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SIMPLE PROBLEMS
1)If 9 men working 6 hours a day can do a work in 88 days. Then 6 men working
8 hours a day can do it in how many days?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so (9*6*88/1)=(6*8*d/1)
on solving, d=99 days.
2)If 34 men completed 2/5th of a work in 8 days working 9 hours a day.How many
more man should be engaged to finish the rest of the work in 6 days working
9 hours a day?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so, (34*8*9/(2/5))=(x*6*9/(3/5))
so x=136 men
number of men to be added to finish the work=136-34=102 men
3)If 5 women or 8 girls can do a work in 84 days. In how many days can 10 women
and 5 girls can do the same work?
Sol: Given that 5 women is equal to 8 girls to complete a work
so, 10 women=16 girls.
Therefore 10women +5girls=16girls+5girls=21girls.
8 girls can do a work in 84 days
then 21 girls ---------------?
answer= (8*84/21)=32days.
Therefore 10 women and 5 girls can a work in 32days
4)Worker A takes 8 hours to do a job. Worker B takes 10hours to do the same
job. How long it take both A & B, working together but independently, to do
the same job?
Sol: A's one hour work=1/8.
B's one hour work=1/10
(A+B)'s one hour work=1/8+1/10 =9/40
Both A & B can finish the work in 40/9 days
5)A can finish a work in 18 days and B can do the same work in half the time
taken by A. Then, working together, what part of the same work they can finish
in a day?
Sol: Given that B alone can complete the same work in days=half the time taken
by A=9days
A's one day work=1/18
B's one day work=1/9
(A+B)'s one day work=1/18+1/9=1/6
6)A is twice as good a workman as B and together they finish a piece of work
in 18 days.In how many days will A alone finish the work.
Sol: if A takes x days to do a work then
B takes 2x days to do the same work
=>1/x+1/2x=1/18
=>3/2x=1/18
=>x=27 days.
Hence, A alone can finish the work in 27 days.
7)A can do a certain work in 12 days. B is 60% more efficient than A. How many
days does B alone take to do the same job?
Sol: Ratio of time taken by A&B=160:100 =8:5
Suppose B alone takes x days to do the job.
Then, 8:5::12:x
=> 8x=5*12
=> x=15/2 days.
8)A can do a piece of work n 7 days of 9 hours each and B alone can do it
in 6 days of 7 hours each. How long will they take to do it working together
8 2/5 hours a day?
Sol: A can complete the work in (7*9)=63 days
B can complete the work in (6*7)=42 days
=> A's one hour's work=1/63 and
B's one hour work=1/42
(A+B)'s one hour work=1/63+1/42=5/126
Therefore, Both can finish the work in 126/5 hours.
Number of days of 8 2/5 hours each=(126*5/(5*42))=3days
9)A takes twice as much time as B or thrice as much time to finish a piece
of work. Working together they can finish the work in 2 days. B can do the
work alone in ?
Sol: Suppose A,B and C take x,x/2 and x/3 hours respectively finish the work
then 1/x+2/x+3/x=1/2
=> 6/x=1/2
=>x=12
So, B takes 6 hours to finish the work.
10)X can do ΒΌ of a work in 10 days, Y can do 40% of work in 40 days and Z can
do 1/3 of work in 13 days. Who will complete the work first?
Sol: Whole work will be done by X in 10*4=40 days.
Whole work will be done by Y in (40*100/40)=100 days.
Whole work will be done by Z in (13*3)=39 days
Therefore,Z will complete the work first.
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