Maths Questions for CTET,KVS Exam : 17 november 2018

Q5. In the given figure, PQRS is a trapezium in which PM || SN, NR = 9 cm, PS = 12 cm, QM = NR and NR= SN. What is the area (in cm²) of trapezium?

(a) 170
(b) 182
(c) 189
(d) 191

Q6. In the given figure, PQR is an equilateral triangle and PS is the angle bisector of ∠P. What is the value of RT:RQ?

(a) 1 : 2
(b) 1 : 1
(c) 2 : 1
(d) 1 : 3

Q7. Two chords of length 20 cm and 24 cm are drawn perpendicular to each other in a circle of radius is 15 cm. What is the distance between the points of intersection of these chords (in cm) from the center of the circle?
(a) √114
(b) √182
(c) √206
(d) √218

Q8. In the given figure, QRTS is a cyclic quadrilateral. If PT = 5 cm, SQ = 4 cm, PS = 6 cm and ∠PQR= 63°, then what is the value (in cm) of TR?

(a) 3
(b) 7
(c) 9
(d) 15

Q9. The compound interest on Rs. 2800 for 18 months at 10 per cent annum is: 
(a) Rs. 441.35
(b) Rs. 436.75
(c) Rs. 434. 00
(d) Rs. 420.00

Q10. Three men or eight boys can do a piece of work in 17 days. In how many days can 2 men and 6 boys complete the same piece of work? 
(a) 20 days
(b) 10 days
(c) 15 days
(d) 12 days

Solutions

S1. Ans.(a)
Sol. According to the question, x^2/yz+y^2/zx+z^2/xy=3
⇒(x^3+y^3+z^3)/xyz=3
⇒ x³ + y³ + z³ = 3xyz
∴ (x + y + z) = 0
Hence, (x + y + z)³ = 0

S2. Ans.(d)
Sol. According to the question,
x^(1/4 )+1/x^(1/4) =2
∴ x = 1
Hence,
x^81+1/x^81 =1+1=2

S3. Ans.(b)
Sol. We know
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
Now,
According to the question,
a² + ab + ca = 45…(i)
ab + b² + bc = 75…(ii)
ca + cb + c² = 105 …(iii)
⇒ Adding (i), (ii) & (iii), we get
a² + b² + c² + 2(ab + bc + ca) = 225
⇒ (a + b + c)² = 225
⇒ (a + b + c) = 15
∴a=45/15=3,b=75/15=5,c=105/15=7
∴ a² + b² + c² = 3² + 5² + 7² = 83

S4. Ans.(c)
Sol. According to the question,
x^2+1/x^2 =1
⇒x+1/x=√3
⇒ cubing both sides
x^3+1/x^3 =(√3)^3-3√3=0
⇒x^6=-1⇒x^6+1=0…(i)
Then,
x^48+x^42+x^36+x^30+x^24+x^18+x^12+x^6+1
=x^42 (x^6+1)+x^30 (x^6+1)+x^18 (x^6+1)+x^6
(x^6+1)+1 ⇒ = 1

S5. Ans.(c)
Sol. Area of trapezium PQRS = 1/2 × (Sum of parallel sides) × height = 1/2 × (12 + 30) × 9 = 189

S6. Ans.(b)
Sol.
Here, PS will be diameter of the circle
∴ ∠PQS = 90° (angle in a semicircle is 90°)
∴ ∠PSQ = 180° – (90° + 30°) = 60°
Now in ∆RTQ
∠RTQ = ∠RQT = 30°
∴ RT = RQ
i.e. RT : RQ = 1 : 1

S8. Ans.(b)
Sol.
We know
PS × PQ = PT × PR
⇒ 6 × (6 + 4) = 5 × (TR + 5)
⇒ TR + 5 = (6 × 10)/5
⇒ TR = 7 cm