M.C.Q (1 Marks) - Maths STD 10 Questions - Vidyadip

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M.C.Q (1 Marks)

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26 questions · auto-graded multiple-choice test.

MCQ 11 Mark

Choose the correct answer from the given four options:
A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought?

A
40
✓
240
C
480
D
750

Answer

Correct option: B.

240

Given, total number of sold tickets = 6000Let she bought x tickets.
Then, probability of her winning the first prize $=\frac{\text{x}}{6000}=0.08$
$\Rightarrow\ \ \text{x}=0.08\times6000$
$\therefore\ \ \text{x}=480$
Hence, she bought 480 tickets.

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MCQ 21 Mark

Choose the correct answer from the given four options:
Consider the following distribution:

Marks obtained	Number of students
More than or equal to 0	63
More than or equal to 10	58
More than or equal to 20	55
More than or equal to 30	51
More than or equal to 40	48
More than or equal to 50	42

the frequency of the class 30-40 is:

A
3
B
4
✓
48
D
51

Answer

Correct option: C.

48

Marks obtained	Number of students
0-10	(63 - 58) = 5
10-20	(58 - 55) = 3
20-30	(55 - 51) = 4
30-40	(51 - 48) = 3
40-50	(48 - 42) = 6
50...	42 = 42

Hence, frequency in the class interval 30-40 is 3

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MCQ 31 Mark

Choose the correct answer from the given four options:
In the formula $\bar{\text{x}}=\text{a}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}},$ for finding the mean of grouped data di’s are deviations from a of:

A
Lower limits of the classes.
B
Upper limits of the classes.
✓
Mid points of the classes.
D
Frequencies of the class marks.

Answer

Correct option: C.

Mid points of the classes.

We know that, $d_i=x_i-a$
i.e., di's are the deviation from a of mid$-$points of the classes.

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MCQ 41 Mark

Choose the correct answer from the given four options:
An event is very unlikely to happen. Its probability is closest to:

✓
0.0001
B
0.001
C
0.01
D
0.1

Answer

Correct option: A.

0.0001

The probability of an event which is very unlikely to happen is closest to zero and from the given options 0.0001 is closest to zero.

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MCQ 51 Mark

Choose the correct answer from the given four options:
In the formula $\bar{\text{x}}=\text{a}+\text{h}\Big(\frac{\sum\text{f}_\text{x}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big),$ for finding the mean of grouped frequency distribution, $u_i=$

A
$\frac{\text{x}_\text{i}+\text{a}}{\text{h}}$
B
$\text{h}(\text{x}_\text{i}-\text{a})$
✓
$\frac{\text{x}_\text{i}-\text{a}}{\text{h}}$
D
$\frac{\text{a}+\text{x}_\text{i}}{\text{h}}$

Answer

Correct option: C.

$\frac{\text{x}_\text{i}-\text{a}}{\text{h}}$

Given, $\bar{\text{x}}=\text{a}+\text{h}\Big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big)$
Above formula is a step deviation formula.
$\text{u}_\text{i}=\frac{\text{x}_\text{i}-\text{a}}{\text{h}}$

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MCQ 61 Mark

Choose the correct answer from the given four options:
The times, in seconds, taken by 150 atheletes to run a 110 m hurdle race are tabulated below:

Class	13.8-14	14-14.2	14.2-14.4	14.4-14.6	14.6-14.8	14.8-15
Frequency	2	4	5	71	48	20

The number of atheletes who completed the race in less then 14.6 seconds is:

A
11
B
71
✓
82
D
130

Answer

Correct option: C.

82

The number of atheletes who completed the race in less than 14.6
= 2 + 4 + 5 + 71 = 82

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MCQ 71 Mark

Choose the correct answer from the given four options:
When a die is thrown, the probability of getting an odd number less than 3 is:

✓
$\frac{1}{6}$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
D
$0$

Answer

Correct option: A.

$\frac{1}{6}$

When a die-is thrown, then total number of outcomes = 6 Odd number less than 3 is 1 only.
Number of possible outcomes = 1
$\therefore\ \text{Required probability}=\frac{1}{6}$

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MCQ 81 Mark

Choose the correct answer from the given four options:
The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its:

A
Mean.
✓
Median.
C
Mode.
D
All the three above.

Answer

Correct option: B.

Median.

Since, the intersection point of less than ogiven and more than ogiven given the median on the abscissa.

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MCQ 91 Mark

Choose the correct answer from the given four options:
If $x_i^{\prime}$ s are the mid points of the class intervals of grouped data, $f _{ i }^{\prime}$ s are the corresponding frequencies and $\overline{ x }$ is the mean, then $\sum\left(f_i-\bar{x}\right)$ is equal to:

✓
$0$
B
$-1$
C
$1$
D
$2$

Answer

Correct option: A.

$0$

$\because\ \ \ \bar{\text{x}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\text{n}}$
$\therefore\ \ \sum(\text{f}_\text{i}\text{ x}_\text{i}-\bar{\text{x}})=\sum\text{f}_\text{i}\text{ x}_\text{i}-\sum\bar{\text{x}}$
$=\text{n}\bar{\text{x}}-\text{n}\bar{\text{x}}$
$=0$

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MCQ 101 Mark

Choose the correct answer from the given four options:
A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. A single student is selected at random to be the class monitor. The probability that the selected student is not from A, B and C is:

A
$\frac{4}{23}$
✓
$\frac{6}{23}$
C
$\frac{8}{23}$
D
$\frac{17}{23}$

Answer

Correct option: B.

$\frac{6}{23}$

Total Number of students = 23
Number of students in house A, B and C = 4 + 8 + 5 = 17
$\therefore$ Remains students = 23 - 17 = 6
So, probability that the selected student is not from A, B and C $=\frac{6}{23}$

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MCQ 111 Mark

Choose the correct answer from the given four options:
One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is:

✓
$\frac{1}{5}$
B
$\frac{3}{5}$
C
$\frac{4}{5}$
D
$\frac{1}{3}$

Answer

Correct option: A.

$\frac{1}{5}$

Number of total outcomes = 40
Multiples of 5 between 1 to 40 = 5, 10, 15, 20, 25, 30, 35, 40
$\therefore$ Total number of possible outcomes = 8
$\therefore\ \ \text{Reqired probability}=\frac{8}{40}=\frac{1}{5}$

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MCQ 121 Mark

Choose the correct answer from the given four options:
Consider the data:

Class	65-85	85-105	105-125	125-145	145-165	165-185	185-205
Frequency	4	5	13	20	14	7	4

The difference of the upper limit of the median class and the lower limit of the modal class is:

A
0
B
19
✓
20
D
38

Answer

Correct option: C.

20

Here,

Class	Frequency	Cumulative frequency
65-85	4	4
85-105	5	9
105-125	13	22
125-145	20	42
145-165	14	56
165-185	7	63
185-205	4	67

Here, $\frac{\text{N}}{2}=\frac{67}{2}=33.5$ which lies in the interval 125-145.
ence, upper limit of median class is 145.
Here, we see that the highest frequency is 20 which lies in 125-145. Hence, the lower limit of modal class is 125.
Required difference = Upper limit of median class - Lower limit of modal class = 145 - 125 = 20

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MCQ 131 Mark

Choose the correct answer from the given four options:
The probability expressed as a percentage of a particular occurrence can never be:

A
Less than 100.
✓
Less than 0.
C
Greater than 1.
D
Anything but a whole number.

Answer

Correct option: B.

Less than 0.

We know that, the probability expressed as a percentage always lie between 0 and 100. So, it cannot be less than 0.

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MCQ 141 Mark

Choose the correct answer from the given four options:
While computing mean of grouped data, we assume that the frequencies are:

A
Evenly distributed over all the classes.
✓
Centred at the classmarks of the classes.
C
Centred at the upper limits of the classes.
D
Centred at the lower limits of the classes.

Answer

Correct option: B.

Centred at the classmarks of the classes.

In computing the mean of grouped data, the frequencies are centred at the class marks of the classes.

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MCQ 151 Mark

Choose the correct answer from the given four options:
A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favourable to E is:

A
4
B
13
C
48
✓
51

Answer

Correct option: D.

51

In a deck of 52 cards, there are 13 cards of heart and 1 is ace of heart.
Hence, the number of outcomes favourable to E = 51

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MCQ 161 Mark

Choose the correct answer from the given four options:
Which of the following cannot be the probability of an event?

A
$\frac{1}{3}$
B
$0.1$
C
$3\%$
✓
$\frac{17}{16}$

Answer

Correct option: D.

$\frac{17}{16}$

Since, probaility of an event always lies between 0 and 1.

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MCQ 171 Mark

Choose the correct answer from the given four options:
If the probability of an event is p, the probability of its complementary event will be:

A
p - 1
B
p
✓
1 - p
D
$1-\frac{1}{\text{p}}$

Answer

Correct option: C.

1 - p

Since, probability of an event + probability of its complementry event = 1
So, probaility of its complementry event = 1 - Probability of an event = 1 - p

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MCQ 181 Mark

Choose the correct answer from the given four options:
If an event cannot occur, then its probability is:

A
$1$
B
$\frac{3}{4}$
C
$\frac{1}{2}$
✓
$0$

Answer

Correct option: D.

$0$

The event which cannot occur is said to be impossible event and probaillity of impossible event is zero.

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MCQ 191 Mark

Choose the correct answer from the given four options:
Consider the following frequency distribution:

Class	0-5	6-11	12-27	18-23	24-29
Frequency	13	10	15	8	11

The upper limit of the median class is:

A
17
✓
17.5
C
18
D
18.5

Answer

Correct option: B.

17.5

Here,

Class	Frequency	Cumulative frequency
-0.5-5.5	13	13
5.5-11.5	10	23
11.5-17.5	15	38
17.5-23.5	8	46
23.5-29.5	11	57

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MCQ 201 Mark

Choose the correct answer from the given four options:
The probability that a non leap year selected at random will contain 53 sundays is:

✓
$\frac{1}{7}$
B
$\frac{2}{7}$
C
$\frac{3}{7}$
D
$\frac{5}{7}$

Answer

Correct option: A.

$\frac{1}{7}$

A non-leap year has 365 days and therefore 52 weeks and 1 day.The 1 day may be Sunday or Manday or Tuesday or Wednesday or Thursday ot Friday or Saturday. Thus, out of 7 possibilities, 1 favourable enent is the event that the one day is Sunday.
$\therefore\ \text{Required probability}=\frac{1}{7}$

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MCQ 211 Mark

Choose the correct answer from the given four options:
Someone is asked to take a number from 1 to 100. The probability that it is a prime is:

A
$\frac{1}{5}$
B
$\frac{6}{25}$
✓
$\frac{1}{4}$
D
$\frac{13}{50}$

Answer

Correct option: C.

$\frac{1}{4}$

Total Number of outcomes = 100
So, the prime number between 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 56, 61, 67, 71, 73, 79, 83, 89, and 97.
$\therefore$ Total number of possible outcomes = 25
$\therefore\ \ \text{Reqired probability}=\frac{25}{100}=\frac{1}{4}$

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MCQ 221 Mark

Choose the correct answer from the given four options:
If P(A) denotes the probability of an event A, then:

A
$\text{P(A)}<0$
B
$\text{P(A)}>1$
✓
$0\leq\text{P(A)}\leq1$
D
$-1\leq\text{P(A)}\leq1$

Answer

Correct option: C.

$0\leq\text{P(A)}\leq1$

Since, probability of an event always lies between 0 and 1.

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MCQ 231 Mark

Choose the correct answer from the given four options:
A card is selected from a deck of 52 cards. The probability of its being a red face card is

A
$\frac{3}{26}$
B
$\frac{3}{13}$
✓
$\frac{2}{13}$
D
$\frac{1}{2}$

Answer

Correct option: C.

$\frac{2}{13}$

In a deck of 52 cards, there are 12 face cards i.e., 6 red and 6 black cards.So, probability of getting a red face card $=\frac{6}{52}=\frac{3}{26}$

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MCQ 241 Mark

Choose the correct answer from the given four options:
The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is:

A
7
✓
14
C
21
D
28

Answer

Correct option: B.

14

Here, total number of eggs = 400
Probability of getting a bad egg = 0.035
$\Rightarrow\ \ \frac{\text{Number of bad eggs}}{\text{Total number of eggs}}=0.035$
$\Rightarrow\ \ \frac{\text{Number of bad eggs}}{400}=0.035$
$\therefore\ \ \text{Number of bad eggs}=0.035\times400=14$

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MCQ 251 Mark

Choose the correct answer from the given four options:
For the following distribution:

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

the sum of lower limits of the median class and modal class is:

A
15
✓
25
C
30
D
35

Answer

Correct option: B.

25

Here,

Class	Frequency	Cumulative frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	66

Now, $\frac{\text{N}}{2}=\frac{66}{2}=33,$ which lies in the interval 10-15. Therefore, lower limit of the median class is 10.
The highest frequency is 20, which lies in the interval 15-20. Therefore, lower limit of modal class is 15.
Hence, required sum is 10 + 15 = 25.

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MCQ 261 Mark

Choose the correct answer from the given four options:
For the following distribution:

Marks	Number of students
Below 10	3
Below 20	12
Below 30	27
Below 40	57
Below 50	75
Below 60	80

the modal class is:

A
10-20
B
20-30
✓
30-40
D
50-60

Answer

Correct option: C.

30-40

Here,

Marks	Number of students	Cumulative frequency
Below 10	3 = 3	3
10-20	(12 - 3) = 9	12
20-30	(27 - 12) = 15	27
30-40	(57 - 27) = 30	57
40-50	(75 - 57) = 18	75
50-60	(80 - 75) = 5	80

Here, we see that the highest frequency is 30. which lies in the inteval 30-40.

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