Q&A - P5 Math

[tianzhu]

999happygolucky:

NYPS 2007 SA2 Qn48


Three schools, School A, School B and School C participated in the Racial Harmony Games Day. There were 68 representatives from Schools B & C. If 32 representatives were not from School B and 44 representatives were not from School C, what was the total number of representatives from the three schools? Ans : 72

Qn 15
Find the total of the number pattern below
81 - 80 + 79 - 78 + 77 - 76 + .... - 2 + 1
What will be the digit in the tens place? Ans : 4

Pls help

Hi

Q48

B + C ------- 68

A + C ------ 32

A + B ------- 44

2*(A + B + C) ------ 144

A + B + C ------ 144/2 ------ 72

Q15

-80+79 = -1

-78+77 = -1
.
.
.

-2 + 1 = -1

There are 40 pairs of (-1).This gives (-40)

81 + (-40) ------- 41

Alternatively

81 – 80 = 1

79 – 78 = 1
.
.
.

+3 – 2 = 1

There are 40 pairs of (1).This gives (40)

40 + 1 = 41

Hence the answer is 4

Best wishes

[simple88]

Hi, I need help with this question. Thanks in advance.


Kelvin had between $30 and $40 in $1 coins and 50cent coins. The ratio of the number of $1 coins to the number of 50cent coins was 1:3. He exchanged some $1 coins for 50cent coins. The ratio of the number of $1 coins to 50cent coins then became 1:5.
Find a) the total number of money that Kelvin had, b) the number of $1 coins which were exchanged for 50cent coins.

[MathIzzzFun]

simple88:

Hi, I need help with this question. Thanks in advance.


Kelvin had between $30 and $40 in $1 coins and 50cent coins. The ratio of the number of $1 coins to the number of 50cent coins was 1:3. He exchanged some $1 coins for 50cent coins. The ratio of the number of $1 coins to 50cent coins then became 1:5.
Find a) the total number of money that Kelvin had, b) the number of $1 coins which were exchanged for 50cent coins.

at first, number of $1 coins : 50-cent coins --> 1 : 3 (total 4)
for every group of 4 coins, total amount = 1 x $1 + 3 x $0.50 = $2.50

in the end, number of $1 : 50-cent coins --> 1 : 5 (total 6)
for every group of 6 coins, total amount = 1 x $1 + 5 x $0.50 = $3.50

The total amount of money at first and in the end remains the same.
So, find LCM of $2.50& $3.50 --> $17.50
Since total amount is between $30 & $40,
total amount of money = $17.50 x 2 = $35 ... with this, you can work out b)

cheers.

[simple88]

Hi MathIzzzFun,

Thank you so much. Amazed how you can present it in such a way that is so much easier to comprehend than what my kid is taught in school. Sorry I have one more for you. Thanks alot.

Q)At first, Ah Jie only had oranges and Ah Siang only had apples. They gave each other half of their fruits. Ah Jie then sold 18 oranges and Ah Siang sold 8 apples. In the end, the ratio of the number of oranges to the number of apples Ah Jie had was 1:4 and the ratio of the number of oranges to the number of apples Ah Siang had was 1:3. How many oranges did Ah Jie have at first?

[MathIzzzFun]

simple88:

Hi MathIzzzFun,

Thank you so much. Amazed how you can present it in such a way that is so much easier to comprehend than what my kid is taught in school. Sorry I have one more for you. Thanks alot.

Q)At first, Ah Jie only had oranges and Ah Siang only had apples. They gave each other half of their fruits. Ah Jie then sold 18 oranges and Ah Siang sold 8 apples. In the end, the ratio of the number of oranges to the number of apples Ah Jie had was 1:4 and the ratio of the number of oranges to the number of apples Ah Siang had was 1:3. How many oranges did Ah Jie have at first?

Hi

you can use MD, UP (unit/part) or cross-multiply method...

After selling 18 oranges,
Jie's oranges : apples --> 1u : 4u
Before selling the oranges, Jie's oranges : apples --> 1u + 18 : 4u

Siang had same number of oranges & apples
ie Siang's oranges : apples --> 1u + 18 : 4u

after selling 8 apples,
Siang's oranges : apples
--> 1u + 18 : 4u - 8
= 1 : 3

equalizing or cross-multiply:
3u + 54 = 4u -8
1u --> 62

number of oranges Jie had at first --> (62 +18)x2 = 160

cheers.

[MathIzzzFun]

Goodluck8:

Water expands 10% when it freezes to ice. Find the volume of the water to which a container of capacity 3000 cm3 is to filled so that the water freezes completely to ice, it will fill the container exactly.

volume of ice : water
= 110 : 100
= 3000 : 3000/11 x 10
= 3000 : 2727 3/11

volume of water --> 2727 3/11 cm³


cheers.

[simple88]

MathIzzzFun:

simple88:

Hi MathIzzzFun,

Thank you so much. Amazed how you can present it in such a way that is so much easier to comprehend than what my kid is taught in school. Sorry I have one more for you. Thanks alot.

Q)At first, Ah Jie only had oranges and Ah Siang only had apples. They gave each other half of their fruits. Ah Jie then sold 18 oranges and Ah Siang sold 8 apples. In the end, the ratio of the number of oranges to the number of apples Ah Jie had was 1:4 and the ratio of the number of oranges to the number of apples Ah Siang had was 1:3. How many oranges did Ah Jie have at first?

Hi

you can use MD, UP (unit/part) or cross-multiply method...

After selling 18 oranges,
Jie's oranges : apples --> 1u : 4u
Before selling the oranges, Jie's oranges : apples --> 1u + 18 : 4u

Siang had same number of oranges & apples
ie Siang's oranges : apples --> 1u + 18 : 4u

after selling 8 apples,
Siang's oranges : apples
--> 1u + 18 : 4u - 8
= 1 : 3

equalizing or cross-multiply:
3u + 54 = 4u -8
1u --> 62

number of oranges Jie had at first --> (62 +18)x2 = 160

cheers.

Hi,
I have problem understanding this part where you mentioned
'Siang had same number of oranges & apples
ie Siang's oranges : apples --> 1u + 18 : 4u'.

How do you know that Siang had same number of oranges & apples to form this ratio?

Thanks.

[MathIzzzFun]

simple88:

MathIzzzFun:

[quote=\"simple88\"]Hi MathIzzzFun,

Thank you so much. Amazed how you can present it in such a way that is so much easier to comprehend than what my kid is taught in school. Sorry I have one more for you. Thanks alot.

Q)At first, Ah Jie only had oranges and Ah Siang only had apples. They gave each other half of their fruits. Ah Jie then sold 18 oranges and Ah Siang sold 8 apples. In the end, the ratio of the number of oranges to the number of apples Ah Jie had was 1:4 and the ratio of the number of oranges to the number of apples Ah Siang had was 1:3. How many oranges did Ah Jie have at first?

Hi

you can use MD, UP (unit/part) or cross-multiply method...

After selling 18 oranges,
Jie's oranges : apples --> 1u : 4u
Before selling the oranges, Jie's oranges : apples --> 1u + 18 : 4u

Siang had same number of oranges & apples
ie Siang's oranges : apples --> 1u + 18 : 4u

after selling 8 apples,
Siang's oranges : apples
--> 1u + 18 : 4u - 8
= 1 : 3

equalizing or cross-multiply:
3u + 54 = 4u -8
1u --> 62

number of oranges Jie had at first --> (62 +18)x2 = 160

cheers.

Hi,
I have problem understanding this part where you mentioned
'Siang had same number of oranges & apples
ie Siang's oranges : apples --> 1u + 18 : 4u'.

How do you know that Siang had same number of oranges & apples to form this ratio?

Thanks.[/quote]\"At first, Ah Jie only had oranges and Ah Siang only had apples. They gave each other half of their fruits.\"

so, each has 1/2 the number of oranges and 1/2 the number of apples

cheers.

[999happygolucky]

Thanks to all the masters who helped to solve all the problems posted. I greatly appreciate the help, especially when my DS seemed so lost in his Maths. MILLIONS OF THANKS

[Udon]

hi..need help :?:


Thank U....

Q1 The total number of balls in Box A and Box B is 165 less than the total number of balls in Box A and Box C. The number of balls in Box C is 4 times the number of balls in Box B. The number of balls in Box D is 3 times the total number of balls in Box A and Box B. If the total number of balls in all the 4 boxes is 752.

a) How many balls are there in Box A?

b) How many balls are there in Box D?