Q1. 8 men working for 9 hours a day complete a piece of work in 20 days. In how many days can 7 men working for 10 hours a day complete the same piece of work?
(a) 21 days
(b) 20 3/5 days
(c) 20 1/2 days
(d) None of these
Sol.
M1×H1×D1=M2×H2×D2
8×9×20=7×10×D2
⇒ D2=(8 × 9 × 20)/(7 × 10)=20 4/7 days
Q2. Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are opened simultaneously, after how much time should B be closed so that the tank is full in 18 minutes?
(a) 2 min
(b) 4 min
(c) 6 min
(d) 8 min
Sol.
Let B be closed after x minutes.
Then, part filled by (A + B) in x min + part filled by A in (18 – x) min = 1
∴x(1/24+1/32)+(18-x)×1/24=1
Or, 7x/96+(18 – x)/24=1 or 7x+4(18-x) =96
Or, 3x =24 ∴ x = 8.
So, B should be closed after 8 min.
Q3. Three pipes A, B and C can fill a cistern in 6 hrs. After working together for 2 hours, C is closed and A and B fill the cistern in 8 hrs. Then find the time in which the cistern can be filled by pipe C.
(a) 6 hrs
(b) 12 hrs
(c) 14 hrs
(d) 20 hrs
Sol.
A + B + C can fill in 1 hr
=1/6 of cistern.
A + B + C can fill in 2 hrs
=2/6=1/3 of cistern.
Unfilled part =(1-1/3)i.e. 2/3 is filled in A + B in 8 hrs.
∴ (A + B) can fill the cistern in (8 × 3)/2 = 12 hrs.
∴ C i.e. (A + B + C) – (A + B) can fill in 1 hr. =1/6-1/12=1/12 part
∴ C Pipe Full the cistern in 12 hrs.
Q4. A train takes 5 seconds to pass an electric pole. If the length of the train is 120 metres, the time taken by it to cross a railway platform 180 metres long (in seconds)-
(a) 10.5 seconds
(b) 12.5 seconds
(c) 14.5 seconds
(d) Can’t be determined
Sol.
Speed of train =120/5= 24 m/s
∴ Time taken by the trains to cross a railway the platform
=(120 + 180)/24=12.5 seconds
Q5. A train overtakes two person who are walking in the same directions in which the train is going, at the rate of 2 kmph and 4 kmph respectively and passed them completely in 9 and 10 seconds respectively. The length of the train is
(a) 50 m
(b) 60 m
(c) 65 m
(d) 70 m
Sol.
Let the speed of train=x km/hr.
And length of train= d metre
According to the questions,
Relative speed with first man = (x-2) km/hr.
Relative speed with second man=(x-4) km/hr
Now, d/(x-2)=9×5/18 …..(i)
d/(x-4)=10×5/18……(ii)
By solving these two equations
x=22km/h
and d= 50 metre
Q6. How many seconds will a train 60 m in length, travelling at the rate of 42 km an hour, take to pass another train 80 m long, proceeding in the same direction at the rate of 30 km an hour?
(a) 41.2 seconds
(b) 42 seconds
(c) 42.3 seconds
(d) 42.5 seconds
(e) None of these
Sol.
Relative speed = 42 – 30
= 12 km/hr.
=12×5/18=10/3 m/s
Time
=(Total length of both the train )/(Relative speed )=(80 + 60)/(10/3)
=(140 × 3)/10 = 42 seconds
Q7. A car travels the first one-third of a certain distance with a speed of 10 km/hr. the next one-third distance with a speed of 20 km/hr. and the last one-third distance with a speed 60 km/hr. Then the average speed of the car for the whole Journey is-
(a) 18 km/hr.
(b) 24 km/hr.
(c) 30 km/hr.
(d) 36 km/hr.
Sol.
Let the distance covered by car be 3x km
Average speed =
(total distance travelled )/(total time taken )
=3x/(x/10 + x/20 + x/60)
=3x/((6x + 3x + x)/60)
=(3x × 60)/10x = 18 km/hr.
Q8. The ratio of income in two consecutive years is 2 : 3 respectively. The ratio of their expenditure is 5 : 9. Income of second year is Rs. 45000 and Expenditure of 1st year is Rs. 25000. Savings in both years together is:
(a) Rs. 4000
(b) Rs. 5000
(c) Rs. 6000
(d) Rs. 7000
Sol.
Income of 2nd year = 3x = 45000
x = 45000 ÷ 3 = 15000
Income of 1st year = 2x
= 2 × 15000 = 30000
Expenditure of 1st year = 5y = 25000
y = 25000 ÷ 5 = 5000
Expenditure of 2nd year = 9y = 45000
Total savings = Total income – total expenditure
= 45000 + 30000 – (25000 + 45000)
= 75000 – 70000 = 5000
Q9. Three friends divide an amount of Rs. 45,000, such that one of them takes 1/4th of the total amount and the other two divide the rest of the amount equally among themselves. What is the ratio of amount taken by one of the friends to the total amount taken by the other two friends together?
(a) 1 : 3
(b) 1 : 2
(c) 3 : 2
(d) 2 : 3
Sol.
Share of two friends =1-1/4=3/4
Required ratio =1/4 ∶3/4=1:3
Q10. A mixture contains milk and water in the ratio of 4 : 3 respectively. If 6 litres of water is added to this mixture, the respective ratio of milk and water becomes 8 : 7. What is the quantity of milk in the original mixture?
(a) 96 litres
(b) 36 litres
(c) 84 litres
(d) 48 litres
Sol.
Let the quantities of milk and water in the original mixture be 4x litres and 3x litres respectively
∴ 4x/(3x + 6)=8/7
⇒ 28x = 24x + 48
⇒ 28x – 24x = 48
⇒ 4x = 48
∴ Amt. of milk = 48 litres.
