Question 55

Question:The mean age of 25 teachers in a school is 40yr. A teacher retires at the age of 60yr and a new teacher is appointed in his place. If the mean age of the teachers in this school now is 39yr, then the age (in years) of the newly appointed teacher is _________.Options:(Numerical/Integer Type - No options provided in source).

Correct Answer:35.

Year:2021, 18 March Shift-I.

Solution (Source): Given, $n = 25$, $\bar{x} = 40$. Total sum $\Sigma x = 25 \times 40 = 1000$. New sum = $1000 - 60 + x = 940 + x$. New average = 39. $39 = (940 + x) / 25$. $975 = 940 + x \Rightarrow x = 35$.

Step Solution:

    1.  Calculate the original total age: $\text{Total Age} = \text{Number of Teachers} \times \text{Mean} = 25 \times 40 = 1000$ years.

    2.  Adjust for the retiring teacher: $\text{Remaining Age Sum} = 1000 - 60 = 940$ years.

    3.  Set up the equation for the new mean: Let $x$ be the age of the new teacher. $\text{New Mean} = \frac{940 + x}{25}$.

    4.  Substitute the new mean (39): $39 = \frac{940 + x}{25}$.

    5.  Solve for $x$:$940 + x = 39 \times 25 = 975 \implies x = 35$.

Difficulty Level:Easy.

Concept Name:Arithmetic Mean (Change in Observation).

Shortcut Solution:$\text{New Age} = \text{Old Age} + (\text{Change in Mean} \times \text{Total Count}) = 60 + ((39 - 40) \times 25) = 60 - 25 = 35$.

Question 67

 Question:The mean of 10 numbers $7 \times 8, 10 \times 10, 13 \times 12, 16 \times 14, \ldots$ is _________.

Options:(Numerical/Integer Type - No options provided in source).

Correct Answer: 398.

Year:2021, 31 Aug. Shift-1.

Solution (Source):General terms are $a_n = 3n + 4$ and $b_n = 2n + 6$. $T_n = (3n + 4)(2n + 6) = 6n^2 + 26n + 24$. Mean = $\frac{\Sigma T_n}{10} = \frac{6 \cdot \frac{10 \cdot 11 \cdot 21}{6} + 26 \cdot \frac{10 \cdot 11}{2} + 24 \cdot 10}{10} = 398$.

Step Solution:

  1.  Identify the general term ($T_n$): The first parts (7, 10, 13...) form an A.P. with $a=7, d=3 \implies 3n+4$. The second parts (8, 10, 12...) form an A.P. with $a=8, d=2 \implies 2n+6$.

   2.  Expand $T_n$:$T_n = (3n + 4)(2n + 6) = 6n^2 + 18n + 8n + 24 = 6n^2 + 26n + 24$.

   3.  Find the sum of 10 terms ($\Sigma_{n=1}^{10} T_n$): $\text{Sum} = 6 \Sigma n^2 + 26 \Sigma n + 24 \Sigma 1$.

   4.  Calculate the sum: $\text{Sum} = 6(\frac{10 \times 11 \times 21}{6}) + 26(\frac{10 \times 11}{2}) + 24(10) = 2310 + 1430 + 240 = 3980$.

   5.  Calculate the Mean: $\text{Mean} = \frac{3980}{10} = 398$.

Difficulty Level:Medium.

Concept Name:Arithmetic Mean of a Series (Summation of $n$ and $n^2$).

Shortcut Solution: Directly calculate the mean of the quadratic expression: $\text{Mean} = 6 \cdot \text{Mean}(n^2) + 26 \cdot \text{Mean}(n) + 24$.

Question 79

Question:Consider the data on $x$ taking the values $0, 2, 4, 8, \dots, 2^n$ with frequencies ${}^nC_0, {}^nC_1, {}^nC_2, \dots, {}^nC_n$ respectively. If the mean of this data is $\frac{728}{2^n}$, then $n$ is equal to _________.

Options:(Numerical/Integer Type - No options provided in source).

Correct Answer:6.

Year: NA Sep. 06, 2020 (II).

Solution (Source):$\text{Mean} = \frac{\Sigma x_i f_i}{\Sigma f_i} = \frac{0 \cdot {}^nC_0 + 2 \cdot {}^nC_1 + 2^2 \cdot {}^nC_2 + \dots + 2^n \cdot {}^nC_n}{{}^nC_0 + {}^nC_1 + \dots + {}^nC_n}$. Denominator $= 2^n$. Numerator $= (1+2)^n - {}^nC_0 = 3^n - 1$. So, $\frac{3^n - 1}{2^n} = \frac{728}{2^n} \implies 3^n = 729 \implies n = 6$.

Step Solution:

 1.  Find the sum of frequencies ($\Sigma f_i$): $\Sigma f_i = {}^nC_0 + {}^nC_1 + \dots + {}^nC_n = 2^n$.

  2.  Find the sum of products ($\Sigma x_i f_i$): $\Sigma x_i f_i = 0 \cdot {}^nC_0 + 2 \cdot {}^nC_1 + 2^2 \cdot {}^nC_2 + \dots + 2^n \cdot {}^nC_n$.

  3.  Apply the Binomial Theorem:The sum is the expansion of $(1+2)^n$ excluding the first term ($1 \cdot {}^nC_0 = 1$), so $\Sigma x_i f_i = 3^n - 1$.

  4.  Equate to the given mean:$\frac{3^n - 1}{2^n} = \frac{728}{2^n}$.

   5.  Solve for $n$: $3^n - 1 = 728 \implies 3^n = 729 \implies 3^n = 3^6 \implies n = 6$.

Difficulty Level:Medium/Hard.

Concept Name:Mean of Frequency Distribution and Binomial Theorem.

Shortcut Solution:Recognize the pattern: the weighted sum is almost exactly $3^n$. Since the $x=0$ term eliminates ${}^nC_0$, the numerator is $3^n - 1$. Then solve $3^n - 1 = 728$

Question 80

Question:The minimum value of $2^{\sin x} + 2^{\cos x}$ is:

Options:  

  A. $2^{1 - \sqrt{2}}$ 

  B. $2^{-1 + \sqrt{2}}$ 

  C. $2^{1 + \sqrt{2}}$ 

  D. $2^{1 - \frac{1}{\sqrt{2}}}$

Correct Answer: D

Year: Sep. 04, 2020 (II)

Solution (Source):$\frac{2^{\sin x} + 2^{\cos x}}{2} \geq (2^{\sin x + \cos x})^{\frac{1}{2}} (\because \text{AM} \geq \text{GM}) \implies 2^{\sin x} + 2^{\cos x} \geq 2 \cdot 2^{\frac{\sin x + \cos x}{2}}$. Since $-\sqrt{2} \leq \sin x + \cos x \leq \sqrt{2}$, the minimum value is $2^{1 - \frac{1}{\sqrt{2}}}$.

Step Solution:

    1.  Apply AM-GM Inequality:For two positive numbers $2^{\sin x}$ and $2^{\cos x}$, $\frac{2^{\sin x} + 2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}}$.

    2.  Simplify the RHS:$\sqrt{2^{\sin x} \cdot 2^{\cos x}} = \sqrt{2^{\sin x + \cos x}} = 2^{\frac{\sin x + \cos x}{2}}$.

    3.  Identify the range of the exponent: The expression $\sin x + \cos x$ has a minimum value of $-\sqrt{2}$.

    4.  Find the lower bound:Substitute the minimum value into the inequality: $2^{\sin x} + 2^{\cos x} \geq 2 \cdot 2^{\frac{-\sqrt{2}}{2}}$.

    5.  Calculate the final result: $\min = 2^1 \cdot 2^{-\frac{1}{\sqrt{2}}} = 2^{1 - \frac{1}{\sqrt{2}}}$.

The Difficulty Level:Medium.

The Concept Name:AM-GM Inequality and Range of Trigonometric Functions.

Short cut solution:The minimum of $2^u + 2^v$ occurs when $u+v$ is at its minimum and $u=v$. Since $\min(\sin x + \cos x) = -\sqrt{2}$, the value is $2 \cdot 2^{-\sqrt{2}/2} = 2^{1 - 1/\sqrt{2}}$.

Question 90

Question:If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is:

Options:  

  A. 30 

  B. 51 

  C. 50 

  D. 31

Correct Answer: D

Year:Jan. 12, 2019 (I)

Solution (Source): Given $\sum_{i=1}^{50} (x_i - 30) = 50 \implies \sum x_i - 50(30) = 50 \implies \sum x_i = 1550$. Mean $\bar{x} = \frac{1550}{50} = 31$.

Step Solution:

  1.  Translate the word Question into an equation: $\sum_{i=1}^{50} (x_i - 30) = 50$.

  2.  Expand the summation: $\sum x_i - \sum 30 = 50$.

  3.  Calculate the constant sum: Since there are 50 terms, $\sum 30 = 50 \times 30 = 1500$.

  4.  Find the total sum of observations: $\sum x_i = 50 + 1500 = 1550$.

 5.  Compute the mean: $\text{Mean} = \frac{\text{Total Sum}}{\text{Number of Observations}} = \frac{1550}{50} = 31$.

The Difficulty Level: Easy.

The Concept Name: Properties of Arithmetic Mean (Sum of Deviations).

Short cut solution: $\text{Actual Mean} = \text{Assumed Mean} + \frac{\text{Sum of Deviations}}{N} = 30 + \frac{50}{50} = 31$.

Question 97

Question:If for some $x \in R$, the frequency distribution of the marks obtained by 20 students in a test is: [Table provided in source: Mark 2, 3, 5, 7 with frequencies $(x+1)^2, (2x-5), (x^2-3x), x$ respectively], then the mean of the marks is:

Options:  

  A. 3.2 

  B. 3.0 

  C. 2.5 

  D. 2.8

Correct Answer: D

Year:April 10, 2019 (I)

Solution (Source):Sum of frequencies is $(x+1)^2 + (2x-5) + (x^2-3x) + x = 20 \implies 2x^2 + 2x - 4 = 20 \implies x^2 + x - 12 = 0 \implies x=3$. Average marks = $\frac{32 + 3 + 0 + 21}{20} = 2.8$.

Step Solution:

    1.  Set up the frequency equation: $(x+1)^2 + (2x-5) + (x^2-3x) + x = 20$.

    2.  Solve the quadratic equation for $x$: $2x^2 + 2x - 4 = 20 \implies 2x^2 + 2x - 24 = 0 \implies x^2 + x - 12 = 0 \implies x=3$ (as $x$ must be positive).

    3.  Find specific frequencies: For $x=3$, $f_1=16, f_2=1, f_3=0, f_4=3$.

    4.  Apply the weighted mean formula: $\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(2 \times 16) + (3 \times 1) + (5 \times 0) + (7 \times 3)}{20}$.

 5.  Final calculation: $\text{Mean} = \frac{32 + 3 + 0 + 21}{20} = \frac{56}{20} = 2.8$.The Difficulty Level: Medium.

The Concept Name: Arithmetic Mean of Frequency Distribution.

Short cut solution: Quickly determine $x=3$ by testing small integers in the frequency sum; once $x=3$ is found, the mean is a simple weighted average: $\frac{56}{20} = 2.8$

Question 108

Question:The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is :

Options:  

    A. 25

    B. 30

    C. 35

    D. 40

Correct Answer:C

Year:Online April 8, 2017

Solution (Source): Let; $\frac{x_1 + x_2 + \ldots + x_{25}}{25} = \bar{x} = 40 \Rightarrow x_1 + x_2 + \dotsc + x_{25} = 1000$. $\therefore x_2 + x_2 + \ldots + x_{25} - 60 + A = 39 \times 25$. Let $A$ be the age of new teacher. $\Rightarrow 1000 - 60 + A = 975 \Rightarrow A = 975 - 940 = 35$.

Step Solution:

    1.  Calculate original total age: $\text{Total} = 25 \times 40 = 1000$ years.

    2.  Adjust for the retiring teacher: $\text{Remaining sum} = 1000 - 60 = 940$ years.

    3.  Establish new total age equation: Let $x$ be the new teacher's age; $\text{New Total} = 940 + x$.

    4.  Relate to the new mean (39): $\frac{940 + x}{25} = 39$.

    5.  Solve for $x$: $940 + x = 39 \times 25 = 975 \implies x = 35$.

Difficulty Level: Easy.

Concept Name: Arithmetic Mean (Change of Observation).

Short cut solution: $\text{New Age} = \text{Old Age} + (\text{Change in Mean} \times \text{Total Count}) = 60 + ((39 - 40) \times 25) = 60 - 25 = 35$.

Question 113

Question:The mean of the data set comprising of 16 observations is 16 . If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data , is:

Options:  

    A. 15.8

    B. 14.0

    C. 16.8

    D. 16.0

Correct Answer:B

Year:2015

Solution (Source): Sum of 16 observations $= 16 \times 16 = 256$. Sum of resultant 18 observations $= 256 - 16 + (3 + 4 + 5) = 252$. Mean of observations $= \frac{252}{18} = 14$.

Step Solution:

    1.  Find the initial sum: $16 \text{ observations} \times \text{mean } 16 = 256$.

    2.  Adjust sum for deletion and additions: $\text{New Sum} = 256 - 16 + (3 + 4 + 5) = 252$.

    3.  Determine the new number of observations ($N$): $N = 16 - 1 (\text{deleted}) + 3 (\text{added}) = 18$.

    4.  Calculate the new mean: $\text{Mean} = \frac{\text{New Sum}}{\text{New Count}} = \frac{252}{18}$.

    5.  Final Division: $\text{Mean} = 14$.

Difficulty Level:Easy.

Concept Name: Arithmetic Mean (Addition and Deletion of Data).

Short cut solution:Use the total sum method: $\frac{(16 \times 16) - 16 + 12}{18} = \frac{252}{18} = 14$.

Question 116

Question:In a set of 2n distinct observations , each of the observations below the median of all the observations is increased by 5 and each of the remaining observations is decreased by 3 . Then the mean of the new set of observations:

Options:  

    A. increases by 1

    B. decreases by 1

    C. decreases by 2

    D. increases by 2

Correct Answer:A

Year: Not specified in source.

Solution (Source):There are $2n$ observations $x_1, x_2, \ldots, x_{2n}$. $\text{Mean} = \sum_{i=1}^{2n} \frac{x_i}{2n}$. Divide into two parts (each size $n$); $1^{\text{st}}$ part 5 is added ($\sum x_i + 5n$), $2^{\text{nd}}$ part 3 is subtracted ($\sum x_i - 3n$). Total new sum $= \sum x_i + 5n - 3n = \sum x_i + 2n$. $\text{New Mean} = \frac{\sum x_i + 2n}{2n} = \text{Old Mean} + 1$.

Step Solution:

    1.  Identify the group sizes: Since the data is split by the median into $2n$ observations, there are $n$ observations below the median and $n$ observations at or above it.

    2.  Calculate the sum change for the first group: $n \times (+5) = +5n$.

    3.  Calculate the sum change for the second group: $n \times (-3) = -3n$.

    4.  Find the net change in the total sum: $\text{Net Change} = 5n - 3n = +2n$.

    5.  Calculate the change in mean: $\Delta\text{Mean} = \frac{\text{Net Change in Sum}}{\text{Total Observations}} = \frac{2n}{2n} = 1$.

Difficulty Level: Medium.

Concept Name: Arithmetic Mean (Transformation of Data).

Short cut solution:Use the formula $\text{Change in Mean} = \frac{\sum (\text{individual changes})}{\text{Total Count}} = \frac{n(5) + n(-3)}{2n} = \frac{2n}{2n} = 1$.