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 By: jan (offline)  Saturday, January 25 2014 @ 10:09 PM CST (Read 1394 times)
jan

1. There are some chocolate and strawberry candies in a container. 2/5 of the chocolate candies and 1/2 of the strawberry candies add up to 138 candies. 2/3 of the chocolate candies and 5/7 of the strawberry candies add up to 210 candies. How many chocolate candies are there?

2. Jo spent 5/8 of his money on a book and 7 pens. The cost of each pen is 1/6 of his remaining money. If the total cost of 7 pens is \$16 more than the cost of the book, how much did Jo have at first?

Newbie

Registered: 05/10/09
Posts: 2

 By: echeewh (offline)  Sunday, January 26 2014 @ 05:08 PM CST
echeewh

Hey <jan>

Appreciate if you can provide the <Answerkey> as well for verification purpose in future. Thank you.

My worked solution is as follows:

Q1.

(2/5)C + (1/2)S = 138 --- (1)
(2/3)C + (5/7)S = 210 --- (2)

(1)x10:
4C + 5S = 1380 --- (3)

(2)x21:
14C + 15S = 4410 --- (4)

Applying Simultaneous method, we Eliminate S,

(3)x3:
12C + 15S = 4140 --- (5)

Comparing (4), (5),
(4)-(5):

14C - 12C = 4410 - 4140
2C = 270
C = 270 ÷ 2 = 135

=========

Q2.

B + 7P = (5/8)total --- (1)
1P = (1/6) × (3/8)total
= (1/16)total
7P = 7 × (1/16)total = (7/16)total

In (1), we have ...

B + (7/16)total = (5/8)total
B = (5/8)total - (7/16)total
= (3/16)total

Given that total cost of 7 pens is \$16 more than the cost of the book,

7P - 1B = 16
(7/16)total - (3/16)total = 16
(4/16)total = 16
Total (at first) = 16 ÷ (4/16)
= 16 × (16/4)
= \$64

========

Trust this helps.

Do let me know again if this is different from your <Answerkey> or if there's further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 By: jan (offline)  Wednesday, January 29 2014 @ 07:53 PM CST
jan

Thanks, Edward.

I'd solved them and same answer as yours.
Thanks again.

Newbie

Registered: 05/10/09
Posts: 2

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