F.1 maths questions

[匿名]

2008-05-19 1:37 am

1. One side of a triangular lot is 10m shorter than twice the length of the
second side. The third side is 9 m shorter than the first. Te perimeter of
the lot is 61m. How long is the shortest side?
Ans: 17m
2. May has 80 coins which consists of 10 cents coins, 20 cents coins and $1
coins. Among the coins, there are 4 more $1 coins than 10 cents coins
and 6 more 20 cents coins than $1 coins . How much money has she?
Ans: $34.6
3. A collection of and $2 coins has a total value of $28.5.
If there are 30 coins in the collection, hoe many coins of each kind are
there in the collection?
Ans: 21 of 50 cents coins,9 of $2 coins

回答 (1)

cliffordchow1225

2008-05-19 2:05 am

✔ 最佳答案

1) Let x be the length of the shortest side
Then length of first side is x + 9
and first side = 2(second side) - 10
second side = [(x+9)+10]/2

x + (x+9) + [(x +9+ 10)]/2 = 61
2x + 2x + 18 +x +19 = 122
5x = 85
x = 17 (m)

Therefore the shortest side is 17 cm.

2) Let y be the number of 10 cents
then number of $1 is y + 4
number of 20cents is (y+4)+6

Total number of coins is 80, then
y + (y+4) + (y+4+6) = 80
3y = 80 -4-10
3y = 66
y = 22
Therefore May has 22 10-cents, 32 20-cents and 26 $1.
Value of coins
= 22 x 0.1 + 32 x 0.2 + 26 x 1
= 2.2 + 6.4 + 26
= 34.6
Therefore May has $34.6

3) Let p be the number of 50cents coins
0.5p + 2(30-p) = 28.5
p +120 - 4p = 57
3p = 63
p = 21
Therefore, there are 21 50-cents, and 9 $2 coins

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