Online Aptitude Test

Formula: (Divisor*Quotient) + Remainder = Dividend.

Soln:

(56*Q)+29 = D -------(1)

D%8 = R -------------(2)

From equation(2),

((56*Q)+29)%8 = R.

=> Assume Q = 1.

=> (56+29)%8 = R.

=> 85%8 = R

=> 5 = R.

Cost Price of 1 toy = Rs.		375		= Rs. 31.25
		12

Selling Price of 1 toy = Rs. 33

So, Gain = Rs. (33 - 31.25) = Rs. 1.75

Profit % =		1.75	x 100	%	=	28	% = 5.6%
		31.25				5

Let the total profit be Rs. 100.

After paying to charity, A's share = Rs.		95 x	3		= Rs. 57.
			5

If A's share is Rs. 57, total profit = Rs. 100.

If A's share Rs. 855, total profit =		100	x 855		= 1500.
		57

Required sum = (2 + 4 + 6 + ... + 30)

This is an A.P. in which a = 2, d = (4 - 2) = 2 and l = 30.

Let the number of terms be n. Then,

t_n = 30     a + (n - 1)d = 30

 2 + (n - 1) x 2 = 30

 n - 1 = 14

 n = 15

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

S.I. = Rs. (15500 - 12500) = Rs. 3000.

Rate =		100 x 3000	%	= 6%
		12500 x 4

Let the ten's digit be x and unit's digit be y.

Then, number = 10x + y.

Number obtained by interchanging the digits = 10y + x.

 (10x + y) + (10y + x) = 11(x + y), which is divisible by 11.

Amount	= Rs. 25000 x 1 + 12 3 100
	= Rs. 25000 x 28 x 28 x 28 25 25 25
	= Rs. 35123.20

 C.I. = Rs. (35123.20 - 25000) = Rs. 10123.20

Average of 20 numbers = 0.

 Sum of 20 numbers (0 x 20) = 0.

It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).

Let man's rate upstream be x kmph.

Then, his rate downstream = 2x kmph.

(Speed in still water) : (Speed of stream) =		2x + x		:		2x - x
		2				2

   = 3 : 1.

(11² + 12² + 13² + ... + 20²) = (1² + 2² + 3² + ... + 20²) - (1² + 2² + 3² + ... + 10²)

	Ref: (12 + 22 + 32 + ... + n2) =	1	n(n + 1)(2n + 1)
		6

=		20 x 21 x 41	-	10 x 11 x 21
		6		6

= (2870 - 385)

= 2485.

(4 + 5 + 2) - (1 + 6 + 3) = 1, not divisible by 11.

(2 + 6 + 4) - (4 + 5 + 2) = 1, not divisible by 11.

(4 + 6 + 1) - (2 + 5 + 3) = 1, not divisible by 11.

(4 + 6 + 1) - (2 + 5 + 4) = 0, So, 415624 is divisible by 11.

Let the number be x. Then

	60	x	3	x x = 36
	100		5

x =		36 x	25		= 100
			9

 Required number = 100

P can complete the work in (12 x 8) hrs. = 96 hrs.

Q can complete the work in (8 x 10) hrs. = 80 hrs.

P's1 hour's work =	1	and Q's 1 hour's work =	1	.
	96		80

(P + Q)'s 1 hour's work =		1	+	1		=	11	.
		96		80			480

So, both P and Q will finish the work in		480		hrs.
		11

Number of days of 8 hours each =		480	x	1		=	60	days = 5	5	days.
		11		8			11		11

Let the number be x and on dividing x by 5, we get k as quotient and 3 as remainder.

    x = 5k + 3

    x² = (5k + 3)²

   = (25k² + 30k + 9)

   = 5(5k² + 6k + 1) + 4

On dividing x² by 5, we get 4 as remainder.

Let the number of students in rooms A and B be x and y respectively.

Then, x - 10 = y + 10      x - y = 20 .... (i)

     and x + 20 = 2(y - 20)      x - 2y = -60 .... (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

 The required answer A = 100.

The year 2006 is an ordinary year. So, it has 1 odd day.

So, the day on 8^th Dec, 2007 will be 1 day beyond the day on 8^th Dec, 2006.

But, 8^th Dec, 2007 is Saturday.

 8^th Dec, 2006 is Friday.

Amount = Rs. (30000 + 4347) = Rs. 34347.

Let the time be n years.

Then, 30000		1 +	7		n	= 34347
			100

		107		n	=	34347	=	11449	=		107		2
		100				30000		10000			100

 n = 2 years.

Clearly, 4864 is divisible by 4.

So, 9P2 must be divisible by 3. So, (9 + P + 2) must be divisible by 3.

 P = 1.