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 By: Kuro (offline)  Sunday, July 06 2014 @ 08:30 AM CDT (Read 740 times)
Kuro

1. Mdm Susan bought 1/3 as many chocolates as sweets. She gave each of her neighbours' children 4 chocolates and 3 sweets, after which she had 6 chocolates and 180 sweets left.
a) How many children received the chocolates and sweets? Ans 18
b) How many sweets did she buy? Ans 234

Junior

Registered: 12/31/06
Posts: 27

 By: echeewh (offline)  Monday, July 07 2014 @ 12:07 AM CDT
echeewh

Hello <Kuro>

Following is my worked solution:

Let the number of children be 1u { since the chocolates(C) and sweets(S) were given to the same number of children }

Number of C given out: 4u
Number of S given out: 3u

Let the initial number of C and S be 1p and 3p respectively.

<C>
1p - 4u = 6 -- (1)

<S>
3p - 3u = 180 -- (2)

Using <simultaneous> method to solve these 2 equations, we eliminate <p> { since we need to know the number of Children }

(1)x3:
3p - 12u = 18 -- (3)

Comparing (3),(2), we take ...

(2)-(3):
-3u - (-12u) = 180 - 18
12u - 3u --> 162
9u --> 162
1u --> 18

(a)
Number of Children = 1u --> 18

(b)
Number of S bought = 3p

in (2):
3p = 180 + 3u --> 180 + (3 × 18)
= 180 + 54
= 234

========

Trust this helps.

Do let me know again if there's further clarification .

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

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