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# Maths Questions for CTET,KVS Exam : 18th 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. Q2. The ratio of the side and height of an equilateral triangle is
(a) 2 : 1
(b) 1 : 1
(c) 2 : √3
(d) √3 : 2

Q3. The number of vertices in a polyhedron that has 22 edges and 10 faces is
(a) 12
(b) 14
(c) 16
(d) 20

Q4. In the figure, ABC is an isosceles triangle with CA = CB and BC is produced to a point D. If CE ⊥ BC, such that ∠D = ½ ∠E = ½ ∠A, then measure of ∠ACD is (a) 110°
(b) 120°
(c) 135°
(d) 140°

Q5. A ray has __________ endpoint(s).
(a) one
(b) two
(c) three
(d) no

Q6. In ∆PQT, PQ = PT. Points R and S are on QT such that PR = PS. If ∠PTS = 62° and ∠RPS = 34°, then measure of ∠ QPR is
(a) 11°
(b) 13°
(c) 15°
(d) 17°

Q7. How many lines of symmetry does a parallelogram have ?
(a) 3
(b) 2
(c) 4
(d) None

Q8. The number of lines of symmetry of the following figure is (a) 1
(b) 2
(c) 4
(d) 8

Q9. A cube has how many faces ?
(a) 4
(b) 6
(c) 8
(d) 12

Q10. A book is an example of the
(a) cube
(b) cuboid
(c) cylinder
(d) square

Solutions

S1. Ans.(b)
Sol. Three or more lines are said to be concurrent if they all intersect at the same point. In option 2, this condition is satisfied.

S2. Ans.(c)
Sol.
Let ABC be an equilateral triangle of side x.
AM ⊥ BC
∴ BM = MC = x/2
In ∆AMC,
AM = √(AC^2–MC^2 )
= √(x^2–(x/2)^2 )
= √(x^2–x^2/4)
= √((3x^2)/4)
= √3/2 x
Ratio of side and height
= x∶√3/2 x
= 1∶√3/2
= 2 : √3

S3. Ans.(b)
Sol.
We know that
F + V = E + 2
Or, 10 + V = 22 + 2
Or, V = 24 – 10
Or, V = 14

S4. Ans.(b)
Sol.
∠D = 1/2∠E …(i)
∠D + ∠E = 90° …(ii)
Solving (i) and (ii), we get
∠D = 30°, ∠E = 60°
Now, 1/2∠E=1/2∠A
⇒ ∠A = ∠E
∴ ∠A = 60°
∵ ∠A = ∠B
(angle opposite to equal sides CA and CB)
∴ ∠B = 60°
∠ACD = ∠A + ∠B
(Exterior angle theorem)
= 60° + 60° = 120°

S5. Ans.(a)
Sol.
A ray has one end point.

S6. Ans.(a)
Sol. We have
PR = PS
∴ ∠PSR = ∠PRS
Now,
∠PSR + ∠PRS + ∠RPS = 180°
⇒ ∠PRS + ∠PRS + 34° = 180°
⇒ 2∠PRS = 146°
⇒ ∠PRS = 73°
Now,
∠PRS + ∠PRQ = 180°
(Angles on one side of a line)
⇒ 73° + ∠PRQ = 180°
⇒ ∠PRQ = 180° – 73°
⇒ ∠PRQ = 107°
Now,
PQ = PT
⇒ ∠ PQT = ∠PTQ
⇒ ∠PQT = ∠PTS = 62°
In triangle PQR,
∠PQR + ∠PRQ + ∠QPR = 180°
⇒ 62° + 107° + ∠QPR = 180°
⇒ 169° + ∠QPR = 180°
⇒ ∠QPR = 180° – 169°
∴ ∠QPR = 11°

S7. Ans.(d)
Sol.
A parallelogram does not have any line of symmetry.

S8. Ans.(c)
Sol.
The “line of symmetry” (shown in the following figure as dotted) is the imaginary line where you can fold the image and have both halves match exactly. Four such lines are possible in the given figure.

S9. Ans.(b)
Sol.
A cube is a three-dimensional object bounded by 6 squares faces.

S10. Ans.(b)
Sol.
Length, breadth and height are usually different of a book.
Therefore, it is a cuboid. Join India's largest learning destination

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