Q&A - P5 Math

[Superbugs]

Nan Hua SA2 Q16 2010

Mr Bond had some pencils, rulers & erasers. 60% of the stationery were pencils, 30% of them were rulers while the rest were erasers. There were 240 more pencils than erasers. After Mr Bond sold some of his pencils, 40% of the stationery left were pencils.
a) How many pencils did Mr Bond have at first?
b) How many pencils were sold?

P-60%
R-30%
E-10%
Difference in % between P&E=50%=240
a) 60%=(60%/50%)x240=288

E+R-60%=(40%/50%)x240=192
P-40%=(40%/60%)x192=128
Therefore, P sold = 288-128=160
However answer sheet shows 60 only. Pls help. Tks

[linden2000]

linden2000:

Chan09:

Need some help on this:

Mrs A made a jar of cookies in the morning. Out of the 2/7 which she took out, 18 were given away and the remaining 52 cookies were eaten by her children. In the evening, she made another lot of cookies and she put 1/3 of them into the same jar as those made in the morning. After this, the total number of cookies in the jar was 204. How many cookies did she made in the evening?
Ans: 87

2/7 of morning cookies = 52+18=70
5/7 of morning cookies (in jar) = 70/2 x 5=175

1/3 of evening cookies = 204-175=29
No. of evening cookies = 29 x 3=87

Hi superbugs. Hope this is useful to you.

[Superbugs]

linden2000:

linden2000:

[quote=\"Chan09\"]Need some help on this:

Mrs A made a jar of cookies in the morning. Out of the 2/7 which she took out, 18 were given away and the remaining 52 cookies were eaten by her children. In the evening, she made another lot of cookies and she put 1/3 of them into the same jar as those made in the morning. After this, the total number of cookies in the jar was 204. How many cookies did she made in the evening?
Ans: 87

2/7 of morning cookies = 52+18=70
5/7 of morning cookies (in jar) = 70/2 x 5=175

1/3 of evening cookies = 204-175=29
No. of evening cookies = 29 x 3=87

Tks a lot, see how careless both me & my daughter are!

Hi superbugs. Hope this is useful to you.[/quote]

[lizawa]

ksmama:

Hi, take a look at this q pls

Average mass of 40 kids in class 43.625kg. Av mass of boys 45.5kg. Av mass of girls 42.5kg. After 4 boys left the class, av mass of boys 45kg.
a) fd no. of boys and girls in class at lst
b) when a girl joined the class, av mass of the girls bcam 43.4kg. What was mass of new girl?
Tks

Use the \"Make a supposition\" method to solve this.

If there are 40 boys, total weight = 45.5 x 40 = 1820 kg.
If there is 1 boy less and 1 girl more, the change in average weight = 45.5 - 42.5 = 3kg.

Total weight = 43.625 x 40 = 1745
Difference ( between 40 boys weight and total weight) = 1820 - 1745 = 75
75 / 3 = 25
=> means that there are 25 boys fewer and 25 girls more.
# of boys = 40 - 25 = 15
# of girls = 25

When a girl join the class, # of girls = 26
Total weight = 26 x 43.4 = 1128.4 kg
Total weight of 25 girls = 25 x 42.5 = 1062.5 kg
Weight of additional girl = 1128.4 - 1062.5 = 65.9kg

PS : I am not sure how the information \"after 4 boys left, the ave mass of the boys is 45kg\" can be used in this question.

[Chan09]

Need some help:

1) 608 children took part in a test. Altogether 110 awards were presented to 1/5 of the boys and 1/7 of the girls. How many boys took part in the test?
Ans: 110

[lizawa]

Chan09:

Need some help:

1) 608 children took part in a test. Altogether 110 awards were presented to 1/5 of the boys and 1/7 of the girls. How many boys took part in the test?
Ans: 110

5u (boys) + 7u (girls) = 608

1u (boys) + 1u (girls) = 110
5u (boys) + 5u (girls) = 110 x 5 = 550
2u (girls) = 608 - 550 = 58
1u (girls) = 58 /2 = 29

1u (boys) = 110 - 29 = 81
Total # of boys = 5u = 5 x 81 = 405

PS : I think the answer given is not correct.

[linden2000]

Superbugs:

Nan Hua SA2 Q16 2010

Mr Bond had some pencils, rulers & erasers. 60% of the stationery were pencils, 30% of them were rulers while the rest were erasers. There were 240 more pencils than erasers. After Mr Bond sold some of his pencils, 40% of the stationery left were pencils.
a) How many pencils did Mr Bond have at first?
b) How many pencils were sold?

P-60%
R-30%
E-10%
Difference in % between P&E=50%=240
a) 60%=(60%/50%)x240=288

E+R-60%=(40%/50%)x240=192
P-40%=(40%/60%)x192=128
Therefore, P sold = 288-128=160
However answer sheet shows 60 only. Pls help. Tks

I also got answer 160. Perhaps typo in ans sheet?

[Chan09]

Not sure whether answers are right for these : Ai Tong 2010

1) 40 children in class sold average of 19 tickets. There were 8 more boys than girls in the class. The ave number of tickets sold by the boys was 15. Find the ave number of tickets sold by girls?
Ans: 25

2) Agnes made some fruit pies. She gave some of them to neighbor and collected $14.40 after selling 3/5 of the remaining pies for $1.20. She then had 1/5 of the original number of pies left. How many pies did she give to neighbor ?
Ans: 20

[ksmama]

Hi Lizawa, tks! I was trying to see how to use d 4 boys’ info to solve the prob but was indeed baffled. Will see if school uses another method when my boy goes back to school, then share info.


Ano prob here
A bag has 200 blue beads and 540 yellow beads. A tin has 850 blue beads and 350 yellow beads. Some blue and yellow beads are transferred from the tin to the bag. When that is done the percentage of blue beads in the bag bcom 50% and that of blue beads in the tin bcom 70%. How many blue and yellow beads are transferred?

[ksmama]

ksmama:

Hi Lizawa, tks! I was trying to see how to use d 4 boys' info to solve the prob but was indeed baffled. Will see if school uses another method when my boy goes back to school, then share info.


Ano prob here
A bag has 200 blue beads and 540 yellow beads. A tin has 850 blue beads and 350 yellow beads. Some blue and yellow beads are transferred from the tin to the bag. When that is done the percentage of blue beads in the bag bcom 50% and that of blue beads in the tin bcom 70%. How many blue and yellow beads are transferred?

Hi, worked out soln to share
7u + 1p =1050--------(1)
3u + 1p =890----------(2)
4u = 160
1u =40
7u=280
3u=120
therefore, no blue beads transf =850-280=570
no yellow beads transf = 350-120=230