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Source: — Data Sufficiency |
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by shovan85 » Tue Oct 05, 2010 3:21 am
IMO A.

1: 10^x=(2^x)*(5^x)
(4^y)(5^z) = 2^2y * 5^z Thus (2^x)*(5^x) = 2^2y * 5^z => x = 2y and z.
If we see x=2y then x is even so sufficient.

2. After solving we get x+5 = 3y+3. Hence x can be odd or even depending on the value of y. so Not Suff.
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by ramannjit » Tue Oct 05, 2010 3:29 am
shovan85 wrote:IMO A.

1: 10^x=(2^x)*(5^x)
(4^y)(5^z) = 2^2y * 5^z Thus (2^x)*(5^x) = 2^2y * 5^z => x = 2y and z.
If we see x=2y then x is even so sufficient.

2. After solving we get x+5 = 3y+3. Hence x can be odd or even depending on the value of y. so Not Suff.
Thanks yes OA is A.

My little mind is confused with (1), I came up with x=2y so it is even, what about x=z. Help me with this, why we will not consider relation between x and z, please.
Ramannjit
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by mj78ind » Tue Oct 05, 2010 4:35 am
ramannjit wrote:
shovan85 wrote:IMO A.

1: 10^x=(2^x)*(5^x)
(4^y)(5^z) = 2^2y * 5^z Thus (2^x)*(5^x) = 2^2y * 5^z => x = 2y and z.
If we see x=2y then x is even so sufficient.

2. After solving we get x+5 = 3y+3. Hence x can be odd or even depending on the value of y. so Not Suff.
Thanks yes OA is A.

My little mind is confused with (1), I came up with x=2y so it is even, what about x=z. Help me with this, why we will not consider relation between x and z, please.
x = z is besides the point, what are we going to do with it......since HAS to 2y hence both x and z HAVE to be even

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