respectively at intervals of 6,7,8,9 and 12 seconds.
How many times they will toll together
in one hour, excluding the one at the start?
L.C.M. of 6,7,8,9 and 12
= 2x2x3x7x2x3 = 504
ie, The bells will toll together after each 504
seconds. In one hour, they will toll together
(60*60)/504= 7 times
working 8 hours a day can do it in how many days?
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.
How many more man should be engaged to finish the rest of the work in 6 days
working 9 hours a day?
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
10 women and 5 girls can do the same work?
Given that 5 women is equal to 8 girls to complete a work. So, 10 women =
16 girls. Therefore 10 women + 5 girls = 16 girls + 5 girls = 21 girls.
8 girls can do a work in 84 days then 21 girls can do a work in (8*84/21) = 32 days.
Therefore 10 women and 5 girls can a work in 32 days
same job. How long it take both A & B, working together but independently, to do
the same job?
As one hour work = 1/8. Bs 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
time taken by A. Then,working together, what part of the same work they can finish
in a day?
Given that B alone can complete the same work in days = half the time
taken by A = 9 days
As one day work = 1/18
Bs one day work = 1/9
(A+B)s one day work = 1/18+1/9 = 1/6
work in 18 days.In how many days will A alone finish the work.
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.
How many days does B alone take to do the same job?
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.
in 6 days of 7 hours each. How long will they take to do it working together 8 2/5
hours a day?
A can complete the work in (7*9) = 63 days
B can complete the work in (6*7) = 42 days
= > As one hours work = 1/63 and
Bs 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)) = 3 days
of work. Working together they can finish the work in 2 days. B can do the work
alone in ?
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.
Z can do 1/3 of work in 13 days. Who will complete the work first?
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.
How much time B will take to complete the work?
Men at work earlier = 10
Men at work later = 5
Ration of workers = a : b = 2 : 1
10 men can perform the complete task = 10 days
5 men will perform the same task in = b : a time = 1 : 2 => 20 days.
absent and so the rest do the work in 12 days. Find the original number of Men.
More men, less days.
= Total men at work =30
days does B alone take to do the same job?
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.
of 7 hours each. How long will they take to do it working together 8 2/5 hours a day?
A can complete the work in (7*9) = 63 days
B can complete the work in (6*7) = 42 days
= > As one hours work = 1/63 and
Bs 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)) = 3 days
When work was scheduled to commence, it was found necessary to send 25
men to another project. How much longer will it take to complete the work?
Before:
One day work = 1 / 20
One mans one day work = 1 / ( 20 * 75)
Now:
No. Of workers = 50
One day work = 50 * 1 / ( 20 * 75)
The total no. of days required to complete the work = (75 * 20) / 50 =
30
wage of Rs. 10 for the day he works, but he have to pay a fine of Rs. 2 for
each day of his absence. If he gets Rs. 216 at the end, he was absent for work
for ... days
The equation portraying the given problem is:
10 * x – 2 * (30 – x) = 216 where x is the number of working days.
Solving this we get x = 23
Number of days he was absent was 7 (30-23) days.
Let 1 mans 1 days work = $ x $ and 1 boys 1 days work = $ y $.
Then, 6$ x $ + 8$ y $ =$ \dfrac{1}{10} $and 26$ x $ + 48$ y $ =$ \dfrac{1}{2} $.
Solving these two equations, we get : $ x $ =$ \dfrac{1}{100} $and $ y $ =$ \dfrac{1}{200} $.
[15 men + 20 boy]s 1 days work =$ \left(\dfrac{15}{100} +\dfrac{20}{200} \right) $=$ \dfrac{1}{4} $.
$\therefore$ 15 men and 20 boys can do the work in 4 days.
1 womans 1 days work =$ \dfrac{1}{70} $
1 childs 1 days work =$ \dfrac{1}{140} $
$\left(5 women + 10 children\right)$s days work =$ \left(\dfrac{5}{70} +\dfrac{10}{140} \right) $=$ \left(\dfrac{1}{14} +\dfrac{1}{14} \right) $=$ \dfrac{1}{7} $
$\therefore$ 5 women and 10 children will complete the work in 7 days.
Bs 10 days work =$ \left(\dfrac{1}{15} \times 10\right) $=$ \dfrac{2}{3} $.
Remaining work =$ \left(1 -\dfrac{2}{3} \right) $=$ \dfrac{1}{3} $.
Now,$ \dfrac{1}{18} $work is done by A in 1 day.
$\therefore$ $ \dfrac{1}{3} $work is done by A in$ \left(18\times\dfrac{1}{3}\right) $= 6 days.