GATE Questions

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SP of each car is Rs. 404415, he gains 15% on first car and losses 15% on second car.

Percentage of students passed in English = 60%

Percentage of students passed in Hindi = 45%

Percentage of students passed in both subjects = 25%

ATQ,

Percentage of students passed in only English = 60 - 25 = 35%

Percentage of students passed in only Hindi = 45 - 25 = 20%

Therefore, Percentage of students failed = 100 - (35 + 25 + 20)

= 100 - 80 = 20%.

The incomes of A and B be 3P and 4P.

Expenditures = Income - Savings

(3P - 100) and (4P - 100)

The ratio of their expenditure = 1:2

(3P - 100):(4P - 100) = 1:2

2P = 100 => P = 50

Their incomes = 150, 200

At 1 pm ... Distance between A and B = 2x3.5 = 7 kms
at 2 pm .....Distance between A and B = 7 + 3.5 - 1 = 9.5 kms
At 3 pm ....Distance between A and B = 9.5 + 3.5 - 2 = 11 kms.
..
..
..
At 7 pm distance between A and B = 3 kms.
From 7 pm to 8 pm, speed of A = 3.5 kmph
speed of B = 8 kmph
Relative speed = 4.5 kmph

Time taken by B to meet A = 3/4.5 hrs = 30/45 hrs = 120/3 mins = 40 mins.

So meeting time is 7 : 40 PM.

If there are total 100 candidates,
40 candidates are really capable and
60 candidates are incapable.
Of the candidates who are really capable = 80%
40*(80/100)=32 pass the test and
of the incapable= 30%
60*(30/100)=18 pass the test.
Total candidates who passed the test = 32+18 = 50
The proportion of the really capable students who can pass the test to the total students who can pass = 32/50 = 16/25.

Total running distance in four weeks = (24 x 240) + (4 x 400)

= 5760 + 1600

= 7360 meters

= 7360/1000

=> 7.36 kms

Given share of L is 2/7 th of profit. Then, their profits are divided in to 5:2 ratio.

Ratio of K and L is

16000×8x×4=52

x = Rs. 12,800.

From above diagram
AC represents the hill and DE represents the tower

Given that AC = 100 m

angleXAD = angleADB = 30° (∵ AX || BD )
angleXAE = angleAEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1)

tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)
∵ BD = CE and Substitute the value of CE from equation 1
100/√3 = 100−h(√3) => h = 66.66 mts

The height of the tower = 66.66 mts.