Time and Work Quiz 1

work done by thomas in day = $\frac{1}{7}$
work done by aashay in a day = $\frac{1}{9}$
work done by both in a day= $\frac{1}{7} + \frac{1}{9} = \frac{16}{63}$
days required by both = $\frac{63}{16}$ =3.9

Let' s say Jack's one day work = $\frac{1}{16}$

Paul's one day work = $\frac{1}{24}$

Hari's one day work = $\frac{1}{z}$

combined one day work of all three together = $\frac{1}{8}$

so $\frac{1}{16}$ + $\frac{1}{24}$ + $\frac{1}{z}$ = $\frac{1}{8}$

$\frac{5}{48} + \frac{1}{z} = \frac{1}{8}$

$\frac{1}{z} = \frac{1}{8} - \frac{5}{48}$

$\frac{1}{z} = \frac{6 - 5}{48} = \frac{1}{48}$

Hari can complete the work in 48 days

Video solution:

A can do the total work in 8 days, and B can do it in 6 days.
Let the total work be 24 units. Now capacities are
A = $\frac{24}{8}$ = 3,
B = $\frac{24}{6}$ = 4,
A + B + C = $\frac{24}{3}$ = 8
So Capacity of C = 1 unit.
In two days C will do 2 units which is $\frac{2}{24}^{th}$ part of the total work. So $\frac{1}{12}^{th}$ part.

George one day work is $\frac{1}{15}$
Marks one day work is $\frac{1}{10}$
$\frac{1}{G+M} = \frac{1}{15} + \frac{1}{10} = \frac{1}{6}$
G+M = 6 days

Babla 10 day ------ now $\frac{30}{10}$ = 3 unit per day

L.C.M = 30 (total work they have to perform)

Ashu 15 day ------- $\frac{30}{15}$ = 2 unit per day

so their ratios of work is 3:2

so Babla will get $\frac{3}{5} \times 5000$ = 3000

Video solution:

A's 2 day's work = $2 \times \frac{1}{20} = \frac{1}{10}$

(A + B + C)'s 1 day's work = $\frac{1}{20} + \frac{1}{30} + \frac{1}{60}  = \frac{1}{10}$

Work done in 3 days = $\frac{1}{10} + \frac{1}{10} = \frac{1}{5}$

Now, $\frac{1}{5}$ work is done in 3 days.

Whole work will be done in (3 x 5) = 15 days.

Video solution:

$m1 \times d1 \times h1 = m2 \times d2 \times h2$
$4 \times 8 \times 5 = 2 \times 20 \times h2$
160 = 40h2
h2 = 4
but because of double papers it becomes $4 \times 2$= 8 hours

Video solution:

Here, we see that George does $\frac{3}{5}^{th}$ of work in 9 days.

So, he will complete the work in $9 \times \frac{5}{3}$ = 15 days.

After 9th day, Paul Comes to join him.

Work left to be done = 1 - $\frac{3}{5} = \frac{2}{5}$

They both complete the work in 4 more days.

Since, it takes george 15 days to complete the work.

In one day, he will complete $\frac{1}{15}^{th}$ of work.

In 4 days, he will do $\frac{4}{15}^{th}$ of work.

Now, work done by Paul in 4 days = $\frac{2}{5} - \frac{4}{15} = \frac{2}{15}$

So,Paul completes $\frac{2}{15}^{th}$ of work in 4 days.

Therefore, In one day he will complete $\frac{2}{60}$ of work or $\frac{1}{30}$ of work.

So, number of days taken by paul to complete the work is 30 days.

as per data.. Y=$\frac{2}{3}$X---- (1)

Z = $\frac{3}{4}$Y----(2)

They all type the same time X+Y+Z=1----(3)

and from equations (1) and (2) substitute interms of Y in eq (3)

$\frac{3}{2}Y + Y + \frac{3}{4}Y = 1$

$\frac{6Y+4Y+3Y}{4}$=1

$\frac{13Y}{4}$ =1

Y = $\frac{4}{13}$

(i) A   :   C

Efficiency      5    :   3

No of days   3x   :  5x

Given that, 5x - 6 =3x

=> x = 3

Number of days taken by A = 9

Number of days taken by C = 15

(ii) B  :  C

Days   2  :  3

Therefore, Number of days taken by B = 10

Work done by B and C in initial 2 days = 2[$\frac{</span><span id="MathJax-Span-6" class="mrow"><span id="MathJax-Span-7" class="mfrac"><span id="MathJax-Span-8" class="mn">1}{</span><span id="MathJax-Span-9" class="mn">10}$ + $\frac{</span><span id="MathJax-Span-11" class="mfrac"><span id="MathJax-Span-12" class="mn">1}{</span><span id="MathJax-Span-13" class="mn">15}</span></span></span><span id="MathJax-Span-14" class="mo">] </span></span></span></span></span>= \frac{1}{3}$