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by raju232007 » Fri Oct 03, 2008 1:23 pm
Apple costs 0.7$/piece & banana costs 0.5$/piece

The eqn is
0.7x+0.5y=6.30....1

7x+5y=63(1*10)

Here x& y are positive integers & the only solution that satisfies the eqn is x=4 & y=7
x+y=11
So the ans must be 11 ...

Let me know if you have any doubts..
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by cramya » Fri Oct 03, 2008 1:55 pm
The only common mutliptle of banana price and apple price is 3.5 (mutlitple of .5 and .7 is 3.5).

To maximize the number of fruits we have to have more bananas

For $3.5 the number of bananas = 7 and we can only guy 4 apples for the remaining (6.3-3.5 = $ 2.8)

Hope this helps!
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by cramya » Fri Oct 03, 2008 1:56 pm
The only common mutliptle of banana price and apple price is 3.5 (mutlitple of .5 and .7 is 3.5).

To maximize the number of fruits we have to have more bananas

For $3.5 the number of bananas = 7 and we can only buy 4 apples for the remaining ( 6.3-3.5 = $ 2.8 )

Hope this helps!
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by vivek.kapoor83 » Sat Oct 04, 2008 12:28 am
that means it was hit & trial but i am looking more convincing solution, dont know whther it exists
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by stop@800 » Sat Oct 04, 2008 3:00 am
vivek.kapoor83 wrote:that means it was hit & trial but i am looking more convincing solution, dont know whther it exists
It can't be.
If have one equation and two unknowns so we will have to take advantage that x and y are integers.

7x + 5y = 63

A multiple of 5 will always end in 0 or 5
to have sum as 63 [with unit digit as 3] the mutiple of 7 must end in 3 or 8.

7*4 = 28
7*9 = 63

so
x = 4, 9
and accordingly
y = 7, 0

Ans will be 11 or 9
so Ans B

Hope this helps.....
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