 # Maths Questions for CTET,KVS Exam : 17 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. 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

S9. Ans.(c)
Sol. Amount = P (1+R/100)^T
=2800(1+10/100)^(3/2)
=2800(1+1/10)(1+1/(2×10))
=2800×11/10×21/20=Rs.3234
∴ Compound interest
= Rs. (3234 – 2800) = Rs. 434

S10. Ans.(d)
Sol. 3 men ≡ 8 boys
∴ 2 men + 6 boys
=(2+9/4)men=17/4 men
∴M_1 D_1=17/4×D_2
⇒3×17=17/4×D_2
⇒D_2=(3×4×17)/17=12 days