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DS- Arithmetics

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by phoenix9801 » Sat Jul 10, 2010 10:53 pm
Hi, can someone please help to explain by step-by-step instructions (in detail) in the most simplest way and clear to understand. Would Greatly appreciate it.

1-

Is 5^k less than 1,000

(1) 5^k+1 > 3,000

(2) 5^k-1 = 5^K - 500


2-

If x is an integer, is 9^x + 9 ^-x = b

(1) 3^x + 3^-x = √ b+c

(2) x > 0
Last edited by phoenix9801 on Mon Jul 12, 2010 1:53 pm, edited 1 time in total.
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Source: — Data Sufficiency |
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by gmatmachoman » Sat Jul 10, 2010 11:01 pm
phoenix9801 wrote:Hi, can someone please help to explain by step-by-step instructions (in detail) in the most simplest way and clear to understand. Would Greatly appreciate it.

1-

Is 5^k less than 1,000

(1) 5^k+1 > 3,000

(2) 5^k-1 = 5^K - 500


st 1:

(1) 5^k+1 > 3,000

( 5^k ) * 5 > 3000

( 5^k ) > 3000/5

= ( 5^k ) > 600

Not sufficient as it can be greater than or less than or equal to 1000.

st 2:

(2) 5^k-1 = 5^K - 500

5^k/5 = 5^K - 500

5^k = 625

So its clear 5^K < 1000

Sufficient

Pick B







2-

If x is an integer, is 9^x + 9 ^-x = b

(1) 3^x + 3^-x = √ b+c

(2) x > 0
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by Rahul@gurome » Sun Jul 11, 2010 5:28 am
phoenix9801 wrote:Hi, can someone please help to explain by step-by-step instructions (in detail) in the most simplest way and clear to understand. Would Greatly appreciate it.

2-

If x is an integer, is 9^x + 9 ^-x = b

(1) 3^x + 3^-x = √ b+c

(2) x > 0
(1) 3^x + 3^-x = √ b+c implies (3^x + 3^-x)^2 = b + c
(3^x + 3^-x)^2 = 3^2x + 3^(-2x) + 2(3^x)(3^-x)
= 3^2x + 3^(-2x) + 2.(3^x-x) = 3^2x + 3^(-2x) + 2 [Since 3^x-x = 3^0 = 1]
So, 3^2x + 3^(-2x) + 2 = b + c
Now, 3^2x = 9^x and 3^-2x = 9^-x
Hence 9^x + 9^-x = b + c - 2, which implies answer to the main question is "no".
So, (1) Is SUFFICIENT to answer the question.

(2) This statement is NOT SUFFICIENT to answer the question as we don't know the relation between x and b.

[spoiler]The correct answer is (A).[/spoiler]
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by debmalya_dutta » Sun Jul 11, 2010 7:00 am
Rahul-
You do not know the value of c.
Here it is asked whether 9^x + 9 ^-x = b

So , what if c-2=0, my point being since you do not know the value of c , is the statement rather insufficient


Rahul@gurome wrote:
phoenix9801 wrote:Hi, can someone please help to explain by step-by-step instructions (in detail) in the most simplest way and clear to understand. Would Greatly appreciate it.



2-

If x is an integer, is 9^x + 9 ^-x = b

(1) 3^x + 3^-x = √ b+c

(2) x > 0
(1) 3^x + 3^-x = √ b+c implies (3^x + 3^-x)^2 = b + c
(3^x + 3^-x)^2 = 3^2x + 3^(-2x) + 2(3^x)(3^-x)
= 3^2x + 3^(-2x) + 2.(3^x-x) = 3^2x + 3^(-2x) + 2 [Since 3^x-x = 3^0 = 1]
So, 3^2x + 3^(-2x) + 2 = b + c
Now, 3^2x = 9^x and 3^-2x = 9^-x
Hence 9^x + 9^-x = b + c - 2, which implies answer to the main question is "no".
So, (1) Is SUFFICIENT to answer the question.

(2) This statement is NOT SUFFICIENT to answer the question as we don't know the relation between x and b.

[spoiler]The correct answer is (A).[/spoiler]
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by Rahul@gurome » Sun Jul 11, 2010 3:17 pm
debmalya_dutta wrote:Rahul-
You do not know the value of c.
Here it is asked whether 9^x + 9 ^-x = b

So , what if c-2=0, my point being since you do not know the value of c , is the statement rather insufficient
That's right, if c = 2 then 9^x + 9 ^-x = b. So, (1) will NOT be SUFFICIENT. Thanks Debmalya!
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