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MATHS MCQ

STATISTICS · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 165 questions.

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MCQ 11 Mark
The wickets taken by a bowler in a one day cricket match are 4, 5, 6, 3, 4, 0, 3, 2, 3, 5. The mode of the data is ________ .
  • ✓
    3
  • B
    4
  • C
    5
  • D
    2
Answer
Correct option: A.
3
Mode of the set of data is the observation which occurs the most.
4, 6 occurs 2 times each, 6, 2 and 00 occurs 11 time each, whereas 3 occurs 3 times.
Thus, the number 33 occurs the maximum number of times i.e., 3. Therefore, mode of the data is 33.
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MCQ 21 Mark
The average of the first five odd prime numbers is:
  • A
    7
  • ✓
    7.8
  • C
    8
  • D
    8.7
Answer
Correct option: B.
7.8
Required average $ = \frac{3 + 5 + 7 + 11 + 13}{5} = \frac{39}{5} =7.8$
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MCQ 31 Mark
Mode of the distribution is that value of the variate for which the_____is_____.
  • ✓
    frequency, maximum
  • B
    Frequency, minimum
  • C
    frequency, arithmetic mean
  • D
    median, mode
Answer
Correct option: A.
frequency, maximum
Mode of the distribution is that value of the variate for which the frequency is maximum.
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MCQ 41 Mark
The average age of 6 students is 11 years. If two more students of age 14 and 16 years join, their average will become
  • A
    13 years
  • ✓
    12 years
  • C
    $ 12\dfrac{1}{2}​\text{ years}$
  • D
    $ 11\dfrac{1}{2}\text{ years}$
Answer
Correct option: B.
12 years
⇒ The average age of 66 students is 11 years.
⇒ Sum of age of 66 students = 6 × 11 = 66
⇒ When two more students of age 14 and 16 added to 6 students then,
total students will become 8 ⇒ Sum of age of 8 students = 66 + 14 + 16 = 96
⇒ Required average $ =\dfrac{96}{8}=12\ \text{years}$




     
  1.  
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MCQ 51 Mark
If the mean of x + 2, 2x+ 3, 3x + 4, 4x + 5 is x + 2 then x is equal to:
  • A
    0
  • B
    1
  • ✓
    -1
  • D
    2
Answer
Correct option: C.
-1
Mean of the given distribution is,
$ =\frac{(\text{x}+2)+(2\text{x}+3)+(3\text{x}+4)+(4\text{x}+5)}{4} = \text{x} +2$
= 4(x + 2) + (2x + 3) + (3x + 4) + (4x + 5) ​= x + 2, (given)
$=\frac{10\text{x}+14}{4} = \text{x}+2$
= 10x + 14 = 4x + 8 ⇒ x = -1
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MCQ 61 Mark
Choose the correct answer. Consider the first 10 positive integers. If we multiply each number by -1 and then add 1 to each number, the variance of the numbers so obtained is:
  • ✓
    8.25
  • B
    6.5
  • C
    3.87
  • D
    2.87
Answer
Correct option: A.
8.25
Since, the first 10 positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
On multiplying each number by -1, we get -1, -2, -3, -4, -5, -6, -7, -8, -9, -10 On adding 1 in each number.
We get 0, -1, -2, -3, -4, -5, -6, -7, -8, -9.
$\therefore\ \sum\text{x}_\text{i}=-\frac{9\times10}{2}=-45$
and $\sum\text{x}^2_\text{i}=0^2+(-1)^2+(-2)^2+\ ....\ +(9)^2=\frac{9\times10\times19}{6}=285$
$\text{SD}=\sqrt{\frac{285}{10}-\Big(\frac{-45}{10}\Big)^2}=\sqrt{\frac{285}{10}-\frac{2025}{100}}$
$=\sqrt{\frac{2850-2025}{100}}=\sqrt{8.25}$
Now, $\text{variance}=(\text{SD})^2=\big(\sqrt{8.25}\big)^2=8.25$
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MCQ 71 Mark
The standard deviation of first 10 natural numbers is:
  • A
    5.5
  • B
    3.87
  • C
    2.97
  • ✓
    2.87
Answer
Correct option: D.
2.87
We know that the standard deviation of first n natural number is $\sqrt{\frac{\text{n}^2-1}{12}}.$
$\therefore$ Standard deviation of first 10 natural numbers
$=\sqrt{\frac{10^2-1}{12}}$
$=\sqrt{\frac{99}{12}}$
$=\sqrt{8.25}$
$=2.87$
Hence, the correct answer is option (d).
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MCQ 81 Mark
Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is:
  • ✓
    s
  • B
    ks
  • C
    s + k
  • D
    $\frac{\text{s}}{\text{k}}$
Answer
Correct option: A.
s
The given observations are a, b, c, d, e.
$\text{Mean}=\text{m}=\frac{\text{a+b+c+d+e}}{5}$
$\Rightarrow\sum\text{x}_\text{i}=\text{a}+\text{b}+\text{c}+\text{d}+\text{e}=5\text{m}\ ...(1)$
Standard deviation, $\text{s}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$
Now, consider the observations a + k, b + k, c + k, d + k, e + k.
New mean $=\frac{(\text{a+k})+(\text{b+k})+(\text{c+k})+(\text{d+k})+(\text{e+k})}{5}$
$=\frac{\text{a+b+c+d+e+5k}}{5}$
$=\frac{5\text{m}+5\text{k}}{5}$
$=\text{m}+\text{k}$
$\therefore$ New standard deviation
$=\sqrt{\frac{\sum(\text{x}_\text{i}+\text{k})^2}{5}-(\text{m}+\text{k})^2}$
$=\sqrt{\frac{\sum(\text{x}_\text{i}^2+\text{k}^2+2\text{x}_\text{i}\text{k})}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}+\frac{\sum\text{k}^2}{5}+\frac{\sum2\text{x}_\text{i}\text{k}}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{5\text{k}^2}{5}-\text{k}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{5}-2\text{mk}}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{2\text{k}\times5\text{m}}{5}-2\text{mk}}$ [Using (1)]
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$
$=\text{s}$
Hence, the correct answer is option (a).
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MCQ 91 Mark
The mean of 9 observations is 36. If the mean of the first 5 observations is 32 and that of the last 5 observations is 39, then the fifth observation is __________.
  • A
    28
  • ✓
    31
  • C
    43
  • D
    37
Answer
Correct option: B.
31
Mean of 9 observations = 36 ⇒ Sum of these 9 observations = 324
Sum of first five observations = 32 × 5 = 160
Sum of last five observations = 39 × 5 = 195
Fifth observation = Sum of first five observations + Sum of last five observations - Sum of all 9 observations
= 160 + 195 - 324 = 31
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MCQ 101 Mark
If n = 10, $\overline{\text{X}}=12$ and $\sum\text{x}_\text{i}^2=1530,$ then the coefficient of variation is:
  • A
    36%
  • B
    41%
  • ✓
    25%
  • D
    None of these
Answer
Correct option: C.
25%
Standard deviation is expressed in the following manner:
$\sigma=\sqrt{\frac{1}{\text{n}}\sum_\text{i}\text{x}_\text{i}^2-(\overline{\text{X}})^2}$
$=\sqrt{\frac{1530}{10}-(12)^2}$
$=\sqrt9$
$=3$
$\text{CV}=\frac{\sigma}{\overline{\text{X}}}\times100$
$=\frac{3}{12}\times100$
$=25%$
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MCQ 111 Mark
The difference between the maximum and the minimum obervations in data is called the ____________:
  • A
    Mean of the data
  • ✓
    Range of the data
  • C
    Mode of the data
  • D
    Median of the data
Answer
Correct option: B.
Range of the data
In arithmetic, the range of a set of data is the difference between the largest and smallest values.
So, difference between minimum and maximum values is called range.
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MCQ 121 Mark
The mean of 8 numbers is 25 if each number is multiplied by 2 the new mean will be:
  • A
    12.5
  • B
    25
  • C
    40
  • ✓
    50
Answer
Correct option: D.
50
Mean of 8 numbers=25
$ \therefore\ \text{A.M}=\frac{\sum \text{x}}{\text{n}}$
​$ \Rightarrow 25=\dfrac{\sum \text{x}}{8}$
$ \Rightarrow∑\text{x}=25×8=200$
If each number is multiply by 2 then new sum
$ =200\times 2=400$
$ ∴ \text{New mean}=8400​=50$
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MCQ 131 Mark
Let $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{\mathrm{n}}$ be n observations. Let $\mathrm{y}_{\mathrm{i}}=\mathrm{a} \mathrm{x}_{\mathrm{i}}+\mathrm{b} \mathrm{y}_{\mathrm{i}}+\mathrm{b}$ for $\mathrm{i}=1,2,3, \ldots, \mathrm{n}$, where a and b are constants. If the mean of $x_i^{\prime} s$ is 48 and their standard deviation is 12 , the mean of $y_i$ 's 55 and standard deviation of $y_i$ 's is 15 , the values of a and b are:
  • ✓
    a = 1.25, b = -5
  • B
    a = -1.25, b = 5
  • C
    a = 2.5, b = -5
  • D
    a = 2.5, b = 5
Answer
Correct option: A.
a = 1.25, b = -5
  1. a = 1.25, b = -5
Solution:
It is given that $y_i = ax_i + b$ for i = 1, 2, 3, ..., n, where a and b are constants.
$\overline{\text{x}_\text{i}}=48$ and $\sigma_{\text{x}_\text{i}}=12$
$\overline{\text{y}_\text{i}}=55$ and $\sigma_{\text{y}_\text{i}}=15$
$\text{y}_\text{i}=\text{ax}_\text{i}+\text{b}$
$\Rightarrow\frac{\sum\text{y}_\text{i}}{\text{n}}=\frac{\sum(\text{ax}_\text{i}+\text{b})}{\text{n}}$
$\Rightarrow\frac{\sum\text{y}_\text{i}}{\text{n}}=\text{a}\frac{\sum\text{x}_\text{i}}{\text{n}}+\frac{\sum\text{b}}{\text{n}}$
$\Rightarrow\overline{\text{y}_\text{i}}=\text{a}\overline{\text{x}_\text{i}}+\text{b}$
$\Rightarrow55=48\text{a}+\text{b}\ ...(1)$
Now,
Standard deviation of $y_i​​​​​​​$ = Standard deviation of $ax_i + b$
$\Rightarrow\sigma_{\text{y}_\text{i}}=\text{a}\times\sigma_{\text{x}_\text{i}}$
$\Rightarrow15=12\text{a}$
$\Rightarrow\text{a}=\frac{15}{12}=1.25$
Putting a = 1.25 in (1), we get
$b = 55 - 48 \times 1.25 = 55 - 60 = -5$
Thus, the values of a and b are 1.25 and -5, respectively.
Hence, the correct answer is option (a).
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MCQ 141 Mark
Let set M = { x, 2x, 4x } for any number x. If the average (arithmetic mean) of the numbers in set M is 14, find the value of x:
  • A
    2
  • ✓
    6
  • C
    7
  • D
    10
Answer
Correct option: B.
6
Given set M = {x, 2x, 4x}
Average (arithmetic mean) of the numbers in set M is 14.
Value of x will be,
$ \Rightarrow \frac{\text{x}\ +\ 2\text{x}\ +\ \text{4x}}{3}$
$ \Rightarrow7\text{x}=42$
$ \Rightarrow\text{x}=6$
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MCQ 151 Mark
Given the list of numbers {1, 6, 3, 9, 16, 11, 2, 9, 5, 712, 13, 8} what is the median?
  • A
    7
  • ✓
    8
  • C
    9
  • D
    11
Answer
Correct option: B.
8
Given list is {1, 6, 3, 9, 16, 11, 2, 9, 5, 712, 13, 8} Arrange given set of integers in ascending order.
Then, we have 1, 2, 3, 5, 6, 8, 9, 11, 13, 16, 712
The middle number of the set is 8. Therefore the median is 8.
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MCQ 161 Mark
The age of 13 school students are listed below. Find the median: 12, 9, 8, 13, 15, 14, 6, 18, 7, 11, 9, 14, 10
  • A
    8
  • B
    14
  • ✓
    11
  • D
    10
Answer
Correct option: C.
11
The median of a set of data is the middlemost number in the set.
So, first arrange the data in order.
6, 7, 8, 9, 10, 10, 11, 12, 13, 14, 14, 15, 18
The median is 11.
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MCQ 171 Mark
The mean of the squares of the first n natural numbers is:
  • A
    $ \displaystyle {\text{n}}^{2}+1\text{n}2+1$
  • B
    $ \displaystyle \frac{\text{n}^{4}+1}{\text{n}}$
  • ✓
    $ \displaystyle \frac{\left ( \text{n}+1 \right )\left ( 2\text{n}+1 \right )}{6}$
  • D
    $ \displaystyle \frac{\left ( \text{n}+1 \right )\left ( \text{n}+2 \right )}{\text{m}}$
Answer
Correct option: C.
$ \displaystyle \frac{\left ( \text{n}+1 \right )\left ( 2\text{n}+1 \right )}{6}$
The first natural numbers are 1,2,3,......nTheir square are1$1 ^2,2^2,3^2......\text{n}^2$
$ \text{Mean}=\dfrac{1^2+2^2+3^2+........+\text{n}^2}{\text{n}}$
$ =\dfrac{\text{n}(\text{n}+1)(2\text{n}+1)}{6\text{n}}$
$\therefore$ Mean=n12 + 22 + 32 +........+ n2
$\therefore$ square of n natural numbers is$ =\dfrac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$ \text{Mean}=\dfrac{(\text{n}+1)(2\text{n}+1)}{6}$
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MCQ 181 Mark
Kavita obtained 16, 14, 18 and 20 marks (out of 25) in maths in weekly test in the month of Jan 2000; then mean marks of Kavita is:
  • A
    18
  • B
    16.5
  • ✓
    17
  • D
    17.5
Answer
Correct option: C.
17
No. of test in the month Jan 2000 = 4 Total Marks obtained in 4 test
= 16 + 14 + 18 + 20
$ 68\therefore \text{A.M}=\frac{\sum \text{x}}{\text{n}}=\frac{68}{4}=17$
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MCQ 191 Mark
Choose the correct answer. Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is:
  • ✓
    s
  • B
    ks
  • C
    s + k
  • D
    $\frac{\text{s}}{\text{k}}$
Answer
Correct option: A.
s
Given observations are a, b, c d and e.
$\text{Mean}=\text{m}=\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}$
$\sum\text{x}_\text{i}={\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}={5}\text{m}$
Now, $\text{mean}=\frac{\text{a}+\text{k}+\text{b}+\text{k}+\text{c}+\text{k}+\text{d}+\text{k}+\text{e}+\text{k}}{5}$
$=\frac{(\text{a}+\text{b}+\text{c}+\text{d}+\text{e})+5\text{k}}{5}=\text{m}+\text{k}$
$\therefore\ \text{SD}=\sqrt{\frac{{\sum(\text{x}_\text{i}^2+\text{k}^2+2\text{k}\text{x}_\text{i})}}{\text{n}}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{\text{n}}-2\text{mk}}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+2\text{km}-2\text{mk}}$ $\Big[\because\ \frac{\sum\text{x}_\text{i}}{\text{n}}=\text{m}\Big]$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2}$
$=\text{s}$
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MCQ 201 Mark
A company produces on an average 4000 items per month for the first 3 months. How many items it must produce on an average per month over the next 9 months, to average 4375 items per month over the whole?
  • ✓
    4500
  • B
    4600
  • C
    4680
  • D
    4710
Answer
Correct option: A.
4500
Total production has to be 4375 × 12 = 52500
First three months production is 4000 × 3 = 12000
Total production has to be in remaining 9 months = 52500 - 12000 = 40500
Average production per month in remaining 9 months $ = \frac{40500}{9} = 4500$
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MCQ 211 Mark
If the standard deviation of a variable X is $\sigma,$ then the standard deviation of variable $\frac{\text{aX+b}}{\text{c}}$ is:
  • A
    $\text{a}\ \sigma$
  • B
    $\frac{\text{a}}{\text{c}}\sigma$
  • ✓
    $\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
  • D
    $\frac{\text{a}\sigma+\text{b}}{\text{c}}$
Answer
Correct option: C.
$\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
$\text{Y}=\frac{\text{aX+b}}{\text{c}}$
$\overline{\text{Y}}=\frac{\sum\text{y}_\text{i}}{\text{n}}=\frac{\frac{\text{a}\sum\text{X}+\text{nb}}{\text{c}}}{\text{n}}$
$=\frac{\text{a}\sum\text{X}}{\text{nc}}+\frac{\text{nb}}{\text{nc}}$
$=\frac{\text{a}\overline{\text{X}}}{\text{c}}+\frac{\text{b}}{\text{c}}$
$\text{Var}(\text{X})=\frac{\sum\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\sigma^2$
$\text{Var}(\text{Y})=\frac{\sum\big(\text{y}_\text{i}-\overline{\text{Y}}\big)^2}{\text{n}}$
$=\frac{\sum\Big(\frac{\text{aX}}{\text{c}}+\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}-\frac{\text{b}}{\text{c}}\Big)}{\text{n}}$
$=\frac{\sum\Big(\frac{\text{aX}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}\Big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\frac{\sum\big(\text{x}_1-\overline{\text{X}}\big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2$
$\text{SD}(\sigma)=\sqrt{\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2}$
$=\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
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MCQ 221 Mark
Choose the correct answer. The standard deviation of some temperature data in °C is 5. If the data were converted into °F, the variance would be:
  • ✓
    81
  • B
    57
  • C
    36
  • D
    25
Answer
Correct option: A.
81
Given that $\sigma_\text{c}=5$
We know that $\text{C}=\frac{5}{9}(\text{F}-32)$
$\Rightarrow\text{F}=\frac{9\text{C}}{5}+32$
$\therefore\ \sigma_\text{F}=\frac{9}{5}\sigma_\text{c}=\frac{9}{5}\times5=9$
$\therefore\ \sigma^2_{\text{F}}=(9)^2=81$
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MCQ 231 Mark
A measure of central location which splits the data set into two equal groups is called the:
  • A
    Mean
  • B
    Mode
  • ✓
    Median
  • D
    Standard deviation
Answer
Correct option: C.
Median
Median is the middle most value of a series. So it divides a series of observations into two equal parts where 50% of the observations are below.
The median value and other 50% are above the median value.
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MCQ 241 Mark
The average of 2, 4, 6, 8, 10 is .................
  • A
    5
  • B
    6
  • ✓
    7
  • D
    8
Answer
Correct option: C.
7
$\text{ Average} = \displaystyle \frac{2 + 4 + 6 + 8 + 10}{5} = 6$
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MCQ 251 Mark
If i < m < n, then the median of the list i, m, n is ____.
  • A
    i
  • B
    n
  • ✓
    m
  • D
    a
Answer
Correct option: C.
m
The median of a set of data is the middlemost number in the set.
So, the median of the list i, m, n is m.
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MCQ 261 Mark
Find the mean of the first five multiples of $7.$
  • ✓
    $21$
  • B
    $20$
  • C
    $15$
  • D
    $24$
Answer
Correct option: A.
$21$
The first five multiples of $7$ are $7, 14, 21, 28$ and $35.$
$ \text{Required mean }= \dfrac{7+14+21+28+35}{7}=\dfrac{105}{7}=21$
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MCQ 271 Mark
Find the mean of: 9, 11, 12, 4 and 7
  • A
    5.3
  • B
    7.1
  • ✓
    8.6
  • D
    9.4
Answer
Correct option: C.
8.6
$\text{ Mean} = \frac{9+11+12+4+7}{5}$
$ ​\text{Mean} = \dfrac{43}{5}= 8.6$
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MCQ 281 Mark
The average of monthly salary of fifteen employees in a company is Rs. 9450. If the supervisors salary is added, the average salary increase by Rs. 650 What is the salary of the supervisor?
  • A
    Rs.19,850
  • ✓
    Rs.20,050
  • C
    Rs. 20,250
  • D
    Rs. 20,450
Answer
Correct option: B.
Rs.20,050
Average salary of 15 employees = Rs. 9450
Sum of the salaries of 15 employees = 15 × 9450 = 141750
New average after adding salary of supervisor = 9450 + 650 = 10100
Sum of salaries of 16 employees = 10100 × 16 = 1616600
Let the salary of the supervisor = x
Thus. x + 141750 = 161600
x = 19,850
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MCQ 291 Mark
Variance of the distribution 73, 77, 81, 85, ...,113 is:
  • A
    10
  • B
    160
  • ✓
    161
  • D
    None of these
Answer
Correct option: C.
161
  1. 161
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MCQ 301 Mark
In a triangle, the side lengths are a = 5, b = 3 and c = 2. Find the length of the median drawn to the side c:
  • ✓
    4
  • B
    3
  • C
    2
  • D
    1
Answer
Correct option: A.
4
Median of length drawn to the side, $\text{c}= \frac{1}{2}\sqrt{\left (2(\text{a}^{2}+\text{b}^{2})-\text{x}\text{c}^{2}\right)}$
$ = \frac{1}{2}\sqrt{\left (2(5^{2}+3^{2})-2^{2}\right)} $
$ = \frac{1}{2}\sqrt{\left (2(34)-2^{2}\right)}$
$ = \frac{1}{2}\sqrt{\left (68-4\right)} $
$ = \frac{1}{2}\sqrt{64} $
$ = \frac{8}{2}$
$=4$
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MCQ 311 Mark
The mean deviation of the series a, a + d, a + 2d, ..., a + 2n from its mean is:
  • A
    $\frac{(\text{n}+1)\text{d}}{2\text{n}+1}$
  • B
    $\frac{\text{n}\text{d}}{2\text{n}+1}$
  • ✓
    $\frac{\text{n}(\text{n}+1)\text{d}}{2\text{n}+1}$
  • D
    $\frac{(2\text{n}+1)\text{d}}{\text{n}(\text{n}+1)}$
Answer
Correct option: C.
$\frac{\text{n}(\text{n}+1)\text{d}}{2\text{n}+1}$
  1. $\frac{\text{n}(\text{n}+1)\text{d}}{2\text{n}+1}$
Solution:
$x_i$ $\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|=\big|\text{x}_\text{i}-(\text{a}+\text{nd})\big|$
a nd
a + d (n - 1)d
a + 2d (n - 2)d
a + 3d (n - 3)d
: :
: :
a + (n + 1)d d
a + nd 0
a + (n + 1)d d
: :
: :
a + 2nd nd
$\sum\text{x}_\text{i}=(2\text{n}+1)(\text{a}+\text{nd})$ $\sum\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|=\text{n}(\text{n}+1)\text{d}$
Therefore are 2n + 1 terms.
⇒ N = 2n + 1
$\sum\text{x}_\text{i}=\text{a}+\text{a}+\text{d}+\text{a}+2\text{d}+\text{a}+3\text{d}+...+\text{a}+2\text{nd}$
$=(2\text{n}+1)\text{a}+\text{d}(1+2+3+...+2\text{n})$ [a + a + a + ...(2n + 1)times = (2n + 1)a]
$=(2\text{n}+1)\text{a}+\frac{2\text{n}(2\text{n}+1)\text{d}}{2}$ $\Big[$Sum of the first n natural numbers is $\frac{\text{n}(\text{n}+1)}{2},$ but here we are considering$\Big]$
$=(2\text{n}+1)\text{a}+(2\text{n}+1)\text{nd}$
$=(2\text{n}+1)(\text{a}+\text{nd})$
$\overline{\text{X}}=\frac{(2\text{n}+1)(\text{a}+\text{nd})}{(2\text{n}+1)}$
$=\text{a}+\text{nd}$
$\sum\big|\text{x}_\text{i}-\overline{\text{X}}\big|=\text{nd}+(\text{n}-1)\text{d}+(\text{n}-2)\text{d}\\+...+\text{d}+0+\text{d}+2\text{d}+3\text{d}+...+\text{nd}$
$=\text{d}(\text{n}+(\text{n}-1)+(\text{n}-2)+...+1)\\ \ +0+\text{d}(1+2+3+....+\text{n})$
$=\frac{\text{dn}(\text{n}+1)}{2}+\frac{\text{dn}(\text{n}+1)}{2}$ $\Big\{\because1+2+3+....+\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big\}$
$=\text{n}(\text{n}+1)\text{d}$
Mean deviation about the mean $=\frac{\sum\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|}{\text{N}}$
$=\frac{\text{n}(\text{n}+1)\text{d}}{(2\text{n}+1)}$
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MCQ 321 Mark
The mean of first five prime numbers is:
  • A
    3
  • B
    3.6
  • C
    7
  • ✓
    5.6
Answer
Correct option: D.
5.6
The first five prime numbers are 2, 3, 5, 7, 11
$\text{Mean}=\frac{\text{sum of the terms}}{\text{no. of terms}}$
$\text{Mean}=\frac{2+3+5+7+11}{5}$
$=\frac{28}{5}=5.6$
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MCQ 331 Mark
Choose the correct answer. The following information relates to a sample of size 60 $\sum\text{x}^2=18000$ and $\sum\text{x}=960,$ then the variance is:
  • A
    6.63
  • B
    16
  • C
    22
  • ✓
    44
Answer
Correct option: D.
44
We know that variance $(\sigma^2)\frac{\sum\text{x}_\text{i}^2}{\text{N}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{N}}\Big)^2$
$=\frac{18000}{60}-\Big(\frac{960}{60}\Big)^2=300-256=44$
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MCQ 341 Mark
In a class of 100 students there are 70 boys whose average marks in a subject are 75 If the average marks of whole class is 72 then what is the average marks of the girls?
  • A
    73
  • ✓
    65
  • C
    68
  • D
    79
Answer
Correct option: B.
65
Total students = 100 Average marks
=72 Total marks of the class = 72 × 100 = 7200
Total marks of the boys = 70 × 75 = 5250
Total marks of the girls = 7200 = 5250 = 1950
Average marks of the girls $ = \dfrac{1950}{30}=65$ hence, option B is correct.
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MCQ 351 Mark
The sum of the squares deviations for 10 observations taken from their mean 50 is 250. The coefficient of variation is:
  • ✓
    10%
  • B
    40%
  • C
    50%
  • D
    None of these
Answer
Correct option: A.
10%
We have:
$\overline{\text{X}}=50,\ \text{n}=10$
$\sum\limits^{10}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2=250$
$\therefore\text{SD}=\sqrt{\text{Variance of X}}$
$=\sqrt{\frac{\sum\limits^{10}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}{\text{n}}}$
$=\sqrt{\frac{250}{10}}$
$=5$
Using $\text{CV}=\frac{\sigma}{\overline{\text{X}}}\times100$
$\Rightarrow\text{CV}=\frac{5}{50}\times100$
$=10\%$
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MCQ 361 Mark
The mean age of 30 student is 9 years. If the age of their teacher is included, it becomes 10 years. The age of teacher ( in years ) is:
  • A
    27
  • B
    31
  • C
    35
  • ✓
    40
Answer
Correct option: D.
40
Given:Average age of 30 students = 9years.
Total age of 3030 students = 9 × 30 = 270 years. Teachers age included.
So, average age of 30 students + one teacher = 10years.
⇒ Total age of 30 students + one teacher = 10 × 31 = 310 years.
$ \therefore$ age of teacher = 310 - 270 = 40 years.
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MCQ 371 Mark
Means of a set of 60 values is 23, if 4 is added to each these values the the new mean is:
  • ✓
    27
  • B
    25
  • C
    64
  • D
    56
Answer
Correct option: A.
27
New mean = x~ = 23 + 4 = 27
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MCQ 381 Mark
The following observations have been arranged in ascending order. If the median of the data is 78, find the value of x. 44, 47, 63, 65, x + 13, 87, 93, 99, 110.
  • ✓
    65
  • B
    68
  • C
    66
  • D
    64
Answer
Correct option: A.
65
The series in ascending order is: 44, 47, 63, 65, x + 13, 87, 93, 99, 110.
The series has 9 observations.
hence, the middle observation will be the median of the series.
Here, x + 13 is the middle observation
Therefore, x + 13 = 78
x = 65
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MCQ 391 Mark
Median of 15, 28, 72, 56, 44, 32, 31, 43 and 51 is 43:
  • ✓
    True
  • B
    False
  • C
    Neither
  • D
    Either
Answer
Correct option: A.
True
The terms are: 15, 28, 72, 56, 44, 32, 31, 43 and 51.
Arranging them in ascending order: 15, 28, 31, 32, 43, 44, 51, 56, 72
Since the total number of terms is odd that is 9, therefore the median will be the middle term that is the 5th term which is 43.
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MCQ 401 Mark
The attendance of a class of 45 boys for 10 days is given as 40, 30, 35, 45, 44, 41, 38, 44 and 41 then the mean attendance of a class is:
  • A
    39
  • ✓
    40
  • C
    41
  • D
    43
Answer
Correct option: B.
40
In this question one day attendance not givenGiven attendance as per Answer.
are 40, 42, 30, 35, 45, 44, 41, 38, 44 and 41 Then mean
$ =\frac{40+42+30+35+45+44+41+38+44+41}{10}$
$ =\frac{400}{10}=40$
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MCQ 411 Mark
The mean of 864, 874, 884, 1000 and 1008 is:
  • A
    928
  • B
    1010
  • ✓
    926
  • D
    None of these
Answer
Correct option: C.
926
Formula,
$ \cfrac{\sum {\text{x }} }{N}=\cfrac{864+874+884+1000+1008}{5}$
$ =\frac{4630}{5}$
$ =926$
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MCQ 421 Mark
The daily sale of milk (in litres) in a ration shop for eight days is as follows-
60, 40, 10, 40, 4, 70, 30 and 10. The average daily sale is:
  • A
    40
  • ✓
    33
  • C
    64
  • D
    24
Answer
Correct option: B.
33
By definition of average,
$ =\cfrac{60+40+10+40+4+70+30+10}{8}$
$ =\cfrac{264}{8}=33$
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MCQ 431 Mark
A set F, which contains the elements 4, 5, 11, 13, 16, 18, and x. If both the median and average (arithmetic mean) of Set F equal 11, what must be the value of x?
  • A
    9
  • ✓
    10
  • C
    11
  • D
    12
Answer
Correct option: B.
10
Given, set $$f = 4, 5, 11, 13, 16, 18, x Average of set
f = 11
$\Rightarrow \dfrac{4+5+11+13+16+18+\text{x}}{7}=11$
$\Rightarrow \frac{67+\text{x}}{7}=11$
$\Rightarrow 67+{\text{x}}=77$
$\Rightarrow \text{x}=77-67=10$
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MCQ 441 Mark
_____ is the most frequently observed data value:
  • A
    Median
  • ✓
    Mode
  • C
    Mean
  • D
    Quartil
Answer
Correct option: B.
Mode
The mode is the value thats repeated the maximum number of times in the data set.
A worked example: Marks obtained in an examination is
given as 5, 9, 7, 12, 15, 7, 5, 7, 7, 8, 7
We identify the number that is repeated the maximum number of times as: 7 (repeated 5 times).
Thus the mode for this data set is 7.
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MCQ 451 Mark
Mean of 10 values is 32.6. If another values is included the mean becomes 31. The included value is:
  • A
    16
  • B
    14
  • ✓
    15
  • D
    19
Answer
Correct option: C.
15
Included value = 31 × 11 − 32.6 × 10 = 15
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MCQ 461 Mark
The mean of the cubes of the first n natural numbers is:
  • A
    $ \displaystyle \frac{\text{n}\left (\text{ n}+1 \right )^{2}}{2}$
  • ✓
    $ \displaystyle \frac{\text{n}\left ( \text{n}+1 \right )^{2}}{4}$
  • C
    $ \displaystyle \frac{\text{n}\left ( \text{n}+1 \right )\left ( \text{n}+2 \right )}{8}$
  • D
    $ \displaystyle {\text{n}}^{2}+\text{n}+1$
Answer
Correct option: B.
$ \displaystyle \frac{\text{n}\left ( \text{n}+1 \right )^{2}}{4}$
First n natural numbers are 1,2,3,4,......,n
$ \therefore \text{Mean}=\dfrac{1^3+2^3+3^3+.......+\text{n}^3}{\text{n}}$
Sum of the cubes of n natural numbers $ =\left(\frac{\text{n}(\text{n}+1)}{2}\right)^2$
$ \therefore \text{Mean}=\left(\dfrac{\text{n}(\text{n}+1)}{2}\right)^2\times\frac{1}{\text{n}}$
$ =\dfrac{\text{n}(\text{n}+1)^2}{4}$
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MCQ 471 Mark
let $x_1, x_2, \ldots, x_n$ be n observations and $\overline{\text{X}}$ be their arithmetic mean. The standard deviation is given by:​​​​​​​
  • A
    $\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
  • B
    $\frac{1}{\text{n}}\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
  • ✓
    $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$
  • D
    $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\text{x}_\text{i}^2-\overline{\text{X}}^2}$
Answer
Correct option: C.
$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$
  1. $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$
Solution:
It is given that $x_1, x_2, \ldots, x_n$ be n observations and $\overline{\text{X}}$ be their arithmetic mean.
The standard deviation is given observations is $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}.$
Also,
$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}=\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\text{x}_\text{i}^2-\overline{\text{X}}^2}$
Hence, the correct answers are options (c) and (d).
Disclaimer: For option (c) to be the only correct answer, option (d) should be different from the given value.
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MCQ 481 Mark
The daily sale of kerosene (in litres) in a ration shop for six days is as follows: 75, 120, 12, 50, 70.5 and 140.5 The average daily sale is:
  • A
    150
  • B
    10
  • C
    142
  • ✓
    78
Answer
Correct option: D.
78
$\text{Mean}=\frac{75+120+12+50+70.5+140.5}{6}$
​= 78 The average daily sale is therefore the mean = 78.
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MCQ 491 Mark
Choose the correct answer. Mean deviation for n observations $x_1, x_2, \ldots \ldots, x_n$ from their mean x is given by:​​​​​​​
  • A
    $\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})$
  • ✓
    $\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
  • C
    $\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2$
  • D
    $\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2$
Answer
Correct option: B.
$\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
  1. $\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
Solution:
$\text{MD}=\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
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MCQ 501 Mark
The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is -
  • A
    40
  • B
    20
  • ✓
    80
  • D
    60
Answer
Correct option: C.
80
Let the number of boys and girls be x and y.
$ \therefore 52\text{x}+42\text{y}=50(\text{x}+{y})$
$ \Rightarrow 52\text{x}+42\text{y}=50\text{x}+50\text{y}$
$ \Rightarrow 2\text{x}=8{\text{y}} = \text{x }{ =4y} $
$ \therefore$Total number of students in class
= x + y = 4y + y = 5y
$ \therefore$ Required % of boys
$=\frac { 4\text{y} }{ 5\text{y} } \times 100=80%$
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MCQ - MATHS STD 11 Science Questions - Vidyadip