Online Aptitude Test

Distance = (240 x 5) = 1200 km.

Speed = Distance/Time

Speed = 1200/(5/3) km/hr.     [We can write 1 hours as 5/3 hours]

Required speed =		1200 x	3	km/hr	= 720 km/hr.
			5

A : B =

4x x 3 +

4x -

1

x 4x

x 7

:

5x x 3 +

5x -

1

x 5x

x 7

4

5

   = (12x + 21x) : (15x + 28x)

   = 33x :43x

   = 33 : 43.

A's share = Rs.		760 x	33		= Rs. 330.
			76

Let the required number of bottles be x.

More machines, More bottles (Direct Proportion)

More minutes, More bottles (Direct Proportion)

Machines	6	:	10		:: 270 : x
Time (in minutes)	1	:	4

 6 x 1 x x = 10 x 4 x 270

x =	(10 x 4 x 270)
	(6)

 x = 1800.

L.C.M. of 21, 36, 66 = 2772.

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11

To make it a perfect square, it must be multiplied by 7 x 11.

So, required number = 2² x 3² x 7² x 11² = 213444

Required number = (L.C.M. of 12, 15, 20, 54) + 8

   = 540 + 8

   = 548.

15^th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days  4 odd days.

Jan.  Feb.   March    April    May  June  July  Aug. 
(31 +  28  +  31   +   30   +  31  + 30  + 31  + 15) = 227 days

 227 days = (32 weeks + 3 days)  3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7  0 odd days.

Given day is Sunday.

23) 1056 (45
      92
      ---
      136
      115
      ---
       21
      ---
     
 Required number = (23 - 21)    
                 = 2.   

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) =	n(E)	=	9	.
	n(S)		20

	5	A +	4	B	=	2		6	A +	8	B
	100		100			3		100		100

	1	A +	1	B	=	1	A +	4	B
	20		25			25		75

		1	-	1	A =		4	-	1	B
		20		25			75		25

	1	A =	1	B
	100		75

 Required ratio = 4 : 3

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

 The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

This is an A.P. in which a = 51, l = 100 and n = 50.

Sum =	n	(a + l)	=	50	x (51 + 100)   = (25 x 151)   = 3775.
	2			2

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

 Required number of numbers = (1 x 5 x 4) = 20.

Let A, B, C represent their respective weights. Then, we have:

A + B + C = (45 x 3) = 135 .... (i)

A + B = (40 x 2) = 80 .... (ii)

B + C = (43 x 2) = 86 ....(iii)

Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)

Subtracting (i) from (iv), we get : B = 31.

 B's weight = 31 kg.

Let C = x.

Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000.

So, x + x + 5000 + x + 9000 = 50000

 3x = 36000

 x = 12000

A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12.

A's share = Rs.		35000 x	21		= Rs. 14,700.
			50

Speed of train relative to man = (60 + 6) km/hr = 66 km/hr.

=		66 x	5	m/sec
			18

=		55	m/sec.
		3

Time taken to pass the man =		110 x	3	sec = 6 sec.
			55