Shantanu's pocket allowance = ₹ 3832.50
Total days in 2011 (general year) = 365 days

Correct Option: B

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

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]

Correct Option: C

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

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)

put the value and solve the equation.

Correct Option: C

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 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

Correct Option: E

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 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),

Correct Option: B

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