# CBSE 10th Maths Exam 2021: Important MCQs from Chapter 14 Statistics with Answers

CBSE Class 10 Maths Chapter 14 - Statistics MCQs form a very good resource that will help students score good marks in the Maths Exam.

Updated: Mar 19, 2021 13:26 IST
CBSE 10th Maths Exam 2020: Important MCQs from Chapter 14 Statistics with Answers

In CBSE Class 10 Maths Exam 2021, Part-A will be composed of objective type questions only. Students can easily score full marks in these questions with a little practice. We are providing here the MCQ questions from Class 10 Maths NCERT Chapter 14 - Statistics. Different types of questions provided here can be used as reference material to revise all important concepts and score good marks in the exam. All the MCQs are thoroughly solved.

Note - Students must note that 'Step deviation Method for finding the mean' has been excluded from the revised CBSE syllabus of Class10 Maths. So, students should prepare according to the new syllabus only.

Check below the solved MCQs from Class 10 Maths Chapter 14 Statistics:

(A) lower limits of the classes

(B) upper limits of the classes

(C) midpoints of the classes

(D) frequencies of the class marks.

Explanation: We know that di = xi – ai. i.edi’s are the deviations from the midpoints of the classes.

Q2. While computing mean of the grouped data, we assume that the frequencies are:

(A) evenly distributed over all the classes

(B) centered at the class marks of the classes

(C) centered at the upper limits of the classes

(D) centered at the lower limits of the classes

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

(A) 0

(B) – 1

(C) 1

(D) 2

Explanation:

Q4. The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency of a grouped data gives its:

(A) Mean

(B) Median

(C) Mode

(D) All of these

Explanation: Since the intersection point of less than type ogive and more than ogive gives the median on the abscissa.

Q5. 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 median class and modal class is:

(A) 15

(B) 25

(C) 30

(D) 35

Explanation:

 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 N/2 = 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

Q6.  If the arithmetic mean of x, x + 3, x + 6, x + 9 and x + 12 is 10, then x = ?

(A) 1

(B) 2

(C) 6

(D) 4

Explanation:

According to question

Q7. If the mean of first n natural numbers is 5n/9, then n =?

(A) 6

(B) 7

(C) 9

(D) 10

Explanation:

But according to question,

Q8. If 35 is removed from the data, 30, 34, 35, 36, 37, 38, 39, 40 then the median increases by:

(A) 2

(B) 1.5

(C) 1

(D) 0.5

Explanation: We have

30, 34, 35, 36, 37, 38, 39, 40

The data has 8 observations, so there are two middle terms, 4th and 5th term i.e. 36 and 37.

The median is the mean of both these terms.

Median = (36 + 37)/2

Median = 36.5

When 35 is removed from given data as 30, 35, 36, 37, 38, 39, 40 then the number of observations becomes 7.

Now the median is the middle most i.e 4th term which is equal to 37.

Therefore median is increased by 37 – 36.5 = 0.5

Q9. The Median when it is given that mode and mean are 8 and 9 respectively, is:

(A) 8.57

(B) 8.67

(C) 8.97

(D) 9.24

Explanation: By Empirical formula:

Mode = 3median – 2 mean

8 = 3medain – 2 X 9

8 = 3median – 18

3median = 8 + 18

Median = 26/3

Median = 8.67

CBSE Class 10 Maths Exam Pattern 2021: Check to Score Good Marks in Exam

(A) 3

(B) 4

(C) 5

(D) 6

Explanation: According to question,

Q11. In a hospital, weights of new born babies were recorded, for one month. Data is as shown:

 Weight of new born baby (in kg) 1.4 – 1.8 1.8 – 2.2 2.2 – 2.6 2.6 – 3.0 No of babies 3 15 6 1

Then the median weight is:

(A) 2kg

(B) 2.03kg

(C) 2.05 kg

(D) 2.08 kg

Explanation: Construct a table as follows:

 Class-interval Frequency (fi) Midpoint (xi) Cumulative Frequency (cf) 1.4-1.8 3 1.6 3 1.8-2.2 15 2 18 2.2-2.6 6 2.4 24 2.6-3.0 1 2.8 25

Since N/2 = 25/2 = 12.5

12.5 is near to cumulative frequency value 18

So median class interval is 1.8 - 2.2

∴Median = l + [(N/2 – cf)/f]/h

Here

Hence median weight is 2.05 kg.

Q12. In a small scale industry, salaries of employees are given in the following distribution table:

 Salary (in Rs.) 4000 - 5000 5000-6000 6000-7000 7000-8000 8000-9000 9000-10000 Number of employees 20 60 100 50 80 90

Then the mean salary of the employee is:

(A) Rs. 7350

(B) Rs.  7400

(C) Rs. 7450

(D) Rs. 7500

Explanation:

Therefore mean is:

CBSE Class 10 Science Important Questions for Board Exam 2021

Q13.  For one term, absentee record of students is given below. If mean is 15.5, then the missing frequencies x and y are:

 Number of days 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40 TOTAL Total Number of students 15 16 x 8 y 8 6 4 70

(A) x = 4 and y = 3

(B) x = 7 and y = 7

(C) x = 3 and y = 4

(D) x = 7 and y = 6

Explanation: Construct a table as follows:

 Class-interval Frequency (fi) Midpoint (xi) fixi 0-5 15 2.5 37.5 5 - 10 16 7.5 120 10 - 15 x 12.5 12.5x 15 - 20 8 17.5 140 20 - 25 y 22.5 22.5y 25 -30 8 27.5 220 30 - 35 6 32.5 195 35 - 40 4 37.5 150 TOTAL 70 12.5x+22.5y+862.5

mean = (12.5x + 22.5y + 862.5)/70

⇒ 15.5 = (12.5x +22.5y + 862.5)/70

⇒ 15.5 X 70 = 12.5x +22.5y + 862.5

⇒ 12.5x + 22.5y = 222.5

⇒ 125x + 225y = 2225

⇒ 5x + 9y = 89                        .....(i)

Also,

x + y + 57 = 70

x + y = 13         ......(ii)

Multiplying equation (ii) by 5 and then subtracting from (i) as,

Substituting the value of y in equation (ii), we get

x + y = 13

⇒ x + 6 = 13

⇒ x = 7

Hence x = 7 and y = 6

Q14. Pocket expenses of a class in a college are shown in the following frequency distribution:

 Pocket expenses 0-200 200-400 400-600 600-800 800-1000 1000-1200 1200-1400 Number of students 33 74 170 88 76 44 25

Then the median for the above data is:

(A) 485.07

(B) 486.01

(C) 487.06

(D) 489.03

Explanation:

 Class-interval Frequency (fi) Midpoint (xi) fixi cf 0-200 33 100 3300 33 200-400 74 300 22200 107 400-600 170 500 85000 277 600-800 88 700 61600 365 800-1000 76 900 68400 441 1000-1200 44 1100 48400 485 1200-1400 25 1300 32500 510 510 321400

Since N/2 = 510/2 = 255

255 is near to cumulative frequency value 277.

So median class interval is 400-600

Here,

l = 400

N/2 = 255

cf = 107

f = 170

h = 100

Therefore,

Explanation: We have

All the above MCQ questions and answers are also available in form of PDF which can be downloaded from the link provided below:

To get the important MCQs from all other chapters of CBSE Class 10 Maths, go to the following link:

Also, check important resources for the preparation of CBSE Class 10 Board Exam 2021:

We are providing below some articles which are quite useful for making effective preparations for the upcoming class 10 board exam. All these important preparation resources have been created after thorough research of the examination trends and latest pattern. Students should go through these articles to remain organised with their preparations and perform well in the exams. Links to all important articles are given below:

रोमांचक गेम्स खेलें और जीतें एक लाख रुपए तक कैश

## Related Categories

Comment (0)

### Post Comment

4 + 7 =
Post
Disclaimer: Comments will be moderated by Jagranjosh editorial team. Comments that are abusive, personal, incendiary or irrelevant will not be published. Please use a genuine email ID and provide your name, to avoid rejection.