A contractor undertook a project to complete it in 20 days which needed 5 workers to work continuously for all the days estimated. But before the start of the work the client wanted to complete it earlier than the scheduled time, so the contractor calculated that he needed to increase 5 additional men every 2 days to complete the work in the time the client wanted it: If the work was further increased by 50% but the contractor continues to increase the 5 workers o every 2 days then how many more days are required over the initial time specified by the client.

Correct Answer: 2 hrs 24 min

Explanation:

Given that three athletes can complete one round around a circular field in 16, 24 and 36 min respectively.

Now, required time after which they met for the first time = LCM of (16, 24 & 36) min

Now, LCM of 16, 24, 36 = 144 minutes = 2 hrs 24 min.

Correct Answer: 56

Explanation:

Days remaining 124 ? 64 = 60 days

Remaining work = 1 - 2/3 = 1/3

Let men required men for working remaining days be 'm'

So men required = (120 x 64)/2 = (m x 60)/1 => m = 64

Men discharge = 120 ? 64 = 56 men.

Correct Answer: 12 days

Explanation:

9M + 12B ----- 12 days ...........(1)
12M + 12B ------- 10 days........(2)
10M + 10B -------?
108M + 144B = 120M +120B
24B = 12M => 1M = 2B............(3)

From (1) & (3)

18B + 12B = 30B ---- 12 days

20B + 10B = 30B -----? => 12 days.

Correct Answer: 13 days

Explanation:

Ratio of times taken by P & Q = 100 : 130 = 10:13

 

Let Q takes x days to do the work

Then, 10:13 :: 23:x

=> x = 23x13/10

=> x = 299/10

 

P's 1 day's work = 1/23

Q's 1 day's work = 10/299

(P+Q)'s 1 day's work = (1/23 + 10/299) = 23/299 = 1/13

Hence, P & Q together can complete the work in 13 days.

Correct Answer: 3:09 pm

Explanation:

Efficiency of P= 100/20= 5% per hour  

Efficiency of Q= 100/25= 4% per hour  

Efficiency of R= 100/40= 2.5% per hour  

Efficiency of S=100/50= 2% per hour 

 

Cistern filled till 10 am by P, Q and R  

Till 10 . 00 am Pipe P filled 20 % Till 10 . 00 am Pipe Q filled 8 % Till 10 . 00 am Pipe R filled 2 . 5 % 30 . 5 %  

Thus, at 10 am pipe P,Q and R filled 30.5% of the cistern. 

Rest of cistern to be filled = 100 - 30.5 = 69.5%  

 

Now, the time taken by P,Q,R and S together to fill the remaining capacity of the cistern 

= 69.5 / (5+4+2.5+2) = 5 Hours and 9 minutes(approx).

Therefore, total time =4 hrs + 5hrs 9 mins = 9 hrs and 9 mins

 

It means cistern will be filled up at 3:09 pm

Correct Answer: 13 days

Explanation:

Now, Total work = LCM(16, 8) = 48

A's one day work = + 48/16 = + 3

B's one day work = - 48/8 = -6

Given A worked for 5 days to build the wall => 5 days work = 5 x 3 = + 15

2days B joined with A in working = 2(3 - 6) = - 6

Remaining Work of building wall = 48 - (15 - 6) = 39

Now this remaining work will be done by A in = 39/3 = 13 days.

Correct Answer: 1232

Explanation:

From the given data,

                       Day      Capacity

A              ->    12          2

B              ->     8           3             2

A + B + C ->     2            12

=> Capacity of C = 12 - 5 = 7

Ratio of capacity of A : B : C = 2 : 3 : 7

Difference of wages of C & B = 4/5 x 1540

= 4 x 308 = Rs. 1232

Correct Answer: 15 days

Explanation:

We know that Time is inversely proportional to Efficiency

Here given time ratio of P & P+Q as

P/P+Q = 150/100 = 3:2

Efficiency ratio of P & Q as

P:Q = 2:1 ....(1)

Given Efficiency ratio of Q & R as

Q:R = 3:1 ....(2)

From (1) & (2), we get

P:Q:R = 6:3:1

P alone can finish the work in

22.5(3+1)/6 = 15 days.

Correct Answer: 25 days

Explanation:

Let Q complete that work in 'L' days

 

=>  1 L + 1 L - 5 = 9 100

 

=>  9 L 2 - 245 L + 500 = 0

 

L = 25 days.

Correct Answer: 1/6

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