
By: jan (offline) Saturday, January 25 2014 @ 10:09 PM CST (Read 1354 times)



jan 
Please help me on these 2 qns. Thank you.
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



