Q1. A motor cyclist goes from Mumbai to Pune, a distance of 192 kms, at an average speed of 32 kmph. Another man starts from Mumbai by car, 2 1/2 hours after the first and reaches Pune half an hour earlier. What is the rate of the speed of the motorcycle and the car?
(a) 1 : 2
(b) 1 : 3
(c) 10 : 27
(d) 5 : 4
Sol.
Speed of the first man = 32 km/hr
Time taken = 192 ÷ 32 = 6 hr
Second man covers 192 km in 3 hr
∴ Speed of the second man
= 192 ÷ 3 = 64 km/hr
Ratio = 32 : 64 or 1 : 2
Q2. A person sets to cover a distance of 12 km in 45 minutes. If he covers 3/4 of the distance in 2/3rd time, what should his speed to cover the remaining distance in the remaining time?
(a) 16 km/hr
(b) 8 km/hr
(c) 12 km/hr
(d) 14 km/hr
Sol.
Distance already covered =3/4×12 = 9 km
Time spent =2/3×45 min = 30 min
Distance left =(12-9)km = 3 km
Time left =(45-30)min = 15 min
∴ Required speed =3/(15/60) km/hr
= 12 km/hr
Q3. Length of a goods train is 287 m and it passes a bridge in 38 sec running at the speed of 90 km/h. What is the length of the bridge?
(a) 665 m
(b) 663 m
(c) 680 m
(d) 580 m
Sol.
Total distance covered =90×5/18×38= 950 m
So, length of bridge = 950 – 287 = 663
Q4. A train running at certain speed crosses a stationary engine in 20 seconds. To find out the speed of the train, which of the following information is necessary?
(a) Only the length of the train
(b) Only the length of the engine
(c) Either the length of the train or the length of the engine
(d) Both the length of the train and the length of the engine
Sol.
Since the sum of the length of the train and the engine is needed, so both the lengths must be known.
Q5. Sixteen men can complete a work in twelve days. Twenty-four children can complete the same work in eighteen days. Twelve men and eight children started working and after eight days three more children joined them. How many days will they now take to complete the remaining work?
(a) 2 days
(b) 4 days
(c) 6 days
(d) 8 days
Sol.
1 man’s 1 day’s work =1/192; 1 child’s 1 day’s work =1/432
Work done in 8 days =8(12/192+8/432)=8(1/16+1/54)=35/54
Remaining work =(1-35/54)=19/54
(12 men + 11 children)’s 1 day’s work =(12/192+11/432)=19/216
Now, 19/216 work is done by them in 1 day.
∴ 19/54 work will be done by them in (216/19×19/54)= 4 days.
Q6. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
(a) 4 days
(b) 5 days
(c) 6 days
(d) 7 days
Sol.
Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y
Then, 6x + 8y =1/10 and 26x + 48y =1/2
Solving these two equations, we get: x=1/100 and y=1/200
(15 men + 20 boys)’s 1 day’s work =(15/100+20/200)=1/4
∴ 15 men and 20 boys can do the work in 4 days.
Q7. A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m^3. The emptying capacity of the tank is 10 m^3 per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?
(a) 50 m^3/min
(b) 60 m^3/min
(c) 72 m^3/min
(d) None of these
Sol.
Let, the filling capacity of the pump be x m^3/min
Then, emptying capacity of the pump =(x+10) m^3/min
So, 2400/x-2400/((x+10) )=8⇒x^2+10x-3000=0
⇒(x-50)(x+60)=0⇒x=50 [Neglecting the -ve value of x)]
Q8. A leak in the bottom of a tank can empty the full tank in 8 hours. An inlet pipe fills water at the rate of 6 litres a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 12 hours. How many litres does the cistern hold?
(a) 7580
(b) 7960
(c) 8290
(d) 8640
Sol.
Work done by the inlet in 1 hour =(1/8-1/12)=1/24
Work done by the inlet in 1 min =(1/24×1/60)=1/1440
∴ Volume of 1/1440 part = 6 litres
∴ Volume of whole = (1440 × 6) litres = 8640 litres.
Q9. A car travels from P to Q at a constant speed. If its sped were increased by 10 km/hr, it would have taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. What is the distance between the two cities?
(a) 420 km
(b) 540 km
(c) 600 km
(d) 650 km
Sol.
Let distance = x km and usual rate = y kmph
Then,
x/y-x/(y+10)=1 or y (y + 10) = 10x …(i)
And, x/y-x/(y+20)=7/4 or y(y + 20) =80x/7 …(ii)
On dividing (i) by (ii), we get y = 60
Substituting y = 60 in (i) we, get : x = 420 km
Q10. Three persons are walking from a place A to another place B. Their speeds are in the ratio of 4 : 3 : 5. The time ratio to reach B by these persons will be:
(a) 4 : 3 : 5
(b) 5 : 3 : 4
(c) 15 : 9 : 20
(d) 15 : 20 : 12
Sol.
Ratio of speeds = 4 : 3 : 5
∴ Ratio of times taken =1/4:1/3:1/5 = 15 : 20 :12
