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Given that b is more than 90 percent of a, is b < 65? (1) a > 75 (

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Given that b is more than 90 percent of a, is b < 65? (1) a > 75 (  [#permalink]

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New post 28 Sep 2015, 13:27
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Given that b is more than 90 percent of a, is b < 65?

(1) a > 75
(2) a - b > 5


How to solve the second one algebraically without plugging in?

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Re: Given that b is more than 90 percent of a, is b < 65? (1) a > 75 (  [#permalink]

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New post 28 Sep 2015, 18:38
1
Given that b is more than 90 percent of a, is b < 65?

(1) a > 75
(2) a - b > 5

For St 2.

in the questions stem given that
b > (90/100 ) a -----1
a - b > 5 -------2
here the inequality sign is same ...so we can add eq. 1 and 2

on adding we get ..

a > 5 + (90/100 ) a

a - (90/100 ) a > 5

( 10a /100 ) > 5
therefore on solving the above inequality

a > 50

Now if i use b > (90/100 ) a from the question stem

We see that b is both less than 65 and greater than 65 .

Hence not sufficient .

Ans is A .

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Re: Given that b is more than 90 percent of a, is b < 65? (1) a > 75 (  [#permalink]

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New post 28 Sep 2015, 18:39
2
reto wrote:
Given that b is more than 90 percent of a, is b < 65?

(1) a > 75
(2) a - b > 5


How to solve the second one algebraically without plugging in?


b > 0.9a,
1st statement a>75,
hence, b> 67.5
hence sufficient.

2nd statement
a-b > 5
lets assume b = 0.9a(lowest possible value of b)
a - 0.9 a > 5
0.1 a >5
a > 50, this conclusion leads to no where, hence insufficient

hence Ans A
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Re: Given that b is more than 90 percent of a, is b < 65? (1) a > 75 (  [#permalink]

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Re: Given that b is more than 90 percent of a, is b < 65? (1) a > 75 ( &nbs [#permalink] 20 Mar 2018, 08:32
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