# Not sure about my answer and the solution in this book

#### LaaLa

##### New member
I'm going through the fifth edition of Anthony Croft and Robert Davison "Foundation of Maths". When coming across a certain problem based on fractions, I was a little skeptical about the answer given at the ends of the book. (Further details: Pg.25 Exercise 2.5 question (h)).

Problem: (2*1/4 / 3/4) x 2
Solution: (Further details: Pg.485 Exercise 2.5. Question (h)).
(2*1/4 / 3/4) x 2 =

(9/4 x 4/3) x 2 =

(3/4 x 4) x 2 =

3 x 2 = 6.

Question: Where did the 4 come from in (3/4 x 4) x 2 ?

(2 1/4 / 3/4) x 2 =

2 1/4
= (2x4+1)/ 4 = 9/4

(9/4 / 3/4 = 9/4 x 4/3 = 36/12 = 6/2 = 3/1) x 2 =
2 x
3/1 = (2x1+3)/1 = 5/1
so 5/1
= 5

A little bit confused here...
How is my answer different to the solution in the book?
Am I missing something here?
Maybe the rules are different when the denominator is a one prehaps?

Your help is appreciated Thank you

#### MarkFL

##### Super Moderator
Staff member
Is this the actual expression?

$$\displaystyle \left(\frac{2\tfrac{1}{4}}{\dfrac{3}{4}}\right)\times2$$

#### mmm4444bot

##### Super Moderator
Staff member
… Problem: (2*1/4 / 3/4) x 2 …
Hello. Did you intend to type 2+1/4 instead of 2*1/4?

#### LaaLa

##### New member
Hi there,
Sorry, I didn't think to put a photo of the page in the book to make things a lot more convenient.
Hopefully this clears the confusion...

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#### mmm4444bot

##### Super Moderator
Staff member
Hello, again. Thank you for posting the image. Yes, those exercises are showing 'mixed' numbers (i.e., combination of a Whole number and fraction).

In (d) think of 1½ as (1+1/2)

In (e) think of 2¼ as (2+1/4)

If they meant 2×¼ then they'd need to show the multiplication explicitly, using × or grouping symbols: (2)(¼).

Cheers

#### Denis

##### Senior Member

[(9/4) / (3/4)] * 2 = 6

#### Jomo

##### Elite Member
I'm going through the fifth edition of Anthony Croft and Robert Davison "Foundation of Maths". When coming across a certain problem based on fractions, I was a little skeptical about the answer given at the ends of the book. (Further details: Pg.25 Exercise 2.5 question (h)).

Problem: (2*1/4 / 3/4) x 2
Solution: (Further details: Pg.485 Exercise 2.5. Question (h)).
(2*1/4 / 3/4) x 2 =

(9/4 x 4/3) x 2 =

(3/4 x 4) x 2 =

3 x 2 = 6.

Question: Where did the 4 come from in (3/4 x 4) x 2 ?

(2 1/4 / 3/4) x 2 =

2 1/4
= (2x4+1)/ 4 = 9/4

(9/4 / 3/4 = 9/4 x 4/3 = 36/12 = 6/2 = 3/1) x 2 =
2 x
3/1 = (2x1+3)/1 = 5/1
so 5/1
= 5

A little bit confused here...
How is my answer different to the solution in the book?
Am I missing something here?
Maybe the rules are different when the denominator is a one prehaps?

Your help is appreciated Thank you
Yes 2 1/4 = 9/4.
(9/4 x 4/3) x 2----> 9/3 reduces to 3. So the 9/4 becomes 3/4 and the 4/3 becomes 4. Personally I would have cancelled out the 4s as well.
=(3/4 x 4)

3 x 2 = 6. Correct.

You are missing something very important. For example 2 1/4 means 2 +1/4 while 2*(1/4) is NOT the same as 2 1/4.

Note that 3/1 = 3 (you did know that 5/1=5). Now 2x(3/1) = 2x3 =6 not 2 x 3/1 = (2x1+3)/1 = 5/1 . You did 2 + 3 = 5 = (2 3/1) (2x1+3)/1 =5/1=5

Again things like 7 1/2 and 7*(1/2) are different. After all 7 1/2 = 7 + 1/2

Last edited:

#### LaaLa

##### New member
Hi again, thanks everyone for your comments. I was only referring to question (h) but its nice to see your keen on helping!

Jomo! Thanks so much! I forgot that 3/1 = 3, makes perfect sense! I should have converted the fraction to a whole and then times it by 2 in the end:
3/1 = 3
(3) x 2 = 6
Sorry, I didn't know that I had to start converting the fractions as it never said to do so with the exercises.
I will now start converting!
Thanks again

#### Denis

##### Senior Member
Oilcan Denis: "Curses...Jomo is the hero again...."