madhavsrinivas wrote:
Emma got a certain amount as pocket money from her father. She went to a book store and bought a book for herself by paying $4 more than half her pocket money. She then went to a shop and bought a pencil by paying $ 4 more than one-third the amount left with her. After this, she bought some chocolates for her younger sister by paying $ 4 more than one-fifth the amount left with her. She could have bought twice the number of chocolates for herself with the amount left with her, but she preferred to save the money. How much more did she pay for the book than for the pencil?
A. $49
B. $48
C. $36
D. $38
E. $37
This question is lengthy rather than tough and the language of the question might confuse you. It can be solved by taking x as the value of pocket money or to solve it faster we can use reverse solving approach.
Reverse solving approach can come into practice only if we understand the problem quickly and then solve a lots of such problems.
So. Lets 1st solve by:
Lengthy process :Let X be the amount of pocket money received by Emma.
Money used to buy Book = 4 + X/2
Money Left = X/2 -4
Money used to buy Pencil = 4 + 1/3*(X/2 -4 )
Money Left = 2/3*(X/2 -4 ) - 4
Money used to buy Chocolate = 4 + 1/5*(2/3*(X/2 -4 ) - 4)
Money Left = [4/5* (2/3*(X/2 -4 ) - 4)-4]
Now money left can buy twice the number of chocolates.
So, [4/5* (2/3*(X/2 -4 ) - 4)-4] = 2 * [ 4 + 1/5*(2/3*(X/2 -4 ) - 4)]
-> 12 = 2/5*(2/3*(X/2 -4 ) - 4)
->30+4 = 2/3*(X/2 -4 )
-> 55= X/2
-> X = 110
So, Money used to buy Book = 4 + X/2 = 59
Money used to buy Pencil = 4 + 1/3*(X/2 -4 ) = 21
Difference = 38
Answer DLet shorten this process by a bit.
Medium Lengthy process or Shorter Approach To solve this process, first check what fractions are used. So here, 1/3rd, 1/5th and 1/2 is used to purchase the items. So we will LCM of denominators of fraction to assume the value of pocket money received by Emma.
LCM of 3,5,2 = 30
Let 30X be the amount of pocket money received by Emma.
Money used to buy Book = 4 + 15X
Money Left = 15X -4
Money used to buy Pencil = 4 + 1/3*(15X-4 ) = 5X +8/3
Money Left = 2/3*(15X -4 ) - 4 = 10X -20/3
Money used to buy Chocolate = 4 + 1/5*(10X-20/3) = 2X + 4 -4/3 = 2X + 8/3
Money Left = [4/5* (10X-20/3)-4] = 8X-16/3 - 4 = 4X-28/3
Now money left can buy twice the number of chocolates.
So, [8X-28/3] = 2 * [2X + 8/3]
-> 4X = 28/3 + 16/3 = 44/3
-> X = 11/3
-> 30X = 110
Pocket Money received = 110
So, Money used to buy Book = 4 + 15X = 59
Money used to buy Pencil = 5X +8/3 = 55/3 +8/3 = 21
Difference = 38
Answer DLets try to solve it by relatively more shorter approach by solving the problem in reverse order.
Shortest Approach Let the amount left after buying pencil be X
So, Money used to buy chocolate = X/5 + 4
Money left after buying chocolate = 4X/5 - 4
Money left after buying chocolate = 2 x Money used to buy chocolate
-> 4X/5 - 4 = 2 (X/5 + 4)
-> 2X/5 = 12
-> X = 30
So, Money left after buying Book = 3/2*(30+4) = 51
Money used to buy Pencil = 4 + 1/3 *51 = 21
Money received by Emma = 2*(51+4) = 110
Money used to buy book = 4+1/2*110 = 59
So, Difference = 59 - 21 = 38
Answer DSo, we can solve by these three methods.
Longer method will take approx 5 minutes.
Shorter Method will take approx 3.5 minutes.
SHortest method will take approx 2 minutes. But it need a lots of practice to start with this approach.
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