8
Q:

# Adam and Smith are working on a project. Adam takes 6 hrs to type 36 pages on a computer, while Smith takes 5 hrs to type 40 pages. How much time will they take, working together on two different computers to type a project of 120 pages?

 A) 8 hrs 45 min B) 8 hrs 42 min C) 8 hrs D) 8 hrs 34 min

Answer:   D) 8 hrs 34 min

Explanation:

Number of pages typed by Adam in 1 hour = $\inline \fn_jvn \small \frac{36}{6}$ = 6
Number of pages typed by Smith in 1 hour = $\inline \fn_jvn \small \frac{40}{5}$ = 8
Number of pages typed by both in 1 hour = (6 + 8) = 14
Time taken by both to type 110 pages = (120 * 1/14) = 8 $\inline \fn_jvn \small \frac{4}{7}$ = 8 hrs 34 min.

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.

8 975
Q:

A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, the rest of the work is finished by C in two more days. If they get Rs. 3000 as wages for the whole work, what are the daily wages of A, B and C respectively (in Rs):

 A) 200, 250, 300 B) 300, 200, 250 C) 200, 300, 400 D) None of these

Explanation:

A's 5 days work = 50%

B's 5 days work = 33.33%

C's 2 days work = 16.66%          [100- (50+33.33)]

Ratio of contribution of work of A, B and C = $\inline \fn_jvn 50:33\frac{1}{3}:16\frac{2}{3}$

= 3 : 2 : 1

A's total share = Rs. 1500

B's total share = Rs. 1000

C's total share = Rs. 500

A's one day's earning = Rs.300

B's one day's earning = Rs.200

C's one day's earning = Rs.250

32 9665
Q:

Amit can do a piece of work in 45 days, but Bharath can do the same work in 5 days less, than Amit, when  working alone. Amit and Bharath both started the work together but Bharath  left after some days and Amit finished the remaining work in 56 days with half of his efficiency but he did the work with Bharath with his complete efficiency. For how many days they had worked together?

 A) 6 B) 8 C) 9 D) 12

Explanation:

$\inline \fn_jvn \begin{matrix} & Amit& Bharath\\ No.of\: days& 45& 40\\ Efficiency& 2.22\: percent\left ( =\frac{1}{45} \right ) &2.5 \: percent\left ( =\frac{1}{40} \right ) \end{matrix}$

Amit did the work in 56 days = $\inline \fn_jvn 56\times \frac{1}{45\times 2}=\frac{28}{45}$

$\inline \fn_jvn \therefore$ Rest work $\inline \fn_jvn \left ( \frac{17}{45} \right )$ was done by Amit and Bharath = $\inline \fn_jvn \frac{17/45}{17/360}$ = 8 days

( since Amit and Bharath  do the work in one day = $\inline \fn_jvn \frac{1}{45}+\frac{1}{40}=\frac{17}{360}$)

3 950
Q:

Two pipes A and B can fill a cistern in 4 minutes and 6 minutes respectively . If these pipes are turned on alternately for 1 minute each how long will it take to the cistern to fill?

As the pipes are operating alternatively, thus their 2 minutes job is =$\inline \frac{1}{4}+\frac{1}{6}=\frac{5}{12}$

In the next 2 minutes the pipes can fill another $\inline \frac{5}{12}$ part of cistern.

$\inline \therefore$ In 4 minutes the two pipes which are operating alternatively will fill $\inline \frac{5}{12}+\frac{5}{12}=\frac{5}{6}$

Remaining part = $\inline 1-\frac{5}{6}=\frac{1}{6}$

Pipe A can fill $\inline \frac{1}{4}$ of the cistern in 1 minute

Pipe A can fill $\inline \frac{1}{6}$ of the cistern in =$\inline 4\times \frac{1}{6}=\frac{2}{3}$  min

$\inline \therefore$Total time taken to fill the Cistern

$\inline 4+\frac{2}{3}=4\frac{2}{3}$ minutes

874
Q:

A contractor undertakes to complete a work in 130 days. He employs 150 men for 25 days and they complete 1/4 of the work . He then reduces the number of men to 100, who work for 60 days, after which there are 10 days holidays.How many men must be employed for the remaining period to finish the work?

150 men in 25 days do = $\inline \frac{1}{4}$ work

1 man in 1 day does = $\inline \frac{1}{4}\times \frac{1}{25}\times \frac{1}{150}$ work

$\inline \therefore$ 100 men in 60 days do = $\inline \frac{1}{4}\times \frac{1}{25}\times \frac{1}{150}\times 100\times 60=\frac{2}{5}$ work

Total work done =$\inline \frac{1}{4}+\frac{2}{5}=\frac{5+8}{20}=\frac{13}{20}$

$\inline \therefore$ Remaining work =$\inline 1-\frac{13}{20}=\frac{7}{20}$

Remaining time = 130 - (25+60+10) = 35 days

$\inline \therefore\: \frac{1}{4}$work is done in 25 days  by 150 men

$\inline \therefore\: \frac{7}{20}$ work is done in 35 days by $\inline \frac{150\times 4\times 25\times 7}{35\times 20}=150$ men