Free Basic Arithmetic - 04 Practice Test - CAT 

A can finish the work in 12 days.

He can finish $100\%$ of the work in 12 days.

In 1 day he can finish  $\left(\frac{100}{12}\right)\%=8.33\%$ of the work.

Similarly B can finish  $\left(\frac{100}{15}\right)\%=6.67\%$  of the work in 1 day.

When they both work together, they can finish  $(8.33+6.67)\%=15\%$  of the work in 1 day

So, to complete  $100\%$  of the work, they will take  $\left(\frac{100}{15}\right)e=6\left(\frac{2}{3}\right)$  days. Hence option (a).

2nd method:- Total work (W) = 12A = 15B ,  $B=\frac{4}{5}A$

n(A+B) = W = 12A or 15B (Here, we take 12A).

$n(A+B)=12A,n(A+\frac{4}{5}A),n=\frac{20}{3}=6\left(\frac{2}{3}\right)$days. Option (a).

Arun can finish  $100\%$  of work in 12 days he can finish  $\left(\frac{100}{12}\right)=8.33\%$  of the work in a day.

Ajit can finish  $100\%$  of work in 15 days he can finish  $\left(\frac{100}{15}\right)=6.67\%$   of the work in a day.

Amit can finish  $100\%$  of the work in 20 days he can finish $\left(\frac{100}{20}\right)=5\%$   of the work in a day.

So, if all three work together, then they can finish  $(8.33+6.67+5)\%=20\%$  of the work in a day.

So, they can complete the work in  $\left(\frac{100}{20}\right)$ = 5 days. Hence option (B).

2nd method:- W= 12A=15B=20C . 

$n(A+B+C)=12A,n(A+\frac{3}{4}A+\frac{3}{5}A)=12A,n=5days$.

Option (2).

In one day Ajit can do  $\left(\frac{100}{24}\right)\%$  of the work.

In one day Bharath can do  $\left(\frac{100}{30}\right)\%$   of the work.

Working together they can do  $\left(\frac{100}{24}\right)\%+\left(\frac{100}{30}\right)\%=7.5\%$  of the work in one day.

Now when Chandra joins them, they complete the work in 12 days Ajit, Bharath and Chandra together complete  $\left(\frac{100}{12}\right)\%$  of the work in a day.

Now, Ajit and Bharath do $7.5\%$ of the work in a day.

Ajit, Bharath and Chandra do  $\left(\frac{100}{12}\right)\%$   of the work in a day.

Chandra can do  $(\frac{100}{12}-7.5)\%=\left(\frac{10}{12}\right)\%$  of the work in a day.

So, to complete $100\%$ of the work, Chandra will take $\left[\frac{100}{\left(\frac{10}{12}\right)}\right]$= 120 days. Hence option (e).

2nd method:- In 12 days ajit completes $\frac{12}{24}$  i.e  $50\%$  and Bharat completes $\frac{12}{30}$ i.e  $40\%$ , then chandra completes $(100-90)=10\%$ work in 12 days. Then he will take 120 days to complete full work.

Let a, b and c be the % of the work done by A, B and C in one day respectively.

$A\&B$  can complete the work in 12 days   $a+b=\frac{100}{12}\%=8.33\%...\left(i\right)$

$B\&C$  can complete the work in 15 days   $b+c=\frac{100}{15}\%=6.67\%....\left(ii\right)$

$A\&C$  can complete the work in 20 days   $a+c=\frac{100}{20}\%=5\%....\left(iii\right)$

Adding (i), (ii) and (iii)

$2(a+b+c)=20\%\Rightarrow (a+b+c)=10\%$

So, working together they finish $10\%$  of work in a day. So, they can complete the work in 10 days.

A can finish the work in 16 days. In one day he can finish  $\left(\frac{100}{16}\right)\%$  of the work.

In 4 days, working alone he will finish   $\left[\right(\frac{100}{16})\times 4]=25\%$  of the work.

Now B joins him.

Amount of work left = 100 – 25 =  $75\%$

B can do  $\left(\frac{100}{8}\right)\%$  of the work.

A and B together can do  $\left(\frac{100}{8}\right)\%+\left(\frac{100}{16}\right)\%=\left(\frac{300}{16}\right)\%$  of the work.

Work left  $=75\%$ . Hence time taken $=\left[\frac{(75\times 16)}{300}\right]=4$ days.

So, total time taken = 4 + 4 = 8 days. 

2nd method:- In 4 days A will finish  $25\%$  of work. Remaining work is  $75\%$, Efficiency of B is double than A i.e  $50\%$  work will done by B and  $25\%$ will done by A. Thus it will take 4 days. Total 8 days.

Method 1: Using man-days

Amount of work in terms of man days  $=24\times 20=480$  man days

So, number of men required to complete the work in 12 days  $=\frac{480}{12}=40$  men.

16 more men are required. Hence option (c)

Method 2: Using constant product rule

Decrease in number of days  $=\frac{1}{\left(\frac{20}{8}\right)}=\frac{1}{2.5}$

Increase in number of men  $=\frac{1}{1.5}=$ 16 men to finish the same work.

Given that, 1 man = 2 child.

So, 6 children + 2 men = 3 men + 2 men = 5 men.So, 5 men complete a job in 6 days.

So, amount of work mandays $=5\times 6=30$ man days.  Now, when 4 men are working, then number of days taken  $=\frac{30}{4}=7.5$  days

Hence option (d)

Total work is constant.
$\therefore$ Total work = (x-2)(x) = $\frac{4}{3}$*(x+7)(x-10)
$\Rightarrow x^2-6x-280=0$
Solving for x; we get x = 20.
Now 18 men takes 20 days to complete work
$\therefore$ 24 men will take $\frac{20\times 18}{24}$ days = 15 days

Alternatively:
$(x-2)$ $\to x$ days
$(x+7)$ $\to (x-10)$ days for ${\frac{3}{4}}^{th}$ of work
$\Rightarrow (x-2)$ will take $\frac{3}{4}x$ days for ${\frac{3}{4}}^{th}$ of work
So $\frac{x-2}{x+7}=\frac{x-10}{3x/4}$
$\Rightarrow x^2-6x-280=0$
So, x = 20
Now 18 men takes 20 days to complete work
$\therefore$ 24 men will take $\frac{20\times 18}{24}$ days = 15 days

In 1 day 4 men and 2 boys can do  $\left[\frac{100}{\left(\frac{20}{3}\right)}\right]\%=15\%$   of the job. ... (i)

In 1 day 3 women and 4 boys can do  $\frac{100}{5}=20\%$  of the job.... (ii)

In 1 day 2 men and 3 women can do  $\frac{100}{4}=25\%$  of the job.... (iii)

So, adding (i), (ii), (iii)

6 men, 6 women and 6 boys can do 15 + 20 + 25 = $60\%$  of the job in 1 day.

So, 1 man, 1 woman and 1 boys can do $10\%$ of the job in a day.

All double their efficiency, they can do $20\%$ of the job in a day.

They can complete the job in  $\frac{100}{20}$ = 5 days.

Pipe A can fill a tank completely in 4 hours, Therefore in 1 hour, it can fill  $\frac{100}{4}=25\%$  of the tank.

Pipe B can empty the tank in 5 hours, Therefore in 1 hour, it can empty  $\frac{100}{5}=20\%$  of the tank.

So, when both are simultaneously working, then in 1 hour  $5\%$  of the tank will be filled.

For filling the tank completely, it will take  $\frac{100}{5}$  = 20 hours. 

2nd method:-  $\left(\frac{1}{4}\right)-\left(\frac{1}{5}\right)=\frac{1}{20}$ . Thus tank will fill in 20 days. 

First mix varieties  $A\&C$   in a ratio to get a mixture at the cost Rs. 141 per litre.

Applying Alligation:

Now mix varieties  $A\&B$   to get a mixture at the cost of Rs. 141 per litre.

So, A: C in ratio 11: 7 and A: B in ratio 1: 7.The required ratio A: B: C = 11: 77: 7. Hence option (c).

2nd method:- By, observation, 141 is nearest to B (144) then A(120) and then C(174). When we mix all three then highest contribution will be according the above, i.e  $B\to A\to C$ , Now only option (c). 

Quantity of milk in the first container = $\frac{1}{20}$

Quantity of milk in the second container = $\frac{1}{2}$

Applying Alligation:

So, mixture should be taken out from the containers in the ratio = 2: 11

Total amount of mixture required = 26 litres. So, quantity required from $1^{st}$ container = 4 litres and from the second container = 22 litres.

Hence option (d)

After these two transfers = milk is left in the container in ratio 9: 25

$\frac{(k-6{)}^2}{k^2}=\frac{9}{25k}=15.$

Hence option (b)

Suppose the total quantity is x litres.

So, the required ratio 2: 3

$\frac{3x-6}{2x+6}=\frac{2}{3}\Rightarrow x=10\Rightarrow 3x=30$

So, the quantity of the mixture was 30 ltr.