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

[apoorva.srivastva]

If d1, d2, d3, d4 are four different integers, whether is 0.d1d2+0.d3d4 greater than 1?

1) The least number is 4
2) (0.d1d2)(0.d3d4)>0.5

Please Explain st. 2

[yogami]

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

[apoorva.srivastva]

thanks mate..

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

thanks and regards,
Apoorva

[nitya34]

in (2)

we need not pick up single digit integer(only)

if we choose double digit what will happen?

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

[pops]

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

[rah_pandey]

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

[apoorva.srivastva]

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

regards,
apoorva

[carpe_diem]

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.