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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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28 Sep 2015, 14:27
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25% (medium)

Question Stats:

82% (01:42) correct 18% (01:54) wrong based on 78 sessions

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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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28 Sep 2015, 19:38
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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28 Sep 2015, 19:39
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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20 Mar 2018, 09:32
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Re: Given that b is more than 90 percent of a, is b < 65? (1) a > 75 (   [#permalink] 20 Mar 2018, 09:32
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