Q&A - PSLE Math
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Thank you very much for the detailed explanations with answer.
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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. -
Hi
Can anyone assist me to solve this problem?
-Ben and Steven have some sweets. If Ben eats 4 sweets and Steven eats 3 sweets each day, Ben will have 156 sweets left when Steven finishes all his sweets. If Ben eats 1 sweet and Steven eats 4 sweets each day, Ben will have 338 sweets left when Steven finishes all his sweets. How many sweets does Ben have ?
- Liow and Thomas were 535km apart. At 0830,Liow started to drive towards Thomas at a constant speed. Half an hour later ,Thomas drove towards Liow at a constant speed of 15km/h slower than Liow’s.They were 125km apart at 1200.
(a) Find Liow’s speed
(b) How much more time would they take to pass each other? -
Hi
Can anyone assist me to solve this problem?
-Ben and Steven have some sweets. If Ben eats 4 sweets and Steven eats 3 sweets each day, Ben will have 156 sweets left when Steven finishes all his sweets. If Ben eats 1 sweet and Steven eats 4 sweets each day, Ben will have 338 sweets left when Steven finishes all his sweets. How many sweets does Ben have ?
- Liow and Thomas were 535km apart. At 0830,Liow started to drive towards Thomas at a constant speed. Half an hour later ,Thomas drove towards Liow at a constant speed of 15km/h slower than Liow’s.They were 125km apart at 1200.
(a) Find Liow’s speed
(b) How much more time would they take to pass each other? -
kiasuaunt:
HiHi - 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.
Thank you.
Q1.
In the end,
Boys : Girls --> 11u : 8u (2/5 of girls)
At first,
Boys : Girls --> 11u + 145 : 20u
Total Boys + Girls ==> 31u + 145 = 610
1u --> 15
At first, total number of boys --> 11 x 15 + 145 = 310
In the end, total number of boys + girls --> 19 x 15 = 285
cheers. -
kiasuaunt:
Hi - nd help with this:
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.
Hi
Geraldine's spending : Amy's spending : --> 4u : 10u
Amount left, Geraldine : Amy --> 4u + 16 : 1u + 4 (**geraline's amount left = 4 x amy's amount left)
Total amount of money both had
19u + 20 = $162.50
1u --> $7.5
Geraldine's spending --> 4 x $ 7.5 = $30
At first, Amy had --> spending + amount left = 11 x $7.5 + 4 = $86.50
cheers. -
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
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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. -
tamunawara:
Hi,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?
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.
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Thank you, MathIzzzFun and 2cor41618 - appreciate your help. I understand now - will get my dd to try out later. God Bless.
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