M.C.Q (1 Marks)

MCQ 511 Mark

If the mean of the following distribution is $2.6 ,$ then the value of $y$ is

Class interval $(x_i)$	$1$	$2$	$3$	$4$	$ 5$
Frequency	$4$	$5$	$y$	$1$	$2$

A
$3$
✓
$8$
C
$13$
D
$24$

Answer

Correct option: B.

$8$

$n=\Sigma f_i=4+5+y+1+2=12+y$.
$\Sigma f_i x_i=4+10+3 y+4+10=28+3 y \text {. }$
Mean $=\frac{\Sigma f_i x_i}{\Sigma f_i} $
$\Rightarrow 2.6=\frac{28+3 y}{12+y}$
$\Rightarrow 26(12+y)=10(28+3 y) $
$\Rightarrow y=8$

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MCQ 521 Mark

The mean of $x_1, x_2, \ldots \ldots, x_n$ is $M$. If $x_i$, $i=1,2, \ldots \ldots, n$ is replaced by $5 x_i$, the mean becomes $M_1$, then $M_1$ is equal to

✓
$5 M$
B
$M+5$
C
$M+100$
D
$10 M$

Answer

Correct option: A.

$5 M$

(a) : Given that
$
\begin{array}{c}
\frac{\sum_{i=1}^n x_i}{n}=M \text { and } \frac{\sum_{i=1}^n 5 x_i}{n}=M_1 \\
\Rightarrow 5 \cdot \frac{\sum_{i=1}^n x_i}{n}=M_1 \Rightarrow 5 M=M_1
\end{array}
$

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MCQ 531 Mark

If mean of ten consecutive odd numbers is $120$, then the mean of first five odd numbers among them is

A
$113$
✓
$115$
C
$114$
D
$116$

Answer

Correct option: B.

$115$

Let ten consecutive odd numbers be $2 x+1$, $2 x+3, \ldots, 2 x+19$
Hence$, (2 x+1)+(2 x+3)+\ldots+(2 x+19)=10 \times 120$
$\Rightarrow 20 x+100=1200$
$\Rightarrow x=\frac{1100}{20}=55$
$\therefore$ Mean of first five odd numbers
$=\frac{(2 x+1)+(2 x+3)+\ldots .+(2 x+9)}{5}$
$=\frac{10 \times 55+25}{5}=\frac{575}{5}=115$
$\text { 38. (c) }: \frac{3(x-2)+5(x+2)+7(x-3)+9(x+3)}{(x-2)+(x+2)+(x-3)+(x+3)}=6.5$
$\Rightarrow \frac{24 x+10}{4 x}=6.5 $
$\Rightarrow x=5$

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MCQ 541 Mark

The numbers $3,5,7$ and 9 have their respectively frequencies $x-2, x+2, x-3$ and $x+3$. If the mean is 6.5 , then the value of $x$ is

Answer

$\begin{aligned} & \text {(c) : } \frac{3(x-2)+5(x+2)+7(x-3)+9(x+3)}{(x-2)+(x+2)+(x-3)+(x+3)}=6.5 \\ & \Rightarrow \quad \frac{24 x+10}{4 x}=6.5 \Rightarrow x=5\end{aligned}$

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MCQ 551 Mark

If the mean and mode of a frequency distribution be $53.4$ and $55.2$ respectively, then find the median.

✓
$54$
B
$52$
C
$55$
D
$53$

Answer

Correct option: A.

$54$

We have$, 3$ Median $=$ Mode $+2$ Mean
$=55.2+2(53.4)=55.2+106.8=162$
$\Rightarrow$ Median $=162 / 3=54$

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MCQ 561 Mark

For a certain frequency distribution, if $\Sigma f_i=50$ and $\Sigma f_i x_i=2550$, then what is the mean of the distribution?

A
53
✓
51
C
50
D
45

Answer

Correct option: B.

(b) : Mean of the distribution $=\frac{\sum f_i x_i}{\sum f_i}=\frac{2550}{50}=51$.

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MCQ 571 Mark

A sequence $a, a+d, a+2 d, \ldots, a+n d$, has odd number of terms. Find its median.

A
$a+\left(\frac{n-1}{2}\right) d$
B
$a+\left(\frac{n}{2}-1\right) d$
✓
$a+\left(\frac{n+2}{2}\right) d$
D
$a+\left(\frac{n}{2}+2\right) d$

Answer

Correct option: C.

$a+\left(\frac{n+2}{2}\right) d$

(c) : Clearly, the given sequence is an A.P. with first term $a$ and common difference is $d$. Given, sequence is $a, a+d, a+2 d, a+3 d, \ldots, a+n d$ As, there are odd number of terms, so the median is
$
\left(\frac{n+1+1}{2}\right)^{\text {th }} \text { term }=\left(\frac{n+2}{2}\right)^{\text {th }} \text { term }=a+\left(\frac{n+2}{2}\right) d
$

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MCQ 581 Mark

$(x+2), x$ and $(x-1)$ are the frequencies of the numbers 12,15 and 20 respectively. If the mean of the distribution is 14.5 , the value of $x$ is

A
2
✓
3
C
4
D
5

Answer

Correct option: B.

(b) : We have, mean $=\frac{\sum f_i x_i}{\sum f_i}$
$ \begin{aligned} & \Rightarrow \frac{12(x+2)+15(x)+20(x-1)}{(x+2)+(x)+(x-1)}=14.5 \\ & \Rightarrow 2(47 x+4)=29(3 x+1) \\ & \Rightarrow 94 x+8=87 x+29 \Rightarrow 7 x=21 \Rightarrow x=3 \end{aligned} $
(43-47) : Let us consider the following table :

Class	Class marks $\left(x_i\right)$	$d_i=x_i-A$	Frequency $\left(f_i\right)$	$f_i d_i$
30-40	35	-20	80	-1600
40-50	45	-10	110	-1100
50-60	55 = A	0	120	0
60-70	65	10	70	700
70-80	75	20	40	800
			$\Sigma f_i=420$	$\sum f_i d_i=1200$

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MCQ 591 Mark

For some data, if mean and median are 21 and 23 respectively then mode = ___________.

✓
27
B
22
C
17
D
23

Answer

Correct option: A.

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Maths M.C.Q (1 Marks)