Tutor MathsGuru: Ask me for your burning Maths questions!

[MathIzzzFun]

kwcllf:

Hi please help with the following question.


Dolly had 80 more stickers than Jenny. Dolly gave 25% of her stickers to Jenny. Jenny in return gave 60% of her stickers to Dolly. In the end, Dolly had 100 stickers more than Jenny. How many stickers did Dolly have at first.

Thanks!

Hi

there are a few methods.. eg using units, model.

At first,
Jenny -> 4 units
Dolly -> 4 units + 80

After Dolly gave 25% ie 1 unit + 20, to Jenny :
Jenny -> 5 units + 20
Dolly -> 3 units + 60

After Jenny gave 60% (3/5) ie 3 units + 12 to Dolly :
Jenny -> 2 units + 8
Dolly -> 6 units + 72

In the end Dolly had 100 more than Jenny:
So, 6 units + 72 = 2 units + 8 + 100
--> 4 units = 108 - 72 = 36, 1 unit --> 9

At first, Dolly had 4 units + 80 --> 116 stickers.

cheers.

[kwcllf]

Thanks again!


I need help on another question.

Jan and Kay had equal number of sweets and chocolate. Jan ate 12 sweets and Kay ate 18 chocolates and ten the ratio of Jan’s sweets to chocolate became 1:7 and the ratio of Kay’s sweets to chocolate became 1:4. How many sweets did Jan have at first?

Thanks!

[MathIzzzFun]

kwcllf:

Thanks again!


I need help on another question.

Jan and Kay had equal number of sweets and chocolate. Jan ate 12 sweets and Kay ate 18 chocolates and ten the ratio of Jan's sweets to chocolate became 1:7 and the ratio of Kay's sweets to chocolate became 1:4. How many sweets did Jan have at first?

Thanks!

hi

u r welcome.

There many examples of similar questions in the forum - you may want to check out the P5 maths/P6 maths threads for the solutions first. Let me know if you still can't sort it out.

cheers.

[kwcllf]

kwcllf:

Thanks again!


Jan and Kay had equal number of sweets and chocolate. Jan ate 12 sweets and Kay ate 18 chocolates and ten the ratio of Jan's sweets to chocolate became 1:7 and the ratio of Kay's sweets to chocolate became 1:4. How many sweets did Jan have at first?

Hi MathIzzzFun,

Not sure if I get this correct.

Jan - 1 : 7 (after eating 12 sweets) ------> (x 4) ------> 4 : 28
Kay - 1 : 4 (after eating 18 chocolates) --> (x 7) ------> 7 : 28

7u - 4u = 3u

3u = 12 sweets
1u = 4 sweets

Therefore, Jan at first has (4u x 4) + 12 sweets = 16 + 12 = 28 sweets

[MathIzzzFun]

kwcllf:

kwcllf:

Thanks again!


Jan and Kay had equal number of sweets and chocolate. Jan ate 12 sweets and Kay ate 18 chocolates and ten the ratio of Jan's sweets to chocolate became 1:7 and the ratio of Kay's sweets to chocolate became 1:4. How many sweets did Jan have at first?

Hi MathIzzzFun,

Not sure if I get this correct.

Jan - 1 : 7 (after eating 12 sweets) ------> (x 4) ------> 4 : 28
Kay - 1 : 4 (after eating 18 chocolates) --> (x 7) ------> 7 : 28

7u - 4u = 3u

3u = 12 sweets
1u = 4 sweets

Therefore, Jan at first has (4u x 4) + 12 sweets = 16 + 12 = 28 sweets

Hi

Note that Jan and Kay had equal number of sweets and equal number of chocolates, so after Kay ate 18 chocolates, the number of chocolates that Jan had and Kay had would be different, so we cannot equalize the two.
Here's one example that you could refer to - http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&start=4050

cheers.

[MathIzzzFun]

kwcllf:

kwcllf:

Thanks again!


Jan and Kay had equal number of sweets and chocolate. Jan ate 12 sweets and Kay ate 18 chocolates and ten the ratio of Jan's sweets to chocolate became 1:7 and the ratio of Kay's sweets to chocolate became 1:4. How many sweets did Jan have at first?

Hi MathIzzzFun,

Not sure if I get this correct.

Jan - 1 : 7 (after eating 12 sweets) ------> (x 4) ------> 4 : 28
Kay - 1 : 4 (after eating 18 chocolates) --> (x 7) ------> 7 : 28

7u - 4u = 3u

3u = 12 sweets
1u = 4 sweets

Therefore, Jan at first has (4u x 4) + 12 sweets = 16 + 12 = 28 sweets

Hi

Here's another approach ...

http://i40.tinypic.com/33yjj7p.jpg\">

cheers.

[violet04]

Hi,

Please help with the following P3 question. Thanks!

Find the smallest number such that it leaves a remainder of 3 when divided by 5 and leaves a remainder of 5 when divided by 7.

[MathIzzzFun]

violet04:

Hi,

Please help with the following P3 question. Thanks!

Find the smallest number such that it leaves a remainder of 3 when divided by 5 and leaves a remainder of 5 when divided by 7.

Hi

the standard method for solving this type of questions is using the listing method.

List out the multiples of 5 & 7 and add the remainder and find the smallest number that matches.

5 --> 5, 10, 15, 20, 25, 30, 35, 40
+3 --> 8, 13, 18, 23, 28, 33, 38, 43

7 --> 7, 14, 21, 28, 35, 42
+5 --> 12, 19, 26, 33, 40, 47

From above, we see that 33 is the lowest number that matches.

A quicker method is to see that by adding 2 to the number it will be divisible by both 5 & 7 (divide by 5 --> remainder 3, so add 2, the number is divisible by 5; divide by 7 --> remainder 5, add 2 and the number is also divisible by 7). So, the smallest number that is divisible by both 5 & 7 is 5 x 7 = 35. Since we have added 2, the number is 35 - 2 = 33.

Using this method, subsequent numbers that meet the above conditions are :
35 x 2 - 2 = 68
35 x 3 - 2 = 103
35 x 4 - 2 = 138
.....

cheers.

[violet04]

MathIzzzFun:

violet04:

Hi,

Please help with the following P3 question. Thanks!

Find the smallest number such that it leaves a remainder of 3 when divided by 5 and leaves a remainder of 5 when divided by 7.

Hi

the standard method for solving this type of questions is using the listing method.

List out the multiples of 5 & 7 and add the remainder and find the smallest number that matches.

5 --> 5, 10, 15, 20, 25, 30, 35, 40
+3 --> 8, 13, 18, 23, 28, 33, 38, 43

7 --> 7, 14, 21, 28, 35, 42
+5 --> 12, 19, 26, 33, 40, 47

From above, we see that 33 is the lowest number that matches.

A quicker method is to see that by adding 2 to the number it will be divisible by both 5 & 7 (divide by 5 --> remainder 3, so add 2, the number is divisible by 5; divide by 7 --> remainder 5, add 2 and the number is also divisible by 7). So, the smallest number that is divisible by both 5 & 7 is 5 x 7 = 35. Since we have added 2, the number is 35 - 2 = 33.

Using this method, subsequent numbers that meet the above conditions are :
35 x 2 - 2 = 68
35 x 3 - 2 = 103
35 x 4 - 2 = 138
.....

cheers.

Hi MathIzzzFun,
Thank you very much of the solution. Cheers!

[MathIzzzFun]

violet04:

MathIzzzFun:

[quote=\"violet04\"]Hi,

Please help with the following P3 question. Thanks!

Find the smallest number such that it leaves a remainder of 3 when divided by 5 and leaves a remainder of 5 when divided by 7.

Hi

the standard method for solving this type of questions is using the listing method.

List out the multiples of 5 & 7 and add the remainder and find the smallest number that matches.

5 --> 5, 10, 15, 20, 25, 30, 35, 40
+3 --> 8, 13, 18, 23, 28, 33, 38, 43

7 --> 7, 14, 21, 28, 35, 42
+5 --> 12, 19, 26, 33, 40, 47

From above, we see that 33 is the lowest number that matches.

A quicker method is to see that by adding 2 to the number it will be divisible by both 5 & 7 (divide by 5 --> remainder 3, so add 2, the number is divisible by 5; divide by 7 --> remainder 5, add 2 and the number is also divisible by 7). So, the smallest number that is divisible by both 5 & 7 is 5 x 7 = 35. Since we have added 2, the number is 35 - 2 = 33.

Using this method, subsequent numbers that meet the above conditions are :
35 x 2 - 2 = 68
35 x 3 - 2 = 103
35 x 4 - 2 = 138
.....

cheers.

Hi MathIzzzFun,
Thank you very much of the solution. Cheers![/quote]u r welcome

cheers!