IBM Ratio and Proportions Quiz – 1

Let small envelopes bought=2x
Large envelopes bought=x
Total No. of envelopes bought=x+2x=3x
Small envelope unused=(5/11)*2x=10x/11
Large envelope unused=(2/3)*x=2x/3
Total envelope unused=(24x/13)+(x/6)=52x/33
Fraction of the total envelope unused=.
(52x/33)/3x=52/99

No. of Employees working per day= 60
Total Employees working in 3 days = 180
Total No. of Employees taking calls = 30+30+40=100
Required Fraction= 100/180= 5/9

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You can also submit your own version of explanations under user explanation section and also view other submitted explanations by other users.

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You can also submit your own version of explanations under user explanation section and also view other submitted explanations by other users.

area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) -tri(ABC)
area of tri(ADE)/area of tri(ABC) = 9/25
area of tri(ABC)/area of tri(ADE) = 25/9
area of tri(ABC) = area of tri(ADE) + area of trap(BCDE)
(area of tri(ADE) + area of trap(BCDE))/area of tri(ADE) = 25/9
1 + area of trap(BCDE)/area of tri(ADE) = 25/9
take1on both sides
area of trap(BCDE)/area of tri(ADE) = (25/9) - 1 = 16/9

since question is asking for area of tri(ADE)/area of trap(BCDE) = 9/16

M s/w has 78 line of code
N s/w has = 78+78*4/13= 102 line of code
P s/w 5% more N' code 102+5.1=107.1
or
107 line of code

20 students average height is 105cms
sum of heights/ 20 =105cms
sum of heights of 20 students= 105*20= 2100

10 students of average height 120 added
so their, sum of 10 students height / 10 = 120

sum of heights of 10 students = 1200cms

By adding heights of 20 and 10 students , 1200+2100 = 3300
average of 30 students = 3300/30 = 110 cms

100*80=40*x
x=200

20 liters

If milk and water in original mixture is 5x and x respectively, then
5x/(x+20) = 5/6, x=4
So milk in original mixture=5*4=20 liters