Exams > Cat > Quantitaitve Aptitude
BASIC ARITHMETIC MCQs
:
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.
:
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.
:
Total work is constant.
∴ Total work = (x-2)(x) = 43*(x+7)(x-10)
⇒x2−6x−280=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:
(x−2) →x days
(x+7) →(x−10) days for 34th of work
⇒(x−2) will take 34x days for 34th of work
So x−2x+7=x−103x/4
⇒x2−6x−280=0
So, x = 20
Now 18 men takes 20 days to complete work
∴ 24 men will take 20×1824 days = 15 days
:
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)
:
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.
:
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.
:
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 B→A→C , Now only option (c).
:
B
After these two transfers = milk is left in the container in ratio 9: 25
(k−6)2k2=925k=15.
Hence option (b)
:
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)
:
D
Suppose the total quantity is x litres.
So, the required ratio 2: 3
3x−62x+6=23⇒x=10⇒3x=30
So, the quantity of the mixture was 30 ltr.