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# Maths Quiz for 2016-17 Exams

Q1. Supposing that telegraph poles on a railway track are 50 metres apart. How many pole will be passed by a train in 2 hours. If the speed of the train is 45 km/hr
(a) 1801
(b) 1800
(c) 1601
(d) 1600
Sol.
Distance travelled by train in 2 hours = 45 × 2 = 90 kms = 90000 m
No. of pole passed = 90000/50+1
= 1800 + 1
= 1801 poles

Q2. A train 150 metre long, passes a pole in 15 seconds and another train of the same length travelling in the opposite direction in 12 seconds. The speed of the second train is
(a) 45 km/hr
(b) 48 km/hr
(c) 52 km/hr
(d) 54 km/hr
Sol.
When distance is same, speed is indirectly proportional to the time
Speed of the first train = 150/15 = 10 m/s
Speed of the second train = 10/12×15
= 150/12 m/s
=150/12×18/5
= 45 km/hr

Q3. A train 100 metres long moving at a speed of 50 kmph crossed a train 120 metres long coming from opposite direction in 6 seconds. The speed of the second train is
(a) 82 km/hr
(b) 84 km/hr
(c) 86 km/hr
(d) Can’t be determined
Sol.
Total speed = (100 + 120)/6
= 220/6 m/sec.
= 220/6×18/5 km/hr.
= 132 km/hr.
Speed of second train = 132 – 50 = 82 km/hr.

Q4. A boat covers 12 km. upstream and 18 km downstream in 3 hours while it covers 36 km upstream and 24 km downstream in 6  hours, what is the velocity of the stream?
(a) 1.5 km/hr
(b) 1 km/hr
(c) 2 km/hr
(d) 2.5 km/hr
Sol.
12/y+18/x = 3   …(i)
36/y+24/x=13/2    …(ii)
To equate (i) & (ii)
Multiply (i) by 3 first
36/y+54/x = 9   …(iii)
36/y+24/x=13/2    …(iv)
Subtracting (iv) from (iii)
30/x=9-13/2=5/2
⇒ x = 12km/hr.
Now, put value of x in equation (i), we get y = 8 km./hr.
∴ Velocity of stream(Sc)=(12 – 8)/2 = 2 km./hr.

Q5. In a flight of 600 kms, an aircraft was slowed down due to bad weather. Its average speed for the trip is reduced by 200 kms/hr. and the time of flight increased by 30 min. The duration of the flight is –
(a) 1 hr.
(b) 2 hrs.
(c) 3 hrs.
(d) 4 hrs.
Sol.
Let the duration of flight be t hours.
S = D/T
S1-S2 = 200 kms/hr.
600/t-600/(t+1/2) = 200
600/t-(2×600)/(2t+1)=200
(2t + 1) 600 – t × 1200 = 200t (2t + 1)
3(2t + 1) – 6t = t (2t + 1)
6t + 3 – 6t = 2t^2 + t
2t^2 + t – 3 = 0
2t^2 + 3t – 2t – 3 = 0
t (2t + 3) – 1(2t + 3)
(2t + 3) (t – 1) = 0
t = 1 hour.

Q6. Without any stoppage a person travels a certain distance at an average speed of 42 km/hr. and with stoppage he covers the same distance at an average speed of 28 km/hr. How many minutes per hour does he stop?
(a) 20 min.
(b) 30 min.
(c) 21 min.
(d) 23 min.
Sol.
Here x = 42 and y = 28
∴ Stoppage time/hr. = (x  –  y)/x
⇒(42 – 28)/42⇒1/3 hr. ⇒ 20 min.

Q7. The speed of a car increases by 2 km/hr after every hour. If the distance travelled in the first one hour was 35 km, the total distance travelled in 12 hours was-
(a) 456 km
(b) 482 km
(c) 552 km
(d) 556 km
Sol.
Distance travelled in first hour = 35 km
Distance travelled in second hour = 37 km
common difference = 2 km
So, distance travelled in 12 hour = 12/2 [2 × 35 + (12 – 1) × 2]
= 12/2 × (70 + 22)
= 12 × 46
= 552 km

Q8. Walking at 6/7th of his usual speed a man is 12 minutes late. The usual time taken by him to cover that distance is-
(a) 1 hr.
(b) 1 hr. 12 min.
(c) 1 hr. 15 min.
(d) 1 hr. 20 min.
Sol.
Reduced in speed = (1-6/7)/1=1/7
So, increase in time 1/(7 – 1)=1/6 time=12
total time = 6 × 12
= 1 hr. 12 min.

Q9. A starts from a place P to go to a place Q. At the same time B starts from Q for P. If after meeting each other A and B took 4 and 9 hours more respectively to reach their destinations, the ratio of their speeds is
(a) 3:2
(b) 5:2
(c) 9:4
(d) 9:13
Sol.
(Speed of A)/(Speed of B)=√(9/4  )
(Speed of A)/(Speed of B)=3/2
Ratio=3:2

Q10. I have to be at a certain place at a certain time and I find that I shall be 15 minutes too late, if I walks at 4 km and hour; and 10 minutes too soon, if I walks at 6 km an hour. How far have I to walk?
(a) 3 km
(b) 5 km
(c) 6 km
(d) 8 km
Sol.
The difference = 15 – (–10) = 25 m
Increase in speed = 4 km to 6 km
= 2/4=1/2
So,
1/(2 + 1) × time = 25 minute
Time = 75 minute
Total walk = 4 × 75/60
= 5 km.