Quantitative Aptitude PYQs

Formula used: 

Selection of r things out of total n things:

\(^nC_r = \frac{n!}{r!\times (n-r)!}\)

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:

Possible pair will be: 

(1, 2), (2, 3), (3, 4), (4, 5)

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

(1, 4), (2, 5)

(1, 5)

⇒ n(S) = 10

To make the unit digit zero in the product, possible pairs will be 

(4, 5), (2, 5)

⇒ n(E) = 2

Hence, the required probability

P(E) = n(E)/n(S) = 2/10

⇒ P(E) =  1/5

∴ Required probability is \(\frac{1}{5}\).

Alternate Method

Given digits are 1, 2, 3,  4, and 5. 

As discussed above we know, that two digits can be chosen out of 5 in ⁵C₂ ways.

⇒ n(S) = ⁵C₂ = \(\frac{5!}{2!\times 3!}\)

 ⇒ n(S) = \(\frac{5 \times 4 \times 3!}{2 \times 1\times 3!} = 10\)

⇒ n(S) = 10

Since the last digit in the product appears as 0 which is only possible when any of 2 or 4 is multiplied by 5 so possible required result,

(E) = {(2, 5), (4, 5)}

⇒ n(E) = 2

Hence, the required probability

P(E) = n(E)/n(S) = 2/10

⇒ P(E) =  1/5

∴ Required probability is \(\frac{1}{5}\).

Concept:

Mean: Mean (average) is the sum of all quantities divided by no of quantities of the data set.

Mode: It is the most repeated value in the data set.

Median: Median is the middlemost value of the data set.

Skewness:

• The skewness is a measure of the asymmetry of the probability distribution assuming a unimodal
• distribution (assuming there is only one peak in the distribution of values). We can say that the
• skewness indicates how much our underlying distribution deviates from the  normal distribution
• since the normal distribution has a skewness of 0. Generally, we have three types of skewness.

Symmetrical: When the skewness is close to 0 and the mean is almost the same as the median

Negative skew:

• When the left tail of the histogram of the distribution is longer and the majority of the observations are concentrated on the right tail.
• In this case, we can use also the term "left-skewed" or "left-tailed" and the median is greater than the mean.

Positive skew:

• When the right tail of the histogram of the distribution is longer and the majority of the observations are concentrated on the left tail.
• In this case, we can use also the term "right-skewed" or "right-tailed" and the median is less than the mean.

Explanation:

• We know that when the left tail of the histogram of the distribution is longer and the majority of the observations are concentrated on the right tail. In this case, we can use also the term "left-skewed" or "left-tailed". and the median is greater than the mean.
• So, the frequency curve corresponding to the data obtained is skewed to the left and the graph obtained is as below:

For the above left-skewed curve, Mode > Median > Mean.

∴ Mode > Median > Mean for left-skewed curve.

Concept:

\(\displaystyle Mean=\bar x=\frac{Sum\ of \ observation\ }{Number\ of \ observation}\)

σ2 ​= \(\displaystyle \frac{∑_{i=1}^5(x_i−\bar x)^2 }{N}\) 

(a + b)² = a² + 2ab + b²

Calculation:

Given: N = 5 and σ² ​= 3.6

Let 'a' be the fifth observation.

We know  that \(\displaystyle Mean=\bar x=\frac{Sum\ of \ observation\ }{Number\ of \ observation}\)

⇒ \(\displaystyle \bar x=\frac{2\ +\ 2\ +\ 4\ +\ 5\ +\ a }{5}\)

⇒ 5x̅ = 13 + a

⇒ a = 5x̅ - 13       ------(i)

Also, we know that, σ2 ​= \(\displaystyle \frac{∑_{i=1}^5(x_i−\bar x)^2 }{N}\) = 3.6

⇒ \(\displaystyle \frac{∑_{i=1}^5(x_i−\bar x)^2 }{N}\) = (3.6)

⇒ (2 − x̅)² + (2 − x̅)² + (4 − x̅)² + (5 − x̅)² + (α − x̅)² = 18

⇒ 4 + x̅² - 4x̅ + 4 + x̅2 - 4x̅ + 16 + x̅2 - 8x̅ + 25 + x̅2 - 10x̅ + a² + x̅2 - 2ax̅ = 18 

⇒ 49 + 5x̅2 - 26x̅ + a2 - 2ax̅ = 18

Putting the value of equation (i),

⇒ 49 + 5x̅2 - 26x̅ + (5x̅ - 13)2 - 2(5x̅ - 13)x̅ = 18

⇒ 49 + 5x̅2 - 26x̅ +  25x̅2 - 130x̅ + 169 - 10x̅2 + 26x̅  =18

⇒ 218 + 20x̅2 - 130x̅  = 18

⇒ 20x̅2 - 130x̅ + 200 = 0

⇒ 2x̅2 - 13x̅ + 20 = 0

⇒ 2x̅2 - 8x̅ - 5x̅ + 20 = 0

⇒ 2x̅(x̅ - 4) - 5(x̅ - 4) = 0

⇒ (x̅ - 4) (2x̅ - 5) = 0

⇒ x̅ = 4, x̅ = 5/2

Putting the value of x̅ in equation (i),we get,

a = 5 × 4 - 13 = 7 and a = 5 × 2.5 - 13 = - 0.5

As the observations are positive, a = - 0.5 is not possible.

∴ The required number is 7.

Formula used:

Sum of n terms of AP = \(\frac{n}{2}[2a + (n-1)d]\)

Where a and d are the first term & common differences of AP.

Arithmetic mean = (Sum of all terms)/Total number of terms

Calculation:

Given that, for an AP

a = 4, d = 4 & n = 50

Using the formula of the sum of AP

S = \(\frac{n}{2}[2a + (n-1)d]\)

⇒ S = \(\frac{50}{2}[2× 4+ (50-1)4]\)

⇒ S = 25(8 + 196) = 25 × 204

Now, arithmetic mean of 50 terms is

A = \(\frac{25\times 204}{50}\) = 102

∴ The required arithmetic mean is 102.

Concept:

Mean deviation from mean = \(\displaystyle \frac{Σ∣x_i​ − x̅ ∣​}{N}\) 

Coefficient of mean deviation = \(\displaystyle \frac{Mean\ Deviation}{\bar x}\)

where N is the number of observation

x̅ is the mean of the data

x is the value of the observation.

Calculation:

\(\displaystyle Mean=\bar x=\frac{Sum\ of \ observation\ }{Number\ of \ observation}\)

\(\displaystyle Mean=\bar x=\frac{21+34+23+39+26+37+40+20+33+27}{10} \)

⇒ x̅ = \(\displaystyle \frac{300}{10}=30\)

 Sum of deviation from mean is given by Σ∣xi​ − x̅∣,

⇒ Σ∣x_i​ − x̅∣ = ∣21 − 30∣ + ∣34 − 30∣ + ∣23 − 30∣ + ∣39 − 30∣ + ∣26 − 30∣ + ∣37 − 30∣ + ∣40 − 30∣ + ∣20 − 30∣ + ∣33 − 30∣ + ∣27 − 30∣

⇒ Σ∣xi​ − x̅∣ = 9 + 4 + 7 + 9 + 4 + 7 + 10 + 10 + 3 + 3

⇒ Σ∣xi​ − x̅∣ = 66

Mean deviation from mean = \(\displaystyle \frac{Σ∣x_i​ − x̅ ∣​}{10}\) = \(\displaystyle \frac{66}{10}\) ​= 6.6

Coefficient of mean deviation = \(\displaystyle \frac{6.6}{30}\) = 0.22

∴ Coefficient of mean deviation = 0.22

Concept:

Mean = (Sum of all the observations/Total number of observations)

Deviation = Σ |X – µ|

Here, 

Σ represents the addition of values

X represents each value in the data set

µ represents the mean of the data set

Calculation:

Sum of the deviations x₁, x₂, x₃ ...... x_n, from 100 is -20.

⇒ (x₁ - 100) + (x₂ - 100) + (x₃ - 100) + ......... +(x_n - 100) = - 20

⇒ x₁ + x₂ + x₃ + .......... + x_n - 100n = - 20

⇒ x1 + x2 + x3 + ........... + xn = 100n - 20      ------(i)

Sum of deviations of the values from 92 is 140

⇒ (x₁ - 92) + (x₂ - 92) + (x₃ - 92) + .......... + (x_n - 92) = 140

⇒ x₁ + x₂ + x₃ + .........  + x_n - 92n = 140

⇒ 100n - 20 - 92n = 140      [Using (1)]

⇒ 8n = 160

⇒ n = 20

Thus the total number of observations is 20.

So, sum of observations = x₁ + x₂ + x₃ + ....... + x_n = 100 x 20 - 20 = 2000 - 20 = 1980

And hence, mean of the data is 1980/20 = 99.

∴ The mean of the values is 99.

Concept:

Mean = (Sum of all the observations/Total number of observations)

Deviation = Σ |X – µ|

Here, 

Σ represents the addition of values

X represents each value in the data set

µ represents the mean of the data set

Calculation:

We know that the sum of observations

x1 + x2 + x3 + ....... + xn = 1980      ----(1)

And the total number of observations

n = 20                   ----(2)

Sum of the deviations x1, x2, x3 ...... xn, from 99 is given by

(x1 - 99) + (x2 - 99) + (x3 - 99) + ......... +(xn - 99)

⇒ x1 + x2 + x3 + .......... + xn - 99n     

Putting the value of equations (1) and (2) in equation (3), we get,

1980 - 99 × 20 = 1980 - 1980 = 0

∴ The Sum of the deviations is 0.

Concept:

Mean = (Sum of all the observations/Total number of observations)

Deviation = Σ |X – µ|

Here, 

Σ represents the addition of values

X represents each value in the data set

µ represents the mean of the data set

Calculation:

We know that the sum of observations

x1 + x2 + x3 + ....... + xn = 1980      ------(1)

And the total number of observations

n = 20                                            -----(2)

Sum of the deviations x1, x2, x3 ...... xn, from y is 180

(x1 - y) + (x2 - y) + (x3 - y) + ......... +(xn - y) = 180

⇒ x1 + x2 + x3 + .......... + xn - ny = 180    

Putting the value of equation (1) and (2) in above equation, we get,

⇒ 1980 - 20 × y = 180

⇒ 1980 - 180 = 20y

⇒ 1800 = 20y

⇒ y = 90

∴ The value of y is 90.

Concept Used:

1). The nth term of AP is given by T_n = a + (n – 1) × d

2). Sum of n terms in AP is given by S_n = \(\frac{n}{2}\)[2a + (n − 1) × d]

3). Sum of n terms in a finite AP with the last term as ‘l’ is given by S = \(\frac{n}{2}\)(a + l),

Where a = First term, n = nth term, d = common difference, l = last term.

4). \(\displaystyle Mean = \frac{Sum \ of\ all \ the\ observations}{Total\ number \ of \ observations}\)

Calculation:

Given:

n = 51, a = 4, d = 3.

Sum of n terms in AP is given by Sn = \(\frac{n}{2}\)[2a + (n − 1) × d]

⇒ Sn = \(\frac{n}{2}\)[2a + (n − 1) × d]

⇒ S51 = \(\frac{51}{2}\)[2 × 4 + (51 − 1) × 3]

⇒ S51 = \(\frac{51}{2}\)[8 + 50 × 3]

⇒ S₅₁ = \(\frac{51}{2}\) × 158

⇒ S₅₁ = 51 × 79

\(\displaystyle Mean = \frac{Sum \ of\ all \ the\ observations}{Total\ number \ of \ observations}\)

⇒ Mean = \(\displaystyle \frac{51 \times 79}{51}\)

⇒ Mean = 79

∴ The mean of the marks is 79.

Concept: 

For an arithmetic progression, the median can be calculated using the formula:

Calculation:

Given that the marks obtained by 51 students in a class are in AP with its first term 4 and common difference 3.

Median = Middle term = The median is the middle number in a sorted, ascending or descending list of numbers =  a + (n-1)/2 * d

a = 4

n = 51

d = 3

Middle term = 4 + (51-1)/2 * 3

= 79

Hence, the median of the marks obtained by 51 students is 79.

Concept:

1). The nth term of AP is given by Tn = a + (n – 1) × d

Where a = First term, n = nth term, d = common difference.

2). Deviation = Σ (X – µ)

Here, Σ represents the addition of values

X represents each value in the data set

µ represents the mean of the data set

Calculation:

The nth term of AP is given by

Tn = a + (n – 1) × d

Tn = 4 + (51 – 1) × 3

Tn = 154       ----(i)

So, A.P must be 4, 7, 10, ......... 151, 154        ----(ii)

Now, Sum of n terms in AP is given by

Sn = \(\frac{n}{2}\)[2a + (n − 1) × d]

⇒ Sn = \(\frac{51}{2}\)[2 × 4 + (51 − 1) × 3]

⇒ Sn = 4029

\(\displaystyle Mean = \frac{Sum \ of\ all \ the\ observations}{Total\ number \ of \ observations}\)

⇒ Mean = \(\displaystyle \frac{4029}{51}\)

⇒ Mean = 79       ----(iii)

Now, Sum of the deviations of the marks measured from median is given by

From (i), (ii) and (iii), we can write

(4 - 79) + (7 - 79) + (10 - 79) + .........+ (151 - 79) + (154 - 79)

⇒ (-75) + (-72) + (-69) + .........+ (72) + (75) = 0

∴ The sum of the deviations measured from the median is 0.

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:

Let, G be the event that the first applicant interviewed is a graduate.

T be the event that the first applicant interviewed has at least 3 years of experience.

G̅  is the event that the first applicant interviewed is not a graduate

T̅  is the event that the first applicant interviewed has less than three years of experience

Now, from the table we have,

Number of applicants who are graduates and have less than three years of experience,

 \(n (G \cap \overline T)\) = 36

Total number of applicants n(S) = 90

So the required probability is given by

\(P (G \cap \overline T)\) = \(\displaystyle \frac{n (G \cap \overline T)}{n(S)}\) = \(\displaystyle \frac{36}{90}=\frac{2}{5}\)

\(P (G \cap \overline T)\) equal to \(\displaystyle \frac{2}{5}\).

Concept:

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

Conditional probability:

of two events is given by P(A|B) = \(\displaystyle \frac{n (A \cap B)}{n(B)}\)

P(A|B) represents the probability of occurrence of A given B has occurred.

Calculation:

Let, G be the event that the first applicant interviewed is a graduate.

T be the event that the first applicant interviewed has at least 3 years of experience.

G̅  is the event that the first applicant interviewed is not a graduate

T̅  is the event that the first applicant interviewed has less than three years of experience

Now, from the table we have,

Number of applicants having less than three years of experience,

 \(n (\overline T)\) = 36 + 27 = 63

Total number of applicants, n(S) = 90

Now, \(\displaystyle P(\overline T)=\frac{n(\overline T)}{n(S)}=\frac{63}{90}=\frac{7}{10}\)

So the required probability is given by:

\(P(G | \overline T) \) = \(\displaystyle \frac{P (G \cap \overline T)}{P(\overline T)}\) = \(\displaystyle \frac{\frac{2}{5}}{\frac{7}{10}}=\frac{4}{7}\)

\(P(G | \overline T) \) equal to \(\displaystyle \frac{4}{7}\).

Concept:

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

Conditional probability:

of two events is given by P(A|B) = \(\displaystyle \frac{n (A \cap B)}{n(B)}\)

P(A|B) represents the probability of occurrence of A given B has occurred.

Calculation:

Let, G be the event that the first applicant interviewed is a graduate.

T be the event that the first applicant interviewed has at least 3 years of experience.

G̅  is the event that the first applicant interviewed is not a graduate

T̅  is the event that the first applicant interviewed has less than three years of experience

Now, from the table we have,

Number of applicants who are not graduates, 

 \(n (\overline G )\) = 27 + 9 = 36

Number of applicants who are not graduates and have less than three years of experience, 

\(n ( \overline T \cap \overline G )\) = 27

Total number of applicants n(S) = 90

So the required probability is given by,

\(P(\overline T | \overline G)\) = \(\displaystyle \frac{P(\overline T \cap \overline G )}{P(\overline G)}\) = \(\displaystyle \frac{\frac{n ( \overline T \cap \overline G )}{n(S)}}{\frac{n (\overline G )}{n(S)}} \)

\(\Rightarrow P(\overline T | \overline G)= \frac{n ( \overline T \cap \overline G )}{n (\overline G )} =\frac{27}{36}=\frac{3}{4}\)

\(P (G \cap \overline T)\) equal to \(\displaystyle \frac{3}{4}\).

Given :

\(33\frac{1}{3}%\) of the workers have a chance of suffering from the disease

Concept:

The formula for binomial distribution:

\(P\left( x \right) = ;{n_{{C_x}}}{p^x}{q^{n - x}}\)

where,

p = probability of success of a trial

q = (1- p) = probability of failure of a trail

n = total no of trial

x = no of a successful trial for the given condition.

\({n_{{C_x}}} = \frac{{n!}}{{\left( {n - x} \right)!x!}}\) = number of combinations of n taken x at a time

P(x) = probability of success of x trial

Calculation:

p = \(\frac{1}{3}\) & q = 1 - p = 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)

Probability, where out of 6, exactly 3  suffer from the disease is

P (X = 3)

⇒ P(x = 3) = \(\displaystyle ^6C_3\ (\frac{1}{3})^3\ (\frac{2}{3})^{6-3}\)

⇒ P(x = 3) = \(\displaystyle ^6C_3\ (\frac{1}{3})^3\ (\frac{2}{3})^3\)

⇒ P(x = 3) = \(\displaystyle \frac{6!}{3!.3!} \frac{{2}^3}{{3}^6}\)

⇒ P(x = 3) = \(\displaystyle \frac{160}{729}\)

∴ The probability that exactly 3 out of 6 workers suffer from a disease is \(\displaystyle \frac{160}{729}\)

Given :

\(33\frac{1}{3}%\) of the workers have a chance of suffering from the disease

Concept:

The formula for binomial distribution:

\(P\left( x \right) = ;{n_{{C_x}}}{p^x}{q^{n - x}}\)

where,

p = probability of success of a trial

q = (1- p) = probability of failure of a trail

n = total no of trial

x = no of a successful trial for the given condition.

\({n_{{C_x}}} = \frac{{n!}}{{\left( {n - x} \right)!x!}}\) = number of combinations of n taken x at a time

P(x) = probability of success of x trial

Calculation:

 p = \(\frac{1}{3}\) & q = 1 - p = 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)

Probability, where out of 6, exactly 3  suffer from the disease is

P (X = 0)

⇒ P(x = 0) = \(\displaystyle ^6C_0\ (\frac{1}{3})^0\ (\frac{2}{3})^{6-0}\)

⇒ P(x = 0) = \(\displaystyle (\frac{2}{3})^6\)

⇒ P(x = 0) = \(\displaystyle \frac{64}{729}\)

∴ The probability that no one out of 6 workers suffers from a disease is \(\displaystyle \frac{64}{729}\)

Given :

\(33\frac{1}{3}%\) of the workers have a chance of suffering from the disease

Concept:

The formula for binomial distribution:

\(P\left( x \right) = ;{n_{{C_x}}}{p^x}{q^{n - x}}\)

where,

p = probability of success of a trial

q = (1- p) = probability of failure of a trail

n = total no of trial

x = no of a successful trial for the given condition.

\({n_{{C_x}}} = \frac{{n!}}{{\left( {n - x} \right)!x!}}\) = number of combinations of n taken x at a time

P(x) = probability of success of x trial

Calculation:

From the previous question,

The probability that no one out of 6 workers suffers from a disease

P(A) = \(\displaystyle \frac{64}{729}\)

Let, P(B) is the probability that at least one out of 6 workers suffer from a disease.

We know that, P(A) + P(B) = 1

⇒ P(B) = 1 - p(A)  = 1 - \(\displaystyle \frac{64}{729}\)

∴  p(B) = \(\displaystyle \frac{665}{729}\)

Alternate Method

p = \(\frac{1}{3}\) &  q = 1 - p = 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)

Probability, where out of 6, exactly 3  suffer from the disease is

P (X ≥ 1)

⇒ P (X ≥ 1) = P(x = 1) + (x = 2) + (x = 3) + (x = 4) + (x = 5) + (x = 6)

⇒ P (X ≥ 1) = \(\displaystyle ^6C_1\ (\frac{1}{3})^1\ (\frac{2}{3})^{6-1}+\ ^6C_2\ (\frac{1}{3})^2\ (\frac{2}{3})^{6-2}+\ ^6 C_3\ (\frac{1}{3})^3\ (\frac{2}{3})^{6-3}+ \ ^6C_4\ (\frac{1}{3})^4\ (\frac{2}{3})^{6-4}+\ ^6C_5\ (\frac{1}{3})^5\ (\frac{2}{3})^{6-5}+\ ^6C_6\ (\frac{1}{3})^6\ (\frac{2}{3})^{6-6}\) 

⇒ P (X ≥ 1) = \(\displaystyle ^6C_1\ \frac{{2}^5}{{3}^6}+\ ^6C_2\ \frac{{2}^4}{{3}^6}+\ ^6 C_3\ \frac{{2}^3}{{3}^6}+ \ ^6C_4\ \frac{{2}^2}{{3}^6}+\ ^6C_5\ \frac{{2}^1}{{3}^6}+\ ^6C_6\ \frac{{2}^0}{{3}^6}\) 

⇒ P (X ≥ 1) = \(\displaystyle \frac{6!}{5!.1!} \frac{{2}^5}{{3}^6}+ \frac{6!}{4!.2!} \frac{{2}^4}{{3}^6}+ \frac{6!}{3!.3!} \frac{{2}^3}{{3}^6}+ \frac{6!}{2!.4!} \frac{{2}^2}{{3}^6}+ \frac{6!}{1!.5!} \frac{{2}^1}{{3}^6}+ \frac{6!}{6!} \frac{{2}^0}{{3}^6}\)

⇒ P (X ≥ 1) = \(\displaystyle \frac{665}{729}\)

∴ The probability that at least one out of 6 workers suffer from a disease is \(\displaystyle \frac{665}{729}\)

Concept:

If x₁, x₂, x₃, …x_n are the number of observations (Or class marks of classes) with respective frequencies f₁, f₂, f₃, … f_n, then

The sum of observations = f₁x₁ + f₂x₂ + f₃x₃ +….+ f_nx_n

The total number of observations = f₁+ f₂ +…..+ f_n

Therefore, the mean of the data, x̄ = (f₁x₁+ f₂x₂ + f₃x₃ + ….+ f_nx_n)/( f₁+f₂+… + f_n).

The mean of the data, x̅ = \(\displaystyle ∑\frac{f_ix_i}{N} \)

Where, x_i is the class marks of classes 

Class mark = (lower limit + upper limit)/2

Calculation:

Total frequency = 120

⇒ 17 + p + q + 32 + p - 3q + 19 = 120

⇒ 2p - 2q = 52

⇒ p - q = 26      ------(1)

We know mean x̅ = \(\displaystyle ∑\frac{f_ix_i}{N} \),

⇒ Class mark of 1^st class intreval (x₁) = \(\displaystyle \frac{x_l + x_u}{2}\) = \(\displaystyle \frac{0 + 20}{2}\) = 10

Similarly, we get

⇒ x₂ = \(\displaystyle \frac{20+40}{2}\) = 30

⇒ x₃ = \(\displaystyle \frac{40 + 60}{2}\) = 50

⇒ x₄ = \(\displaystyle \frac{60 + 80}{2}\) = 70

⇒ x₅ = \(\displaystyle \frac{80 + 100}{2}\) = 90

⇒ x̅ = \(\displaystyle ∑\frac{f_ix_i}{N} \)

⇒ x̅ = \(\displaystyle \frac{f_1x_1+f_2x_2+f_3x_3+f_4x_4+f_5x_5}{N}\)

⇒ 50 = \(\displaystyle \frac{17×10+(p+q)×30+32×50+(p-3q)×70+19×90}{120}\)

⇒ 6000 = 170 +30p + 30q + 1600 + 70p - 210 q + 1710

⇒ 6000 = 100p - 180q + 3480

⇒ 2520 = 100p - 180q

⇒ 252 = 10p - 18q      ------(2)

Solving (1) and (2) we get

⇒ p = 27.

∴ The value of p = 27.

Concept:

If x1, x2, x3, …xn are the number of observations (Or class marks of classes) with respective frequencies f1, f2, f3, … fn, then

The sum of observations = f1x1 + f2x2 + f3x3 +….+ fnxn

The total number of observations = f1+ f2 +…..+ fn

Therefore, the mean of the data, x̄ = (f1x1+ f2x2 + f3x3 + ….+ fnxn)/( f1+f2+… + fn).

The mean of the data, x̅ = \(\displaystyle ∑\frac{f_ix_i}{N} \)

Where, xi is the class marks of classes 

Class mark = (lower limit + upper limit)/2

Calculation:

Total frequency = 120

⇒ 17 + p + q + 32 + p - 3q + 19 = 120

⇒ 2p - 2q = 52

⇒ p - q = 26      ------(1)

We know mean x̅ = \(\displaystyle ∑\frac{f_ix_i}{N} \),

⇒ Class mark of 1^st class intreval (x1) = \(\displaystyle \frac{x_l + x_u}{2}\) = \(\displaystyle \frac{0 + 20}{2}\) = 10

Similarly, we get

⇒ x2 = \(\displaystyle \frac{20+40}{2}\) = 30

⇒ x3 = \(\displaystyle \frac{40 + 60}{2}\) = 50

⇒ x4 = \(\displaystyle \frac{60 + 80}{2}\) = 70

⇒ x5 = \(\displaystyle \frac{80 + 100}{2}\) = 90

⇒ x̅ = \(\displaystyle ∑\frac{f_ix_i}{N} \)

⇒ x̅ = \(\displaystyle \frac{f_1x_1+f_2x_2+f_3x_3+f_4x_4+f_5x_5}{N}\)

⇒ 50 = \(\displaystyle \frac{17×10+(p+q)×30+32×50+(p-3q)×70+19×90}{120}\)

⇒ 6000 = 170 +30p + 30q + 1600 + 70p - 210 q + 1710

⇒ 6000 = 100p - 180q + 3480

⇒ 2520 = 100p - 180q

⇒ 252 = 10p - 18q      ------(2)

Solving (1) and (2) we get

⇒ q = 1

∴ The value of q = 1.

Concept:

If x1, x2, x3, …xn are the number of observations with respective frequencies f1, f2, f₃, … fn, then

The sum of observations = f1x1 + f2x2 + f3x3 + ….+ fnxn.

The total number of observations = f1+ f2+… + fn.

Therefore, the mean of the data, x̄ = (f1x1+ f2x2 + f3x3 + ….+ fnxn) / ( f1+ f2+… + fn).

The mean of the data, x̅ = \(\displaystyle ∑\frac{f_ix_i}{N} \)

Calculation:

If the frequency of each class is doubled, then

Total frequency = 2 ×120 = 240

x̅ = \(\displaystyle ∑\frac{f_ix_i}{N} \)

⇒ x̅ = \(\displaystyle \frac{f_1x_1+f_2x_2+f_3x_3+f_4x_4+f_5x_5}{N}\)

⇒ x̅ = \(\displaystyle \frac{2×17×10+2×(p+q)×30+2×32×50+2×(p-3q)×70+2×19×90}{2×120}\)

⇒ x̅ = \(\displaystyle \frac{17×10+(p+q)×30+32×50+(p-3q)×70+19×90}{120}\)

⇒ x̅ = \(\displaystyle \frac{100p \ - \ 180q \ +\ 3480}{120}\)

Substituting the value of p and q in the above equation.

⇒ x̅ = \(\displaystyle \frac{100\times27 \ - \ 180\times 1 \ +\ 3480}{120}\)

⇒ x̅ = 50 

∴ The mean remains the same i.e. 50.