Same Denominator in Fraction Questions

I have received some feedback from one visitor regarding my previous post. She asked me to post an example of same denominator question. Here it is.

Question:

Adam and Ben had same amount of money. Adam spent 1/4 of his money and Ben spent 2/5 of his money. Adam spent $300 less than Ben. How much did the boys have altogether at first?

Solution:

Adam spent 1/4

Ben spent 2/5

Adam and Ben had same amount of money --> Adam's 4u has to be same as Ben's 5u

Adam 1/4 (multiply both numerator and denominator by 5) --> 5/20

Ben 2/5 (multiply both numerator and denominator by 4) --> 8/20

Adam spent 5u and Ben spent 8u

8u - 5u = 3u --> $300

1u --> $100

The boys had 16u altogether at first --> 16 x $100 = $1600

Same Numerator in Fraction Questions

Adam and Ben had $3400 altogether. After Adam spent 3/4 of his money and Ben spent 7/9 of his money, they had the same amount of money left. How much money did each boy have at first?

Solution:

Normally students tend to use model to solve this question. However, it is actually unnecessary to draw model. It can be solved just using the fraction itself.

Fraction of what Adam left --> 1/4

Fraction of what Ben left --> 2/9

They had the same amount of money left, so Adam's 1 unit needs to be changed to 2 units so as to be the same as Ben's units.

1/4 --> 2/8

At first Adam had 8 units, Ben had 9 units.
17u --> $3400
1u --> $200

Adam 8u --> $1600
Ben 9u --> $1800

Note:
The key for this kind of questions is to change either numerator or denominator to be the same depends on the questions. 

Working Backwards

A fraction question sent in by Mr Chiow

Solution:

In the end: Z = L à 174 

After L gave 2/5, she had 3 units left.

So 3u à 174

1u à 58

Z received 2u from L, so Z originally had 174 – 2u à 174 – 116 = 58

Before L gave 2/5, she had 5 units. 5u à 290

Z had 58 after he gave 1/3, 2u à 58, 1u à 29 (given to L)

L had 290 after she received 29

L at first à 290 – 29 = 261 

Cross Method for Fraction / Ratio Questions

Here is a question from visitor Belnanna:

There were 3/5 as many children as adults on a bus. At the next stop, 3 children boarded and 2 adults alighted from the bus.Then, there were 5/6 as many children as adults on the bus. How many children were there on the bus at first.

Solution:

Ratio, Ratio and Ratio!!

A very common Ration question!

Solution:

The importance of 'same' in Fraction questions

When it comes to fraction questions, pay attention to the 'same' word if it appears in the question. It will be the key to solve the question. Here is an example.

Question:

Adam and Bella had $5100 altogether. After Adam spent 4/5 of his money and Bella spent 5/7 of her money, they had the same amount of money left. How much money did each of them have at first?
 
Solution:

Key info: 'same amount of money left' 

Adam left 1 unit and Bella left 2 units. It means Adam's 1 unit is equal to Bella's 2 units. 

Change Adam's units to Bella's Units as shown in the table below:

17 units -> $5100
1 unit -> $300
Adam 10 units -> $3000
Bella 7 units -> $2100 

Another Question from Kiasuparents

In class 5A, there were 1/2 as many girls as boys. In class 5B , there were 1/3 as many boys as girls. The number of boys in class 5B was 2/3 as many as the number of girls in class 5A. There were 32 pupils in class 5B.

A) express the number of pupils in 5B as a fraction of the number of pupils in 5A

B) after some boys were transferred from 5A to 5B , there were 1/2 as many of 5B boys as the number of 5B girls. How many pupils were there in class 5B after the transfer ?

Solution:

5B: Boys : Girls = 1 :3 so 4units -> 32, 1unit -> 8. Boys is 8 and girls is 24
5B Boys : 5A girls = 2:3, so 5A girls is 12
5A boys: 12x2=24 
5A total pupils is 36

A) 32/36=8/9

B) After transfer
No. of 5B girls remains the same -> 24
No. of 5B boys --> half of 5B girls -> 12
Total no. of pupils in 5B -> 24 + 12 = 36