AIEEE Questions

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The Dronacharya Award, formally known as Dronacharya Award for Outstanding Coaches in Sports and Games, is sports training honour of the Republic of India. The award is named after Drona, often referred as "Dronacharya" or "Guru Drona", a character from the Sanskrit epic Mahabharata of ancient India.

Let the sum be Rs. S.
(S x 18 x 2)/100 - (S x 12 x 2)/100 = 840 => 36S/100 - 24S/100 = 840

=> 12S/100 = 840 => S = 7000.

For solving this problem first we would break the whole range in 5 sections

1) From 1 to 9
Total number of zero in this range = 0

2) From 10 to 99
Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer)

3) From 100 to 999 - three type of numbers are there in this range
a) x00 b) x0x c) xx0 (here x represents a non zero number)
Total possibilities
for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18
for x0x = 9*1*9 = 81, hence total zeros = 81
similarly for xx0 = 81
total zeros in three digit numbers = 18 + 81 +81 = 180

4) From 1000 to 9999 - seven type of numbers are there in this range
a)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xx
Total possibilities
for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27
for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162
for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162
for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162
for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729
for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729
for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729
total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700
thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)
= 2893

n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways.
The letter pattern 'MESMERISE' consists of 10 letters of which there are 2M's, 3E's, 2S's and 1I and 1R.
Number of arrangements = 9!(2!)2×3!

The given letter equation puzzle means that 

7 W of the W = 7 Wonders of the World.

Angle between hands of a clock

When the minute hand is behind the hour hand, the angle between the two hands at M minutes past H 'o clock

=> 30H-M5+M2 degrees

Here H = 4, M = 15 and the minute hand is behind the hour hand.

Hence the angle

 = 30[4-(15/5)]+15/2 = 30(1)+7.5 = 37.5 degrees

Part filled in 4 minutes = 4(1/15 + 1/20) = 7/15

Remaining part = 1 - 7/15 = 8/15

Part filled by B in 1 minute = 1/20

1/20 : 8/15 :: 1 : k

k = (8/15 )x 1 x 20 = 10( 2/3) min = 10 min 40 sec.

The tank will be full in (4 min. + 10 min. 40 sec) = 14 min 40 sec.

Speed of the current = 24-16/2

= 8/2

= 4 km/hr.