Q&A - PSLE Math

[2cor41618]

kiasuaunt:

Hi - nd help with this:


1. There were 610 pupils in a school. After 3/5 of the girls and 145 boys left the school, the new ratio of the number of boys to the girls became 11:8.
a.Find the number of boys in the school at first.
b.Find the total number of pupils in the school in the end.

2.Geraldine and Amy went shopping for groceries together. They had $162.50 with them. Amy spent 2.5 times as much money as Geraldine. The amount of money Geraldine had left was $16 more than what she had spent. She had 4 times as much money left as Amy.

a.How much money did Geraldine spent?

b.How much money did Amy have at first?

Thank you.

1) Since 2/5 of the girls equals to 8u, thus the total number of units for girls at first must be 8/2 x 5 = 20u

Therefore, girls at first---20u, boys--11u + 145

31u + 145 --- 610
1u --- (610 -145)/31 =15

(a) No. of boys at first---15 x 11 + 145 = 310

(b) No. of pupils in the end--- 15 x 19 = 285


2) Ratio for money spent---A : G = 5u : 2u
Geraldine left---2u + $16
Amy left--- (2u + $16)/4 = 0.5u + $4

Geraldine at first--- 2u + 2u + $16 = 4u + $16
Amy at first----5u + 0.5u +$4 = 5.5u + $4

Thus, 4u + $16 + 5.5u + $4 --- $162.50
9.5 u ---$$142.50
1u ---$15

a) Geraldine spent--2 x $15 = $30

b) Amy at first ---5.5 x $15 + $4 = $86.50

Hope this helps

[tutorsg_punggol]

Hihi,


by using algebra method,
Qn 1, Ben has 380 sweets.
Qn 2, Liow's speed is 70km/h and they will take one hour to pass each other.

As I'm not good in using all the formats here, I will explain w/o using models. Hopefully you can understand.

Qn 1: Let the 1st example be x days. This will means Ben has (4x + 156) sweets and Steven has (3x) sweets.

Then we let 2nd example be y days. So this will means Ben has (y + 338) sweets and Steven has (4y) sweets.

Therefore (4x + 156) = (y + 338) and 3x = 4y.
since 3x = 4y, it means y = 3/4 x.
So (4x + 156) = (3/4x + 338)
4x - 3/4 x = 338 -156
3 1/4x = 182
x = 56.
Ben has (4x + 156), so (4 X 56) + 156 = 380

Qn 2:
( Sorry that i do not know how to draw the diagram here. If you want to, let me know. I will try to snap a photo of my drawing.)

Let speed of Liow be SL. Since Thomas is slower by 15km/hr, speed of Thomas will be ( SL - 15)

Distance is speed X time.
Liow started at 830am. So time until 12pm will be 3.5 hours.
Thomas started half hour later, so his time will be 3 hours.
And please note that they were 125 km apart at 12pm.

Total distance is 535km.

535km = (3.5hrs X SL) + 125km + 3hrs X (SL - 15)
535km = 3.5 SL + 125km + 3 SL - 45km
535km - 125km + 45km = 6.5 SL
6.5 SL = 455
So SL = 70km/hr.

Thomas's speed will be 55km/hr.

Hence for (b), for every hour they travel, it will be (70 + 55)km/h.
This means they travel 125 km every hr.

Therefore they will only take one hour to pass each other.
Please note that the question asks how much more time, and not what time.


Hopefully this helps.

[speedmaths.012624com]

tamunawara:

hi can someone help me with this question


Andy had only $2 notes and Bryan had only $5 notes. the number of notes Andy had was 80% of Bryan's notes. When Bryan gave Andy $550, the number of notes Andy had became thrice that of Bryan.

a)How many notes did Bryan have at first?
b) How much money did Andy have in the end?

Hi,

Perhaps the question is not clear.

I interpret the question as:

\"Andy had only $2 notes and Bryan had only $5 notes (at first).
The number of notes Andy had was 80% of Bryan's notes.
When Bryan gave Andy $550 (in $5 notes),
the number of ($2 and $5) notes Andy had became thrice that of Bryan.\"

So the solution could start with:

....Andy : Bryan
16 Boxes: 20 Boxes (at first)

27 Boxes: 9 Boxes (finally)

Bryan gave 20 - 9 = 11 Boxes of notes to Andy
11 Boxes → 110 $5 notes (worth $550)
1 Box → 10 notes

Can you continue from here?

Cheers


speedmaths.com


PS.
We will teach P6 students how to answer more of such questions at our PSLE Maths Camps; please see our thread in the Happenings forum.
.

[kiasuaunt]

Thank you, MathIzzzFun and 2cor41618 - appreciate your help. I understand now - will get my dd to try out later. God Bless.

[Chan09]

Need help

There were twice as many red pens as blue pens.he sold the pens in bundles of 2 red and 3 blue.
After selling all the blue pens, he still had 120 red pens left. How many penns did he have at first?
Ans: 270

[tianzhu]

Chan09:

Need help

There were twice as many red pens as blue pens.he sold the pens in bundles of 2 red and 3 blue.
After selling all the blue pens, he still had 120 red pens left. How many penns did he have at first?
Ans: 270

Hi

You may use MD or UM.

Red : Blue ------ 2:1 -------- 6:3

He sold 2 units of red pens and 3 units of blue pens.

Hence, 4 units ------ 120

This gives 1 unit ------ 30

Number of pen@first ------ 9 units ------- 270

Best wishes

[isetan]

Hi, need help to cross check the ans to this question:


Two boxes, A and B, contain some red and blue beads. In box A, the ratio of the no. Of red beads to the no. of blue beads is 2:1. In box B, the ratio of the no. Of red beads to blue beads is 1:3. Box A has thrice as many beads as Box B.

A) find the ratio of the no. of Red beads in box A to the no. of blue beads in box B.

B) if 105 red beads are taken out from box A and placed in box B, the ratio of the no. of the red beads to the no. of blue beads in box B becomes 3:2. What is the total no. of beads in the 2 boxes?

Our working leads ans A is – 8:1, ans B is – 480.

Appreciate anyone helps if the ans is correct. Many thanks.

[isetan]

Hi, another question:


Tan, Lim and Wong had some pens to sell. Tan had the least no. of pens. Wong’s no. of pens was 30% less than Lim’s. After Tan and Lim had each sold 50% of their pens, Lim had 95 more pens than Tan. If the three men had a total of 330 pens left, what is the total no. of pens they had at first?

[ksoh3000]

tks

[2cor41618]

isetan:

Hi, need help to cross check the ans to this question:


Two boxes, A and B, contain some red and blue beads. In box A, the ratio of the no. Of red beads to the no. of blue beads is 2:1. In box B, the ratio of the no. Of red beads to blue beads is 1:3. Box A has thrice as many beads as Box B.

A) find the ratio of the no. of Red beads in box A to the no. of blue beads in box B.

B) if 105 red beads are taken out from box A and placed in box B, the ratio of the no. of the red beads to the no. of blue beads in box B becomes 3:2. What is the total no. of beads in the 2 boxes?

Our working leads ans A is -- 8:1, ans B is -- 480.

Appreciate anyone helps if the ans is correct. Many thanks.

I think he answer to part a should be 8:3. Part b is correct