Fractions
 A has twice as much money as B. They play together and at the end of the first game B wins onethird 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 ?

 1/3
 1/5
 1/4
 1/10

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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 500/3Correct Option: D
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.
 Sum of three fractions is 2^{11}/_{24}. If the greatest fraction is divided by the smallest fraction, the result is 7/6, which is greater than the middle fraction by 1/3. find all the three fractions ?

 3/5, 4/7, 2/3
 7/8, 5/6, 3/4
 7/9, 2/3, 3/5
 7/8, 7/9, 7/10

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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 = 72/6 = 5/6
Now, x + y + z = 2^{11}/_{24}Correct Option: B
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 = 72/6 = 5/6
Now, x + y + z = 2^{11}/_{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.
 Sum of three fraction is 2^{11}/_{24} . If the greatest fraction is divided by the smallest fraction the result is 7/6, which is greater than the middle fraction by 1/3. Find all the three fractions.

 3/5, 4/7, 2/3
 7/8, 5/6, 3/4
 7/9, 2/3, 3/5
 7/8, 7/9, 7/10
 None of the above

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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 = 2^{11}/_{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 = 2^{11}/_{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
 In the year 2011, Shantanu gets ₹ 3832.5 as his pocket allowance. Find his pocket allowance per day.

 ₹ 9.5
 ₹ 10.5
 ₹ 12.5
 ₹ 11.5
 None of the above

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Shantanu's pocket allowance = ₹ 3832.50
Total days in 2011 (general year) = 365 daysCorrect Option: B
Shantanu's pocket allowance = ₹ 3832.50
Total days in 2011 (general year) = 365 days
Allowance per day = 3832.5/365 = ₹ 10.5
 1/8 part of a pencil is black and 1/2 part of the remaining is white. If the remaining part is blue and length of this blue part is 3^{1}/_{2} cm, then find the length of the pencil.

 6 cm
 7 cm
 8 cm
 9 cm
 None of the above

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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 = 3^{1}/_{2}
7x/16 = 3^{1}/_{2}
⇒ 7x/16 = 7/2
⇒ x/16 = 1/2
⇒ x = 16/2
∴ x = 8 cm