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# Maths Questions for CTET,KVS Exam : 6th october 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. The sum of mean, mode and median of the data 6, 3, 9, 5, 1, 2, 3, 6, 5, 1, 3 is
(a) 13
(b) 10
(c) 11
(d) 12

Q2. The ages in years of 10 teachers of a school are 34, 42, 27, 41, 36, 28, 22, 31, 37, 39. What is the age of the youngest teacher?
(a) 41
(b) 22
(c) 27
(d) 28

Q3. The base of an isosceles ∆ABC is 48 cm and its area is 168 cm². The length of one of its equal sides is
(a) 8 cm
(b) 15 cm
(c) 17 cm
(d) 25 cm

Q4. The following monthly expenditure values of 15 families: Exp (Rs.): 220, 310, 110, 300, 500, 270, 160, 200, 310, 400, 420, 320, 340, 140, 100 Calculate the arithmetic mean.
(a) Rs. 876.35
(b) Rs. 345.77
(c) Rs. 273.33
(d) Rs. 373.55

Q5. A square and a circle have equal perimeters. The ratio of the area of the square to the area of the circle is
(a) 1 : 1
(b) 1 : 4
(c) π : 2
(d) π : 4

Q6. A room is in the form of a cuboid. Its length, breadth and height are 9 m, 8 m and 4 m respectively. What is the total area of all four walls?
(a) 120 m²
(b) 64 m²
(c) 136 m²
(d) 70 m²

Q7. Sum of mean and median of the number 5.02, 5.18, 5.12, 5.007 and 5.018 is
(a) 10.089
(b) 10.73
(c) 10.71
(d) 10.89

Q8. The diameter of a hemisphere is 21 cm. What is its total surface area?
(a) 1031.5 cm²
(b) 1039.5 cm²
(c) 519.75 cm²
(d) 312.41 cm²

Q9. The product of the mean and median of the numbers 8, 11, 6, 9, 16 is
(a) 60
(b) 64
(c) 80
(d) 90

Q10. The radius of a roller is 1.75 m, and it is 2 m long. If it takes 50 revolutions to level a field, then the area of the field is
(a) 1100 m²
(b) 2200 m²
(c) 550 m²
(d) 225 m²

Solutions

S1. Ans.(b)
Sol. The given numbers are 6, 3, 9, 5, 1, 2, 3, 6, 5, 1, 3
After arranging the numbers in ascending order, we get
1, 1, 2, 3, 3, 3, 5, 5, 6, 6, 9
Number of numbers = 11
Mean
=(1+1+2+3+3+3+5+5+6+6+9)/11
=44/11=4
Here, the number of terms is odd.
Median = (n + 1)/2th term
=(11+1)/2 th term
= 6th term
= 3
Now, 1 is twice, 2 is once, 3 is thrice, 5 is twice, 6 is twice and 9 is once. As 3 occurs the most of the given data is 3.
Sum of mean, mode and median = 4 + 3 + 3 = 10

S2. Ans.(b)
Sol. After arranging all the ages in increasing order, we get
22, 27, 28, 31, 34, 36, 37, 39, 41, 42
Thus, the age of the youngest teacher is 22.   S6. Ans.(c)
Sol. Sum of area of walls = 2 (length × height + breadth × height)
= 2 (9 × 4 + 8 × 4) = 136 m²

S7. Ans.(a)
Sol. Mean = (Sum of numbers)/(Numbers of numbers)
=(5.02+5.18+5.12+5.007+5.018)/5
=25.345/5=5.069
The median, in case of an odd number of numbers, is the middle term of the sequence when the numbers are arranged in ascending order.
Ascending order of numbers
= 5.007, 5.018, 5.02, 5.12, 5.18
Middle term when numbers are arranged in ascending order = 5.02
∴ Median = 5.02
Sum of mean and median
= 5.069 + 5.02 = 10.089

S8. Ans.(b)
Sol. Diameter of hemisphere = 21 cm
= 21/2 = 10.5 cm
Total surface area of he misphere = 3πr²
=3×22/7×(10.5)^2
= 1039.5 cm²

S9. Ans.(d)
Sol. Mean = (Sum of numbers)/(Number of numbers)
=(8+11+6+9+16)/5
=50/5=10
The median of an odd number of numbers is the middle term when the numbers are arranged in ascending order.
6, 8, 9, 11, 16
Median = 9
Product of mean and median
= 10 × 9 = 90

S10. Ans.(a)
Sol. Area covered in one revolution = Curved surface area of the roller = 2πrh
=2×22/7×1.75×2=22 m^2
Therefore, the area of the field
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