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November 17th, 2015, 06:48 AM  #1 
Newbie Joined: Nov 2015 From: UK Posts: 2 Thanks: 0  Find the missing number 10  6 * 15 / ? = 20
Good day all Can someone please explain how this problem is to be tackled? I have no issue with the problem above, however, I'm being faced with much more complex problems where the interger value is higher and there is more multiplication/division. Is there a rule to follow to make something like this more simple? Thanks 
November 17th, 2015, 07:13 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,164 Thanks: 736 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Personally, I would replace the ? with a variable (like $\displaystyle x$) and use basic algebra to make that variable ($\displaystyle x$) the subject. That would be, imho, the clearest, easiest way to figure out the answer. Equation: $\displaystyle 10  \frac{6 \times 15}{x} = 20$ Evaluate $\displaystyle 6 \times 15$: $\displaystyle 10  \frac{90}{x} = 20$ Subtract 10 from both sides: $\displaystyle  \frac{90}{x} = 10$ Multiply by 1 on both sides: $\displaystyle \frac{90}{x} = 10$ Multiply both sides by x: $\displaystyle 90 = 10x$ Divide both sides by 10: $\displaystyle x = \frac{90}{10} = 9$ However, if you haven't covered algebra yet, then you can solve simple problems by looking at each operation step by step and ask yourself what numbers you would need at each step. For example... i) What would the second term need to be to make 20? $\displaystyle 10  ... = 20$ It would have to be $\displaystyle 10$, so you get $\displaystyle 10  10 = 10+10 = 20$ ii) How do you make $\displaystyle 10$ from $\displaystyle \frac{6 \times 15 }{?}$... .. well... $\displaystyle 6 \times 15 = 90$, so you would have to divide that number by $\displaystyle 9$ to make $\displaystyle 10$. Therefore the $\displaystyle ?$ must be $\displaystyle 9$. Last edited by Benit13; November 17th, 2015 at 07:16 AM. 
November 17th, 2015, 11:33 AM  #3 
Newbie Joined: Nov 2015 From: UK Posts: 2 Thanks: 0 
Okay, thanks for your help In the first method, does the rearrangement follow BODMAS? Thanks again 

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