A computer routine was developed to generate two numbers, ( x,y), the

Question

A computer routine was developed to generate two numbers, ( x,y), the first being a random number between 0 and 100 inclusive, and the second being less than or equal to the square root of the first. Each of the following pairs satisfies this routine EXCEPT:

(A) (99, 10)
(B) (85, 9)
(C) (50, 7)
(D) (1 , 1)
(E) (1, 0)

Sol-
A doesnt satisfy one of the conditions mentioned
10^2 is not 100.
Ans- (A)

(A) (99, 10)
(B) (85, 9)
(C) (50, 7)
(D) (1 , 1)
(E) (1, 0)

nkin · Answer

10^2 = 100 which is more than 99, hence the pair (99,10) does not satisfy the given condition.

Option A

dave13 · Answer

the quant became like walking on clouds

HKD1710 your quant posts boosted my confidence in quant

10^2 = 100 so option A is doesnt satidfy conditon

amhill4 · Answer

99 is slightly less than 100, which has a square root of 10. Square root of 99 is therefore less than 10, so answer is  B

nkin · Answer

Hey  amhill4 , did you mean to write A instead of B?

amhill4 · Answer

Whoops, yeah I meant to write A I swear!

Gmat2Cracker · Answer

A computer routine was developed to generate two numbers, ( x,y), the first being a random number between 0 and 100 inclusive, and the second being less than or equal to the square root of the first. Each of the following pairs satisfies this routine EXCEPT:

(A) (99, 10)
(B) (85, 9)
(C) (50, 7)
(D) (1 , 1)
(E) (1, 0)

Sol-
A doesnt satisfy one of the conditions mentioned
10^2 is not 100.
Ans- (A)

(A) (99, 10) 
(B) (85, 9)
(C) (50, 7)
(D) (1 , 1)
(E) (1, 0)

Sol- 
A doesnt satisfy one of the conditions mentioned
10^2 is not 100.
Ans- (A)

Ana4001 · Answer

I need help as when i read the question , the second should be less than or equal to square of first.

so if its (X,Y) then y<=X2???

KanishkM · Answer

Hi  Ana4001

It is "second being less than or equal to the  square root of the first ", i suppose you missed out on the  bold   part

it means y <=  \sqrt{x}

ScottTargetTestPrep · Answer

Looking at the answer choices, we see that since √100 = 10, square root of 99 is less than 10, so (99,10) cannot be the pair of numbers chosen.

Answer: A

CrackverbalGMAT · Answer

Check by squaring the second numbers of the options

A. 10^2 = 100 (so this is where we can say for certain that the second number is not less than or equal to the square root of the first (99))

B.(85,9)- 9^2 = 81. Eliminate. So square root of 85 is > 9 or second number 9 is less than first number 85.

C.(50,7)- 7^2 = 49. Eliminate

D.(1,1) - 1^2 =1 .Eliminate

E. (1,0) - 1^2 =1 which is >0 and hence eliminate.

(option a)

D.S
GMAT SME

kungfury42 · Answer

Since 10 >  \sqrt{99} , statement A is a violation of the given rule.
We can check all other options satisfy the given rule.
Hence, A.

bumpbot · Answer

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