answer to 13

Q13. Joe and Fred are linemen. Joe can string 10 miles of line in 3 days while together they can string it in 1 day. How long would it take Fred alone to string the same line?

ANS:

Joe in 1 day : 10/3

Together in 1 day: 10

So Fred in 1 Day : 10 - 10/3 = 20/3

To string 10 miles Fred alone 'll take 10/(20/3) = 1.5 Days

Q12 answer

Q12. If a certain number of workmen can do a piece of work in 25 days, in what time will another set of an equal number of men do a piece of work twice as great, supposing that 2 of the first set can do as much work in an hour as 3 of the second set can do in an hour?
(a)60 days
(b)75 days
(c)90 days
(d)105 days

Ans:

2 women = 3 men

=> 1 man = 2/3 of 1 woman

if 1 woman take 25 days then 1 man would take 3/2 x 25 days

and to complete the work twice the man would take 3/2 x 25 x 2 days

= 75 days

Q9 Answer

Q9. A pipe can fill a tank in X hours and another can empty it in Y hours. If the tank is 1/3rd full then the number of hours in which they will together fill it in is

Ans:

pipe 1 can fill a tank in X hours

pipe 1 does 1/X job in 1 hour

pipe 2 can fill a tank in Y hours

pipe 2 does 1/Y job in 1 hour

Total job done in 1 hr = 1/X + 1/Y

The tank is 1/3rd full, So to full another 2/3rd part time taken 'll be

    2/3     

(1/x + 1/y)

=2xy / 3(x+y)

1) Let in 1 min. the amount

1) Let in 1 min. the amount filled is "x" m^3.
So, in 1 min. the amount emptied is (X+10) m^3.

The  time required to fill the tank of 3600 capacity is 3600/x
& to empty of the same capacity is 3600/(x+10)

The difference b/w the two time is 12 mins.
So, 3600/(x+10)-3600/x =12.

Solving,will give value as 50 m^3 /mins.

Ans:(a)

Ans: (3) 12.5 km   (4) 30

Ans: (3) 12.5 km

  (4) 30 men.

10) 4/5 th of a cistern can

10)

4/5 th of a cistern can be fill in 1 min.
So,1/5 th of the cistern will be filled in 1/4 min.
=14x60=15 seconds.

so, Ans: (b)

9)The tank is 1/3 rd full so

9)
The tank is 1/3 rd full so we just have to full the rst of 2/3 rd tank.
Time taken to fill the tank is X hrs . and to empty is Y  hrs.
Let the time required to fill the rest of th tank be "h" hrs. so,
h(1/X-1/Y)=2/3.
so,h=2XY/3(Y-X). 
 

Let A left after x

Let A left after x days.
So,
x/21+(x+9)/24=1
Solving, x=7
So,Ans: (b)

(b)

it is asking for emptying capacity= X+10=60

Ans b Bcoz it asked the

Ans b Bcoz it asked the Empting capacity

My answer to Q5:

Q5. A can do a piece of work in 20 days. He works at it for 5 days and then B finishes it in 10 more days. In how many days will A and B together finish the work?

Sol:
Let the total work be 1 unit.
Work done by A in 5 days=5/20.
Let the time taken by B to do the work alone be "a" days.
So,work done by B in 10 days =10/b.
According to question,
5/20+10/b=1
Solving will give the value of b=40/3.
Let the time taken to complete the work if both A & B work together=y.
So, y(1/20+3/40)=1
Solving will give the value of y=8 days.
Ans: 8 days.
 

My Solution to Q4:

Q4. If 8 men, working 9 hours a day can build as wall 18 metres long, 2 metres broad and 12 metres high in 10 days, how many men will be required to build a 32 metres long, 3 metres broad and 9 metres high, by working 6 hours a day, in 8 days?

Sol: Let the number of men required to build the wall 32m long, 3 metres broad and 9 metres high=a.

So, 8x9x10/(18x2x12)=6x8xa/(32x3x9)

Solving, a= 10.
So,number of men required =10.
 

Solution to Q8:

Q8.A and B can do a piece of work in 21 and 24 days respectively. They started the work together and after some days A leaves the work and B completes the remaining work in 9 days. After how many days did A leave?
(a)5
(b)7
(c)8
(d)6

Let A leaves after a days.
So, a/21+(a+9)/24=1

Solving will give a=7.
Answer:(b) 7 days.

Q10. If (4/5)th of a cistern

Q10. If (4/5)th of a cistern is filled in 1 minute, how much more time will be required to fill the rest of it?
(a)20 seconds
(b)15 seconds
(c)12 seconds
(d) 10 seconds

Sol: 4/5 th of the cistern is filled in 1 minute.
So,the cistern will be full in 5/4 minutes.

4/5 of the cistern is full.So, 1/5 of the cistern will be full in 1/5x5/4=1/4 minute.
1/4x60=15 seconds.

Answer: (b) 15 seconds

Solution to Q11:

Q11. It takes bill 9 hours to paint a house and jim 7 hours to do the same job. how long will it take both people working together?

Sol:Let it take a hours if both of tehm work together.
So,a(1/7+1/9)=1
a=63/16 =3.94 hours.

Ans: 3.94 hours

Solution to Q14:

Q14. There are 12 pipes that are connected to a tank. Some of them are fill pipes and the others are drain pipes. Each of the fill pipes can fill the tank in 8 hours and each of the drain pipes can drain the tank completely in 6 hours. If all the fill pipes and drain pipes are kept open, an empty tank gets filled in 24 hours. How many of the 12 pipes are fill pipes?
(a)6
(b)8
(c)7
(d)5

Sol:
Let the number of fill pipes =a .
So number of empty pipes=(12-a)
According to question,
24[(a/8)-(12-a)/6]=1

Solving,a=7
So number of fill pipes=7.
Ans: (c) 7.

Correct me if i m wrong:

Q2. A men employs 10 laborers for 3 weeks for wages amounting to Rs. 1470. What would be total wages payable if 2 laboures do not turn up for the last 2 weeks?
All laborers are paid equal wages.

Sol: 10 labour for 3 weeks will take Rs.1470
So 10 labour for 1 week will take Rs.490
1 labour for 1 week will take Rs.49

According to question, 10 labours work for 1 week and then 8 labours work for 2 week.

Total amount=10x1x49+8x2x49= 1274.

Ans: Rs.1274
 

work may be completed ..

Q7. A contract is to be completed in 56 days and 104 men were set to work, each working 8 hours a day. After 30 days 2/5 of the work is completed. How many additional men may be employed, so that the work may be completed in time, each man now working 9 hours a day?

Sol: Let the amount of total work=a units.
According to question104x8x30=(2/5)a
a=520x120.
Let the additional men = b.
And the rest work=3/5 .
So,(104+b)x26x9=(3/5)x520x120.
b=42.

Answer: 42 men.