# Free Basic Arithmetic - 04 Practice Test - CAT

A man can finish a work in 12 days while B can finish the same work in 15 days. If both work together, then calculate the time taken to complete the work.

A. 6(23) days
B. 6 days
C. 8 days
D. 6(34) days

#### SOLUTION

Solution : A

A can finish the work in 12 days.

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

In 1 day he can finish  (10012)%=8.33% of the work.

Similarly B can finish  (10015)%=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  (10015)e=6(23)  days. Hence option (a).

2nd method:- Total work (W) = 12A = 15B ,  B=45A

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

n(A+B)=12A , n(A+45A) , n=203=6(23)days. Option (a).

Arun can finish a work in 12 days; Ajit can finish the same work in 15 days while Amit can finish the same work in 20 days. Find the total time taken when all three work together to complete the work.

A. 6 days
B. 5 days
C. 4 days
D. 3 days

#### SOLUTION

Solution : B

Arun can finish  100%  of work in 12 days he can finish  (10012)=8.33%  of the work in a day.

Ajit can finish  100%  of work in 15 days he can finish  (10015)=6.67%   of the work in a day.

Amit can finish  100%  of the work in 20 days he can finish (10020)=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  (10020) = 5 days. Hence option (B).

2nd method:- W= 12A=15B=20C .

n(A+B+C)=12A , n(A+34A+35A)=12A , n=5 days.

Option (2).

Ajit can do a certain work in 24 days. Bharath can do the same work in 30 days. When Chandra joins them and they do the work together, they complete the work in 12 days. Find out the time required by Chandra to complete the work alone

A.

100 days

B.

125 days

C.

130 days

D.

120 days

#### SOLUTION

Solution : D

In one day Ajit can do  (10024)%  of the work.

In one day Bharath can do  (10030)%   of the work.

Working together they can do  (10024)%+(10030)%=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  (10012)%  of the work in a day.

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

Ajit, Bharath and Chandra do  (10012)%   of the work in a day.

Chandra can do  (100127.5)%=(1012)%  of the work in a day.

So, to complete 100% of the work, Chandra will take [100(1012)]= 120 days. Hence option (e).

2nd method:- In 12 days ajit completes 1224  i.e  50%  and Bharat completes 1230 i.e  40% , then chandra completes (10090)=10% work in 12 days. Then he will take 120 days to complete full work.

As a worker, Ajit is thrice as good as Dev. If both working together can finish the work in 5 days, then determine the time taken by Ajit to complete the work alone.

A. 103 days
B. 5 days
C. 203 days
D. 7 days

#### SOLUTION

Solution : C

Suppose Ajit can finish the work in ‘x’ days. In one day he can do  (100x)%  of the work.
As Ajit is thrice as good as Dev, Dev will do 13rd of what Ajit can do in a day.
So, Dev can do  (1003x)%  of work in a day.
Now they complete the work in 5 days, so in 1 day they must be doing 1005=20%  of the work.
So, (100x)+(1003x)=20
x=(203)
So, Ajit can complete the work alone in  (203) days. Hence option (d).
2nd method:- Efficiency  A=3D , 5(A+D)=nA  (i.e.,)5×43A=nA , n=203 days.

A and B can do a work in 12 days, B and C can do the same work in 15 days and A and C can complete the work in 20 days.Days required by A, B and C working together is (in days) ___

#### SOLUTION

Solution :

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=10012%=8.33%...(i)

B & C  can complete the work in 15 days   b+c=10015%=6.67%....(ii)

A & C  can complete the work in 20 days   a+c=10020%=5%....(iii)

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

2(a+b+c)=20%(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 a work in 16 days while B can finish the work in 8 days. After A started the work alone, B joined him after 4 days. The total time taken to finish the work is___

#### SOLUTION

Solution :

A can finish the work in 16 days. In one day he can finish  (10016)%  of the work.

In 4 days, working alone he will finish   [(10016)×4]=25%  of the work.

Now B joins him.

Amount of work left = 100 – 25 =  75%

B can do  (1008)%  of the work.

A and B together can do  (1008)%+(10016)%=(30016)%  of the work.

Work left  =75% . Hence time taken =[(75×16)300]=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.

24 men can complete a work in 20 days. How many more men are required to complete the work in 12 days?

A. 40
B. 12
C. 16
D. 18

#### SOLUTION

Solution : C

Method 1: Using man-days

Amount of work in terms of man days  =24×20=480  man days

So, number of men required to complete the work in 12 days  =48012=40  men.

16 more men are required. Hence option (c)

Method 2: Using constant product rule

Decrease in number of days  =1(208)=12.5

Increase in number of men  =11.5= 16 men to finish the same work.

6 children and 2 men complete a job in 6 days. Each child takes twice as much time a man takes to complete the same amount of job. In how many days will 4 men complete the same job.

A. 7
B. 8
C. 6
D. 8.5
E. 7.5

#### SOLUTION

Solution : E

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 =5×6=30 man days.  Now, when 4 men are working, then number of days taken  =304=7.5  days

Hence option (d)

(x-2) men can do a job in x days and (x+7) men can do a job in  34th  of the same work in (x-10) days. The no. of days in which (x+4) men can finish the job is___

#### SOLUTION

Solution :

Total work is constant.
Total work = (x-2)(x) = 43*(x+7)(x-10)
x26x280=0
Solving for x; we get x = 20.
Now 18 men takes 20 days to complete work
24 men will take 20×1824 days = 15 days

Alternatively:
(x2) x days
(x+7) (x10) days for 34th of work
(x2) will take 34x days for 34th of work
So x2x+7=x103x/4
x26x280=0
So, x = 20
Now 18 men takes 20 days to complete work
24 men will take 20×1824 days = 15 days

4 men and 2 boys can do a job in  623  days. 3 women and 4 boys can finish the same job in 5 days. Also 2 men and 3 women can finish the same job in 4 days. In how many days can 1 man, 1 woman and 1 boy finish the work at their double efficiency___

#### SOLUTION

Solution :

In 1 day 4 men and 2 boys can do  [100(203)]%=15%   of the job. ... (i)

In 1 day 3 women and 4 boys can do  1005=20%  of the job.... (ii)

In 1 day 2 men and 3 women can do  1004=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  10020 = 5 days.

Pipe A can fill a tank in 4 hours while Pipe B can empty the same tank completely in 5 hours. If both the pipes are working simultaneously, then determine the time in which an empty tank will be completely filled.

A. 6
B. 9
C. 20
D. 13

#### SOLUTION

Solution : C

Pipe A can fill a tank completely in 4 hours, Therefore in 1 hour, it can fill  1004=25%  of the tank.

Pipe B can empty the tank in 5 hours, Therefore in 1 hour, it can empty  1005=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  1005  = 20 hours.

2nd method:-  (14)(15)=120 . Thus tank will fill in 20 days.

In what ratio should three varieties of chemical A, B and C at the rate Rs. 120, Rs. 144 and Rs. 174 per litre be mixed to come up with a mixture worth Rs. 141 per litre?

A. 1: 2: 3
B. 11: 7: 77
C. 11: 77: 7
D. 11: 7: 1

#### SOLUTION

Solution : C

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  BAC , Now only option (c).

A milk container contains  5%  milk & rest water while another contains  50%   milk & rest water. What amount of mixture should be taken from both the containers if we require 26 litres mixture with ratio of milk & water as 3: 2?

A. 2, 24
B. 1, 25
C. 8, 18
D. 4, 22

#### SOLUTION

Solution : D

Quantity of milk in the first container = 120

Quantity of milk in the second container = 12

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 1st container = 4 litres and from the second container = 22 litres.

Hence option (d)

From a container 6 litres milk was drawn out and was replaced by water. Again 6 litres of mixture was drawn out and replaced by water. Thus the quantity of milk in the container after these two operations is 9: 16. Calculate the quantity of the milk initially in the container:

A. 12 ltr
B. 15 ltr
C. 9 ltr
D. 6 ltr
E. 18 ltr

#### SOLUTION

Solution : B

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

(k6)2k2=925k=15.

Hence option (b)

The ratio of milk and water in the container is 3: 2 and when 10 litres of the mixture is taken out and is replaced by water the ratio becomes 2: 3. The total quantity of the mixture in the container is.

A. 15 ltr
B. 10 ltr
C. 12 ltr
D. 30 ltr

#### SOLUTION

Solution : D

Suppose the total quantity is x litres.

So, the required ratio 2: 3

3x62x+6=23x=103x=30

So, the quantity of the mixture was 30 ltr.