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Roy is now 4 years older than Erik and half of that amount older than Iris. If in 2 years, roy will be twice as old as Erik, then in 2 years what would be Roy's age multiplied by Iris's age?

Is there any general method or are they all so confusing?

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You can try forming some systems of linear equations to solve them. –  user130512 Mar 7 at 20:23
    
Write the associated equation. –  jinawee Mar 7 at 20:23
    
that's the way to go^^ –  Max Mar 7 at 20:24

2 Answers 2

up vote 0 down vote accepted

Roy is now 4 years older than Erik

$$ R = E + 4 $$

and half of that amount older than Iris.

$$ R = I + 2 $$

If in 2 years, roy will be twice as old as Erik

$$ (R + 2) = 2(E + 2) $$

then in 2 years what would be Roy's age multiplied by Iris's age?

$$ (R + 2)(I+2) = \;\; ? $$

Then you just solve the equations. Hopefully you can see that this method will apply to any similar problem equally well.

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This cleared the concept for me! Thanks! –  Prakash Wadhwani Mar 8 at 5:02
1  
The answers is 48! –  Prakash Wadhwani Mar 8 at 5:04

Hint: Set $E$ the age of Erik, $I$ the age of Iris and $R$ the age of Roy today. You are given that $$R=4+E \tag{1}$$ and $$R=2+I \tag{2}$$ if I understood correctly this one (I am not sure what "that amount" is in your formulation) and $$(R+2)=2(E+2) \tag{3}$$ and you want to find $$(R+2)\times (I+2)$$ (if I understood that correctly too). Now you can solve the system of the three equations (1), (2), (3) (please check if they are formulated correctly) in three unknowns $E, I$ and $R$. That is the general method, with the difficulty being in the correct formulation of the given relations.

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I think the second equation should be $R=I+2$ –  user130512 Mar 7 at 20:27
    
@user130512 So with "this amount" is meant this 4? It is not clear –  Stef Mar 7 at 20:30
    
The question definitely should have been worded better, but I think that because this is a precalculus algebra question, they will not be asking him to solve for R as a function of both E and I –  user130512 Mar 7 at 20:32
    
@user130512 Ok, thanks, I corrected it. Seems more legitimate –  Stef Mar 7 at 20:34

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