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Speed in still water =	1	(11 + 5) kmph = 8 kmph.
	2

  I. gives, S.I for 3 years = Rs. 4500.

 II. gives, Rate = 10% p.a.

III. gives, (C.I.) - (S.I.) = Rs. 465.

Each number is the proceeding number multiplied by -2.

So, the required number is -128.

Number of bricks =	Volume of the wall	=		800 x 600 x 22.5		= 6400.
	Volume of 1 brick			25 x 11.25 x 6

Let the number of students be x. Then,

Number of students above 8 years of age = (100 - 20)% of x = 80% of x.

80% of x = 48 +	2	of 48
	3

	80	x = 80
	100

x = 100.

Let 40% of A =	2	B
	3

Then,	40A	=	2B
	100		3

	2A	=	2B
	5		3

	A	=		2	x	5		=	5
	B			3		2			3

A : B = 5 : 3.

P.W. = Rs.		100 x 2310		= Rs. 1680.
		100 + 15 x 5 2

Go on subtracting 3 and dividing the result by 2 to obtain the next number.

Clearly, 46 is wrong.

(A + B)'s 1 day's work =		1	+	1		=	1	.
		15		10			6

Work done by A and B in 2 days =		1	x 2		=	1	.
		6				3

Remaining work =		1 -	1		=	2	.
			3			3

Now,	1	work is done by A in 1 day.
	15

	2	work will be done by a in		15 x	2		= 10 days.
	3				3

Hence, the total time taken = (10 + 2) = 12 days.

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 speed of the train be x metres/sec.

Time taken to cross a signal pole =	Length of the train
	Speed of the train

Time taken to cross a platform =	(Length of the train + Length of the Platform)
	Speed of the train

Length of train = 330 m.

I and III give, 18 =	330	x =	330	m/sec =	55	m/sec.
	x		18		3

II and III give, 36 =	2 x 330	x =	660	m/sec =	55	m/sec.
	x		36		3

Correct answer is (D).

Let the length of the train be x metres and its speed by y m/sec.

Then,	x	= 8         x = 8y
	y

Now,	x + 264	= y
	20

8y + 264 = 20y

y = 22.

Speed = 22 m/sec =		22 x	18		km/hr = 79.2 km/hr.
			5

Let there be x pupils in the class.

Total increase in marks =		x x	1		=	x
			2			2

	x	= (83 - 63)	x	= 20      x= 40.
	2		2

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S)	= Number of ways of drawing 2 balls out of 7
	= 7C2 `
	= (7 x 6) (2 x 1)
	= 21.

Let E = Event of drawing 2 balls, none of which is blue.

n(E)	= Number of ways of drawing 2 balls out of (2 + 3) balls.
	= 5C2
	= (5 x 4) (2 x 1)
	= 10.

P(E) =	n(E)	=	10	.
	n(S)		21

Gain in 2 years	= Rs. 5000 x 25 x 2 - 5000 x 4 x 2 4 100 100
	= Rs. (625 - 400)
	= Rs. 225.

Gain in 1 year = Rs.		225		= Rs. 112.50
		2

Part filled in 4 minutes = 4		1	+	1		=	7	.
		15		20			15

Remaining part =		1 -	7		=	8	.
			15			15

Part filled by B in 1 minute =	1
	20

	1	:	8	:: 1 : x
	20		15

x =		8	x 1 x 20		= 10	2	min = 10 min. 40 sec.
		15				3

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour =		4 x	1		=	2	.
			6			3

Remaining part =		1 -	1		=	1	.
			2			2

	2	:	1	:: 1 : x
	3		2

x =		1	x 1 x	3		=	3	hours i.e., 45 mins.
		2		2			4

So, total time taken = 3 hrs. 45 mins.

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.

Let the required number of working hours per day be x.

More pumps, Less working hours per day (Indirect Proportion)

Less days, More working hours per day (Indirect Proportion)

Pumps	4	:	3		:: 8 : x
Days	1	:	2

4 x 1 x x = 3 x 2 x 8

x =	(3 x 2 x 8)
	(4)

x = 12.

Gain = 25% of C.P.

In order to find gain, we must know the sale price of each article and the number of articles sold.

Correct answer is (D).