Maths Questions for CTET,KVS Exam : 1st october 2018_00.1
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Maths Questions for CTET,KVS Exam : 1st october 2018

Maths Questions for CTET,KVS Exam : 1st october 2018_40.1
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 total surface area of a solid cube is 24 cm². The volume of the cube is

(a) 4 cm³
(b) 8 cm³
(c) 24 cm³
(d) 27 cm³

Q2. What is the range of 22, 5, 6, 15, 3, 4, 9, 35?
(a) 22
(b) 25
(c) 32
(d) 15

Q3. The mean of median and mode of the data 7, 6, 7, 9, 8, 8, 10, 8 is
(a) 5.5
(b) 8
(c) 8.5
(d) 9

Q4. The radius of a cylinder is 7 cm and height is 20 cm. the volume of the cylinder is
(a) 3080 cm³
(b) 1540 cm³
(c) 2310 cm³
(d) 3850 cm³

Q5. The ratio of the bases of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is
(a) 7 : 6
(b) 4 : 9
(c) 20 : 27
(d) 10 : 9

Q6. The following table given the monthly income of 10 families in a city: Income (Rs.): 4600, 5560, 6440, 4530, 7670, 6850, 6750, 7910, 5490, 6800. Calculate the arithmetic mean.
(a) 4897
(b) 6542
(c) 6260
(d) 7260

Q7. The mean of range, mode and median of the data 4, 3, 2, 2, 7, 2, 2, 0, 3, 4, 4 is
(a) 4
(b) 5
(c) 2
(d) 3

Q8. The volume of a cube is 125 cm³. What will be its total surface area?
(a) 100 cm³
(b) 125 cm³
(c) 150 cm³
(d) 200 cm³

Q9. The area of a square is 16/π of the area of a circle. The ratio of the side of the square to the diameter of the circle is
(a) 3 : 1
(b) 2 : 1
(c) π : 1
(d) √2 π : 1

Q10. If the side of a square is increased by 20%, then by what percentage will the area be increased?
(a) 20%
(b) 22%
(c) 40%
(d) 44%

Solutions

S1. Ans.(b)
Sol. Total surface area of cube = 6a²
Where ‘a’ is the side of the cube
Now, 6a² = 24
⇒a^2=24/6=4⇒a=2
Volume of cube = a³ = 2³ = 8 cm³

S2. Ans.(c)
Sol. After arranging all the numbers in ascending order, we get 3, 4, 5, 6, 92, 15, 22 and 35
Range = Highest value – Lowest value
= 35 – 3 = 32

S3. Ans.(b)
Sol. After arranging 7, 6, 7, 9, 8, 8, 10, 8 in ascending order, we get 6, 7, 7, 8, 8, 8, 9, 10
Here, 6 is once, 7 is twice, 8 is thrice, 9 is once and 10 is once.
As 8 occurs the maximum number of times, i.e., three times, the mode is 8.
Here, the number of terms, n = 8
As the number of terms is an even number,
Median = 1/2
[(n/2)th term+(n/2+1)th term]
=1/2 [4th term+(4+1)th term]
=1/2 [4th term+5th term]
=18 [8+8]
= 8
∴ Mean of median and mode = (8 + 8)/2=8

S4. Ans.(a)
Sol. Volume of cylinder = πr²h
=22/7×7^2×20
= 3080 cm³

S5. Ans.(c)
Sol. Let the radii of two cylinders be 2r and 3r and heights be 5h and 3h respectively.
Volume of first cylinder = π (2r)² (5h) = 20π r²h
Volume of second cylinder = π (3r)² (3h) = 27 π r²h
Required ratio
=(20πr^2 h)/(27πr^2 h)
=20/27=20∶27

S6. Ans.(c)
Sol. Mean =
(Sum of all observations)/(Total number of observations)
=((4600+5560 . . . . . . . . 6800))/10
=62600/10=6260

S7. Ans.(a)
Sol. After arranging the given numbers in ascending order, we get 0, 2, 2, 2, 2, 3, 3, 4, 4, 4, 7
Range = Maximum value – Minimum value = 7 – 0 = 7
Here, the number of terms = 11
The median for an odd number of terms is the middle term when the terms are arranged in ascending or descending order.
When the number of terms ‘n’ is odd, then the middle term is ((n + 1))/2 th term.
∴ Median = ((11 + 1)/2)^th term
= 6th term
= 3
Here, 2 is getting repeated the maximum number of times, i.e., 4 times.
Therefore, the mode = 2
Mean of range, mode and median
=(7+2+3)/3=12/3=4

S8. Ans.(c)
Sol. Volume of a cube = (side)³
Let ‘a’ be the side of the cube.
∴ a³ = 125 ⇒ a = 5 cm
Therefore, the total surface area
= 6 × (a)² = 6 × (5)² = 150 cm²


S9. Ans.(b)
Sol. Let the diameter of the circle be d and the side of the square be s.
Area of circle = π (d/2)^2=(πd^2)/4
Area of square = s²
According to the question,
s^2=(πd^2)/4×16/π
⇒s^2/d^2 =4
⇒s/d=2
∴ s : d = 2 : 1

S10. Ans.(d)
Sol. Let the initial side of the square be x. Then, the initial area of the square = x²
Side of the square after 20% increase = x (1 + 20%) = 1.2x
Area of the square after 20% increase in the side = (1.2x)² = 1.44x²
Increase in area = 1.44x² – x²
= 0.44x²
Percentage increase in area
=(0.44x^2)/x^2 ×100%=44%

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