Let x and y be real nos such that
[x] + 3{y}=3.9 and
{x} + 3[y]=3.4
[z] and {z} have their usual meanings.
Find the number of ordered pairs (x,y).
a) 0 b) 1 c) 2 d) 3 e) none of these
i guess there is no such pair :|
ReplyDeleteNow adding the given equations, We get x+3y = 7.3
ReplyDeleteNow [x] can be 1,2 or 3 and [y] should be equal to 1 as -1<{x}<1
Now suppose [x] =1,
we get {y} as 29/30 using 1st equation and hence y = 59/30 (Since [y]=1) and substituting in the equation, we get x=1.4
Similarly when [x] = 2, we get x = 2.4 and y= 49/30
when [x] = 3 we get x = 3.4 and y = 1.3
So total three solutions, answer choice (d)..Really good questions Implex, keep 'em coming:)
Rightly done Bhaskar
ReplyDelete@ Nidhi, Have a look at the solution above, if you have doubts, we can handle it !