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 Forum Index >  Test Paper Related >  Primary 6 Matters
 Math Problem Sum
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By: 2000123 (offline)  Thursday, August 09 2012 @ 09:57 PM CDT (Read 1235 times)  
2000123

Q1.A group of children read an average of 48 books in 3months. The table below show the number of boys and girls and the average number of books each child read. How many girls were there?

Number Average no. of books

Boys 16 40
Girls ? 64

Q2.Genie paid a total of $258.40 for 2 similar dresses and 2 skirts after a discount of 15% was given on the price of each item.If the original price of each dress was 11/9 times the price of the skirt,find the price of each dress before the discount.

Q3.Mrs Lee bought some stationery. 1/4 of the stationery were pens which cost $1.50 each,2/3 of them were files which cost $1.20 each and the rest were erasers which cost $0.75 each.She spent $238.05 on the files and erasers.How much more money did she spend on files than pens?

Q4.An egg-seller had chicken eggs,duck eggs and quail eggs. He had 225 more quails eggs than duck eggs. After he had sold 60% of the chicken eggs,2/3 of the duck eggs and 3 times as many quail eggs as duck eggs sold, he found that he had 25% of his quail eggs and a total of 205 eggs left.Find the total number of eggs he had at first.

Q5.Mdm Ravi packed a sack of salt into packets of 150g and 500g. The number of 150g packets to the number of the 500g packets was 3:1. After selling 30 packets of 150g and 6 packets of 500g, the number of 500g packets was 60% of the number of 150g packets. What was the mass of salt Mdm Ravi had at first?

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By: jo sarah (offline)  Saturday, August 11 2012 @ 09:38 AM CDT  
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Q1. [i take an algebraic approach]
Let the no. of girls be n
Then, total read by boys = 16 x 40 = 640
total read by girls = 64n
total of all children = 640 + 64n = 64(10 + n)
From the fact that the group read an average of 48 books,
the total read by the group = 48(16 + n)

So that, 64(10 + n) = 48(16 + n)
i.e., 4(10 + n) = 3(16 + n)
40 + 4n = 48 + 3n
thus, n = 8
that is to say, there were 8 girls.

Q2. Original price of dress to skirt = 11 : 9
If price of a dress is 11u and of a skirt 9u
After the discount, the prices are respectively 9.35u and 7.65 u
so that the total = 17u

Now, since 2 dresses and 2 skirts cost $258.40
so that, 1 dress and 1 skirt cost $129.20
thus, 17u = 129.2
and so, 1u = 7.6
Thus, the price of the dress before discount was $83.60

Q3. Let the total stationery be 12u
so, she bought 3u pens, 8u files and therefore, 1u erasers.
the cost of files and erasers = $(8u x 1.2 + 1u x 0.75)
i.e., 10.35u = 238.05
and so, 1u = 23
thus, she spent $(8 x 23 x 1.2 - 3 x 23 x 1.5) = $117.30 more on files than on pens.

then, 1 dress and 1 skirt cost $258.40/2 = $129.20

therefore, 17u = 129.2
and so, 1u = 7.6

The price of each dress before discount = 11u = $83.60.

Q3. Let the total stationery be 12u
then, 3u were pens, 8u were files and so, 1u erasers.
total for files and erasers = 8u x 1.20 + 1u x 0.75 = 10.35u
and this is $238.05
thus, 1u = 23
Money spend on files more than on pens = $(8 x 23 x 1.2 - 3 x 23 x 1.5)
= $117.30.

Q4.

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By: jo sarah (offline)  Saturday, August 11 2012 @ 10:03 AM CDT  
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Q4. Let the seller had 1p chicken eggs (C), 1u duck eggs (D) and (1u + 225) quail eggs (Q) at first.
tabulate: ....C....................D.....................Q
at first:.......1p..................1u...............1u + 225
change:..-0.6p.............-(2/3)u...............(-2u)
in end:.....0.4p..............(1/3)u............(225 - 1u)

he had 1/4 Q left, that means 4(225 - 1u) = 1u + 225
so that, 5u = 3 x 225
and, 1u = 135

total no. of eggs left in end was 205.
that is, 0.4p + (1/3)u + 225 - 1u = 205
0.4p = 205 + (2/3)u - 225 = (2/3) x 135 - 20 = 70
and 1p = 70 / 0.4 = 175

So, he had a total of 175 + 2 x 135 + 225 = 670 eggs at first.

Q5........tabulate:
...................150g packets.........500g packets
at first:...............3u............................1u
sold:.................-30............................-6
left:................3u - 30.....................1u - 6
in ratio: 5 : 3

thus, 3 x (3u - 30) = 5 x (1u - 6)
i.e., 9u - 90 = 5u - 30
4u = 60
1u = 15

Mass of salt she had at first = (3 x 15 x 150 + 1 x 15 x 500) g = 14250 g = 14.25 kg

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By: 2000123 (offline)  Saturday, August 11 2012 @ 09:22 PM CDT  
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Thanks Smile

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