M.C.Q (1 Marks)

MCQ 2011 Mark

The mean of the following data is

class interval	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequemcy	$3$	$5$	$9$	$5$	$3$

✓
$20$
B
$24$
C
$22$
D
$25$

Answer

Correct option: A.

$20$

The frequency distribution table from the given data can be drawn as :

Daily allowance $($in $₹)$	Class mark $\left(x_i\right)$	Number of children $\left(f_i\right)$	$f_i x_i$
$11-13$	$12$	$7$	$84$
$13-15$	$14$	$6$	$84$
$15-17$	$16$	$9$	$144$
$17-19$	$18$	$13$	$234$
$19-21$	$20$	$f$	$20f$
$21-23$	$22$	$5$	$110$
$23-25$	$24$	$4$	$96$
		$\sum f_i=f+44$	$\Sigma f_i x_i=$ $20 f+752$

Now, mean, $\bar{x}=\frac{\sum f_i x_i}{\sum f_i}$
$\Rightarrow 18=\frac{20 f+752}{f+44}$
$\Rightarrow 20 f+752=18 f+792$
$\Rightarrow 20 f-18 f=792-752$
$\Rightarrow 2 f=40$
$\Rightarrow f=\frac{40}{2}=20$
Hence, the fourth frequency, $f$ is $20 .$

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MCQ 2021 Mark

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is $₹18.$ Find the missing frequency $f.$

Daily pocket allowance $($in $₹)$	$11-13$	$13-15$	$15-17$	$17-19$	$19-21$	$21-23$	$23-25$
Frequemcy	$7$	$6$	$9$	$13$	$f$	$5$	$4$

A
$18$
✓
$20$
C
$22$
D
$19$

Answer

Correct option: B.

$20$

The frequency distribution table from the given data can be drawn as :

Daily allowance $($in $₹)$	Class mark $\left(x_i\right)$	Number of children $\left(f_i\right)$	$f_i x_i$
$11-13$	$12$	$7$	$84$
$13-15$	$14$	$6$	$84$
$15-17$	$16$	$9$	$144$
$17-19$	$18$	$13$	$234$
$19-21$	$20$	$f$	$20f$
$21-23$	$22$	$5$	$110$
$23-25$	$24$	$4$	$96$
		$\sum f_i=f+44$	$\Sigma f_i x_i=$ $20 f+752$

Now, mean, $\bar{x}=\frac{\sum f_i x_i}{\sum f_i}$
$\Rightarrow 18=\frac{20 f+752}{f+44}$
$\Rightarrow 20 f+752=18 f+792$
$\Rightarrow 20 f-18 f=792-752$
$\Rightarrow 2 f=40$
$\Rightarrow f=\frac{40}{2}=20$
Hence, the fourth frequency, $f$ is $20 .$

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MCQ 2031 Mark

Four observations are 2, 4,6 and 8. The frequencies of the first three observations are 3,2 and 1 respectively. If the mean of the observations is 4 , then find the frequency of the fourth observation.

A
8
B
4
✓
1
D
2

Answer

Correct option: C.

1

(c) : Let the frequency of the fourth observation be $f$.
$
\begin{aligned}
& \text { Mean, } \bar{x}=\frac{x_1 f_1+x_2 f_2+x_3 f_3+x_4 f_4}{f_1+f_2+f_3+f_4} \\
\Rightarrow & 4=\frac{2 \times 3+4 \times 2+6 \times 1+8 \times f}{3+2+1+f}=\frac{6+8+6+8 f}{6+f} \\
\Rightarrow & 24+4 f=20+8 f \\
\Rightarrow & 4 f=4 \Rightarrow f=1
\end{aligned}
$

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MCQ 2041 Mark

If the mean of $x, x+3, x+6, x+9$ and $x+12$ is 10 , then $x$ equals

A
1
B
2
✓
4
D
6

Answer

Correct option: C.

4

(c) : Mean, $\bar{x}=\frac{\text { Sum of all observations }}{\text { Total number of observations }}$
$
\begin{array}{ll}
\Rightarrow & 10=\frac{x+(x+3)+(x+6)+(x+9)+(x+12)}{5} \quad \text { [Given] } \\
\Rightarrow & 50=5 x+30 \\
\Rightarrow & 5 x=20 \\
\Rightarrow & x=4
\end{array}
$

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MCQ 2051 Mark

If the mean of first $n$ natural numbers is $\frac{5 n}{9}$, then $n$ is equal to

A
$5$
✓
$9$
C
$10$
D
$11$

Answer

Correct option: B.

$9$

Mean of first $n$ natural numbers
$=\frac{1+2+3+\ldots+n}{n}$
$\Rightarrow \frac{5 n}{9}=\frac{n(n+1)}{2 n}$
$[\because$ Sum of first $n$ natural numbers $=\frac{n(n+1)}{2}]$
$\Rightarrow 10 n^2=9 n^2+9 n$
$\Rightarrow n^2-9 n=0$
$\Rightarrow n(n-9)=0$
$\Rightarrow n=9$
$(\because n \neq 0)$

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MCQ 2061 Mark

The mean of first ten odd natural numbers is

A
5
✓
10
C
20
D
19

Answer

Correct option: B.

10

(b) : First ten odd natural numbers are 1, 3, 5, 7, 9, $11,13,15,17$ and 19 .
$\therefore \quad$ Required mean
$
=\frac{1+3+5+7+9+11+13+15+17+19}{10}=\frac{100}{10}=10
$

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MCQ 2071 Mark

If $d_i=x_i-13, \Sigma f_i d_i=30$ and $\Sigma f_i=120$, then mean $(\bar{x})$ is equal to

A
13
B
12.75
✓
13.25
D
14.25

Answer

Correct option: C.

13.25

(c) : Here, $d_i=x_i-13 \Rightarrow a=13 \quad\left[\because d_i=x_i-a\right]$ Also, $\sum f_i d_i=30$ and $\sum f_i=120$ Now, mean, $\bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}=13+\frac{30}{120}=13.25$

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MCQ 2081 Mark

Find the class marks of classes $10-20$ and $35-55$.

A
10,35
B
20,55
✓
15,45
D
17.5,45

Answer

Correct option: C.

15,45

(c) : Class mark of class $10-20=\frac{10+20}{2}=15$ Class mark of class 35-55 $=\frac{35+55}{2}=45$

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MATHS M.C.Q (1 Marks)