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SAT PREP 2015

Twice the sum of two consecutive integers is 10 less than 5 times the smaller integer. Find the greater integer.

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PLZ note that most people choose to let X be the smaller integer and X+1 to be the greater integer.
 
But the problem specifically asks us to find the value of the larger integer, so why don't we call it X and solve for X ??
 
That simply means that the smaller integer is X-1 (that's not too hard to understand, is it?).
 
Now, 2*((X-1) + X)) = 5*(X-1) - 10
         2X - 2 + 2X  = 5X - 5 - 10
                     13 = X

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Andrew D. | Harvard Grad offers Test Preparation and TutoringHarvard Grad offers Test Preparation and...
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Call the smaller consecutive integer X
Then the greater is    X+1
 
The sum of these integers is X+(X+1)  =  2X +1
Twice the sum is               2(2X+1)      = 4X+2
 
10 less than 5 times smaller                 = 5X-10
 
So                                 5X-10            =   4X  +2
                                                    X    =    12
 
Therefore the greater integer    X+1      =   13
 
 
Jon P. | Knowledgeable Math, Science, SAT, ACT tutor - Harvard honors gradKnowledgeable Math, Science, SAT, ACT tu...
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Let x be the smaller integer.
 
Since they're consecutive, the larger one must be x + 1.
 
"Twice the sum of two consecutive integers" is 2(x + x + 1) = 2(2x + 1) = 4x + 2
 
"10 less than 5 times the smaller integer" is 5x - 10
 
So...
 
4x + 2 = 5x - 10
2 = x - 10
12 = x
 
So 12 is the smaller integer, which means that 13 is the larger one.
 

Comments

And, checking,
Twice the sum of 12 and 13  (2*25 = 50)  is 10 less than 5*12, which is 60-10=50
  they match!