Quantitative Aptitude PYQs

Concept:

• Mean of x₁, x₂, ....,x_n is \(x_1 + x_2 +...+ x_n \over n\)
• 1 + 2 + 3 + ....+ n = \(n(n+1) \over 2\)
• 1² + 2² + 3² + ....+ n² = \(n(n+1)(2n+1) \over 6\)

Calculation:

Given: x is the mean of squares of first n natural numbers and y is the square of mean of first n natural numbers

⇒ \(x = {1^2 +2^2+3^2+...+ n^2 \over n}\) and  \(y = ({1 +2+3+...+ n \over n})^2\)

⇒  \(x= {{n(n+1)(2n+1) \over 6} \over n}\) and \(y = ({{n(n+1) \over 2} \over n})^2\)

⇒ \(y = {{(n+1)(2n+1) \over 6} }\) and  \(y = {{(n+1)^2 \over 4} }\) 

If \(\frac{\text{x}}{\text{y}} = \frac{55}{42}\), 

⇒ \({ x \over y} = {{ {(n+1)(2n+1) \over 6}\over {(n+1)^2 \over 4} }} = {55 \over 42}\)

⇒ \( { {4(n+1)(2n+1) }\over 6(n+1)^2} = {55 \over 42}\)

⇒ \( { {2(2n+1) }\over 3(n+1)} = {55 \over 42}\)

⇒ 2 × 42(2n + 1) = 3 × 55(n + 1)

⇒ 168n + 84 = 165n + 165

⇒ 3n = 81

⇒ n = 27

∴ The correct option is (3).

Concept:

Probability of an event = \(\text {Favourable outcome } \over \text { Total outcomes}\)

Explanation:

Prime numbers between 1 to 50 are given by,

{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}

⇒ There are 15 prime numbers between 1 to 50

And 1 is neither composite nor prime.

⇒ Number of composite numbers between 1 to 50 = 50 - 15 - 1 = 34

⇒ Favourable outcome = 34

and total outcome = 50

⇒ The probability of getting a composite number in the list of natural numbers from 1 to 50 

= \(\text {Favourable outcome } \over \text { Total outcomes}\)= \(34 \over 50\)

⇒ The probability of getting a composite number in the list of natural numbers from 1 to 50 = \(\frac{17}{25}\)

∴ The correct option is (2).

Concept:

Probability of an event = \(\frac{favourable~ number ~of ~outcomes}{total ~no. ~of ~outcomes}\)

C(n, r) = \(\frac{n!}{(n-r)!~r!}\)(where n! = factorial of n)

Calculation:

In the Given question, we need to find the probability that C(n, 7) is a multiple of 7, where n > 7.

So, Finding out favourable number of outcomes i.e.

For n > 7 ,C(n, 7) must be divisible by 7.

Considering n = 8, We get, C(8, 7) = 8 which is not divisible by 7.

Considering n = 9, We get, C(9, 7) = 36 which is not divisible by 7.

Considering n = 10, We get, C(10, 7) = 240 which is not divisible by 7.

Considering n = 11, We get, C(11, 7) = 1980 which is not divisible by 7.

..

..

..

Considering n = 48, We get ,C(48, 7) = which is not divisible by 7.

BUT , On Considering n = 49 , we get C(49, 7) which is divisible by 7 

⇒Before n = 49, C(n, 7) is not divisible by 7.

Thus, favourable number of outcomes = 1.

Total number of outcomes = C(8, 7) + C(9, 7) + ....

Thus, Probability that C(n,7) is divisible by 7 for n>7 = \(\frac{1}{C(8,7)+C(9,7)...}\)= \(\frac{1}{7}\)

Concept:

Probability of an event = \(\frac{favourable~ number ~of ~outcomes}{total ~no. ~of ~outcomes}\)

Calculation:

The set of first ten natural numbers can be visualized as:

A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Since, We want to find the probability that (x + y) is divisible by 4,

So, for that,

Total number of cases = \(_{}^{10}\textrm{C} \)₂ 

⇒ Total number of cases = 45

Now,

Number of favourable cases will be the pairs of (x, y) such that (x + y) is divisible by 4. i.e.

⇒Total number of Pairs formed = 1 + 3 + 4 + 2 = 10

Thus,

Probability that (x + y) is divisible by 4 = \(\frac{favourable~ number ~of ~pairs}{total ~no. ~of ~pairs}\)

⇒\(\frac{10}{45}\)

⇒\(\frac{2}{9}\)

Thus, the Probability that (x + y) is divisible by 4 = \(\frac{2}{9}\)

Concept:

Probability of an event = \(\text {Favourable number of outcomes } \over \text {Total number of outcomes}\) 

Calculation:

Given,  x + \(\frac{1}{\text{x}}\) > 2

⇒ x² - 2x + 1 > 0

⇒ (x - 1)² > 0

For any x ≠ 1, this inequality is satisfied.

As x is chosen from first n natural numbers.

⇒ Favourable number of choices for x = n - 1

And the total number of choices = n

⇒ Probability for x = \(\frac{(\text{n}−1)}{\text{n}}\)

∴ The correct option is (3).

Concept:

If three numbers are in A.P., then they are of the form a - d, a, a + d, where d is the common difference.

Probability of an event = \(\text {Favourable number of outcomes } \over \text {Total number of outcomes}\) 

Calculation:

Possible numbers in dice = {1, 2, 3, 4, 5, 6}

Total possible in tossing three dice = 6 × 6 × 6 = 216

Case 1: If the common difference AP is 1 or -1

The AP will be (a - 1, a, a + 1)

AP with common difference 1 are (1,2,3), (2,3,4), (3,4,5), (4,5,6), (3,2,1), (4,3,2), (5,4,3), (6,5,4)

So, there can be 4 ways of forming AP with common difference of 4

Case 3: If the common difference AP is 2 or -2

The AP will be (a - 2, a, a + 2)

AP with common difference 2 or -2 are (1,3,5), (2,4,6),(5,3,1), (6,4,2)

So, there can be 2 ways of forming AP with common difference of 2

And the common difference cannot be greater than 2 for forming an AP of 3 numbers.

Favourable outcome (numbers are in AP) = 8 + 4 = 12

⇒ Probability that they show different numbers that are in AP = \(\text {Favourable number of outcomes } \over \text {Total number of outcomes}\)

⇒ Probability that they show different numbers that are in AP = \(12 \over 216\) = \(\frac{1}{18}\)

∴ The correct option is (2).

Concept:

• P(A') = 1 - P(A)
• P(A' ∩ B') = P((A ∪ B)') {Probability of not A and not B}
• P(A and not B) = P(A ∩ B') = P(A) - P(A ∩ B)
• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation:

Given:  P(A) = 0.5, P(B) = 0.7 and P(A ∩ B) = 0.3, 

⇒  P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

⇒  P(A ∪ B) = 0.5 + 0.7 - 0.3 = 0.9__(i)

Now the probability of not A and not B,

P(A' ∩ B') = P((A ∪ B)') = 1 - P(A ∪ B) 

⇒ P(A' ∩ B') = 1 - 0.9 { From (i)}

⇒ P(A' ∩ B') = 0.1 __(ii)

The probability of B and not A,

P(A' ∩ B) = P(B) - P(A ∩ B) 

⇒ P(A' ∩ B) = 0.7 - 0.3 = 0.4 __(iii)

The probability of A and not B,

P(A ∩ B') = P(A) - P(A ∩ B) 

⇒ P(A' ∩ B) = 0.5 - 0.3 = 0.2 __(iv)

From (ii), (iii) and (iv),

P(A' ∩ B') + P(A' ∩ B) + P(A ∩ B') = 0.1 + 0.4 + 0.2 = 0.7

∴ The correct option is (2).

Concept:

If A and B are independent, then P(A and B) = P(A).P(B)

Explanation:

Probability of getting atmost four tails in five tosses = 1 - Probability of getting all five tails

⇒ Probability of getting atmost four tails in five tosses = 1 - P(T,T,T,T,T)

In one toss, P(H) = \(1 \over 2\) and P(T) = \(1 \over 2\)

Since all five tosses are independent,

⇒ P(T,T,T,T,T) = \(1 \over 2\) × \(1 \over 2\) × \(1 \over 2\) × \(1 \over 2\) × \(1 \over 2\) = \(1 \over 32\)

⇒ Probability of getting atmost four tails in five tosses = 1 - ​​\(1 \over 32\) = ​​​\(31 \over 32\)​

∴ The correct option is (1).

Concept:

Probability of an event = \(\text {Favourable number of outcomes } \over \text {Total number of outcomes}\) 

Calculation:

On tossing a dice, there can be following outcomes = {1,2,3,4,5,6}

If three fair dices are thrown, the sum can be greater than or equal to 15 in the following ways:

(Sum = 15, 16, 17 or 18}

 

⇒ Number of Favorable outcomes = 20

And the total number of choices when three dices are thrown = 6 × 6 × 6 = 216

⇒ The probability of getting a total greater than or equal to 15 = \(20 \over 216\) = \(\frac{5}{54}\)

∴ The correct option is (4).

Concept:

If A and B are independent, then P(A and B) = P(A).P(B)

Explanation:

Given that: Probability that a person hits a target is 0.5

⇒ Probability of not hitting the target = P(not hit) = 1 - 0.5 = 0.5

Now, the probability of at least one hit in 4 shots = 1 - Probability of no hit in four shots

Since the four shots are independent 

⇒ Probability of no hit in four shots = P(not hit) × P(not hit) × P(not hit) × P(not hit) 

⇒ Probability of no hit in four shots = 0.5 × 0.5 × 0.5 × 0,5 = \(( {1 \over 2})^4 = {1 \over 16}\)

⇒ The probability of at least one hit in 4 shots = 1 - \(1 \over 16\) = \(\frac{15}{16}\)

∴ The correct option is (3).

Concept:

Number of ways of choosing r objects out of n objects = \(^nC_r = {n! \over r! (n-r)!}\)

Explanation:

Given that: A box contains 2 white balls, 3 black balls, and 4 red balls

⇒ Box contains total 9 balls

3 balls from the box with at least one black ball can be drawn in the following ways

 

Case 1: 1 black ball 

Number of ways of choosing 1 black ball from 3 black balls and 2 balls from remaining 6 balls = 3C1 × 6C2 = 3 × 15 = 45

Case 2: 2 black balls

Number of ways of choosing 2 black balls from 3 black balls and 1 ball from remaining 6 balls = 3C2 × 6C1 = 3 × 6 = 18

Case 3: 3 black balls

Number of ways of choosing 3 black balls from 3 black balls and 0 ball from remaining 6 balls = 3C3 × 6C0 = 1 × 1 = 1

⇒ Total number of required ways = 45 + 18 + 1 = 64

∴ The correct option is (3).

Concept:

Binomial distribution

In n independent trials, if p and q are the probability of success and failure respectively then the probability of exactly r success = ⁿC_r p^r (q)^n-r

Explanation:

Given, During war, one ship out of 5 was sunk on an average in making a certain voyage

⇒ P(sinking) = \(1 \over 5\)  {failure}

⇒ P(safe arriving) =  P(not sinking) = 1 - \(1 \over 5\) = \(4 \over 5\)  {success}

Now the probability that exactly 3 out of 5 ships would arrive safely = Probability of 3 success

 n = 5, p = \(4 \over 5\), q = \(1 \over 5\), r = 3

⇒ The probability that exactly 3 out of 5 ships would arrive safely =  5C3 (\(4 \over 5\))3 (\(1 \over 5\))5-3 

⇒ The probability that exactly 3 out of 5 ships would arrive safely = \({5! \over 3! 2!} ({64\over 125}).({1 \over 25})\) 

⇒ The probability that exactly 3 out of 5 ships would arrive safely = \(\frac{128}{625}\) 

∴ The correct option is (4).

Concept:

• Odds against an event = Number of unfavorable choices : Number of favorable choices
• n(A or B) = n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Calculation:

Given: A card is drawn from a pack of 52 cards. A gambler bets that it is either a spade or an ace. 

Number of favourable events in winning = n(spade or ace)

⇒ Number of favourable events in winning = n(spade) + n(ace) - n(spade ∩ ace)

∵ in the pack of 52 cards, 1 card is both spade and ace ⇒ n(spade ∩ ace) = 1

⇒ Number of favourable events in winning = 13 + 4 - 1 = 16

⇒ Number of unfavourable events in winning = Total - favourable

⇒ Number of unfavourable events in winning = 52 - 16 = 36

⇒  The odds against winning = Number of unfavorable choices : Number of favorable choices

⇒  The odds against winning = 36 : 16 = 9 : 4

∴ The correct option is (1).

Concept:

The correlation coefficient is statistical measure of the strength of the relationship between the relative moments of two variables.

The values range between  -1.0  and  1.0.

It is determined by dividing the covariance by the product of the two variables standard deviation. Thus,

Coefficient of correlation Does not depends on scale and origin.

Calculation:

In the given question, we are given that Coefficient of correlation between the ages of husband and wife at the time of marriage for a set of 100 couples was 0.7

and According to the question, If couples survive to celebrate their silver jubilee, then 

Since, Coefficient of correlation doesn't changes on scale and time

Thus, the Coefficient of correlation at that point will be same as it was earlier 

⇒The Coefficient of correlation = 0.7

Concept:

• P(A|B) is the probability of A after B happens.
• P(A ∩ B) + P(A ∩ B̅) = P(A)
• P(A|B) = \(P(A \cap B)\over P(B)\)
• P(A̅ ) = 1 - P(A)

Calculation:

Let A be the event that the construction job is completed on time and B be the event that there is strike.

Given:

The probability of strike is 0.6 ⇒ P(B) = 0.6  __(i)

The probability that the construction job gets completed on time if there is no strike is 0.85

⇒ P(A| B̅) = 0.85  __(ii)

The probability that the construction job gets completed on time if there is a strike is 0.35.

⇒ P(A| B) = 0.35 __(iii)

Find: The probability that the construction job will not be completed on time = 1 - P(A)

Solution:

From (ii), 

 P(A| B̅) = 0.85 

⇒ \({P(A \cap \overline B)\over P(\overline B) }= 0.85\)

⇒ \({P(A \cap \overline B) }= 0.85 P(\overline B)\)

⇒ \({P(A \cap \overline B) }= 0.85 (1 -P(B))\) 

⇒ \({P(A \cap \overline B) }= 0.85 (1 - 0.6)\)  {From (i)}

⇒ \({P(A \cap \overline B) }= 0.34\)  __(iv)

Similarly from (iii),

 P(A| B) = 0.35

⇒ \({P(A \cap B)\over P( B) }= 0.35\)

⇒ \({P(A \cap B) }= 0.35 P( B)\)  

⇒ \({P(A \cap B) }= 0.35 (0.6)\)  {From (i)}

⇒ \({P(A \cap B) }= 0.21\)  __(v)

Adding (iv) and (v)

⇒ \({P(A \cap \overline B) }+ {P(A \cap B) }= 0.34 + 0.21\)

⇒ P(A) = 0.55

⇒ The probability that the construction job is not completed on time = 1 - P(A) = 1 - 0.55

⇒ The probability that the construction job is not completed on time = 0.45

∴ The correct option is (2).

Concept:

The data with the greatest coefficient of variation has the highest variability.

Coefficient of variation = \({S.D. \over \text {Mean}} \times 100\)

where S.D. is the standard deviation of the data.

Calculation:

Given:

 

So, the coefficient of variation for Mathematics = \({15 \over \text 40} \times 100 = {75 \over 2} \)

⇒ The coefficient of variation for Mathematics = 37.5  __(i)

The coefficient of variation for Physics = \({12 \over \text 28} \times 100 = 42.85\) __(ii)

 The coefficient of variation for Chemistry = \({14 \over \text 38} \times 100 = 36.84\)____(iii)

The coefficient of variation for Biology = \({16 \over \text 36} \times 100 = 44.44\)____(iv)

Comparing (i), (ii), (iii) and (iv),

The subject Biology has the greatest Coefficient of Variation

⇒ Biology shows the highest variability.

The correct option is (4).

Concept:

The data with the greatest coefficient of variation has the highest variability.

Coefficient of variation = \({S.D. \over \text {Mean}} \times 100\)

where S.D. is the standard deviation of the data.

Calculation:

Given:

 

⇒ S.D. of mathematics = 15 and Mean for Mathematics = 40

So, the coefficient of variation for Mathematics =  \({15 \over \text 40} \times 100 = {75 \over 2} \)

⇒ The coefficient of variation for Mathematics = 37.5  

∴ The correct option is (1).

Concept:

For a distribution, Median = \(L + {{N \over 2} - CF \over f} \times h\)

where, L = lower class limit of median class

N = Sum of frequencies

CF = cumulative frequency of class preceding the median class

f = frequency of median class

h = class length of median class.

Calculation:

 

∑f = N = 20

⇒ N/2 = 10 

As 7 < 10 < 13, N/2 lies in the class length 30 - 40 by comparing cumulative frequency.

Median class = 30 - 40

⇒ L = 30, N = 60, CF = 7, f = 6, h = 40 - 30 - 10

⇒ Median = \(L + {{N \over 2} - CF \over f} \times h = 30 + {{20 \over 2} - 7 \over 6} \times 10\)

⇒ Median = \(30+ {3 \over 6} \times 10\)= 35

∴ The correct option is (3).

Concept:

For a distribution, Median =  \(L + {{N \over 2} - CF \over f} \times h\)

where, L = lower class limit of median class

N = Sum of frequencies

CF = cumulative frequency of class preceding the median class

f = frequency of median class

h = class length of median class.

Mean deviation about median = \({∑ f_i|x_i -M | \over N}\)

where M is median, N is sum of frequencies, x_i are the mid points of the class and f_i are the frequencies

Calculation:

 

∑f = N = 20

⇒ N/2 = 10 

As 7 < 10 < 13, N/2 lies in the class length 30 - 40 by comparing cumulative frequency.

Median class = 30 - 40

⇒ L = 30, N = 60, CF = 7, f = 6, h = 40 - 30 - 10

⇒ Median = \(L + {{N \over 2} - CF \over f} \times h = 30 + {{20 \over 2} - 7 \over 6} \times 10\)

⇒ Median = \(30+ {3 \over 6} \times 10\)= 35

Now calculating mean deviation about median,

 

⇒ Mean deviation about median = \({∑ f_i|x_i -M | \over N}\) = \(210\over 20 \) = 10.5

∴ The correct option is (4).

Concept:

• Mean of a distribution = \(∑ f_i x_i \over ∑ f_i\)
• Mean deviation about mean x̅ = \({∑ f_i|x_i -\overline x | \over ∑ f_i}\)

where fi are frequencies and xi are the mid points of class

Calculation:

Calculating mean

 

⇒ Mean = \(∑ f_i x_i \over ∑ f_i\) = \(690 \over 20\) = 34.5

 

Calculating the mean deviation about mean

 

⇒ Mean deviation about mean x̅ = \({∑ f_i|x_i -\overline x | \over ∑ f_i}\) = \(213 \over 20\) = 10.65

∴ The correct option is (2).