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

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

thanks mate..

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

thanks and regards,
Apoorva

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

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

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

great reasoning Rahul and Pops...thanks for your inputs

regards,
apoorva

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.