M.C.Q (1 Marks)

MCQ 1011 Mark

The distribution below gives the arks obtained by 80 students on a test.

Marks:	Less than 10	Less than 20	Less than 30	Less than 40	Less than 50	Less than 60
Nuber of students:	3	12	27	57	75	80

The odal class of this distribution is

A
10-20
B
20-30
✓
30-40
D
50-60

Answer

Correct option: C.

30-40

Marks:	0-10	10-20	20-30	30-40	40-50	50-60
Number of students:	3	9	15	30	18	5

We find that the class with maximum frequency is 30-40. It is the modal class.

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MCQ 1021 Mark

Mode for the following distribution is 22. If x < y < 10 then the value of y is

Class interval:	5	8	10	x	y	30
Frequency (f):	5	8	10	x	y	30

A
2
✓
5
C
3
D
4

Answer

Correct option: B.

(b)
It is given that x < y < 10. So, 20-30 is the modal class such that $l=20, f=10, f_1=8$ $f_2=x$ and $h=10$
$\therefore \quad$ Mode $=l+\frac{f-f_1}{2 f-f_1-f_2} \times h \Rightarrow 22=20+\frac{10-8}{20-8-x} \times 10 \Rightarrow 2=\frac{2}{12-x} \times 10 \Rightarrow 12-x=10 \Rightarrow x=2$
We have,
$N=30 \Rightarrow 5+8+10+x+y=30 \Rightarrow x+y=7 \Rightarrow y=5 \quad[\because x=2]$

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MCQ 1031 Mark

If the sum of 15 observations of a data is (434 + x) and the mean of the observation is x then x =

A
25
B
27
✓
31
D
33

Answer

Correct option: C.

(c)
It is given that the sum of 15 observations is (434 + x) and their mean is x.
∴ $x=\frac{434+x}{15} \Rightarrow 15 x=434+x \Rightarrow 14 x=434 \Rightarrow x=31$.

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MCQ 1041 Mark

If the difference of mode and median of a data is 24, then the difference of median and mean is

✓
12
B
24
C
8
D
36

Answer

Correct option: A.

(a)
We know that
Mode 3 Median - 2 Mean
⇒ Mode - Median = 2 (Median-Mean)
⇒ 24=2 (Median - Mean) [∴ Mode - Median = 24 (given)]
⇒ Median - Mean = 12

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MCQ 1051 Mark

If the mean of the following frequency distribution is 2.6, then the value of p is

Variate (X):	1	2	3	4	5
Frequency (f):	4	5	y	1	2

A
3
✓
8
C
13
D
24

Answer

Correct option: B.

(b)
It is given that
Mean = 2.6
$\Rightarrow \frac{1 \times 4+2 \times 5+3 \times y+4 \times 1+5 \times 2}{4+5+y+1+2}=2.6 \Rightarrow \frac{3 y+28}{12+y}=2.6 \Rightarrow 3 y+28=31.2+2.6 y \Rightarrow 0.4 y=3.2 \Rightarrow y=8$

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MCQ 1061 Mark

If the median of the data $\frac{x}{5}, x, \frac{x}{3}, \frac{2 x}{3}, \frac{x}{4}, \frac{2 x}{5}, \frac{3 x}{4}, x>0$ is 4 , then $x=$

A
5
✓
10
C
8
D
7

Answer

Correct option: B.

(b)
Arranging the given data in ascending order, we obtain: $\frac{x}{5}, \frac{x}{4}, \frac{x}{3}, \frac{2 x}{5}, \frac{2 x}{3}, \frac{3 x}{4}, x$
There are 7 observations. Therefore,
Median = Value of $\left(\frac{7+1}{2}\right)^{\text {th }}$ i.e. $4^{\text {th }}$ observation $\Rightarrow 4=\frac{2 x}{5} \Rightarrow x=10$

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MCQ 1071 Mark

If n is an even positive integer and $x>0, \quad x \neq 1$, then the median of the data a, $a, a x, a x^2, a x^3, \ldots, a x^n$ is

A
$a x^{n-1}$
B
$a x^{\frac{n}{2}-1}$
✓
$a x^{\frac{n}{2}}$
D
$a x^{\frac{n}{2}+1}$

Answer

Correct option: C.

$a x^{\frac{n}{2}}$

(c)
Given observations are in ascending or descending order of magnitude according as $x>1$ or $x<1$. If n is even there are (n + 1) an odd number of observations. The median is the value of middle i.e.,$\left(\frac{n+1+1}{2}\right)^{t/h}=\left(\frac{n}{2}+1\right)^{th}$
∴ Median $=a x^{\frac{n}{2}+1-1}=a x^{\frac{n}{2}}$

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MCQ 1081 Mark

If for a data, Mean:Median = 9:8, then Median: Mode =

A
8:9
✓
4:3
C
7:6
D
5:4

Answer

Correct option: B.

4:3

(b)
We have, $\frac{\text { Mean }}{\text { Median }}+\frac{9}{8}=$ Mean $=\frac{9}{8}$ Median ...(i)
We know that
Mode $=3$ Median -2 Mean
⇒ Mode $=3$ Median $-\frac{9}{4}$ Median $=\frac{3}{4}$ Median ...[Using (i)]
Median : Mode $=4: 3$

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MCQ 1091 Mark

If a variable takes discrete values $x+4, x-\frac{7}{2}, x-\frac{5}{2}, x-3, x-2, x+\frac{1}{2}, x-\frac{1}{2}, x+5$,$x>0$,

A
$x-\frac{5}{2}$
B
$x-\frac{5}{3}$
✓
$x-\frac{5}{4}$
D
$x-\frac{5}{6}$

Answer

Correct option: C.

$x-\frac{5}{4}$

(c)
Arranging the values of the variable in ascending order, we obtain
$x-\frac{7}{2}, x-3, x-\frac{5}{2}, x-2, x-\frac{1}{2}, x+\frac{1}{2}, x+4, x+5$
These are 8 observations.
Median = AM of 4th and 5th observation $=\frac{x-2+x-\frac{1}{2}}{2}=x-\frac{5}{4}$

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MCQ 1101 Mark

The arithmetic mean of x, x + 3, x + 6, x + 9 and x + 12 is

✓
x + 6
B
x + 5
C
x + 7
D
x + 8

Answer

Correct option: A.

x + 6

(a)
Let $\bar{X}$ be the arithmetic mean of x, x + 3 x + 6 x + 9 and x + 12.
Then, $\bar{X}=\frac{x+(x+3)+(x+6)+(x+9)+(x+12)}{5}=\frac{5 x+30}{5}=x+6$.

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MATHS M.C.Q (1 Marks)