A man reads 3/8 of a book on a day and 4/5 of the remainder on the second day. If the number of pages still unread are 40, then how many pages did the book contain?

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

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

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

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 A has ? 200 at the starting of game, then B has ? 100 at the starting of game.
After the game, money left with A = 200 - 200/3 = ? 400/3
? Total money with B = 100 + 200/3 = ? 500/3
Let the fraction of money lost by B = N
Then, 500/3 - N x 500/3 = 400/3 + N x 500/3
? 500/3 - 400/3 = 500N/3 + 500N/3
? 100/3 = 1000N/3
? N = (100/3) x (3/1000) = 1/10
so, the fraction of money lost is 1/10.

Let the original fraction be p/q.
Numerator is increased by 200%.
? Numerator = p + 200% of p
= p + 200p/100
= p + 2p
= 3p
and denominator of the fraction is increased by 150%.
Denominator = q + 150q/100
= (100q + 150q)/100
= 250q/100
= 5q/2
Then, according to the question,
3p/5q/2 = 9/35
3p x 2/5q = 9/35
6p/5q = 9/35
? p/q = 9 x 5/35 x 6
p/q = 3/14

Let A, B, C and D pay ?x, ?y, ?z and ?a
According to the question,
x = 1/2(y + z + a)
2x = (y + z + a) ..........(i)
y = 1/3(x + z + a)
3y = (x + z + a) ...........(ii)
z = 1/4(x + y + a)
4z = (x + y + a) ..........(iii)
Also, x + y + z + a = 60 ...............(iv)
Now, put the value of x + y + a = 4z from (iii) in (iv)
Then, 4z + z = 60 ? 5z = 60
? z = 12
Similarly, on putting the value of x + z + a = 3y from (ii) in (iv), we get
3y + y = 60 = ? 4y = 60
? y = 15
Again, on putting the value of (y + z + a) = 2x from (i) in (iv), we get
2x + x = 60 ? 3x = 60
&ther4; x = 20
Now, x + y + z + a = 60
On putting the value of x, y and z, we get
12 + 15 + 20 + a = 60
? a = 60 - 47 = ? 13

Let denominator of fraction = x
Then, numerator = x - 4
? Fraction = x - 4/x
Now, according to the question,
8[(x - 4) - 2] = (x + 1)
(x -4) - 2= (x + 1)/8
? x - 6 = x + 1/8
? 8x - 48 = x + 1
? 8x - x = 48 + 1
? 7x = 49
? x = 49/7
? x = 7
? Fraction = (x - 4)/x
Put the value of x.
? Fraction = (7 - 4)/7 =3/7

? = 783/9 x 0.75 = 116

Let us assume length of pole is L.
Length of the pole in Mud is = L x 4/7 = 4L/7
Length of the Pole which is pulled out = L x 1/3 = L/3
Length of the pole is still in mud = Length of the pole in Mud - Length of the Pole which is pulled out
Length of the pole is still in mud = 4L/7 - L/3
According to question,
Length of the pole is still in mud = 250
4L/7 - L/3 = 250
(4L x 3 - L x 7)/7 x 3 = 250
12L - 7L = 250 x 21
5L = 250 x 21
L = 250 x 21/5
L = 50 x 21
L = 1050