1. Since its a right triangle 1/2 * b*h= Area of triangle. Now we have x<y<z which implies z greatest i.e. the hypotenuse. We also have A=1=1/2*x*y
Therefore xy=2. Since y>x y=2 and x=1. So answer A.) y> square root 2

3. From the question,
J/A=5/3 --->1
and J=5/9(T) + 50 ---> 2

Equation 1 implies that J=5/8(T) where T=Total ---> 3 Solving 2 and 3 we get T=720 and hence A=3/8(T)= 270


- Deepak

Q1.

Since our triangle is right and x<y<z, this means that z, which is the longest of all the sides, must be the hypotenuse. This in turn means that x and y are both catheti, so the area of our triangle is x*y/2 = 1. So this leads to x*y = 2. Now suppose that x = y. This would mean that x = y = sqrt(2), but since y > x, then y has to be greater than sqrt(2).

Q2. Say x = hundreds digit for m
y = tens digit for m
z = units digit for m.

Then say a = hundreds digit for v
b = tens digit for v
c = units digit for v.

So you have that:
2x^2*3y^2*5z^2 = 9*2a^2*3b^2*5c^2. You can notice that you can immediately eliminate 2, 3 and 5, so you get that:
x^2*y^2*z^2 = 9*a^2*b^2*c^2. This is equivalent to
(x^2*y^2*z^2 - 9*a^2*b^2*c^2) = 0 or (xyz - 3abc)(xyz + 3abc) = 0. But since xyz + 3abc cannot be 0, then xyz = 3abc. I don't know how to continue from here on, unfortunately, because any of x, y, z could equal 3a, 3b or 3c if a, b, care either 1, 2 or 3.
I might have misunderstood the original formula or maybe you wrote it down incorrectly....

Q3. Say j = amount of money john got
a = amount of money Ann got.

We have that: a + j = x
j/a = 5/3, so a = 3/5j
5/9*x = j - 50.

Substitute a in a + j = x with a = 3/5 j and you get that 8/5j = x. This means that j = 5/8x.
Now substitute that in 5/9x = j - 50 and you get that 5/9x = 5/8x - 50 or that 5/72x = 50 or that x/72 = 10. This means that x = 720. Since a = 3/5j = 3/5*5/8x, a = 270.

For Q2 refer this link

https://www.beatthegmat.com/what-s-the-f ... 19361.html

Oh ok then.... I thought it was weird when I first saw it...