9
Q:

# After working for 8 days, Arun finds that only $\inline \fn_jvn \small \frac{1}{3}$ rd of the work has been done. He employs Akhil who is 60% as efficient as Arun. How many days more would Akhil take to complete the work?

 A) 24.5 days B) 26.6 days C) 25 days D) 20 days

Explanation:

Arun has completed $\inline \fn_jvn \small \frac{1}{3}$ rd of the work in 8 days
Then he can complete the total work in
$\inline \fn_jvn \small \frac{1}{3}$ ---- 8
1 ---- ?
= 24 days
But given Akhil is only 60% as efficient as Arun
Akhil = $\inline \fn_jvn \small \frac{1}{24}\times \frac{60}{100}=\frac{1}{40}$
Akhil can complete the total work in 40 days
Now, remaining 2/3rd of work can be completed in
1 ------   40
$\inline \fn_jvn \small \frac{2}{3}$  ------   ?   $\fn_jvn&space;\small&space;\Rightarrow$ 26.66 days.

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

1 15
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

8 248
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

12 187
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.

7 366
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.