Arithmetic Grade Six

Lesson One Hundred-ten: Cross Reducing II


You can cross-reduce just as easily as you can reduce one of the fractions in the problem. Here's an example:


	 3	  2	
	---   *  --- =
	 8	  9

	Since the 3 and the 9 are both divisible by
	3, we can rewrite the problem like this:

	 1	 2
	---  *  ---  =
	 8	 3

	Since the 2 and the 8 are both divisible by
	2, we can rewrite the problem like this:
	
	 1	 1
	---  *  ---  =
	 4	 3

	And, obviously the answer is 1/12

	If we had simply multiplied 3/8 and 2/9 without
	cross-reducing we would have gotten 6/72. As you
	might guess, 6/72 can be reduced to 1/12.

Here's another example:

	 10	 14
	----  * ----  =
	 21	 35

	Since 10 and 35 are both divisible by
	5, we can rewrite the problem like this:

	  2	 14
	----  * ----  =
	 21	  7

	Since 14 and 21 are both divisible by
	7, we can rewrite the problem like this:
	
	 2	 2
	---  *  --- =
	 3	 7

	Multiplying 2/3 and 2/7 we get 4/21. 

	If we had multiplied 10/21 and 14/35 without reducing
	we would have gotten 140/735 which reduces to 4/21 since
	both 140 and 735 are divisible by 35.


Now you get to try a few for yourself.

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