Online Aptitude Test - Aptitude Test - Random

I gives, 2R = 44.

II gives, H = 12.

A = 2RH = (44 x 12).

Cost of painting = Rs. (44 x 12 x 20).

Thus, I and II together give the answer.

Correct answer is (E).

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

   = 540 + 8

   = 548.

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) =	8	=	4	.
	14		7

Whole work is done by A in		20 x	5		= 25 days.
			4

Now,		1 -	4		i.e.,	1	work is done by A and B in 3 days.
			5			5

Whole work will be done by A and B in (3 x 5) = 15 days.

A's 1 day's work =	1	, (A + B)'s 1 day's work =	1	.
	25		15

B's 1 day's work =		1	-	1		=	4	=	2	.
		15		25			150		75

So, B alone would do the work in	75	= 37	1	days.
	2		2

Arun : Kamal : Vinay = (8,000 x 6) : (4,000 x 8) : (8,000 x 8)

= 48 : 32 : 64

= 3 : 2 : 4.

Kamal's share = Rs.		4005 x	2		= Rs. 890.
			9

100 years contain 5 odd days.

Last day of 1^st century is Friday.

200 years contain (5 x 2) 3 odd days.

Last day of 2^nd century is Wednesday.

300 years contain (5 x 3) = 15 1 odd day.

Last day of 3^rd century is Monday.

400 years contain 0 odd day.

Last day of 4^th century is Sunday.

This cycle is repeated.

Last day of a century cannot be Tuesday or Thursday or Saturday.

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Video Explanation: https://youtu.be/WCEF3iW3H2c

From I and II, we can find the length and breadth of the rectangle and therefore the area can be obtained.

So, III is redundant.

Also, from II and III, we can find the length and breadth and therefore the area can be obtained.

So, I is redundant.

Correct answer is "II and either I or III".

Let the required number days be x.

Less spiders, More days (Indirect Proportion)

Less webs, Less days (Direct Proportion)

Spiders	1	:	7		:: 7 : x
Webs	7	:	1

1 x 7 x x = 7 x 1 x 7

x = 7.

B.G. =	(T.D.)2	= Rs.		160 x 160		= Rs. 16.
	P.W.			1600

Let the tens and units digit be x and y respectively.

  I. (10x + y) - (10y + x)       x - y = 3.

 II. x - y = 3.

III. x - y = 3.

Thus, even all the given three statements together do not give the answer.

Correct answer is (E).

Volume of the large cube = (3³ + 4³ + 5³) = 216 cm³.

Let the edge of the large cube be a.

So, a³ = 216         a = 6 cm.

Required ratio =		6 x (32 + 42 + 52)		=	50	= 25 : 18.
		6 x 62			36

Let the amount taxable purchases be Rs. x.

Then, 6% of x =	30
	100

x =		30	x	100	= 5.
		100		6

Cost of tax free items = Rs. [25 - (5 + 0.30)] = Rs. 19.70

	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

Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,

4x + 6 = 26       4x = 20

x = 5.

Deepak's age = 3x = 15 years.