Quantitative Aptitude PYQs

Given:

1. P(E ∩ F) ≥ P(E) + P(F) - 1
2. P(E ∪ F) = P(E) + P(F) + P(E ∩ F)
3. P(E ∪ F) ≤ P(E) + P(F)

Calculation:

(1) P(E ∩ F) ≥ P(E) + P(F) - 1

The above inequality is the lower bound of the intersection of two events E and F. This relation is called the Principle of Inclusion-Exclusion (for two events).

If you think of this intuitively, the left-hand side P(E ∩ F) represents the probability of both events E and F happening together. In the worst case, this can't be higher than the sum of the probabilities of E and F occurring individually (i.e., P(E) + P(F)), less the probability of at least one of them happening (which is 1). So, the relation holds true that the intersection should be greater than or equal to the sum of probabilities of two events E and F, minus 1. (2) P(E ∪ F) = P(E) + P(F) + P(E ∩ F)

The correct formula -

P(E ∪ F) = P(E) + P(F) - P(E ∩ F)

The above equation is the formula for the probability of the union of two events E and F. The original relation stated was incorrect because the plus sign before the last term P(E ∩ F) should be a minus sign.

This formula tells us the total probability of either event E, event F, or both happening is equal to the sum of their individual probabilities minus the probability of them both occurring (since we double-counted that in the sum). (3) P(E ∪ F) ≤ P(E) + P(F)

The above inequality is the upper bound on the union of two events E and F. Essentially, this is stating that the probability of either E happening, or F happening, or both, is naturally less than or equal to the sum of their individual probabilities.

If E,F are mutually exclusive (cannot occur at the same time), the sign becomes equality instead of inequality.

So, statement 1 and 3 are correct. Hence option (3) is correct.

Formula used:

P(A|B) = \(\rm \frac{P(A ∩ B)}{P(B)}\)

P(B|A) = \(\rm \frac{P(A ∩ B)}{P(A)}\)

Explanation:

Given that,

P(A|B) < P(A)

⇒ \(\rm \frac{P(A ∩ B)}{P(B)}\) < P(A)       [∵ P(A|B) = \(\rm \frac{P(A ∩ B)}{P(B)}\)]

⇒ P(A ∩ B) < P(A).P(B)

⇒ \(\rm \frac{P(A ∩ B)}{P(A)}\) < P(B)

⇒ P(B|A) < P(B)       [∵ P(B|A) = \(\rm \frac{P(A ∩ B)}{P(A)}\)]

∴  P(B|A) < P(B)

Dispersion is the extent to which values in a distribution differ from the average of the distribution. To quantify the extent of the variation, there are certain measures namely:

• Measures of Dispersion From Average:
• ​The mean deviation (MD) of a series is the arithmetic average of the deviation of various items from a measure of central tendency (mean, median, and mode).
• Standard Deviation (S. D.): One of the most stable measures of variability, it is the most important and commonly used measure of dispersion. It measures the absolute dispersion or variability of a distribution. Standard deviation is the positive square root of the mean of the squared deviations of observations from their mean. It is denoted by S.D. or σx.
• Measures Based Upon Spread of Values:
• Range: It is the simplest measure of variability both from the point of view of calculation and understanding. It is the difference between the largest and the smallest observations in the data. For example, the range of scores 2, 5, 6, 4, 12, 10, 9 and 8 is 12-2 = 10.
• Interquartile range (IQR) is a measure of variability, based on dividing a data set into quartiles. It divides a rank-ordered data set into four equal parts. Q1 is the mid-value in the first half of the rank-ordered data, Q2 is the score that has exactly two quarters (median). Q3 is the mid-value in the second half of the rank-ordered data. 

For example, consider the number 3, 4, 6, 9, 11, 12, 14, 15.

Q1 is the middle value in the first half of the data set.

The middle value is the average of the two middle values; that is, Q1 = \(4+6\over 2\) = 5.

Q3 is the middle value in the second half of the data set.

i. e. Q3 = \(12+14\over 2\) = 13

Semi Quartile Range = \(Q3 - Q1\over 2\) = \(13 - 5 \over 2\) = 4

Formula used:

Probability: The probability of an event can be calculated by simply

dividing the favorable number of outcomes by the total number

of possible outcomes.

The value of the probability of an event to happen can lie between 0 and 1 i.e. P(̅{E})

P(E)+P(̅{E}) = 1

\(\rm P(E) + P(\overline{E}) = 1\)

The event \(\rm P(\overline{E}) \) means ‘Not E’.

Calculation:

P(problem is solved) = 1 - P(problem is not solved)

⇒ P(problem is solved) = 1 - P(A̅) × P(B̅) P(C̅)

⇒ \(\rm (\frac{29}{32}) = 1-(1 - \frac{1}{2})(1 - \frac{3}{4})(1 - p)\)

⇒ \(\rm (1 - \frac{29}{32}) = (1 - \frac{1}{2})(1 - \frac{3}{4})(1 - p)\)

⇒ \(\rm \frac{3}{32} = \frac{1}{2}× \frac{1}{4} × (1 - p)\)

⇒ 1 - p = \(\rm \frac{3}{4}\)

⇒ P = 1 - \(\rm \frac{3}{4}\)

⇒ P = \(\rm \frac{1}{4}\)  

∴ The value of p is \(\rm \frac{1}{4}\).

Formula used:

Probability of occurrence of the event:

P(E) =  \(\frac{n(E)}{n(S)}\)

Where,

n(E) = Number of favorable outcome

n(S) = Number of possible outcome

Calculation:

A batsman hits sixes 8 times out of 60 balls then

Probability of getting sixes

P(A) = \(\rm \frac{8}{60}\) = \(\rm \frac{2}{15}\)​

The probability that he does not hit a six

P(A̅) = 1 − \(\rm \frac{2}{15}\) 

⇒ P(A̅) = \(\rm \frac{13}{15}\)

∴ Required probability is \(\rm \frac{13}{15}\).

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:

We have,

3x - 4y + 8 = 0

⇒ 4y = 3x + 8

⇒ \(y = \frac{3}{4}x+2\)

Also,

4x - 3y - 1 = 0

⇒ 4x = 3y + 1

⇒ x = \(\rm \frac{3y}{4} + \frac{1}{4}\)

∴ Both 1 and 2 statements are correct.

Formula used:

Coefficient of correlations r = \(\rm ± \sqrt{b_{xy}b_{yx}}\)

The regression coefficient b_xy & byx are the change occurring in x for a

unit change in y and y for a unit change in x respectively.

Calculation:

3x - 4y + 8 = 0  

⇒ 4y = 3x + 8   

⇒ y = \(\rm \frac{3}{4}\)x + 2     ----(1)

⇒  b_yx = \(\rm \frac{3}{4}\)

Similarly,

4x - 3y - 1 = 0   -----(2)

b_xy = \(\rm \frac{3}{4}\)

We know that,

r = \(\pm\sqrt{\frac{3}{4}\times \frac{3}{4}}\)

⇒ r = \(\rm \frac{3}{4}\)

Hence, statement 1 is correct.

Put the value of y from the equation (1) in (2), 

4x - 3(​\(\frac{3}{4}\)x + 2) - 1 = 0 

⇒ 16x - 9x - 28 = 0

⇒ 7x = 28

⇒ x = 4

Hence, from equation (2),

4(4) - 3y - 1 = 0

⇒ 3y = 15

⇒ y = 5

Hence, statement 2 is incorrect.

Formula used:

Median of grouped data:

Median = \(\rm l + \frac{\frac{N}{2} - C}{f} × h\)

Where,

l = lower limit of the median class

N = number of observations

f = Frequency of median class

h = Class size

C = Cumulative frequency of class preceding the median class.

To find the median class:  If N is odd, the median is the [(N+1)/2].

If n is even, then the median will be the average of the n/2^th and

the (n/2 +1)^th observation.

Calculation:

 

N = 60, \(\rm \frac{N}{2} = \frac{60}{2} = 30\)

Median class is 35 - 40

l = 35, f = 5, h = 5 and C = 26

⇒ Median = 35 + \(\rm \frac{30 -26}{5} × 5\)

⇒ Median = 35 + 4 = 39

∴ Required median is 39.

Formula used:

Mode = L +  \([ \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_2}]\times h\)

Where,

l = lower limit of the modal class

h = size of the class interval

f₁ = frequency of the modal class

f₀ = frequency of the class preceding the modal class

f2 = frequency of the class succeeding the modal class

Let us take an example to understand this clearly.

 

Calculation:

From the table, we can see

Class with the highest frequency is 25 - 30 = modal class

Hence,

L = 25, f₁ = 11, f₀ =  5, f₂ = 6, h = 5

Therefore,

Mode = 25 + \([\frac{11 - 5}{2\times 11 - 5 - 6}]\times 5\)

⇒ Mode = 27.73

∴ The required mode is 27.73

Concept:

Sum of n terms of AP with first and last term a & d respectively is

S = \(\frac{n}{2}(a +l)\)

Mean: It is the ratio of the sum of all observations and the total

number of observations. 

Calculation:

The given interval is [15, 64].

Hence, numbers in the interval are

15, 16, ......63, 64 which is AP.

a = 15, l = 64

Let there are n numbers in the series.

⇒ Sum = \(\frac{n}{2}(a +l)\)

⇒ S = \(\frac{n}{2}(15 +64)=\frac{79n}{2}\)

Hence, 

Mean = \(\frac{1}{n}( \frac{79n}{2})\)

⇒ Mean = 39.5

∴ The mean of natural numbers in the interval [15, 64] is 39.5.

Formula used:

Mean: It is the ratio of the sum of all observations and the total

number of observations.

Mode: The mode is the value that is repeatedly occurring in a given set.

Relation b/w mean, mode and median:

Mode = 3(Median) - 2(Mean)

Calculation:

Given set of numbers are

x, x, x + 2, x + 3, x + 10 

According to queston, x is a natural number, then put x = 1

Set of numbers are

1, 1, 3, 4, 11

Mean = \(\rm \frac{1 + 1 + 3 + 4 + 11}{5}\) 

Mean = 4

Mode = 1

Mean > Mode

Hence,

Statement 1 is correct.

We know that,

Mode = 3(Median) - 2(Mean)

⇒ 1 = 3(Median) - 2(4)

⇒ 3(Median) = 9

⇒ Median = 3

Median < Mean

∴ Statement 2 is  incorrect.

Formula used:

Mean: It is the ratio of the sum of all observations and the total

number of observations.

Property of mean:

Let mean of  a1, a2,  ......a_n  = A, then

• Mean of  a1 + k,a2,  + k  ......a_n + k = A + k
• Mean of  a1 - k, a2  - k  ......an  - k = A - k
• Mean of  ka1, ka2,  ......kan  = kA
• Mean of a1/k  a2/k  ......an/k  = A/k

 

Calculation:

Given that, a mean of 10 observations 

 A = 5.5

According to the question,  each observation is multiplied by 4 and

subtracted from 44. Let new mean is A', then using the property of

mean,

⇒ A' = 44 - 4A

⇒ A' = 44 - 4 × 5.5      (∵ A = 5.5)

⇒ A' = 44 - 22  

⇒ A' = 22

∴  The required mean is 22.

Alternate Method

Given that,

Mean of 10 observation = 5.5

Let a_1, a_2,  ......a₁₀  are 10 observations.

⇒ (a1 + a2 +   ......a₁₀)/10 = 5.5

⇒ (a1 + a2 +   ......a10) = 55     ------(1)

According to the question, 

Each observation is multiplied by 4 and subtracted from 44. Then,

Mean = [10 ×44 - 4(a1 + a2 +   ......a10)]/10

Using equation (1)

Mean = [10 ×44 - 4 × 55]/10

⇒ Mean = (440 - 220)/10 

⇒ Mean = 220/10 = 22    

∴ The new mean is 22.

Formula used:

For n terms,  a₁, a₂, a₃, ……., a_n

Geometric mean = \(\rm ({a_{1}.a_{2}.a_{3}....a_{n}})^{\frac{1}{n}}\)

Calculation:

g is the geometric mean of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024

⇒ g = \((2 × 4 × 8 × 16 × 32 × 64 × 128 × 256 × 512 × 1024)^{\frac{1}{10}}\)  

⇒ g = (2 × 2² × 2³ × 2⁴ × 2⁵ × 2⁶ × 2⁷ × 2⁸ × 2⁹ × 2^{10\()^{\frac{1}{10}}\)}

⇒ g = \(\rm 2^{5.5}\)

⇒ 2⁵ < g < 2⁶

∴ 32 < g < 64 is correct.

Formula used:

For n terms,  a₁, a₂, a₃, ……., a_n.

Harmonic mean = \(\rm \frac{n}{\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + ....... \frac{1}{a_{n}} }\)

If b is the Harmonic mean of a and b then 

b = 2ab/(a + b)

Calculation:

Given, The harmonic mean of 60 and x is 48

According to the formula used 

\(\rm \frac{2\times 60x}{x + 60} = 48\)

⇒ 120x = 48(x + 60)

⇒120x = 48x + 2880

⇒ 120x - 48x = 2880

⇒ x = 40

∴ The value of x is 40.

Concept:

Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

Where,

\(\bar{X}= \frac{∑ x}{n}\) = mean, 

∑x = sum of all observations,

n = number of observations. 

Calculation:

First 10 even natural numbers are

2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Mean, 

X̅  = \(\rm \frac{2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20}{10}\) 

⇒ X̅ = 11

Therefore, the mean value is 11

Now, subtract each mean from the first 10 even natural numbers, and ignore the minus symbol if any

⇒ Mean deviation = \(\frac{1}{n}|x_i-\bar{X}|\)

⇒ Mean deviation \(=\frac{1}{10}(|2 - 11|+ |4 - 11| + |6-11|+|8-11|+|10-11|+\|12-11|+|14-11|+|16-11|+|18-11|+|20-11|\)

⇒  \(\rm\frac{9 + 7 + 5 + 3 + 1 + 1 + 3 + 5 + 7 + 9}{10}\) = \(\frac{50}{10}\)

∴ The mean deviation of the first 10 even natural numbers is 5.

Given:

\(\rm \displaystyle \sum_{i = 1}^{10}x_i = 110\)

\(\rm \displaystyle \sum_{i = 1}^{10}x_i^2 = 1540\)

Formula used:

Variance:

\(V(X)=\rm \frac{ \displaystyle \sum_{i = 1}^{n}x_i^2}{n} -\frac{ (\displaystyle \sum_{i = 1}^{n}x_i)^{2}}{n^2} \)

Calculation:

We know that, variance,

\(V(X)=\rm \frac{ \displaystyle \sum_{i = 1}^{n}x_i^2}{n} -\frac{ (\displaystyle \sum_{i = 1}^{n}x_i)^{2}}{n^2} \)

⇒ V(X) = \(\rm \frac{1540}{10} - (\frac{110}{10})^2\)

⇒ V(X) = 154 - 121

⇒ V(X) = 33

∴ The variance is 33.

Concept:

• Divisibility rule of 3 : A number is completely divisible by 3 if the sum of its digits is divisible by 3
• ⁿP_r = n!/(n - r)!

Calculation:

Total number of three-digit numbers = 3P2 = 3!/(3 - 2)! = 6

Total number of three-digit numbers are 137, 173, 371, 317, 713, 731 

Total number of outcomes = 6 

Since Sum of digits is 11 which is not a multiple of 3 

So, The favorable outcome is 0

**∴ The probability that the number is divisible by 3 is 0/6 = 0.**​

Concept:

Consider the quadratic equation ax2 + bx + c = 0

Discriminant (D): D = b² - 4ac

For real root: D ≥ 0

b² - 4ac ≥ 0

Calculation:

Given equation 

x2 + x + n = 0

Finding discriminant: (1)² - 4n ≥ 0

1  ≥  4n

n ≤ \(\rm \frac{1}{4}\)

There is no value which belongs in 1, 2, 3, ..... (i.e. natural number)

This is possible for n = 0

∴ The probability of the required event is zero.

Given:

P(not A) = \(\rm \frac{7}{10}\), and P(not B) = \(\rm \frac{3}{10}\)

P(A|B) = \(\rm \frac{3}{14}\)

Formula used:

• \(\rm P(B|A) = \frac{P(A\cap B)}{P(A)}\)
• \(\rm P(A|B) = \frac{P(A\cap B)}{P(B)}\)
• P(A) = 1 - P(not A)
• P(B) = 1 - P(not B)

 

Calculation:

We have,

\(\rm P(\bar{B}) = 0.3, P(\bar{A}) = 0.7\)

⇒ P(B) = 1 - 0.3 = 0.7 and

P(A) = 1 - 0.7 = 0.3

⇒ P(A) = 0.3      ----(1)

We know that,

\(\rm P(A|B) = \frac{P(A\cap B)}{P(B)}\)

⇒ \(\frac{3}{14} \) = \(\frac{P(A\cap B)}{0.7}\)       (∵ P(A|B) = \(\rm \frac{3}{14}\))

⇒ \(\rm P(A\cap B)\) = 0.15      ----(2)   

We know that,

\(\rm P(B|A) = \frac{P(A\cap B)}{P(A)}\)

⇒ P(B|A) = \(\rm \frac{0.15}{0.3}\)        [From equation (1)& (2)]

⇒ P(B|A) = 0.5 = \(\rm \frac{1}{2}\)

∴  P(B|A) is equal to \(\rm \frac{1}{2}\)

Concept:

Probability:

• The probability of an event can be calculated by simply dividing the favorable number of outcomes by the total number of possible outcomes.
• The value of the probability of an event to happen can lie between 0 and 1.

Combination Formula:

 ⁿC_r = \(\rm \frac{n!}{r!(n-r)!}\)

nCr = number of combinations

n = total number of objects in the set

r =number of choosing objects from the set

Calculation:

The number of ways of arranging 10 balls in a row = ¹⁰ P₁₀.​=10!

Now, the condition is to not arrange two black balls adjacently.

So, first, arrange the 7 while balls first and arrange these 3 black balls in between the created 8 places between each white ball.

So, number of favorable outcomes of arranging 3 black balls and 7 white balls

= 8 P3 × 7!

Thus, the probability that no two black balls are placed adjacently is

P = (8 P3 × 7!)​ / 10 P10

⇒ P = \(\rm \frac{7}{15}\)

∴ The required probability is \(\rm \frac{7}{15}\).