IBM Percentage Quiz -1

Based on all given information, a number of students
who never played specific sports among 200 students
conducted the surveys is as below:

Foot ball: 30
Basket ball : 40
Hockey: 50

We could Assume that all 140 students who play the Tennis
which has the minimum number of all 4 games
(; 200 x 70% = 140) are the first candidate who must
have played all four games.

In order to get the minimum number of students who
must have played all four games:
1. Among these 140 students who play tennis as a hobby,
we should subtract the number of students who never
had played other 3 sports each in case they are the one.
140 - 30 - 40 - 50 = 20

Given 60% of 5 employees including 2 who are capable of doing task.

60% of 5 employeees
= 60/100 * 5
= 3 employees

===> 2 employees who are capable of doing the task and one
employee who is not capable.

percentage of employees assigned who are not capable
= 1/3 * 100
= 33.33%

Let there is x Kgs of alloy.
Cooper present in mixture= 15/100 of x
Lead present in mixture = 25/100 of x= 5 Kgs(Given)
=> By solving above expression : x= 20Kgs
Silver present in mixture = 60/100 of x= (60/100) * 20 = 12 Kgs

total hrs= 120
Divide by 4= each get 30 hr
so, each one of three worker got 750 (30*25)each and contractor
got 300( 30*10).
Total cost= 750*3+300=2550.

Solution Let assume x% of people attend the party
Out of x
35% is men so, x*35/100=7x/20 men and remaining
13x/20 is women
Now out of 7x/20 20% men are attend the party so,
(7x/20)*(20/100)=7x/100
Now out of 13x/20 40% women are attend the party so,
(13x/20)*(40/100)=13x/50

Total people attend the party: Men+ Women
7x/100+13x/50=33x/100
% of people attend the party : (33x/100)*(1/x)(100)=33%

53-25=28
25-22=3
so the liters to be used from the container that is mixed at
53% strength is (12*3)/31=36/31=1 5/31 litres

time taken to complete one record is 75*2=150,150/60=2.5hrs
so 2.5*20=50 full records
3 hrs to take onr hr broadcast ,so 50*3=150

1.5%+6.5% = 8% ( rate increase every year)
25*8% = 2/- 25+2= 27
27*8% = 2.16/- 27+2.16 = 29.16 /-
ans is (b) 29.16

New networks= 8
Already employed networks= 20
Total networks= 28

computer advt. from uni.A= 28*(75/100) =21
out
28 networks 21 came from uni.A
among
these 21 we know 8 are newly hired
so 21-8=13 networks among previously employed networks
came from uni.A

Answer is : 13/20

Taking A= x, B=2x, and C=3x.
(x+2x+3x)/3=600
x=300

so, the numbers are 300,600,900

new average= 105%of 600= 630

so (110% of 300 + 120% of 600 + x)/3 = 630

330 + 720 + x = 1890
x= 1890 - 1050
x=840

so C should be reduced by 60