If 7^(1/2)

Question

If  \sqrt{7} <x<\sqrt{37}  and x is an integer, then x can have how many different values?

(A) Three 
(B) Four 
(C) Five 
(D) Eight 
(E) Ten

elevateV · Answer

7^(1/2) is somewhere between 2 and 3 (as 2^2=4 and 3^2=9),
37^(1/2) is somewhere near 6.1 (as 6^2=36)
Therefore integers between these two values are 3,4,5,6

Answer (B) Four

MathRevolution · Answer

'x' is between 2.xxx and 6.xx and since it is an integer, then x = 3,4,5, and 6.

Four values.

Answer B

GMATWhizTeam · Answer

Solution :

We know,  \sqrt{4} < \sqrt{7} < \sqrt{9} 
   ⟹ 2 <  \sqrt{7}   < 3   
Thus,  \sqrt{7}  lies between 2 and 3.

Also,  \sqrt{36} < \sqrt{37} < \sqrt{49} 
  ⟹  6 < \sqrt{37} < 7   
Thus,  \sqrt{37}  lies between 6 and 7

By the problem, 
   \sqrt{7} <x<\sqrt{37}   
  ⟹ 2.## < x < 6.##  
Therefore, x = 3, 4, 5, 6 (since x is an integer)

Hence, the correct answer is   Option B  .

zoeyl · Answer

given x>0, we can get 7<x^2<37, then x can be 3 (with x^2=9), 4 (with x^2=16), 5 (with x^2=25) and 6 (with x^2=36)

bumpbot · Answer

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If 7^(1/2) < x < 37^(1/2) and x is an integer, then x can have how man

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