A has twice as much money as B. They play together and at the end of the first game B wins one-third of A's money from A, what fraction of the sum that B now has, must A win black in the second game, so that they may have exactly equal money?

Let the weight of vesel = x kg
Then, 1/4 of volume of water = (5.25 - x)kg
Full volume of water = (16.5 - x)
? 4(5.25 - x) = (16.5 - x)
? 21 - 4x = 16.5 - x
? 21 - 16.5 = 3x
? x = 4.5/3 = 1.5 kg

Let the total property = N
Then, property of his daughter = P/7
? Remaining property = P - P/7 = 6P/7
Now, this property is divided among his sons, so that each son gets twice the property of his daughter.
Let number of sons be K.
Then, 6N/7 x K = 2N/7
? K = 3

Let the fraction be N.
Then, N / (13/14) - (13/14) x N = 3/65
? 14N/13 - 13N/14 = 3/65
? (196N - 169N)/182 = 3/65
? 27N/182 = 3/65
? N = (3 x 182) / (27 x 65) = 14/45
So, correct answer will be (13/14) x (14/45) = 13/45

0.3+ 0.6 + 0.7 + 0.8 = x,
? x = 3/9 + 6/9 + 7/9 + 8/9 = 24/9 = 8/3 = 2²/₃

Length of each piece = 3¹/₄ = 13/4 m
Maximum pieces that can be cut = 30/(13/4) = 9
? Rope left = 30 - 13/4 x 9 = 120 - 117/4 = 3/4 m
? Required fraction = 3/4/30 = 1/4

Let the fraction be x, y and z respectively in decreasing order.
Then, according to the question,
x/z = 7/6
? x = 7z/6

And y = 7/6 - 1/3 = 7-2/6 = 5/6

Now, x + y + z = 2¹¹/₂₄
? 7z/6 + 5/6 + z = 59/24
? (13z + 5)/ 6 = 59/24
? 13z = 59/4 - 5 =39/4
? z = 3/4
? x =7z/6 = 7/8
Hence, the fraction are 7/8, 5/6, 3/4.

Let the greatest, middle and smallest fraction be P, Q and R respectively in decreasing order.

According to the question.
(Greatest fraction) / (Smallest fraction) = 7/6
? P/R = 7/6
? P = 7R/6 ....(i)
and Q = 7/6 - 1/3 = 7 - 2/6 = 5/6 (ii)

Now, P + Q + R = 2¹¹/₂₄ (iii )

Substitute the value of P and Q from Eqs. (i), (ii) and put in Eqs (iii), we get
? 7R/6 + 5/6 + R = 59/24
? (7R + 5 + 6R)/6 = 59/24
? (13R + 5) / 6 = 59/24
? 13R = (59/4) - 5 = 39/4
? R = 39 / (4 x 13)
? R = 3/4

On putting the value of R in Eq. (i), we get
P = 7R/6 = (7/6) x (3/4) = 7/8

Shantanu's pocket allowance = ? 3832.50
Total days in 2011 (general year) = 365 days
Allowance per day = 3832.5/365 = ? 10.5

Let total length of pencil be x cm.
Then, black part = x/8
Remaining part pencil after black part = x - x/8 = 7x/8
then White part = 1/2 (7x/8)
= 7x/16
Now Remaining part of pencil is blue = total length of pencil - (length of black part + length of white part)
length of blue part of pencil = x - (x/8 + 7x/16)
length of blue part of pencil = x - (2x + 7x)/16
length of blue part of pencil = x - 9x/16
length of blue part of pencil = (16 x - 9x)/16
? Length of blue part = 7x/16;

According to question
Length of blue part = 3¹/₂
7x/16 = 3¹/₂
? 7x/16 = 7/2
? x/16 = 1/2
? x = 16/2
? x = 8 cm

Let the number of pages be p.
Number of pages read on first day = 3p/8
Remaining pages = p - 3p/8 = (8p - 3p)/8 = 5p/8
Number of pages read on second day = 4/5 x 5p/8 = p/2
Now , remaining pages = p - [ 3p/8 + p/2]
= p - [(3p + 4p)/8] = [p- 7p/ 8]
= [(8p - 7p)/ 8] = p/8
According to the question
p/8 = 40
? p = 40 x 8
? p = 320