Quantitative Aptitude PYQs

Calculation:

Given, bag contains 5 black and 4 white balls.

A man selects two balls at random.

∴ P(both are of same colour)

= P(first ball is black)P(second ball is black) + P(first ball is white)P(seconf ball is white)

= \(\frac{5}{9}\times\frac{4}{8}+\frac{4}{9}\times\frac{3}{8}\) 

= \(\frac{32}{72}\)

= \(\frac{4}{9}\)

∴ The probability that both of these are of the same colour is \(\frac{4}{9}\).

The correct answer is Option 3.

Concept:

Let p be the probability of success and q be the probability of failure of a random variable x of sample size n then 

mean (μ) = np and variance (σ²) = npq where q = 1 - p

Also, P(X = x) = ⁿC_xp^xq^n-x

Explanation:

Here mean

np = 5...(i)

and variance

npq = 4...(ii)

So, \({npq\over np}=\frac45\) ⇒ q = \(\frac45\)

Then  p = 1 - q = 1 - \(\frac45\) = \(\frac15\)

Putting in (i) we get

n = 25

Given 

523P(X = 3) = λ4λ

⇒ 523  25C3(\(\frac15\))³(\(\frac45\))22 = λ4λ

⇒ 523  25C3\(\frac{4^{22}}{5^{25}}\) = λ4λ

⇒ \({25!\over 3!.22!}.\frac1{25}\).4²² = λ4λ

⇒ \({25.24.23.22!\over 3.2.1.22!}.\frac1{25}\).422 = λ4λ

⇒ 23 × 4 × 4²² = λ4λ

⇒ 23 × 423 = λ4λ

⇒ λ = 23

Option (3) is true.

Concept:

The regressing line of y on x is given by y - y̅ = byx(x - x̅), where

\(b_{yx}=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}\)

The regressing line of x on y is given by x - x̅ = bxy(y - y̅), where

\({b_{xy}} = \frac{{n\sum xy - \left( {\sum x} \right)\left( {\sum y} \right)}}{{n\sum {y^2} - {{\left( {\sum y} \right)}^2}}}\)

Calculation:

Here, n = 4

∴ \(b_{yx}=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}\)

= \(\frac{4(11)-0\cdot11}{4\cdot30-0^2}\)

= \(\frac{11}{30}\)

x̅ = \(\frac{\sum x}{n}\) = 0

y̅ = \(\frac{\sum y}{n}\) = \(\frac{11}{4}\) = 2.75

∴ Required line, y - y̅ = b_yx(x - x̅)

⇒ y - 2.75 = \(\frac{11}{30}\)(x - 0)

⇒ y = 2.75 + \(\frac{11}{30}x\)

Comparing with y = a + bx, a = 2.75 and b = \(\frac{11}{30}\)

⇒ 2a + 15b = 5.5 + 5.5 = 11

∴ The value of 2a + 15b is 11.

The correct answer is Option 2.

Given:

Let x + 2y + 1 = 0 and 2x + 3y + 4 = 0 be two lines of regression. If θ is the acute angle between them.

Concept Used:

The angle θ between two lines given by \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) is given by:

\(\tan θ = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|\)

Where m₁ and m₂ are the slopes of the lines.

Explanation:

For line x + 2y + 1 = 0 :

⇒ \(y = -\frac{1}{2}x - \frac{1}{2} \implies m_1 = -\frac{1}{2}\)

For line 2x + 3y + 4 = 0 :

y = \(-\frac{2}{3}x - \frac{4}{3} \implies m_2 = -\frac{2}{3}\)

\(\tan θ = \left|\frac{-\frac{1}{2} + \frac{2}{3}}{1 + \left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)}\right|\)

= \(\left|\frac{-\frac{1}{2} + \frac{2}{3}}{1 + \frac{1}{3}}\right| = \left|\frac{\frac{-3+4}{6}}{\frac{4}{3}}\right| = \left|\frac{\frac{1}{6}}{\frac{4}{3}}\right| = \left|\frac{1}{8}\right| = \frac{1}{8}\)  

Using the triple angle formula for tangent 

\(\tan(3θ) = \frac{3\tanθ - \tan^3θ}{1 - 3\tan^2θ}\)

Substitute tan θ  = 1/8 

\(\tan(3θ) = \frac{3 \cdot \frac{1}{8} - \left(\frac{1}{8}\right)^3}{1 - 3 \left(\frac{1}{8}\right)^2}\)

= \(\frac{\frac{3}{8} - \frac{1}{512}}{1 - \frac{3}{64}}\) = \(\frac{191}{64} \times \frac{8}{61}\) = \(\frac{191}{8} \times \frac{1}{61}\)

Now we multiply this value by 488:

488 tan(3θ) = 488 x  \(\frac{191}{8} \times \frac{1}{61}\) = 191

Thus, the value of 488 tan(3θ) is 191.

Given:

If two random variables X and Y are connected by the relation

\(\frac{2X - 3Y}{5X + 4Y} = 4\) and X follows a Binomial distribution with parameters

n = 10 and \(p = \frac{1}{2} \)

Concept Used:

For Binomial distribution X ∼ B(n, p) ,

E(X) = np and Var(X) = np(1-p)

Explanation:

Given relation: 

\(\frac{2X - 3Y}{5X + 4Y} = 4 \)

⇒ \( 2X - 3Y = 4(5X + 4Y)\)

⇒ \(2X - 3Y = 20X + 16Y\)

⇒ -18X = 19Y

⇒ Y = \(\frac{-18X}{19}\)

Since X ∼ B(10, 1/2) :

E(X) = \(10 \cdot \frac{1}{2} = 5\)

Var(X) = \(10 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{5}{2}\)

Now, finding Var(Y) :

Y = \(\frac{-18X}{19}\)

The variance of Y :

\(\text{Var}(Y) = \left(\frac{-18}{19}\right)^2 \text{Var}(X) = \frac{324}{361} \cdot \frac52 = \frac{810}{361}\)

Thus, the variance of Y is 810/361.

Explanation:

a, b, c are in HP so

\({2\over b}=\frac1a+\frac1c\)...(i)

⇒ \({2\over b}=\frac{a+c}{ac}\)

⇒ b = \(2ac\over a+c\)....(ii)

∴ \(\frac{1}{b-a}+\frac{1}{b-c}\)

= \(\frac{2b-(c+a)}{(b-a)(b-c)}\)

= \(\frac{2b-(c+a)}{b^2-(a+c)b+ac}\)

= \(\frac{2b-{2ac\over b}}{b^2-2ac+ac}\)

= \(2b^2-2ac\over b(b^2-ac)\) = \(2\over b\) = \(\frac1a+\frac1c\)

1 and 2 both are correct.

Calculation:

Given the rates are 150, 200, 250, and 300 rupees per litre, the formula for the harmonic mean (H) for (n) values is:

\(H = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \frac{1}{x_4}}\)

where x₁, x₂ , x₃, and x₄ are the given rates.

Substituting the given values:

\(H = \frac{4}{\frac{1}{150} + \frac{1}{200} + \frac{1}{250} + \frac{1}{300}}\)

First, compute the sum of the reciprocals of the rates:

1/150 = 0.006667

1/200 = 0.005

1/250 = 0.004

1/300 = 0.003333

Sum these values:

0.006667 + 0.005 + 0.004 + 0.003333 = 0.019

Now, calculate the harmonic mean:

H = 4 / 0.019 ≈ 210.526

Rounded to the nearest whole number, the average price of oil per litre is approximately 210.

Explanation -

First, let's calculate the total number of ways to arrange the letters in "TIRUPATI."

The word "TIRUPATI" consists of 8 letters where the letter T is repeated twice.

The total number of permutations of the letters is given by: \(\frac{8!}{2!}\)

where 8! is the factorial of 8, and 2! accounts for the repetition of the letter T.

8! = 40320 and 2! = 2

⇒ \(\frac{8!}{2!} = \frac{40320}{2} = 20160\)

Next, we calculate the number of ways to arrange the letters such that the two Ts are always consecutive.

We can treat the two Ts as a single "super T" unit. This reduces the problem to arranging 7 units: (TT), I, R, U, P, A, I.

The number of ways to arrange these 7 units is \(\frac{7!}{1!}\)

where 7! is the factorial of 7, and 1! accounts for the "super T" being treated as a single unit.

7! = 5040

Since there are no repetitions other than the original T, we don't need to divide by any other factorials.

Therefore, the probability that the two Ts are always consecutive is the ratio of the number of favorable outcomes to the total number of outcomes:

5040 / 20160 = 1/4

So, the correct answer is 1/4.

Given:

Let m = 77ⁿ. The index n is given a positive integral value at random.

Concept Used:

To determine the units place of a number resulting from a multiplication, we only need to consider the units place of the numbers being multiplied.

Let's analyze the units place of 77 raised to various powers:

• 77 has a units place of 7
• 77² = 5929 has a units place of 9
• 77³ = 456533 has a units place of 3
• 77⁴ = 35153041 has a units place of 1

We observe that the units digit of 77ⁿ repeats every 4 cycles: 7, 9, 3, 1.

Calculation:

We need to find the probability that the units place of m (i.e., 77ⁿ) will be 1.

From the above pattern, we see that the units place is 1 whenever n is a multiple of 4 (i.e., n = 4k for some integer k).

Given that n is a positive integer, the probability that n is a multiple of 4 among all positive integers is:

Number of favourable outcomes (multiples of 4) = 1, 2, 3, 4, ...

Total number of outcomes = n

Probability  = 1/4 

∴ The probability that the value of m will have 1 in the units place is 1/4.

Given:

Three different numbers are selected at random from the first 15 natural numbers.

Formula Used:

Probability = \(\frac{Number\ of\ Favourable\ Outcomes}{Total\ Number\ of\ Outcomes}\)

Calculation:

Total number of outcomes = \(\binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455\)  

Let's find the number of favorable outcomes where the product of first two of the numbers is equal to the third number.

All the numbers are distinct numbers from 1 to 15.

The possible sets are:

• (1, 2, 2) - Invalid since numbers must be distinct.
• (1, 3, 3) - Invalid since numbers must be distinct.
• (1, 4, 4) - Invalid since numbers must be distinct.
• (1, 5, 5) - Invalid since numbers must be distinct.
• (1, 6, 6) - Invalid since numbers must be distinct.
• (1, 7, 7) - Invalid since numbers must be distinct.
• (1, 8, 8) - Invalid since numbers must be distinct.
• (1, 9, 9) - Invalid since numbers must be distinct.
• (1, 10, 10) - Invalid since numbers must be distinct.
• (1, 11, 11) - Invalid since numbers must be distinct.
• (1, 12, 12) - Invalid since numbers must be distinct.
• (1, 13, 13) - Invalid since numbers must be distinct.
• (1, 14, 14) - Invalid since numbers must be distinct.
• (1, 15, 15) - Invalid since numbers must be distinct.
• (2, 3, 6) - Valid
• (3, 4, 12) - Valid
• (2, 4, 8) - Valid
• (3, 5, 15) - Valid
• (2, 5, 10) - Valid
• (2, 6, 12) - Valid
• (2, 7, 14) - Valid

 

Total number of favorable outcomes = 7

Therefore, the probability = 7/ 455 = 1/65

∴ The probability that the product of two of the numbers is equal to the third number is 1/65.

Concept:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation:

Given, P(A ∪ B) ≥ 0.75 and 0.125 ≤ P(A ∩ B) ≤ 0.375.

Now, P(A) + P(B) - P(A ∩ B) = P(A ∪ B)

⇒ P(A) + P(B) - P(A ∩ B) ≥ 0.75

To find the minimum value of P(A) + P(B), we need the smallest possible value for P(A ∩ B), which is 0.125.

⇒ P(A) + P(B) ≥ 0.75 + 0.125

⇒ P(A) + P(B) ≥ 0.875

∴ Minimum value of P(A) + P(B) is 0.875

The correct answer is Option 4.

Concept:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation:

Given, P(A ∪ B) ≥ 0.75 and 0.125 ≤ P(A ∩ B) ≤ 0.375.

Now, P(A) + P(B) - P(A ∩ B) = P(A ∪ B)

⇒ P(A) + P(B) - P(A ∩ B) ≥ 0.75

To find the maximum value of P(A) + P(B), we need the largest possible value for P(A ∩ B), which is 0.375.

⇒ P(A) + P(B) ≤  0.75 + 0.375

⇒ P(A) + P(B) ≤ 1.125

∴ Maximum value of P(A) + P(B) is 1.125.

The correct answer is Option 2.

Concept:

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B ∩ C)

Calculation:

Given, P(A) = 0.6, P(B) = 0.4, P(C) = 0.5

P(A ∪ B) = 0.8, P(A ∩ C) = 0.3 and P(A ∩ B ∩ C) = 0.2 and P(A ∪ B ∪ C) ≥ 0.85

Now, P(A ∪ B) = 0.8

⇒ P(A) + P(B) - P(A ∩ B) = 0.8

⇒ 0.6 + 0.4 - P(A ∩ B) = 0.8

⇒ P(A ∩ B) = 0.2 ⋯ (i)

∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B ∩ C)

According to the question, P(A ∪ B ∪ C) ≥ 0.85

⇒ P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B ∩ C) ≥ 0.85

⇒ 0.6 + 0.4 + 0.5 - 0.2 - P(B ∩ C) - 0.3 + 0.2 ≥ 0.85

⇒ 1.2 - P(B ∩ C) ≥ 0.85

⇒ - P(B ∩ C) ≥ - 0.35

⇒ P(B ∩ C) ≤ 0.35

So maximum value of P(B ∩ C) is 0.35. But we need the minimum value.

Let’s consider what values we cannot go below.

We are also given:

P(A ∩ C) = 0.3

P(A ∩ B ∩ C) = 0.2

So the part of C that overlaps with both A and B is already 0.2.

So, P(B ∩ C) ≥ P(A ∩ B ∩ C) = 0.2

Hence, the Minimum value of P(B ∩ C) is 0.2​

The correct answer is Option 2.

Concept:

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B ∩ C)

Calculation:

Given, P(A) = 0.6, P(B) = 0.4, P(C) = 0.5

P(A ∪ B) = 0.8, P(A ∩ C) = 0.3 and P(A ∩ B ∩ C) = 0.2 and P(A ∪ B ∪ C) ≥ 0.85

Now, P(A ∪ B) = 0.8

⇒ P(A) + P(B) - P(A ∩ B) = 0.8

⇒ 0.6 + 0.4 - P(A ∩ B) = 0.8

⇒ P(A ∩ B) = 0.2 ⋯ (i)

∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B ∩ C)

According to the question, P(A ∪ B ∪ C) ≥ 0.85

⇒ P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B ∩ C) ≥ 0.85

⇒ 0.6 + 0.4 + 0.5 - 0.2 - P(B ∩ C) - 0.3 + 0.2 ≥ 0.85

⇒ 1.2 - P(B ∩ C) ≥ 0.85

⇒ - P(B ∩ C) ≥ - 0.35

⇒ P(B ∩ C) ≤ 0.35

∴ The maximum value of P(B ∩ C) is 0.35.

The correct answer is Option 3.

Given: 

An unbiased coin is tossed n times. The probability of getting at least one tail is p and the probability of getting at least two tails is q . It is given that p - q = 5/32

Explanation -

Probability of at least one tail p:

p = 1 - Probability of getting no tail

The probability of getting no tail in n tosses = \( \left( \frac{1}{2} \right)^n \)

Therefore, p = 1 - \( \left( \frac{1}{2} \right)^n \)  

Probability of at least two tails q:

q = 1 - Probability of getting no tail + Probability of getting exactly one tail

The probability of getting exactly one tail in n tosses is 

\(\binom{n}{1} \left( \frac{1}{2} \right)^1 \left( \frac{1}{2} \right)^{n-1} = n \left( \frac{1}{2} \right)^n\)

Therefore,

\(q = 1 - \left( \frac{1}{2} \right)^n - n \left( \frac{1}{2} \right)^n = 1 - \left( \frac{1+n}{2^n} \right)\)

It is given that p - q = 5/32

\(\left(1 - \left( \frac{1}{2} \right)^n\right) - \left(1 - \left( \frac{1+n}{2^n} \right)\right) = \frac{5}{32}\)

Simplifying this, we get:

\( \left(1 - \left( \frac{1}{2} \right)^n\right) - \left(1 - \left( \frac{1+n}{2^n} \right)\right) = \frac{5}{32}\)

\(\frac{1+n}{2^n} - \left( \frac{1}{2} \right)^n = \frac{5}{32}\)

\(\frac{1+n - 1}{2^n} = \frac{5}{32}\)

\(\frac{n}{2^n} = \frac{5}{32}\)

This implies that:

n = 5

Thus, the value of n is 5.

Explanation -

Given:

p = \(1 - \left( \frac{1}{2} \right)^5\)  and q = \(1 - \left( \frac{1+5}{2^5} \right)\)

Now 

\(p = 1 - \left( \frac{1}{2} \right)^5 = 1 - \frac{1}{32} = \frac{31}{32} \)   

And 

\(q = 1 - \left( \frac{1+5}{2^5} \right) = 1 - \left( \frac{6}{32} \right) = 1 - \frac{3}{16} = \frac{13}{16} \)   

⇒ \(p + q = \frac{31}{32} + \frac{13}{16} \)

⇒ p + q = 57/32

Thus, the value of p + q is 57/32.

Concept:

Sum of a n terms GP a, ar, ar², ⋯ , ar^n-1 is given by:

S = \(\frac{a(r^n-1)}{r-1}\), if r > 1

else, S = \(\frac{a(1-r^n)}{1-r}\), if r < 1

Calculation:

Given, 

 

∴ \(\sum_i^n x_i f_i\)

= \(\frac{1}{1}+\frac{2}{2}+\frac{3}{2^2}+⋯+\frac{n}{2^{n-1}}\)

Let S = \(\frac{1}{1}+\frac{2}{2}+\frac{3}{2^2}+⋯+\frac{n}{2^{n-1}}\) ⋯ (i)

⇒ \(\frac{1}{2}\)S =       \(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+⋯+\frac{n}{2^n}\)⋯ (ii)

(i) - (ii)

⇒ \(\frac{1}{2}\)S = \(1+\frac{1}{2}+\frac{1}{2^2}+⋯+\frac{1}{2^{n-1}}-\frac{n}{2^n}\)

⇒ \(\frac{1}{2}\)S = \(\frac{1(1-\frac{1}{2^n})}{1-\frac{1}{2}}-\frac{n}{2^n}\)

⇒ \(\frac{1}{2}\)S = \(2(1-\frac{1}{2^n})-\frac{n}{2^n}\)

⇒ \(\frac{1}{2}\)S = \(2-\frac{1}{2^{n-1}}-\frac{n}{2^n}\)

⇒ S = \(2^2-\frac{1}{2^{n-2}}-\frac{n}{2^{n-1}}\)

⇒ S = \(\frac{2^n-1}{2^{n-2}}-\frac{n}{2^{n-1}}\)

⇒ S = \(\frac{2^{n+1}-2}{2^{n-1}}-\frac{n}{2^{n-1}}\)

⇒ S = \(\frac{2^{n+1}-n-2}{2^{n-1}}\)

∴ The value of \(\sum_i^n x_i f_i\) is equal to \(\frac{2^{n+1}-n-2}{2^{n-1}}\).

The correct answer is Option 2.

Concept:

The mean of a distribution is given by, x̅ = \(\frac{\sum_i^n x_i f_i}{\sum_i^nf_i}\)

Calculation:

Given, 

 

∴ \(\sum_i^n x_i f_i\)

= \(\frac{1}{1}+\frac{2}{2}+\frac{3}{2^2}+⋯+\frac{n}{2^{n-1}}\)

Let S = \(\frac{1}{1}+\frac{2}{2}+\frac{3}{2^2}+⋯+\frac{n}{2^{n-1}}\) ⋯ (i)

⇒ \(\frac{1}{2}\)S =       \(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+⋯+\frac{n}{2^n}\)⋯ (ii)

(i) - (ii)

⇒ \(\frac{1}{2}\)S = \(1+\frac{1}{2}+\frac{1}{2^2}+⋯+\frac{1}{2^{n-1}}-\frac{n}{2^n}\)

⇒ \(\frac{1}{2}\)S = \(\frac{1(1-\frac{1}{2^n})}{1-\frac{1}{2}}-\frac{n}{2^n}\)

⇒ \(\frac{1}{2}\)S = \(2(1-\frac{1}{2^n})-\frac{n}{2^n}\)

⇒ \(\frac{1}{2}\)S = \(2-\frac{1}{2^{n-1}}-\frac{n}{2^n}\)

⇒ S = \(2^2-\frac{1}{2^{n-2}}-\frac{n}{2^{n-1}}\)

⇒ S = \(\frac{2^n-1}{2^{n-2}}-\frac{n}{2^{n-1}}\)

⇒ S = \(\frac{2^{n+1}-2}{2^{n-1}}-\frac{n}{2^{n-1}}\)

⇒ S = \(\frac{2^{n+1}-n-2}{2^{n-1}}\)

∴ \(\sum_i^nf_i\) = \(1+\frac{1}{2}+\frac{1}{2^2}+⋯+\frac{1}{2^{n-1}}\)

= \(\frac{1(1-\frac{1}{2^n})}{1-\frac{1}{2}}\)

= \(2-\frac{1}{2^{n-1}}\)

= \(\frac{2^n-1}{2^{n-1}}\)

⇒ x̅ = \(\frac{\sum_i^n x_i f_i}{\sum_i^nf_i}\) = \(\frac{\frac{2^{n+1}-n-2}{2^{n-1}}}{\frac{2^n-1}{2^{n-1}}}\) = \(\frac{2^{n+1}-n-2}{2^{n}-1}\)

∴ The mean of the distribution is \(\frac{2^{n+1}-n-2}{2^{n}-1}\).

The correct answer is Option 3.

Explanation -

The marks obtained by 10 students in a Statistics test are:

24, 47, 18, 32, 19, 15, 21, 35, 50, and 41.

First, we sort these marks in ascending order:

15, 18, 19, 21, 24, 32, 35, 41, 47, 50

The largest five observations are:

32, 35, 41, 47, 50

Mean Deviation of the Largest Five Observations

The Mean:

\(\text{Mean} = \frac{32 + 35 + 41 + 47 + 50}{5} = \frac{205}{5} = 41 \)

The Mean Deviation:

\(\text{Mean Deviation} = \frac{|32 - 41| + |35 - 41| + |41 - 41| + |47 - 41| + |50 - 41|}{5} \)

= \(\frac{9 + 6 + 0 + 6 + 9}{5}\) = 30/5 = 6

The mean deviation of the largest five observations is 6.

Explanation -

The marks obtained by 10 students in a Statistics test are:

24, 47, 18, 32, 19, 15, 21, 35, 50, and 41.

First, we sort these marks in ascending order:

15, 18, 19, 21, 24, 32, 35, 41, 47, 50

The largest five observations are:

32, 35, 41, 47, 50

Variance of the Largest Five Observations

The Variance:

we already have the mean μ = 41.

The Sum of Squared Deviations:

\(\text{Variance} = \frac{(32 - 41)^2 + (35 - 41)^2 + (41 - 41)^2 + (47 - 41)^2 + (50 - 41)^2}{5} \)

= \(\frac{(-9)^2 + (-6)^2 + (0)^2 + (6)^2 + (9)^2}{5}\)

=\( \frac{81 + 36 + 0 + 36 + 81}{5}\) = 234/5 = 46.8

The variance of the largest five observations is 46.8.