GRE Questions

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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.

The total angle traced by the hour hand is the angle traced in 7 hours and 10 minutes.
We know that the angle traced by the hour hand in one hour is 30º and in one minute is 1/2º.
Therefore, (30º x 7) + (10 x 1/2º) = 215º is the angle traced by the hour hand.

Let 'd' be the distance between A and B
K -time = d/10 + d/9 = 19d/90 hours
L -time = 2d/12 = d/6 hours
We know that, 10 min = 1/6 hours
Thus, time difference between K and L is given as 10 minutes.
=> 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
Thus,
d= 15/4 km = 3.75 km.

hence the distance between A and B is 3.75 km.

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.
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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.

Given 

1 = 1 => 1 x 11 = 11

2 = 22 => 2 x 11 = 22

3 = 33 => 3 x 11 = 33

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Similarly,

11 => 11 x 11 = 121.

An icycle, a hanging spike of ice formed by the freezing of dripping water forms due to cool in winter and dies in summer due to hot.

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.

From the assumption and trial and error method,

last three digits sum must leave the same digit as the answer. It is possible only if the answer must be units digit number leaving tens digit as carry.

And the possible 3rd digit is 5. Since 5 + 5 + 5 = 15 

Now, 2nd step is  1 + x + x + x = 5

=> 3x = 4

It is possible with x = 8. 8 + 8 + 8 + 1 = 25 leaving 2 as carry. 

Now, 3rd step is 2 + x + x + x = 5 => x = 1

Hence, the three dgits are 1, 8, 5.

185 + 185 + 185 = 555.