Sunday, March 28, 2010

Decimal P5 2009 SA1 P2 MGS Q13

13. Arnold has some 50-cent coins and Weili has some 20-cent coins. Arnold has 5 fewer coins than Weili. However, the amount of money Arnold has is $2.60 more than the amount of money Weili has. [P5 2009 SA1 P2 MGS Q13]
a) How much money does Arnold have? [3]
b) How many 20-cent coins does Weili have? [1]


Arnold has 5 fewer coins than Weili. However, the amount of money Arnold has is $2.60 more than the amount of money Weili has.
Given the above information, we need to remove the 5 additional 20-cent coins to make the number of 20-cent and 50-cent coins the same, so as to group them as a set.
5 x $0.20 = $1.00

By removing the 5 additional 20-cent coins, the difference in value between the 20-cent coins and 50-cent coins will increase.
1.00 + 2.60 = $3.60

With the total difference in value and the 1-to-1 difference, we will be able to find the number of sets.
0.50 – 0.20 = $0.30
3.60 ÷ 0.30 = 12 sets

12 x 0.50 = $6.00

a) Arnold has $6.00.


To get the number of 20-cent coins, we need to put back the 5 20-cent coins removed from the sets.
12 + 5 = 17 20-cent coins


b) Weili has 17 20-cent coins.


Alternative method:

5 x 0.20 = $1.00

Number
W 20¢ [1 u ][5]
A  50¢ [1 u ]
-------------------
Amount
W 20¢ [ 10 u ][ 10 u ][100]<       260        >
A  50¢ [         ][         ][         ][         ][         ]

30 u -- > 100 + 260
1 u -- > 360 ÷ 30 = 12

A 50¢ -- > 12 x 50 = 600¢
                              = $6.00

a) Arnold has $6.00.


W 20¢ -- > 12 + 5 = 17 20¢ coins

b) Weili has 17 20-cent coins.

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