Exercise 10

Solve 1
Number of days required by 6 men to complete the work = 7
Number of days required by 1 man to complete the work = 7 × 6
Number of days required by 21 men to complete the work = 42/21 = 2
Hence, 2 days is required by 21 men to complete the work

Solve 2
Shashi weaves one basket in 35/25 =7/5 days
shashi weaves 110 baskets in (110*7)/5=154 days

Solve 3
Suppose two persons are A and B.
A can do 1 work in 4 days, so in 1/4 work in 1 day. Similarly B can do 1/6 work in 1 day.
If they work together, then they can do 1/4 + 1/6 work in 1 day.
A+B = 1/4 + 1/6 work = 5/12 work in 1 day.
Now, in 1 day , 5/12 work
Therefore, in 12/5 days 5/12 * 12/5 = 1 work.
Total work done together is 1/(5/12) = 12/5 days
Answer: 12/5 days.

Solve 4
1st man 1 hour work = 1/3 hour
2nd man 1 hour work = 1/2
Now if they work together
Let the time be x hours
1x/3+1x/2=1
5x/6=1
X =6/5 hours

Solve 5
A and B can do a piece of work in 10 days.
Therefore A and B together can do 1/10 of work in one day.
A + B = 1/10…Eq..1
A alone can complete this work in 15 days. Thus A alone can do this work in 1/15.
Substituting the value of work done by A in Eq..1
A + B = 1/10
1/15 + B = 1/10
B = 1/10 - 1/15
B = (3 - 2)/30
B = 1/30
Thus, B can accomplish 1/30 of work in one day.
Accordingly, B alone could complete this work in 30 days.
Answer B alone could complete this work in 30 days.

Solve 6

Solve 7
A, B and C can cultivate a field in 10, 12 and 15 days respectively if they work together in how many days will they finish the work and what fraction of the work will each of them do

A, B और C क्रमशः 10, 12 और 15 दिनों में एक खेत में खेती कर सकते हैं यदि वे एक साथ कितने दिनों में काम करते हैं तो वे काम को कितने दिनों में पूरा करेंगे और उनमें से प्रत्येक का क्या अंश काम करेगा




Solve 8
A, B and C can do piece of work in 10 days
so they can do 1/10 part of work in a day
A alone can do same work in 40 days
so A can do 1/40 part of work in a day
B alone can do it in 30 days
so B can do 1/30 part of work in a day
let the days taken by C be x
1/10=1/40+1/30+1/x
1/10=7/120+1/x
1/10-7/120=1/x
12-7/120=1/x
5/120=1/x
1/24=1/x
therefore C can do 1/24 in a day
so C can do whole piece of work in 24 days

Solve 9
One day work of (A+B) = 1/8.
One day work of (B+C)= 1/12.
One day work of (A+B+C) =1/6.
Therefore one day work of C= 1/6–1/8 = 1/24.
And one day work of A = 1/6–1/12 = 1/12.
Thus , one day work of (A+C) = 1/12+1/24 = 1/8.
1/8 work A and C can complete in= 1 day.
1. work A and C can complete in= 1/(1/8).= 8 days. Answer.

Solve 10
Time taken by tap 1 to fill the cistern = 4h
work done in 1h = 1/4
time taken by tap 2 to fill the cistern = 3h
work done in 1 h = 1/3
one hour work of both taps = 1/4+1/3 =7/12
time taken by both taps to fill cistern =1÷7/12
=12/7 hours

Solve 12
first tap filled=1/12 min
second tap filled= 1/15 min
third tap =-1/10min
bath be filled if both taps are turned on = 1/12+1/15-1/10
=(5+4-6)60
=20 minutes

Solve 13
Let us call the pipe Pipe A.
Pipe A fills the cistern in 6 hrs
so in one hr Pipe A fills 1/6th of the cistern.
However with the leak it takes 7 hrs for Pipe A to fill the cistern.
Hence in one hour the cistern will have 1/7th left in.
So the difference between not leaking and leaking is 1/6 - 1/7 = 1/42 th of the cistern
this is amount of leak per hour.
Hence it will take 42 hours to fully empty the cistern
full-width