<p>The average (arithmetic mean) of 4 different integers is 75. If the largest integer is 90, what is the least possible value of the smallest integer?
A. 1
B. 19
C. 29
D. 30
E. 33
The answer is E, but I don't understand why it's not A?</p>
<p>I don’t know if it is right, but this is what I did: </p>
<p>The 4 different integers are 90, 89, 88, x
(Since the three numbers besides x have to be the highest possible values to make the x as small as it can be) </p>
<p>(90+89+88+x)/4 = 75</p>
<p>And solve for x, which gives you x=33 </p>
<p>Hope that helps!</p>
<p>Let’s do it this way … 75x4=300</p>
<p>300-90=210 </p>
<p>So the sum of the remaining 3 numbers is 210. From this point onwards, I would start plugging in </p>
<p>If one of the numbers has a value of 1, the remaining two numbers add up to 209 which is not possible as then they would have to be integers greater than 90</p>
<p>The same is true for the remaining incorrect options of B,C,D and E</p>
<p>Slightly tedious method but I can’t think of a better way. Desigirl’s method is good as well although it did require some ‘out of the box’ thinking on her part in the context of the SAT instead of using the conventional Plug In </p>
<p>Anyways hope this helped :)</p>
<p>Sorry kinda messed up my post a bit … The only incorrect options are B,C and D :P</p>
<p>I assumed 1 is the smallest integer because it was the smallest among the choices, thus this is what I did:
300 - (90+1) = 209 then I assumed the other two integers were 105 and 104… So in the end it’s (90+1+105+104)/4 = 75
But now I see what I did wrong. 90 is the greatest integer, so it must be 33 </p>
<p>Thank you!</p>