Quantitative Aptitude PYQs

Given: 

The sum of deviations of n numbers from 10 = p

The sum of deviations of n numbers from 20 = q

(p - q)2 = 10000

Calculation:

p = \(\displaystyle \sum {(x_i-10)} = \displaystyle \sum {x_i-10n}\)

q = \(\displaystyle \sum {(x_i-20)} = \displaystyle \sum {x_i-20n}\)

As, (p - q)2 = 10000

\((\displaystyle \sum {x_i-10n} - \displaystyle \sum {x_i+20n})^2 = 10000\)

\((10n)^2 = 10000\)

n = 10

Given:

Mean (X̅) = 20, No. of observations = 10

Calculation:

Let

S = \(\displaystyle \sum_{i=1}^{10}\left(\frac{3 x_i-4}{5}\right)\)

S = \(\displaystyle \sum_{i=1}^{10}\left(\frac{3 x_i}{5}\right) -{4\over 5} \times \displaystyle \sum_{i=1}^{10} 1\)

S = \({3\over 5} \displaystyle \sum_{i=1}^{10}( x_i) -{4\over 5} \times 10\)           -----(1)

As, \({{\displaystyle \sum_{i=1}^{10}( x_i)}\over 10} = 20\)

\({\displaystyle \sum_{i=1}^{10}( x_i)} = 200\)

Now, from equation (1)

S = \({3\over 5} (200) -{4\over 5}(10)\)

S = \({600-40\over 5}\)

S = 112

∴ \(\displaystyle \sum_{i=1}^{10}\left(\frac{3 x_i-4}{5}\right)\) = 112

Formula Used:

Standard Deviation = \(\sigma = \sqrt{({\sum x_i\over n})^2 -({\sum x^2_i\over n}) }\)

Given:

Mean = 40, Sum of squares = 16160, No. of observations (n) = 10

\({\sum x_i\over n} = 40\), \(\sum x^2_i = 16160\)

Calculation:

\(\sigma = \sqrt{({\sum x^2_i\over n}) -({\sum x_i\over n})^2}\)

\(\sigma = \sqrt{{16160\over 10} - 40^2}\)

\(\sigma = \sqrt{{1616} - 1600}\)

\(\sigma = \sqrt{16}\)

\(\sigma = 4\)

Concept:

Perfect Squares = 1, 4, 9, 16, 25, 36

Among all these, we can get perfect squares 4, 9, 16

Calculation:

The total number of outcomes = 6³ = 216

Total ways for an outcome of 4,

(1, 1, 2) in 3 ways

Total ways for an outcome of 9,

(1, 2, 6) in 3! ways

(1, 3, 5) in 3! ways

(2, 3, 4) in 3! ways

(3, 3, 3) in 1! ways

(2, 2, 5) in 3 ways

(1, 4, 4) in 3 ways

Total Ways =  3! + 3! + 3! + 1! + 3 + 3 = 6 + 6 + 6 + 1 + 3 + 3 = 25

Total ways for an outcome of 16,

(4, 6, 6) in 3 ways

(5, 5, 6) in 3 ways

Total ways = 3 + 3 = 6

Number of favorable outcomes = 3 + 25 + 6 = 34

The probability of getting a sum which is a perfect square,

= 34/216

= 17/108

Concept: 

Mutually exclusive events are those events that do not occur at the same time.

For example, when a coin is tossed then the result will be either head or tail, but we cannot get both the results. 

Sum of mutually exclusive events is equals to 1.

Calculation:

A, B, C and D are mutually exclusive and exhaustive events. Therefore,

P(A) + P(B) + P(C) + P(D) = 1         -----(1)

According to the question,

2P(A) = 3P(B) = 4P(C) = 5P(D) = k (Let)

P(A) = k/2, P(B) = k/3, P(C) = k/4, P(D) = k/5

Put the values in equation (1),

P(A) + P(B) + P(C) + P(D) = 1

\({k\over 2} + {k\over 3} + {k\over 4} + {k\over 5} = 1\)

\({30k + 20k+15k+12k\over 60}= 1\)

\({77k}= 60\)

Put k = 2 P(A),

77 P(A) = 30

Hence, 77P(A) equals to 30.

Given:

Two distinct natural numbers from 1 to 9 are picked at random.

Calculation:

Total number of ways in which two distinct natural numbers from 1 to 9 are picked at random = ⁹C₂

9C2 = \({{9!}\over {{(9-2)!}2!}} \) = 36

Favourable pairs are 

 (7, 3)

Total favourable pairs = 1

Required probability = 1/36

Calculation:

Sample Space:

(1,1); (1,2); (1,3); (1,4); (1,5); (1,6);

(2,1); (2,2); (2,3); (2,4); (2,5); (2,6);

(3,1); (3,2); (3,3); (3,4); (3,5); (3,6);

(4,1); (4,2); (4,3); (4,4); (4,5); (4,6);

(5,1); (5,2); (5,3); (5,4); (5,5); (5,6);

(6,1); (6,2); (6,3); (6,4); (6,5); (6,6);

From the above sample space, there are 14 possibilities for the difference of numbers between them is 2 or 3

Total number of Sample Space = 36

the probability that the difference of numbers on them is 2 or 3,

= 14/36

= 7/18

Given:

Frequencies (f_i) = 9C0, 9C1, 9C2 ..., 9C9

Numbers (x_i) = 1, 2, 3, ... 10

Formula Used:

\(Mean = {{\sum f_ix_i}\over {\sum f_i}}\)

\(:^nC_0+: 2\times^nC_1 + 3\times:^nC_2 + 4\times:^nC_3 .......+^nC_n (n) = (n+2)(2)^{n-1}\)

\(:^nC_0+:^nC_1 + :^nC_2 .......+^nC_n = (2)^n\)

Calculation:

\(Mean = {{1 \times^9C_0 + 2 \times^9C_1+3 \times^9C_2.....10 \times^9C_9}\over {^9C_0 + ^9C_1+^9C_2.....^9C_9}}\)

\(Mean = {{(9+2)\times {2^{9-1}}}\over {2^9}}\)

\(Mean = {{(11)\times {2^{8}}}\over {2^9}}\)

Mean = 5.5

Concept: 

Binomial distribution is the discrete probability distribution that gives only two possible results in an experiment, either Success or Failure. For example, if we toss a coin, there could be only two possible outcomes: heads or tails, and if any test is taken, then there could be only two results: pass or fail. This distribution is also called a binomial probability distribution.

P (x) = \(^{n}C_x \times [P(Success)]^x \times [P(Failure)]^{n-x}\)

Given:

Success: The probability that a person recovers from a disease is 0.8. 

Failure: The probability that a person does not recover from a disease = 1 - 0.8 = 0.2 

Calculation:

P (Success) = 0.8, P(Failure) = 0.2, n = 5, x = 2

P (x = 2) = \(^{5}C_2 \times (0.8)^2 \times(0.2)^{5-2}\)

P (x = 2) = \({{5!}\over {{(5-2)!}2!}} \times (0.64) \times(0.008)\)

P (x = 2) = \(10 \times (0.00512)\)

P (x = 2) = \(0.0512\)

Concept:

Bayes theorem: It is used to determine the conditional probability of event A when event B has already happened. The general statement of Bayes’ theorem is “The conditional probability of an event A, given the occurrence of another event B, is equal to the product of the event of B, given A and the probability of A divided by the probability of event B.” i.e.

P(A|B) = \({{P(A) \times P(B|A)}\over {P(B)}}\)

Given:

Event A: The chance that the design is faulty.

Event B: The chance that the design is not faulty.

Event C: The chance that the building collapses

P(A) = 10%, P(B) = 90%, P(A|C) = 95%, P(B|C) = 45%

Calculation:

P(C|A) = \({{P(A) \times P(C|A)}\over {P(A) \times P(C|A)}+{P(B) \times P(C|B)}}\)

P(C|A) = \({{10 \times 95}\over {10 \times 95}+{90 \times 45}}\)

P(C|A) = \({{950}\over {950}+{4050}}\)

P(C|A) = \({{950}\over {5000}}\)

P(C|A) = 0.19

Given: The correlation coefficient between x and y is r.

Concept:

The correlation coefficient measures the strength and direction of the linear relationship between two variables. 

Calculation:

The transformation of y involves multiplication by -3. Multiplying one of the variables by a negative value changes the direction of the correlation.

= - r 

Hence, the correlation coefficient between ( 3x + 4 ) and ( -3y + 3 ) is  - r

Concept:

\(P(E) = {{n(E)} \over {n(S)}}\)

where n(E) = No. of favourable cases for event A and n(S) = cardinality of sample space.

If a coin is tossed then,

P(getting head) = \(1 \over 2\) P(getting tail) = \(1 \over 2\)

Calculation:

A fair coin is tossed repeatedly. Sixth toss is different from those obtained in the first five tosses.

There are two cases of getting the result as:

Case 1: Getting a tail is a result

Case 2: Getting a head is a result

The possibility is that in the first five tosses, we get 5 heads and 1 tail or five tails and one head,

⇒ \({1 \over 2} × {1 \over 2} × {1 \over 2} × {1 \over 2} × {1 \over 2} + {1 \over 2} ×{1 \over 2} ×{1 \over 2} ×{1 \over 2} ×{1 \over 2} \)

⇒ \({1 \over 64} + {1 \over 64}\)

⇒ \({2 \over 64}\) = \({1 \over 32}\)

Given:

If H is the Harmonic Mean of three numbers 10C4, 10C5, and 10C6.

Concept: 

The harmonic mean is the reciprocal of the arithmetic mean of the values. It is calculated by dividing the number of observations by the reciprocal of each number in the series. 

\(^{n}C_k = {{n!}\over (n-k)! k!}\)

Calculation:

\(^{10}C_4 = {{10!}\over (10-4)! 4!} = 210\)

\(^{10}C_5 = {{10!}\over (10-5)! 5!} = 252\)

\(^{10}C_6 = {{10!}\over (10-6)! 6!} = 210\)

Harmonic mean (H) = \({{3} \over {{1\over210}+{1\over252}+{1\over210}}}\)

Harmonic mean (H) = \({{3} \over {{6+5+6\over1260}}}\)

Harmonic mean (H) = \({{3 \times 1260} \over {17}}\)

\(\frac{270}{H}\) = \({{17 \times270}\over {3 \times 1260}} = {17\over 14}\)

Given: In a class, there are n students including the students P and Q.

Calculation:

Total number of students in the class = n

The probability of P and Q sitting together:

Treat the P and Q as one unit (X).

Now, we have (n - 2, X) to arrange. This can be done in (n - 2 + 1)! ways.

Within the P and Q, they can arrange themselves in 2! ways.

Therefore, the total number of favorable outcomes in this case is (n - 1)! × 2!.

n people can be arranged in a row in n! ways.

\(Probability = {Favorable: outcomes \over Total :possible :outcomes}\)

Probability \(= {{((n - 1)! × 2!)} \over n!}\)

Probability \(= {{2} \over n}\)

Concept: 

Binomial distribution is the discrete probability distribution that gives only two possible results in an experiment, either Success or Failure. For example, if we toss a coin, there could be only two possible outcomes: heads or tails, and if any test is taken, then there could be only two results: pass or fail. This distribution is also called a binomial probability distribution.

B(n, p) = \(^{n}C_x \times [P(Success)]^x \times [P(Failure)]^{n-x}\)

Given:

n = 6

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

Calculation:

Let the success be p

Let the failure be 1 - p

P (x = 2) = \(^{6}C_2 \times (p)^2 \times(1-p)^{6-2}\)

P (x = 2) = \({{6!}\over {{(6-2)!}2!}} \times (p)^2 \times(1-p)^{4}\)

P (x = 2) = \(15 \times (p)^2 \times(1-p)^{4}\)

P (x = 4) = \(^{6}C_4 \times (p)^4 \times(1-p)^{6-4}\)

P (x = 4) = \({{6!}\over {{(6-4)!}4!}} \times (p)^4 \times(1-p)^{2}\)

P (x = 4) = \(15 \times (p)^4 \times(1-p)^{2}\)

As, 9P(X = 4) = P(X = 2)

\(9\times 15 \times (p)^4 \times(1-p)^{2} = ​​15 \times (p)^2 \times(1-p)^{4}\)

\(9\times (p)^2 = (1-p)^{2}\)

\(9 = ({{1-p}\over p})^2\)

\(3 = {{1-p}\over p}\)

3p = 1 - p

p = \( {{1}\over 4}\)

Given: Three boys P, Q, R and three girls S, T, U are to be arranged in a row for a group photograph.

Calculation:

The probability of all three boys sitting together:

Treat the boys as one unit (B).

Now, we have four units (B, S, T, U) to arrange. This can be done in 4! ways.

Within the boys' block (B), the three boys can arrange themselves in 3! ways.

Therefore, the total number of favorable outcomes in this case is 4! × 3!.

6 people can be arranged in a row in 6! ways.

\(Probability = {Favorable: outcomes \over Total :possible :outcomes}\)

Probability \(= {{(4! × 3!)} \over 6!}\)

Probability \(= {{(24 × 6)} \over 720}\)

Probability = 1/5

Therefore, the probability that all three boys sit together is 1/5.

Given: Three boys P, Q, R and three girls S, T, U are to be arranged in a row for a group photograph.

Case 1: Starting with a Girl

In this case, boys can be arranged in this way: B_B_B

Boys can be arranged in this manner in 3! ways.

Girls can be arranged in this manner in 3! ways.

Total Favorable outcomes = 3! × 3! = 6 × 6 = 36 ways

Case 2: Starting with a Boy

In this case, boys can be arranged in this way: B_B_B_

Boys can be arranged in this manner in 3! ways.

Girls can be arranged in this manner in 3! ways.

Total Favorable outcomes = 3! × 3! = 6 × 6 = 36 ways

Total favorable outcomes (Case 1 + Case 2) = 36 + 36 = 72.

Total possible arrangements for 6 people: 6! = 720

The probability that boys and girls sit alternatively,

\(Probability = {Favorable: outcomes \over Total :possible :outcomes}\)

Probability \(= {{(72)} \over 720}\)

Probability = 1/10

Therefore, the probability that boys and girls sit alternatively is 1/10.

Given: Three boys P, Q, R and three girls S, T, U are to be arranged in a row for a group photograph.

Calculation:

Boys can be arranged in this way: B_B_B

Boys can be arranged in this manner in 3! ways.

Girls can be arranged in this manner in 3! × ⁴P₃ ways.

Total Favorable outcomes = 3! × 3! × 4P3 = 6 × 6 × 4! = 144 ways

The probability that no two girls sit together,

6 people can be arranged in a row in 6! i.e. 720 ways.

\(Probability = {Favorable: outcomes \over Total :possible :outcomes}\)

Probability \(= {{(144)} \over 720}\)

Probability = 1/5

Therefore, the probability that no two girls sit together is 1/5.

Given: Three boys P, Q, R and three girls S, T, U are to be arranged in a row for a group photograph.

Calculation:

P and Q take the two end positions: P_ _ _ _ Q or Q_ _ _ _ P

4 people arrange between P and Q in 4! ways.

Total Favorable outcomes = 2 × 4! = 2 × 24 = 48 ways

The probability that P and Q take the two end positions,

6 people can be arranged in a row in 6! ways.

\(Probability = {Favorable: outcomes \over Total :possible :outcomes}\)

Probability \(= {{(48)} \over 720}\)

Probability = 1/15

Therefore, the probability that P and Q take the two end positions is 1/15.

Given: Three boys P, Q, R and three girls S, T, U are to be arranged in a row for a group photograph.

Calculation:

The probability that Q and U sit together.

Treat the Q and U as one unit (X).

Now, we have five units (X, P, R, S, T) to arrange. This can be done in 5! ways.

Within Q and U, they can arrange themselves in 2! ways.

Therefore, the total number of favorable outcomes in this case is 5! × 2!.

6 people can be arranged in a row in 6! ways.

\(Probability = {Favorable: outcomes \over Total :possible :outcomes}\)

Probability \(= {{(5! × 2!)} \over 6!}\)

Probability \(= {{(120 × 2)} \over 720}\)

Probability = 1/3

Therefore, the probability that all three boys sit together is 1/3.