11
Q:

# There are three boats B1, B2 and B3 working together they carry 60 people in each trip. One day an early morning B1 carried 50 people in few trips alone. When it stopped carrying the passengers B2 and B3 started carrying the people together. It took a total of 10 trips to carry 300 people by B1, B2 and B3. It is known that each day on an average 300 people cross the river using only one of the 3 boats B1, B2 and B3. How many trips it would take to B1, to carry 150 passengers alone?

 A) 15 B) 30 C) 25 D) 10

Explanation:

Combined efficiency of all the three boats = 60 passenger/trip

Now, consider option(a)

15 trips and 150 passengers means efficiency  of B1 = $\inline \fn_jvn 10\frac{p}{t}$

which means in carrying 50 passengers B1 must has taken 5 trips. So the rest trips equal to 5 (10-5 = 5) in which B2 and B3 together carried remaining 250 (300 - 50 = 250) Passengers.

Therefore the efficiency of B2 and B3 = $\inline \fn_jvn \frac{250}{5}=50\frac{p}{t}$

Since, the combined efficiency of B1, B2 and B3 is 60. Which is same as given in the first statement hence option(a) is correct.

Q:

 A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left?

 A) 8/15 B) 7/9 C) 6/13 D) 4/11

Explanation:
 P's 1 day's work = 1 15
 Q's 1 day's work = 1 20
 (P + Q)'s 1 day's work = 1 + 1 = 7 15 20 60
 (P + Q)'s 4 day's work = 7 x 4 = 7 60 15
 Therefore, Remaining work = 1 - 7 = 8 . 15 15

2 51
Q:

X can complete the work in 10 days, Y can do the same in 15 days. If they are hired for 5 days to do the work together, what is the work that left unfinished?

 A) 1/3 B) 2/3 C) 1/6 D) 5/6

Explanation:

Given X can do in 10 days

=> 1 day work of X = 1/10

Y can do in 15 days

=> 1 day work of Y = 1/15

1day work of (X + Y) = 1/10 + 1/15 = 1/6

Given they are hired for 5 days

=> 5 days work of (X + Y) = 5 x 1/6 = 5/6

Therefore, Unfinished work = 1 - 5/6 = 1/6

8 256
Q:

If 10 men take 15 days to complete a work. In how many days will 37 men complete the work?

 A) 3 days B) 4 days C) 5 days D) 6 days

Explanation:

Given 10 men take 15 days to complete a work

=> Total mandays = 15 x 10 = 150

Let the work be 150 mandays.

=> Now 37 men can do 150 mandays in 150/37 =~ 4 days

12 199
Q:

Three taps P, Q and R can fill a tank in 12 hrs, 15 hrs and 20 hrs respectively. If P is open all the time and Q and R are open for one hour each alternately, starting with Q, then the tank will be full in how many hours ?

 A) 9 hrs B) 7 hrs C) 13 hrs D) 11 hrs

Explanation:

Given,

P can fill in 12 hrs

Q can fill in 15 hrs

R can fill in 20 hrs

=> Volume of tank = LCM of 12, 15, 20 = 60 lit

=> P alone can fill the tank in 60/12 = 5 hrs

=> Q alone can fill the tank in 60/15 = 4 hrs

=> R alone can fill the tank in 60/20 = 3 hrs

Tank can be filled in the way that

(P+Q) + (P+R) + (P+Q) + (P+R) + ....

=> Tank filled in 2 hrs = (5+4) + (5+3) = 9 + 8 = 17 lit

=> In 6 hrs = 17 x 6/2 = 51 lit

=> In 7th hr = 51 + (5+4) = 51 + 9 = 60 lit

=> So, total tank will be filled in 7 hrs.

7 369
Q:

P and Q can complete a job in 24 days working together. P alone can complete it in 32 days. Both of them worked together for 8 days and then P left. The number of days Q will take to complete the remaining work is ?

 A) 56 days B) 54 days C) 60 days D) 64 days

Explanation:

(P+Q)'s 1 day work = 1/24

P's 1 day work = 1/32

=> Q's 1 day work = 1/24 - 1/32 = 1/96

Work done by (P+Q) in 8 days = 8/24 = 1/3

Remainining work = 1 - 1/3 = 2/3

Time taken by Q to complete the remaining work = 2/3 x 96 = 64 days.