Online Aptitude Test

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left =		7x -	7	x 9		litres =		7x -	21	litres.
			12						4

Quantity of B in mixture left =		5x -	5	x 9		litres =		5x -	15	litres.
			12						4

	7x - 21 4	=	7
	5x - 15  + 9 4		9

	28x - 21	=	7
	20x + 21		9

 252x - 189 = 140x + 147

 112x = 336

 x = 3.

So, the can contained 21 litres of A.

Clearly, (2272 - 875) = 1397, is exactly divisible by N.

Now, 1397 = 11 x 127

 The required 3-digit number is 127, the sum of whose digits is 10.

(3²⁵ + 3²⁶ + 3²⁷ + 3²⁸) = 3²⁵ x (1 + 3 + 3² + 3³) = 3²⁵ x 40

     = 3²⁴ x 3 x 4 x 10

     = (3²⁴ x 4 x 30), which is divisible by30.

132 = 4 x 3 x 11

So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.

264  11,3,4 (/)

396  11,3,4 (/)

462  11,3 (X)

792  11,3,4 (/)

968  11,4 (X)

2178  11,3 (X)

5184  3,4 (X)

6336  11,3,4 (/)

Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.

Required number of number = 4.

The year 2004 is a leap year. It has 2 odd days.

 The day on 8^th Feb, 2004 is 2 days before the day on 8^th Feb, 2005.

Hence, this day is Sunday.

Number of shares =		4455		= 540.
		8.25

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

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

The year 2008 is a leap year. So, it has 2 odd days.

1^st day of the year 2008 is Tuesday (Given)

So, 1^st day of the year 2009 is 2 days beyond Tuesday.

Hence, it will be Thursday.

Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.

 10(6x + 10) = 11(5x + 10)

 5x = 10

 x = 2.

 Sagar's present age = (5x + 6) = 16 years.

Let S_n = (2 + 4 + 6 + ... + 30). This is an A.P. in which a = 2, d = 2 and l = 30

Let the number of terms be n. Then,

a + (n - 1)d = 30

     2 + (n - 1) x 2 = 30

    n = 15.

Sn =	n	(a + l)	=	15	x (2 + 30) = (15 x 16) = 240.
	2			2

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

 Required number = (90 x 4) + 4   = 364.

Ratio of rates of working of A and B = 2 : 1.

So, ratio of times taken = 1 : 2.

A's 1 day's work =	1	; (2 times of B's work)
	6

(A + B)'s 1 day's work =		1	+	1		=	3	=	1	.
		6		12			12		4

So, A and B together can finish the work in 4 days.

Required numbers are 102, 108, 114, ... , 996

This is an A.P. in which a = 102, d = 6 and l = 996

Let the number of terms be n. Then,

a + (n - 1)d = 996

   102 + (n - 1) x 6 = 996

   6 x (n - 1) = 894

   (n - 1) = 149

   n = 150.

Unit digit in 7⁹⁵ = Unit digit in [(7⁴)²³ x 7³]
= Unit digit in [(Unit digit in(2401))²³ x (343)]
= Unit digit in (1²³ x 343)
= Unit digit in (343)
= 3

Unit digit in 3⁵⁸ = Unit digit in [(3⁴)¹⁴ x 3²]
= Unit digit in [Unit digit in (81)¹⁴ x 3²]
= Unit digit in [(1)¹⁴ x 3²]
= Unit digit in (1 x 9)
= Unit digit in (9)
= 9

Unit digit in (7⁹⁵ - 3⁵⁸) = Unit digit in (343 - 9) = Unit digit in (334) = 4.

So, Option B is the answer.

Number = (12 x 35)

Correct Quotient = 420 � 21 = 20