Maths Questions for CTET,KVS Exam : 29 october 2018_00.1
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Maths Questions for CTET,KVS Exam : 29 october 2018

Maths Questions for CTET,KVS Exam : 29 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. In what time will a 100 metre long train running with a speed of 50 km/hr cross a pillar?
(a) 7.0 sec
(b) 72 sec
(c) 7.2 sec
(d) 70 sec

Maths Questions for CTET,KVS Exam : 29 october 2018_50.1

Q3. In a triangle PQR, the side QR is extended to S. ∠QPR = 72° and ∠PRS = 110°, then the value of ∠PQR is:
(a) 38°
(b) 32°
(c) 25°
(d) 29°

Maths Questions for CTET,KVS Exam : 29 october 2018_60.1

Q5. In ∆ABC, ∠B = 70° and ∠C = 60°. The internal bisectors of the two smallest angles of ∆ABC meet at O. The angle so formed at O is:
(a) 125°
(b) 120°
(c) 115°
(d) 110°

Q6. 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


Q7. A cylindrical container of 32 cm height and 18 cm radius is filled with sand. Now all this sand is used to form a conical heap of sand. If the height of the conical heap is 24 cm, what is the radius of its base?
(a) 12 cm
(b) 24 cm
(c) 36 cm
(d) 32 cm

Q8. 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

Q9. 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

Q10. The ratio of the radius 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

SOLUTIONS

S1. Ans.(c)
Sol.
time = (100 meter)/(50×5/18 m/sec) = 7.2 sec.

S2. Ans.(c)
Sol.
(l + m + n)2 = l2+ m2 + n2 + 2(lm + mn + nl)
81 = 31 + 2 (lm + mn + nl)
(lm + mn + nl) = 25

S3. Ans.(a)
Sol.
∠PQR = (∠PRS – ∠QPR)
= 110° – 72° = 38°

S4. Ans.(a)
Sol.
x+1/x=√3
cubing both sides,
x^3+1/x^3 +3(x+1/x)=3√3
x^3+1/x^3 =3√3-3√3=0

S5. Ans.(a)
Sol
AOC = 90+(∠B)/2
= 90 + 35 = 125°

S6. 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

S7. Ans.(c)
Sol.
Volume of cylinder and cone are equal.
so, πr_1^2 h_1=1/3 πr_2^2 h_2
so, (3×18×18×32)/24=r_2^2
r_2=√((3×18×18×32)/24)=36 cm

S8. Ans.(c)
Sol. Mean =
(Sum of all observations)/(Total number of observations)
=((4600+5560 . . . . . . . . 6800))/10
=62600/10=6260


S9. Ans.(a)
Sol. Volume of cylinder = πr²h
=22/7×7^2×20
= 3080 cm³

S10. 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

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