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is 0.d1d2+0.d3d4 greater than 1?

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by yogami » Tue Jun 23, 2009 1:10 pm
I think the answer is D

(1) is obvious because if least is 4 then you know that least value of other digits are 5,6,7,8 and that guarantees a sum > 1.0
(2) says that the product is > 0.5. Lets try different minimum values for d1,d2,d3 and d4
any value less than 5 for d1 or d3 will mean that either d3 or d1 need to be really high to push the product > 0.5 So intuitively both d1 and d3 need to be greater than 5 to push the product to . 0.5 and hence the sum of 0.d1d2 and 0.d3d4 is guaranteed to be . 1.0
200 or 800. It don't matter no more.
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by apoorva.srivastva » Tue Jun 23, 2009 1:43 pm
thanks mate..

The OA is D...could you please elaborate on the your explanation

thanks and regards,
Apoorva
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by nitya34 » Tue Jun 23, 2009 9:18 pm
in (2)

we need not pick up single digit integer(only)

if we choose double digit what will happen?
yogami wrote:I think the answer is D

(1) is obvious because if least is 4 then you know that least value of other digits are 5,6,7,8 and that guarantees a sum > 1.0
(2) says that the product is > 0.5. Lets try different minimum values for d1,d2,d3 and d4
any value less than 5 for d1 or d3 will mean that either d3 or d1 need to be really high to push the product > 0.5 So intuitively both d1 and d3 need to be greater than 5 to push the product to . 0.5 and hence the sum of 0.d1d2 and 0.d3d4 is guaranteed to be . 1.0
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Re: is 0.d1d2+0.d3d4 greater than 1?

by pops » Tue Jun 23, 2009 10:00 pm
statement 1: straighforward and same as explained above that if least number is 4 then definitely 0.d1d2+0.d3d4 > 1

statement 2: take this
since (0.d1d2)(0.d3d4)>0.5
=> 2 * 0.d1d2 > 1 / 0.d3d4
now 1 / 0.d3d4 is greater than 1 (1 over a number less than 1)
so 2 * 0.d1d2 > 1
that means 0.d1d2 has to be greater than 0.5

similarly 0.d3d4 has to be greater than 0.5
and hence 0.d1d2+0.d3d4 > 1
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by rah_pandey » Tue Jun 23, 2009 10:02 pm
Using statement 2
we get

(d1/10+d2/100)*(d3/10+d4/100)>1/2
d1d3/100+d1d4/1000+d2d3/1000+d2d4/10000>1/2
=>d1d3/100+(d1d4+d2d3)/1000+d2d4/10000>0.500000

from above d1d3/100 should be major contributor since d1,d2,d3,d4 are all single digit. This has to be assumed otherwise .d1d2+.d3d4 will be nonsensical.

max((d1d4+d2d3)/1000)=.1620 (Assume d1,d2,d3,d4=9
max(d2d4/10000)=.0081

thus (d1d3)/100>.3299 => (d1,d3)~(5,7)
thus .d1d2+.d3d4>1
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by apoorva.srivastva » Tue Jun 23, 2009 11:00 pm
great reasoning Rahul and Pops...thanks for your inputs

regards,
apoorva
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by carpe_diem » Thu Jun 25, 2009 9:59 am
i guess for statement 2, in spite of going for long calculation we can take the general property of multiplication here.

0.d1d2 * 0.d3d4 > 1/2

this means that both the numbers should be greater than 1/2 in order to make the product greater than 1/2.

and therefore 0.d1d2 + 0.d3d4 > 1.
Hence sufficient.
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