Online Aptitude Test - Aptitude Test 4

Let the total number of shots be x. Then,

Shots fired by A =	5	x
	8

Shots fired by B =	3	x
	8

Killing shots by A =	1	of	5	x	=	5	x
	3		8			24

Shots missed by B =	1	of	3	x	=	3	x
	2		8			16

	3x	= 27 or x =		27 x 16		= 144.
	16			3

Birds killed by A =	5x	=		5	x 144		= 30.
	24			24

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

Then,	(6x + 6) + 4	=	11
	(5x + 6) + 4		10

10(6x + 10) = 11(5x + 10)

5x = 10

x = 2.

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

Given that:

1. The difference of age b/w R and Q = The difference of age b/w Q and T.

2. Sum of age of R and T is 50 i.e. (R + T) = 50.

Question: R - Q = ?.

Explanation:

R - Q = Q - T

(R + T) = 2Q

Now given that, (R + T) = 50

So, 50 = 2Q and therefore Q = 25.

Question is (R - Q) = ?

Here we know the value(age) of Q (25), but we don't know the age of R.

Therefore, (R-Q) cannot be determined.

1	+	1	=	1  +  1 1 + an am 1 + am an
1 + a(n - m)		1 + a(m - n)

=	am	+	an
	(am + an)		(am + an)

=	(am + an)
	(am + an)

   = 1.

Given Exp. =

1  +  1  +  1 1 + xb + xc xa xa 1 + xa + xc xb xb 1 + xb + xa xc xc

=	xa	+	xb	+	xc
	(xa + xb + xc)		(xa + xb + xc)		(xa + xb + xc)

=	(xa + xb + xc)
	(xa + xb + xc)

   = 1.

Simran : Nanda = (50000 x 36) : (80000 x 30) = 3 : 4.

Simran's share = Rs.		24500 x	3		= Rs. 10,500.
			7

A's 2 day's work =		1	x 2		=	1	.
		20				10

(A + B + C)'s 1 day's work =		1	+	1	+	1		=	6	=	1	.
		20		30		60			60		10

Work done in 3 days =		1	+	1		=	1	.
		10		10			5

Now,	1	work is done in 3 days.
	5

Whole work will be done in (3 x 5) = 15 days.

Let the distance travelled by x km.

Then,	x	-	x	= 2
	10		15

3x - 2x = 60

x = 60 km.

Time taken to travel 60 km at 10 km/hr =		60	hrs	= 6 hrs.
		10

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.

Required speed =		60	kmph.	= 12 kmph.
		5

Suppose he move 4 km downstream in x hours. Then,

Speed downstream =		4		km/hr.
		x

Speed upstream =		3		km/hr.
		x

	48	+	48	= 14 or x =	1	.
	(4/x)		(3/x)		2

So, Speed downstream = 8 km/hr, Speed upstream = 6 km/hr.

Rate of the stream =	1	(8 - 6) km/hr = 1 km/hr.
	2

By the rule of alligation:

Cost of 1 kg pulses of 1st kind Cost of 1 kg pulses of 2nd kind
Rs. 15	Mean Price Rs. 16.50	Rs. 20
3.50		1.50

Required rate = 3.50 : 1.50 = 7 : 3.

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

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

So, a + b = 1.

A : B = 200 : 169.

A : C = 200 : 182.

C

=

C

x

A

=

182

x

200

= 182 : 169.

B

A

B

200

169

When C covers 182 m, B covers 169 m.

When C covers 350 m, B covers		169	x 350	m	= 325 m.
		182

Therefore, C beats B by (350 - 325) m = 25 m.

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.

The day on 6^th March, 2005 will be 1 day beyond the day on 6^th March, 2004.

Given that, 6^th March, 2005 is Monday.

6^th March, 2004 is Sunday (1 day before to 6^th March, 2005).

Angle traced by hour hand in 12 hrs = 360°.

Angle traced by hour hand in 5 hrs 10 min. i.e.,	31	hrs =		360	x	31		°	= 155°.
	6			12		6

Angle traced by the hour hand in 6 hours =		360	x 6		°	= 180°.
		12

For an income of Rs. 756, investment = Rs. 9000.

For an income of Rs.	21	, investment = Rs.		9000	x	21		= Rs. 125.
	2			756		2

For a Rs. 100 stock, investment = Rs. 125.

Market value of Rs. 100 stock = Rs.		125 -	1		= Rs. 124.75
			4

Clearly, there are 52 cards, out of which there are 12 face cards.

P (getting a face card) =	12	=	3	.
	52		13

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

P.W. = Rs. (2562 - 122) = Rs. 2440.

S.I. on Rs. 2440 for 4 months is Rs. 122.

Rate =		100 x 122	%	= 15%.
		2440 x 1 3

Each of the number except 279 is a multiple of 11.