \cdot (1 - k)} = \frac{1}{2} \Rightarrow 4k = 1 - k \Rightarrow 5k = 1 \Rightarrow k = \frac{1}{5} = 0.2 \)

Hence, the correct answer is Option 1.

Calculation:

The given data is a grouped frequency distribution.

 

Compute the totals:

⇒ \( \sum f_i = 20+30+30+20 = 100 \)

⇒ \( \sum x_i f_i = 200+600+900+800 = 2500 \)

Mean of the distribution is given by:

⇒ \( \text{Mean} = \frac{\sum x_i f_i}{\sum f_i} \)

Substituting the values:

⇒ \( \text{Mean} = \frac{2500}{100} = 25 \)

Hence, the correct answer is Option 2.

Calculation:

Given:

 P(A) = 0.3 , P(B) = 0.4 and P( A |B)=0.5

Using the formula of conditional probability:

\( P!\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)} \)

Substitute the given values:

\( 0.5=\frac{P(A\cap B)}{0.4} \)

Solving for \(P(A\cap B)\):

\( P(A\cap B)=0.5\times 0.4=0.2 \)

Now find \(P!\left(\frac{B}{A}\right)\):

\( P!\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)} \)

Substitute the values:

\( P!\left(\frac{B}{A}\right)=\frac{0.2}{0.3}=\frac{2}{3}\approx 0.667 \)

Hence, the correct answer is Option 4.

Calculation:

Given:

\( P(A)=\frac{1}{3},\quad P(B)=\frac{1}{2},\quad P(A\cap B)=\frac{1}{4} \)

We know that:

⇒ \( P(\bar{A})=1-P(A) \)

Using the formula:

⇒ \( P(\bar{A}\cup B)=P(\bar{A})+P(B)-P(\bar{A}\cap B) \)

Substitute \(P(\bar{A})=1-P(A)\):

⇒ \( P(\bar{A}\cup B)=1-P(A)+P(B)-[P(B)-P(A\cap B)] \)

Substitute the given values:

\( =1-\frac{1}{3}+\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{4}\right) \)

Simplifying:

\( =\frac{2}{3}+\frac{1}{2}-\frac{1}{4} \)

\( =\frac{8+6-3}{12}=\frac{11}{12} \)

Hence, the correct answer is Option 4.

Calculation:

Statement I: Mean and variance have the same unit of measurement.

• Mean has the same unit as the original data.
• Variance is the square of the unit of the data (since deviations are squared).
• Therefore, mean and variance do NOT have the same unit.
• Statement I is incorrect.

Statement II: Mean deviation and standard deviation have the same unit of measurement.

• Mean deviation is based on absolute deviations, so it has the same unit as the data.
• Standard deviation is the square root of variance, hence it also has the same unit as the data.
• Therefore, Statement II is correct.

Hence, the correct answer is Option 2.