Objective type questions in CBSE Class 10 Maths Exam 2020 will be asked for 20 marks. 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.
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.
Answer: (C)
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
Answer: (B)
Explanation: In computing the mean of grouped data, the frequencies are centred at the class marks of the classes
#Important MCQs Class 10 Science for Board Exam 2020
(A) 0
(B) – 1
(C) 1
(D) 2
Answer: (A)
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
Answer: (B)
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
Answer: (B)
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
Answer: (D)
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
Answer: (C)
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
Answer: (D)
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
Answer: (B)
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
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(A) 3
(B) 4
(C) 5
(D) 6
Answer: (D)
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
Answer: (C)
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
Answer: (C)
Explanation:
Therefore mean is:
CBSE Class 10 Science Important Questions for Board Exam 2020
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
Answer: (D)
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
Answer: (C)
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,
Answer: (B)
Explanation: We have
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| CBSE Class 10 Maths Important MCQ Questions & Answers: All Chapters |
Also check important resources for the preparation of CBSE Class 10 Board Exam 2020:
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- CBSE Class 10 Exam Pattern 2020: All Subjects
- CBSE Class 10 Syllabus for Board Exam 2020: All Subjects
- CBSE Class 10 Sample Papers with Marking Scheme 2020: All Subjects
- CBSE Class 10 Question Papers 2019 with Solutions
- CBSE Class 10 NCERT Exemplar Problems, Solutions for Maths & Science
- CBSE Class 10 NCERT New Textbooks & NCERT Solutions: All Subjects