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4a² - 4a + 1 + 3a = (1)² + (2a)² - 2 x 1 x 2a + 3a

   = (1 - 2a)² + 3a

   = (1 - 2a) + 3a

   = (1 + a)

   = (1 + 0.1039)

   = 1.1039

P.W. = Rs. (540 - 90) = Rs. 450.

S.I. on Rs. 450 = Rs. 90.

S.I. on Rs. 540 = Rs.		90	x 540		= Rs. 108.
		450

B.D. = Rs. 108.

Let speed of the car be x kmph.

Then, speed of the train =	150	x	=		3	x	kmph.
	100				2

	75	-	75	=	125
	x		(3/2)x		10 x 60

	75	-	50	=	5
	x		x		24

x =		25 x24		= 120 kmph.
		5

(1/2)x	+	(1/2)x	= 10
21		24

	x	+	x	= 20
	21		24

15x = 168 x 20

x =		168 x 20		= 224 km.
		15

Speed =		300		m/sec =	50	m/sec.
		18			3

Let the length of the platform be x metres.

Then,		x + 300		=	50
		39			3

3(x + 300) = 1950

x = 350 m.

A : B = 100 : 90.

A : C = 100 : 72.

B : C =	B	x	A	=	90	x	100	=	90	.
	A		C		100		72		72

When B runs 90 m, C runs 72 m.

When B runs 100 m, C runs		72	x 100	m	= 80 m.
		90

B can give C 20 m.

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.

36 = 2² x 3²

84 = 2² x 3 x 7

H.C.F. = 2² x 3 = 12.

Given numbers are 1.08, 0.36 and 0.90.   H.C.F. of 108, 36 and 90 is 18,

H.C.F. of given numbers = 0.18.

Number of shares =		4455		= 540.
		8.25

Face value = Rs. (540 x 10) = Rs. 5400.

Annual income = Rs.		12	x 5400		= Rs. 648.
		100

log	a	+ log	b	= log (a + b)
	b		a

log (a + b) = log		a	x	b		= log 1.
		b		a

So, a + b = 1.

Let total number of children be x.

Then, x x	1	x =	x	x 16     x = 64.
	8		2

Number of notebooks =	1	x2 =		1	x 64 x 64		= 512.
	8			8

Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y.

Then, 6x + 8y =	1	and 26x + 48y =	1	.
	10		2

Solving these two equations, we get : x =	1	and y =	1	.
	100		200

(15 men + 20 boy)'s 1 day's work =		15	+	20		=	1	.
		100		200			4

15 men and 20 boys can do the work in 4 days.

(B + C)'s 1 day's work =		1	+	1		=	7	.
		9		12			36

Work done by B and C in 3 days =		7	x 3		=	7	.
		36				12

Remaining work =		1 -	7		=	5	.
			12			12

Now,	1	work is done by A in 1 day.
	24

So,	5	work is done by A in		24 x	5		= 10 days.
	12				12

Let S be the sample space.

Then, n(S) = 52C2 =	(52 x 51)	= 1326.
	(2 x 1)

Let E = event of getting 2 kings out of 4.

n(E) = 4C2 =	(4 x 3)	= 6.
	(2 x 1)

P(E) =	n(E)	=	6	=	1	.
	n(S)		1326		221

	x	-	1		2	= x +	1	- 2
			x				x

= (3 + 22) +	1	- 2
	(3 + 22)

= (3 + 22) +	1	x	(3 - 22)	- 2
	(3 + 22)		(3 - 22)

   = (3 + 22) + (3 - 22) - 2

   = 4.

		x	-	1		= 2.
				x

Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.

C.P. of 30 articles = Rs.		5	x 30		= Rs. 25.
		6

S.P. of 30 articles = Rs.		6	x 30		= Rs. 36.
		5

Gain % =		11	x 100	% = 44%.
		25