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If 9/10 < a/b < 10/11, where a and b are positive integers, then find the smallest possible value of b.

 Dec 10, 2019
 #1
avatar+19851 
0

put the fractions in a common denominator  (110)

 

99/110 < a/b < 100/110     

  does not help much, huh? There are no integers between 99 and 100   sooooo: Let's double everything (the next common denominator)

198/220 < a/b < 200/220          Now we might guess b to be least at 220  (and a to be 199)

 Dec 10, 2019
 #2
avatar+69 
0

Doing some bashing we get the answer a/b as 19/21 which leaves the answer of b as 21.

 Dec 12, 2019
 #3
avatar+106933 
+1

This is how Gor.. came up with this answer.

 

9/10 < a/b < 10/11

so

18/20 < a/b < 20/22

 

There is a fair bet that 19/21 will be between the 2.  (Maybe it is certainty, but I wouldn't want to say that without proving it)

 

Check by using decimals

9/10 = 0.9

10/11 = 0.909090....

 

19/21 = 0.90476

 

And it is between  the 2 as predicted.

 

The smallest value of b is 21

Melody  Dec 12, 2019

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