Quantitative Aptitude PYQs

Calculation:

To find the other two observations, let's denote them as x and y. We have the following information:

The sum of observations:

11 + 16 + 20 + x + y = 47 + x + y

The mean of observations:

(11 + 16 + 20 + x + y) / 5 = (47 + x + y) / 5 = 14

From the equation above, we can solve for x + y:

47 + x + y = 14 * 5

47 + x + y = 70

x + y = 70 - 47

x + y = 23

Variance:

The variance formula is given by:

Variance = [(Observation 1 - Mean)² + (Observation 2 - Mean)² + ... + (Observation 5 - Mean)²] / Number of Observations

Putting the values of observations and mean, we get:

13.2 = [(11 - 14)² + (16 - 14)² + (20 - 14)² + (x - 14)² + (y - 14)²] / 5

13.2 = [9 + 4 + 36 + (x - 14)² + (y - 14)²] / 5

13.2 = (49 + (x - 14)² + (y - 14)²) / 5

66 = 49 + (x - 14)² + (y - 14)²

17 = (x - 14)² + (y - 14)²

We have the system of equations:

x + y = 23

(x - 14)² + (y - 14)² = 17

Now, we can solve this system of equations. There are multiple possible solutions, but let's consider one possible solution:

x + y = 23

x = 23 - y

Substituting this into the second equation:

(23 - y - 14)² + (y - 14)² = 17

(9 - y)² + (y - 14)² = 17

81 - 18y + y² + y² - 28y + 196 = 17

2y² - 46y + 260 = 0

y² - 23y + 130 = 0

(y - 10)(y - 13) = 0

Therefore, we have two possible values for y: y = 10 or y = 13.

Substituting these values back into x + y = 23, we find the corresponding values for x:

For y = 10:

x + 10 = 23

x = 13

For y = 13:

x + 13 = 23

x = 10

So, the two possible sets of values for the other two observations are

(x, y) = (13, 10) or (x, y) = (10, 13).

The correct answer is option (3)

Calculation:

Let's break down the given information and use the properties of probability to find the value of k.

We are given:

P(A̅) = 0.7

P(B̅) = k

P(A ∪ B) = 0.8

First, let's consider the complement of the union of A and B, which is (A ∪ B)'

Using the property

P((A ∪ B)' = 1 - P(A ∪ B),

we can write:

P((A ∪ B))' = 1 - P(A ∪ B)

P((A ∪ B))' = 1 - 0.8

P((A ∪ B))'  = 0.2

Next, we can express (A ∪ B)' in terms of A̅ and B̅ using the De Morgan's law:

(A ∪ B)' = A̅ ∩ B̅

Now, we can equate this expression with the probability P((A ∪ B)':

A̅ ∩ B̅ = 0.2

Since A and B are independent events, the probability of their intersection is the product of their individual probabilities:

P(A̅ ∩ B̅) = P(A̅) * P(B̅)

Substituting the given values:

0.2 = 0.7 * k

To solve for k, divide both sides of the equation by 0.7:

0.2 / 0.7 = k

k = 2/7

The correct answer is option (3)

Calculation:

• The probability of getting a tail on each toss is equal to 1 minus the probability of getting a head.
• Given that the probability of getting a head is 1/4, the probability of getting a tail on each toss is 1 - 1/4 = 3/4.
• Since each toss is independent, the probability of getting a tail on the first four tosses followed by a head on the fifth toss can be calculated by multiplying the probabilities of each individual toss.

Therefore, the probability of getting a tail in all the first four tosses followed by a head is:

(3/4) × (3/4) × (3/4) × (3/4) × (1/4) = (3/4)⁴× (1/4)

Simplifying the expression:

(3/4)⁴× (1/4) = 81/256 × 1/4 = 81/1024

Therefore, the probability of getting a tail in all the first four tosses followed by a head is 81/1024.

The correct answer is option (2)

Concept:

If A, B and C are 3 independent events, then P (A ∩ B ∩ C) = P(A) × P(B) × P(C)

Calculation:

Given coin is biased heads come up thrice as likely as tails.

\( \Rightarrow P\left( H \right) = \frac{3}{4};and;P\left( T \right) = \frac{1}{4}\)

Let, T be a random variable for the no. of tails for 3 independent tosses.

To get exactly the number of three heads, the number of tails must be zero.

\( \Rightarrow P\left( {T = 0} \right) = P\left( {HHH} \right) = {\left( {\frac{3}{4}} \right)^3} = \frac{{27}}{{64}} \)

The  probability of getting exactly three heads = 27/64

The correct answer is option (2)

Calculation:

To find the variance of Y,

we can use the properties of the binomial distribution and the fact that \

X + Y = 100.

In a binomial distribution with parameters n and p, the mean (μ) is given by

μ = np,

and the variance (σ²) is given by

σ² = np(1 - p).

Since X follows a binomial distribution with

n = 100 and p = 4/5,

we have μ(X) = np = 100 × 4/5 = 80

and

σ²(X) = np(1 - p) = 100 × 4/5 × (1 - 4/5) = 16.

Now, let's solve for the variance of Y.

We know that X + Y = 100.

Rearranging the equation,

We have Y = 100 - X.

The variance of Y can be calculated using the property:

σ²(aX + b) = a²σ²(X), where a and b are constants.

Here,

a = -1 and b = 100.

So, we have:

σ²(Y) = σ²(-X + 100) = (-1)²σ²(X) = σ²(X) = 16.

Therefore, the variance of Y is 16.

The correct answer is option (3)

Concept:

The regression line of y on x: y = b₁x + b₀

The regression line of x on y: x = a₁y + a₀

where b₀, a₀ is constant and b₁, a₁ is the regression coefficient.

Calculation:

Given lines are

x + 4y + 1 = 0    ...(i)

4x + 9y + 7 = 0   ...(ii)

Let line x + 4y + 1 = 0 be the line of regression x on y, then other be the line of regression y on x.

From (i),

Line of regression x on y we have x = - 4y -1

∴ b_xy ​= Regression coefficient x on y = - 4

From (ii), 

Line of regression y on x we have y = \(\frac{-4}{9}\)y - \(\frac{7}{9}\)

∴ b_yx = Regression coefficient y on x = -4/9

∴ r² =b_yxb_xy ​= (- 4)(- 4/9) = 16/9 

⇒ | r | = \(\frac{4}{{3}}\) ​ which is not possible as 0 ≤ r² ≤ 1.

Hence, our assumption is wrong.

Line x + 4y + 1 = 0 is the line of regression y on x, and 

Line 4x + 9y + 7 = 0 is the regression x on y.

∴ 4x + 9y + 7 = 0

⇒ 4x + 9(- 3) + 7 = 0

⇒ 4x - 27 + 7 = 0

⇒ 4x - 20 = 0

⇒ x = 5

∴ The value of x when y = -3 is 5.

Concept:

Since we are dealing with a pie chart, the sum of all the central angles of the sectors in the circle is 360 degrees.

Calculation:

Let's assign variables to the central angles of the sectors:

Given that p, q, r, and s represent the central angles of the four sectors.

According to the given relation:

9p = 3q = 2r = 6s,

we can express these angles in terms of a common value, let's say k.

9p = 3q = 2r = 6s = k

p = k/9

q = k/3

r = k/2

s = k/6

Now, we know that the sum of the central angles is 360 degrees:

k/9 + k/3 + K/2 + K/6 = 360

k = 324

Substituting the values of p, q, r, and s in terms of k, we have:

Putting the value of p in terms of k in equation 4p - q, we get,

p = k/9

q = k/3

4p - q = 4k/9 - k/3

= k/9

= 324/9

= 36

The correct answer is option (4)

Calculation:

To find the value of (2m + M), where m is the mean of the lowest 8 observations and M is the mean of the highest 4 observations, we first need to determine the values of m and M.

Given the observations:

4, 1, 4, 3, 6, 2, 1, 3, 4, 5, 1, 6.

To find m, the mean of the lowest 8 observations,

We arrange the observations in ascending order and calculate the mean:

1, 1, 1, 2, 3, 3, 4, 4.

mean (m)

= (1 + 1 + 1 + 2 + 3 + 3 + 4 + 4) / 8

= 19 / 8

To find M, the mean of the highest 4 observations,

we arrange the observations in descending order and calculate the mean:

6, 6, 5, 4.

M = (6 + 6 + 5 + 4) / 4

= 21 / 4

Then (2m + M):

= 2×19/8 + 21/4

= 10

Therefore, (2m + M) is equal to 10.

The correct answer is option 1

Concept:

Equation of line joining the points (x1, y1) and (x2, y2) is given by: \(\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\)

Calculation:

Given, data points are  (-1, 1) = (x₁, y₁) and (3, 2) = (x₂, y₂).

For two data points, the line of regression will be same as line joining the two points.

∴ Equation of line: \(\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\)

⇒ \(\frac{y-1}{x-(-1)}=\frac{2-1}{3-(-1)}\)

⇒ \(\frac{y-1}{x+1}=\frac{1}{4}\)

⇒ 4y - 4 = x + 1

⇒ x - 4y + 5 = 0

∴ The line of regression of y on x is given by x - 4y + 5 = 0.

Calculation:

To find the mode of the data obtained from throwing a die 10 times,

We need to determine the number that appears most frequently in the data set.

The given data set is:

1, 2, 1, 1, 2, 1, 4, 6, 5, 4.

Counting the frequency of each number:

1 appears 4 times

2 appears 2 times

4 appears 2 times

6 appears 1 time

5 appears 1 time

From the frequency counts, we can see that 1 appears most frequently, with a count of 4.

Therefore, the mode of the data set is 1.

The correct answer is option (4)

Calculation:

• To find the median of a frequency distribution, we need to determine the middle value of the data set.
• In this case, we have a frequency distribution with values (1, 2, 3, 5) and corresponding frequencies (4, 6, 9, 7).

To find the median, we need to arrange the values in ascending order and find the middle value or values if there is an even number of observations.

Arranging the values in ascending order:

1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5

Now we count the total number of observations, which is the sum of the frequencies:

4 + 6 + 9 + 7 = 26

Since we have an even number of observations (26), the median will be the average of 13^th and 14^th term, i.e.,

= (3 + 3)/2

=  3

​Therefore, the median of the distribution is 3.

The correct answer is option (3)

Calculation:

Given the data: -

-1, 1, 4, 3, 8, 12, 17, 19, 9, 11.

Arranging the terms in ascending order::

-1, 1, 3, 4, 8,9,11,12,17,19

The median is the middle value, which is 3.

Therefore, M = 3.

To find N, the median of the last five observations, we arrange the last five numbers in ascending order:

 9, 11, 12, 17.,19

The median is the middle value, which is 12.

Therefore, N = 12.

Now, we can calculate the value of 4M - N:

4M - N = 4 × 3 - 11

= 12 - 12

= 0

Therefore, the value of 4M - N is 0

The correct answer is option (4)

Calculation:

Given that:

Let

\(5 P=4 Q=\frac{R}{2}\) = K

So,

P = K/5

Q = K/4

R = 2K

Putting the values of P , Q,  and R in terms of K in \(\frac{P+Q}{2 P+0.7 R}\) , we get

\(= \frac{\frac{k}{5}+ \frac{k}{4}}{2\times\frac{k}{5}+0.7\times(2k)}\)

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

The correct answer is option (4)

Calculation:

The given sequence is 1, 2, 4, 8, ..., 2^n-1,

We need to find the geometric mean (G) of this sequence.

G.P:

The geometric mean (G) is calculated by taking the nth root of the product of all the terms. In this case, we have n terms.

The product of all the terms can be expressed as:

1 × 2 × 4 × 8 × ... × 2^n-1 = 2^{( 0 + 1 + 2 + ......(n-1))}

= 2^(n-1)n/2

To find G, we take the nth root of this product:

G = 2(n-1)n/2×1/n  

= 2^(n-1)/2

Now, let's simplify the expression 1 + 2log₂(G):

1 + 2log2(G) = 1 + 2log₂( 2(n-1)/2)).

Using the logarithmic property log_ab^c = c log_ab,

we can rewrite this as:

1 + 2 × ((n-1)/2) × log₂(2).

Since log₂(2) = 1, the expression simplifies to:

1 + 2 ×  ((n-1)/2) = 1 + (n-1) = n.

Therefore, the value of 1 + 2log2(G) is  n

The correct answer is option (4)

Calculation:

The harmonic mean of the the x₁ , x₂ , x₃ ......x_n is given by 

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

n/H = \(\frac{1}{x_1} + \frac{1}{x_2}+ \frac{1}{x_3} + .....\frac{1}{x_n}\)

n/H = \(\frac{1}{1} + \frac{1}{2}+ \frac{1}{2^2} + .....\frac{1}{2^(n-1)}\)

The above is geometric mean of common ratio 1/2 and first term 1. The sum of above G.P would be

= \(\frac{a(1-r^n)}{(1-r)}\)

= \(\frac{1(1-(1/2^n))}{(1-1/2)}\)

= \(2-\frac{1}{2^{n-1}}\)

The correct answer is option (2)

Concept:

• Mean is the average of all of the numbers.
• Median is the middle number, when in order.
• Mode is the most common number.

Calculation:

Given, P be the median, Q be the mean and R be the mode of observations x1, x2, x3, .....xn

Q = \(\frac{\sum_{i=1}^n x_i}{n}\)          -----(1)

Given, \(S=\sum_{i=1}^n\left(2 x_i-a\right)^2\)

For minima, dS/dx = 0

⇒  \(2\times 2\sum_{i=1}^n\left(2 x_i-a\right)\) = 0

⇒ \(\sum_{i=1}^n\left(2 x_i-a\right)\) = 0

⇒ \(2\sum_{i=1}^n x_i-na\) = 0

⇒ a = \(\frac{2\sum_{i=1}^n x_i}{n}\) 

From equation (1)

a = 2Q

∴ S takes minimum value at a = 2Q.

Given:

Bag 1: 3 white balls, 2 black balls

Bag 2: 2 white balls, 3 black balls

2 balls are drawn from Bag 1 and put into Bag 2, then a ball is drawn from Bag 2.

Concept:

This problem involves conditional probability and the concept of combinations

in probability.

Calculation:

       3W          2W

       2B           3B

    _____      ____

       (I)             (II)

There are three possible scenarios:

Draw two black balls (BB) ⇒ 2W, 5B

Draw two white balls (WW) ⇒ 4W, 3B

Draw one white ball and one black ball (WB or BW) ⇒ 3W, 4B

Required probability

\(\frac{^{2}C_2}{^5C_2} \times \frac{2}{7}+\frac{^{3}C_2}{^5C_2} \times \frac{4}{7}+\frac{^{2}C_1 \times ^3C_1}{^5C_2} \times \frac{3}{7}\)

\(\frac{2+18+12}{70} = \frac{32}{70}\)

Therefore, the probability that the ball drawn from the second bag is white is 32/70.

Calculation:

• To calculate the probability that each face of three dice shows only multiples of 3, we need to determine the total number of favorable outcomes and divide it by the total number of possible outcomes.
• If we go through each die individually.
• Each die has six faces, and out of those six faces, there are two multiples of 3: 3 and 6.
• Therefore, the probability of getting a multiple of 3 on one die is \(\frac{2}{6}\) or \(\frac{1}{3}\).

Since the three dice are independent, we need to multiply the probabilities together to find the probability of all three dice showing multiples of 3.

\(\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3} =\frac{1}{27}\)

The correct answer is option  (3)

Calculation:

In the month of December 31 day, we have 31 days i.e.

28 + 3 days

for 28 days there would be 4 Sundays.

for the remaining 3 days we need to consider.

if we consider the first 3 days of the month of December, and to get 5 Sundays in the month of December, Sunday should be in the first three days.

i.e, it could be 

1^st December Sunday

or 2^nd December be Sunday

or 3^rd December be Sunday.

We know that the Total days in a week = 7

The probability of getting 1 December be Sunday = 1/7

Similarly, 2^nd and 3^rd to be Sunday = 1/7 and 1/7 respectively.

The total probability of getting 5 Sundays in the month of December would be 

= \(\frac{1}{7}\) + \(\frac{1}{7}\) + \(\frac{1}{7}\)

= \(\frac{3}{7}\)

or

if we consider the last 3 days

i.e., 28 days + 3 days

The possible outcome for the last 3 days would be

• The last three days fall on Monday, Tuesday, and Wednesday.
• The last three days fall on Tuesday, Wednesday, and Thursday.
• The last three days fall on Wednesday, Thursday, and Friday.
• The last three days fall on Thursday, Friday, and Saturday.
• The last three days fall on Friday, Saturday, and Sunday.
• The last three days fall on Saturday, Sunday, and Monday.
• The last three days fall on Sunday, Monday, and Tuesday.

There are a total of 7 outcomes out of which Sundays will fall on

Friday, Saturday, and Sunday.

Saturday, Sunday, and Monday.

Sunday, Monday, and Tuesday.

So, the probability of getting 5 Sundays = 3/7

​The correct answer is option (3)

Calculation:

The first 50 natural numbers start at 1 and end at 50.

For n = 1,n = 49, and for n = 50, the output is greater than 50 and for all other values of n, the output is less than 50.

So, the probability that \(n+\frac{50}{n}<50 \)  would be

= 1 - {probability of getting output greater than 50)

= 1 - { 3/50)

= 1 - 3/50

= 47/50

The correct answer is option (2)