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# Maths Questions for CTET,KVS Exam : 20 november 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.

(a) – 9
(b) – 81
(c) + 9
(d) + 81

Q2. If (√(5 + x)+√(5-x))/(√(5 + x)-√(5-x))=3, then what is the value of x?
(a) 5/2
(b) 25/3
(c) 4
(d) 3

Q3. If (x + y + z) = 12, xy + yz + zx = 44 and xyz = 48, then what is the value of x^3 + y^3 + z^3?
(a) 104
(b) 144
(c) 196
(d) 288

Q4. If x=(4√ab)/(√a+√b), then what is the value of (x+2√a)/(x-2√a)+(x+2√b)/(x-2√b) (when a ≠ b)?
(a) 0
(b) 2
(c) 4
(d) (√a + √b)/( √a –√b)

Q5. In a triangle PQR, ∠Q = 90°. If PQ = 12 cm and QR = 5 cm, then what is the radius (in cm) of the circumcircle of the triangle?
(a) 5
(b) 6
(c) 6.5
(d) 6√2

Q6. If a chord of a circle subtends an angle of 30° at the circumference of the circle, then what is the ratio of the radius of the circle and the length of the chord respectively?
(a) 1 : 1
(b) 2 : 1
(c) 3 : 1
(d) √2 : 1

Q7. The tangents drawn at points A and B of a circle with centre O, meet at P. If ∠AOB = 120° and AP = 6 cm, then what is the area of triangle (in cm2) APB?
(a) 6√3
(b) 8√3
(c) 9
(d) 9√3

Q8. P is a point outside the circle at distance of 6.5 cm from centre O of the circle. PR be a secant such that it intersects the circle at Q and R. If PQ = 4.5 cm and QR = 3.5 cm, then what is the radius (in cm) of the circle?
(a) 1.5
(b) 2
(c) 2.5
(d) 3

Q9. What will Rs 12500 amounts to, in 2 years at the rate of 20% p.a. if interest is compounded yearly?
(a) Rs 18500
(b) Rs 19000
(c) Rs 18000
(d) Rs 19500

Q10. Simplify (4/5+9/2)÷(1/2-1/5)
(a) 53/3
(b) 52/3
(c) 50/3
(d) 47/3

Solution

S1. Ans.(d); For expression x⁶ – 18x³ + K to be a perfect square, expression should be in the form of a² + b² – 2ab
On comparing
a² + b² – 2ab = (x³)² + K – 2 × 9 × x³
so here b = 9
so K = b² = 81

S2.Ans.(d); (√(5+x)+√(5–x))/(√(5+x) –√(5–x))=3
Applying componendo and dividendo
(2√(5+x))/(2√(5–x))=2
(5+x)/(5 –x)=4
5 + x = 20 – 4x
5x = 15
x = 3

S3.Ans.(d); (x^3+y^3+z^3–3xyz)/(x^2+y^2+z^2–xy –yz –zx)=x+y+z
Also, (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
⇒12² = x² + y² + z² + 2 × 44
⇒x² + y² + z² = 144 – 88
⇒x² + y² + z² = 56
Now, (x^3+y^3+z^3–3×48)/((56 –44) )=12
⇒ x³ + y³ + z³ – 144 = 144
⇒ x³ + y³ + z³ = 288

S4. Ans.(b); x=(4√ab)/(√a+√b)
x/(2√a)=(2√b)/(√a+√b) …(i)
And
x/(2√b)=(2√a)/(√a+√b) ……..(ii)
Applying componendo and dividend in (i) and (ii)
(x+2√a)/(x –2√a)=(2√b+√a+√b)/(2√b –√a–√b)=(3√b+√a)/(√b–√a)
And, (x+2√b)/(x –2√b)=(2√a+√a+√b)/(2√a –√a–√b)=(3√a+√b)/(√a –√b)
so,(x+2√a)/(x–2√a)+(x+2√b)/(x–2√b)=(3√b+√a)/(√b–√a)+(3√a+√b)/(√a–√b)
=(3√a+√b –3√b –√a)/(√a–√b) =2

S5. Ans.(c);

Radius of circumcircle of right angle triangle is half of length of hypotenuse
PR = 13 cm

S7. Ans.(d);

Join AB and OP
∠APB = 60ᵒ
OP divides ∠APB equally
So ∠APO = 30°
If OP and AB intersect at D then
DP/AP=cos⁡30
DP/6=√3/2
DP = 3√3
Area of ∆APB=1/2×6×3√3=9√3 cm^2

S8. Ans.(c);
Let PO is produced such that it touches the circle at M
Since
QP × PR = PN × PM
(3.5 + 4.5) × 4.5 = (6.5 – r) (6.5 + r)
36 = 42.25 – r²
r² = 6.25
r = 2.5 cm

S9. Ans.(c)
Sol.
Amount = P[1+R/100]^2
=12500 [ 1+20/100]^2=Rs 18000

Q10. Ans.(a)
Sol.
(4/5+9/2)÷(1/2-1/5)
=((8+45)/10)÷((5-2)/10)
=53/10×10/3=53/3