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# Maths Questions for CTET,KVS Exam : 19th september 2018

Dear Students!!! There is most general as well as a scoring section in all the competitive entrance examinations in the teaching field i.e “Mathematics”.Because in this section only one thing is work i.e your accuracy and that could be nourished with the daily practice. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2018, DSSSB ,KVS,STET Exam.

Q1. Find the median of the following values:
30, 20, 15, 10, 25, 35, 18, 21, 28, 40, 36
(a) 25
(b) 39
(c) 42.7
(d) 35

Q6. Find the mode of the following data:
110, 120, 130, 120, 110, 140, 130, 120, 140, 120
(a) 240
(b) 120
(c) 320
(d) 110

Q7. Find the mode of the following series:
2.5, 2.3, 2.2, 2.4, 2.7, 2.7, 2.5, 2.3, 2.2, 2.6, 2.2
(a) 2.2
(b) 6.5
(c) 8.9
(d) 2.7

Q8. The ages of 15 students of a class are reported below.
Find the modal age (in year). Age (years): 24 24 17 18 17 19 18 21 20 21 20 23 22 22 22
(a) 50
(b) 22
(c) 92
(d) 17

Q10. Below presented is the death rate of population across different countries. Find the mode.
Death rate (per thousand): 11.1, 10.9, 10.7, 11.1, 10.6, 11.3, 10.6, 10.7, 10.6, 10.9, 10.6, 10.5, 10.4, 10.6
(a) 10.6
(b) 11
(c) 13.9
(d) 11.1

Solutions

S1. Ans.(a)
Sol. 30, 20, 15, 10, 25, 35, 18, 21, 28, 40, 36 in ascending order can be written as 10, 15, 18, 20, 21, 25, 28, 30, 35, 36, 40
Here, N = 11, which is odd
∴ Median = ((11 + 1)/2)th term = 6th term, which is 25
Therefore, the median is 25.

S2. Ans.(b)
Sol. Here, N = 122
∴N/2 = 61
The cumulative frequency just greater than N/2, i.e., 61, is 72 and value of the income corresponding to 72 is 6000.
∴ The median is 6000.

S3. Ans.(a)
Sol. Here, N = 75
∴ N/2 = 37.5
The cumulative frequency just greater than N/2, i.e., 37.5, is 51 and value of the size corresponding to 51 is 7.
∴ The median is 7.

S4. Ans.(c)
Sol. Here, N = 150
∴ N/2 = 75
The cumulative frequency just greater than N/2, i.e., 75, is 95 and value of the size corresponding to 95 is 35.
∴ The median is 35.

S5. Ans.(a)
Sol. Here, N = 200
∴ N/2 = 100
The cumulative frequency just greater than N/2, i.e., 100, is 135 and value of the size corresponding to 135 is 48.
∴ The median is Rs. 48.

S8. Ans.(b)
Sol. Arrange the series in ascending order as: Age (years): 17  17  18  18  19  20  20  21  21  22  22  22  23  24  24
An inspection of the series shows that 22 occurs most frequently.
So, mode = 22

S9. Ans.(c)
Sol. Since the frequency of 140 is the maximum.
∴ Mode = Rs. 140.