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Let the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

	1	+	1	=	1
	x		(x + 6)		4

	x + 6 + x	=	1
	x(x + 6)		4

x² - 2x - 24 = 0

(x -6)(x + 4) = 0

x = 6.     [neglecting the negative value of x]

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27x metres,

and length of the second train = 17y metres.

	27x + 17y	= 23
	x+ y

27x + 17y = 23x + 23y

4x = 6y

	x	=	3	.
	y		2

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E

= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),      (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),      (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(E) = 27.

P(E) =	n(E)	=	27	=	3	.
	n(S)		36		4

  I. Let C.P. be Rs. x.

Then, M.P. = 130% of x = Rs.		13x	.
		10

III. S.P. = 90% of M.P.

Thus, I and III give, S.P. = Rs.		90	x	13x		= Rs.		117x
		100		10				100

Gain = Rs.		117x	- x		= Rs.	17x
		100				100

Thus, from I and III, gain % can be obtained.

Clearly, II is redundant.

Originally, let the number of boys and girls in the college be 7x and 8x respectively.

Their increased number is (120% of 7x) and (110% of 8x).

		120	x 7x		and		110	x 8x
		100					100

	42x	and	44x
	5		5

The required ratio =		42x	:	44x		= 21 : 22.
		5		5

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

Numbers are 1³, 2³, 3³, 4³, 5³, 6³.

So, the missing number is 7³ = 343.

29.94	=	299.4
1.45		14.5

=		2994	x	1		[ Here, Substitute 172 in the place of 2994/14.5 ]
		14.5		10

=	172
	10

= 17.2

Gain = 20%

I. Profit = (S.P.) - (C.P.) = Rs. 40.

Thus, I give the answer. But, II does not give the answer.

Correct answer is (A).

Ratio of times taken by A and B = 1 : 3.

The time difference is (3 - 1) 2 days while B take 3 days and A takes 1 day.

If difference of time is 2 days, B takes 3 days.

If difference of time is 60 days, B takes		3	x 60		= 90 days.
		2

So, A takes 30 days to do the work.

A's 1 day's work =	1
	30

B's 1 day's work =	1
	90

(A + B)'s 1 day's work =		1	+	1		=	4	=	2
		30		90			90		45

A and B together can do the work in	45	= 22	1	days.
	2		2

(A's 1 day's work) : (B's 1 day's work) =	7	: 1   =   7 : 4.
	4

Let A's and B's 1 day's work be 7x and 4x respectively.

Then, 7x + 4x =	1	11x =	1	x =	1	.
	7		7		77

A's 1 day's work =		1	x 7		=	1	.
		77				11

1 woman's 1 day's work =	1
	70

1 child's 1 day's work =	1
	140

(5 women + 10 children)'s day's work =		5	+	10		=		1	+	1		=	1
		70		140				14		14			7

5 women and 10 children will complete the work in 7 days.

Let the number be x.

Then, x + 17 =	60
	x

x² + 17x - 60 = 0

(x + 20)(x - 3) = 0

x = 3.

Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years.

(2x + 2) + 2x + x = 27

5x = 25

x = 5.

Hence, B's age = 2x = 10 years.

By the rule of alligation, we have:

Profit on 1st part Profit on 2nd part
8%	Mean Profit 14%	18%
4		6

Ration of 1^st and 2^nd parts = 4 : 6 = 2 : 3

Quantity of 2nd kind =		3	x 1000	kg	= 600 kg.
		5

S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%.

C.P. of 1 kg of the mixture = Rs.		100	x 68.20		= Rs. 62.
		110

By the rule of alligation, we have:

Cost of 1 kg tea of 1st kind. Cost of 1 kg tea of 2nd kind.
Rs. 60	Mean Price Rs. 62	Rs. 65
3		2

Required ratio = 3 : 2.

Let original length = x and original breadth = y.

Original area = xy.

New length =	x	.
	2

New breadth = 3y.

New area =		x	x 3y		=	3	xy.
		2				2

Increase % =		1	xy x	1	x 100	%	= 50%.
		2		xy

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

   = 1008 + 7

   = 1015

T.D. =		B.G. x 100		= Rs.		270 x 100		= Rs. 750.
		R x T				12 x 3

B.D. = Rs.(750 + 270) = Rs. 1020.