Q&A - P5 Math

[COOKIE64]

Hi please help .

Classes 5A and 5B have the same number of pupils. The ratio of the number of boys in 5A to the number of boys in 5B is 2:3 The ratio of the number of girls in 5A to the number of girls in 5B is 7:4

(A) find the ratio of the number of boys to the number of girls in 5A

(B) if there are 15 fewer girls than boys in 5B, how many pupils are there in 5B.


Thanks.

[MathIzzzFun]

Hi please help .

Classes 5A and 5B have the same number of pupils. The ratio of the number of boys in 5A to the number of boys in 5B is 2:3 The ratio of the number of girls in 5A to the number of girls in 5B is 7:4

(A) find the ratio of the number of boys to the number of girls in 5A

(B) if there are 15 fewer girls than boys in 5B, how many pupils are there in 5B.


Thanks.

There are equal number of students in 5A and 5B --> difference in number of boys in 5A and 5 B will be 5he same as the difference in the number of girls in 5B and 5A for eg. if there are 6 more boys in 5B than in 5A, then there will be 6 more girls in 5A than in 5B.

So, make the difference in ratio parts the same.

Girls, 5A:5B = 7u : 4u ...difference of 3u
Boys, 5A:5B = 2:3 = 6u :9u ...dIfference of 3u.

a) ratio of boys :girls in 5A = 6:7
b) 15 fewer girls than boys in 5B, so
9u - 4u = 5u --> 15
u --> 3

Number of pupils in 5B = (4+9) x 3 = 39

cheers.

[belnanna]

Please help on this question..
thanks in advance

[COOKIE64]

Hi please help .

Classes 5A and 5B have the same number of pupils. The ratio of the number of boys in 5A to the number of boys in 5B is 2:3 The ratio of the number of girls in 5A to the number of girls in 5B is 7:4

(A) find the ratio of the number of boys to the number of girls in 5A

(B) if there are 15 fewer girls than boys in 5B, how many pupils are there in 5B.


Thanks.

There are equal number of students in 5A and 5B --> difference in number





of boys in 5A and 5 B will be 5he same as the difference in the number of girls in 5B and 5A for eg. if there are 6 more boys in 5B than in 5A, then there will be 6 more girls in 5A than in 5B.

So, make the difference in ratio parts the same.

Girls, 5A:5B = 7u : 4u ...difference of 3u
Boys, 5A:5B = 2:3 = 6u :9u ...dIfference of 3u.

a) ratio of boys :girls in 5A = 6:7
b) 15 fewer girls than boys in 5B, so
9u - 4u = 5u --> 15
u --> 3

Number of pupils in 5B = (4+9) x 3 = 39


cheers.


Hi, why do we have to make the difference in part ratio the same. Can you please help me , I cannot explain the same to my DS
Thanks

[MathIzzzFun]

Hi please help .

Classes 5A and 5B have the same number of pupils. The ratio of the number of boys in 5A to the number of boys in 5B is 2:3 The ratio of the number of girls in 5A to the number of girls in 5B is 7:4

(A) find the ratio of the number of boys to the number of girls in 5A

(B) if there are 15 fewer girls than boys in 5B, how many pupils are there in 5B.


Thanks.

There are equal number of students in 5A and 5B --> difference in number





of boys in 5A and 5 B will be 5he same as the difference in the number of girls in 5B and 5A for eg. if there are 6 more boys in 5B than in 5A, then there will be 6 more girls in 5A than in 5B.

So, make the difference in ratio parts the same.

Girls, 5A:5B = 7u : 4u ...difference of 3u
Boys, 5A:5B = 2:3 = 6u :9u ...dIfference of 3u.

a) ratio of boys :girls in 5A = 6:7
b) 15 fewer girls than boys in 5B, so
9u - 4u = 5u --> 15
u --> 3

Number of pupils in 5B = (4+9) x 3 = 39


cheers.


Hi, why do we have to make the difference in part ratio the same. Can you please help me , I cannot explain the same to my DS
Thanks

https://www.flickr.com/photos/mathizzzfun/5760364658/

https://www.flickr.com/photos/mathizzzfun/5703951396/

cheers.

[MathIzzzFun]

Please help on this question..
thanks in advance

Cut the shaded area into 3 triangles :
Triangle AEB --> base 5 cm, height 7 cm
Triangle BCE --> base 4 cm,height 7 cm
Triangle CDE --> base 3 cm, height 3 cm
and add the areas.

cheers.

[bb_ mom]

Please help on this question..
thanks in advance

Cut the shaded area into 3 triangles :
Triangle AEB --> base 5 cm, height 7 cm
Triangle BCE --> base 4 cm,height 7 cm
Triangle CDE --> base 3 cm, height 3 cm
and add the areas.

cheers.

Thanks! Never thought of cutting the triangles that way plus got misled by rectangular looking squares.

[belnanna]

Please help on this question..
thanks in advance

Cut the shaded area into 3 triangles :
Triangle AEB --> base 5 cm, height 7 cm
Triangle BCE --> base 4 cm,height 7 cm
Triangle CDE --> base 3 cm, height 3 cm
and add the areas.

cheers.

Thanks mathizzzfun
Thanks to all the kind souls here, you guys will have good karma.. god bless all of you and your family . Thanks sincerely from my heart.

[leng2005]

I post this the second time.Please help me.

[speedmaths.com]

I post this the second time.Please help me.

Hi leng2005,

Let's try; one possible method:


At first:

Let Farhan have 1 unit (or 1 u) of cards
Let Michelle have 2 units (or 2 u) of cards


Changes:

Farhan bought another 33 cards
Michelle lost 1 u of cards


Finally:

Let Farhan have (1 u + 33) of cards left
Let Michelle have 1 u of cards left

Let Eliza have (1 u + 33) - 6 = (1 u + 27) of cards left


Total cards left → (1 u + 33) + 1 u + (1 u + 27)
Total cards left → 3 u + 60

3 u + 60 → 150
3 u → 90
1 u → 30


Finally:

Farhan have 1 u + 33 → 63 cards
Michelle have 1 u → 30 cards
Eliza have 1 u + 27 → 57 cards


At first:

Farhan have 1 u → 30 cards

Michelle have 2 u → 60 cards

Eliza have 57 + 14 → 71 cards


Hope this helps.

Cheers



speedmaths
.