Let total (t) = a +b + c + d. We know that t = 9900, but lets keep it t for now a = (b+c+d)/2 = (t-a)/2, so a = t/3 Similarly b = (t-b)/4, so b = t/5 c = (t-c)x 2/3 = 2t/5 Putting it back in original equation, we have d = t - (a + b + c) = t - (t/3 + t/5 + 2t/5) = 9900 - 3300 - 1980 - 3960 = 9900 - ...

Ah, thanks guys.. looks like I didn't consider all the possibilities

for option A ) cant we do squaring on both sides and arrive at x-5 > 0 and hence X > 5 ? so that the answer is D No, since this is an inequality. Think of it this way, first condition is satisfied with both x = 4 and x = 6, but in one case the value of x is < 5, and in the other its > 5

The circle is immaterial in this one, instead think of this question as how can you fill 4 boxes with 5 colours, with no 2 adjacent boxes with the same color. Here's a (bad) representation of the boxes |5 |4 |4 |4 | Now, first box can be filled in 5 ways, next one in 4 ways (since same colors cant b...

Rate = 5280 feet per 100x24 hours

So, feet per hour = 5280 / (100 x 24) = 11/5

Should be (B)

I think this is where you are wrong: y^2 + 3 > y^2 The inequality sign would change if y is negative, so you can't really say the above is true, unless you know for a fact that y > 0 Thanks, but how can an inequality sign changes. Rules are if you multiply or divide an inequality by a negative numb...

I think this is where you are wrong:
y^2 + 3 > y^2

The inequality sign would change if y is negative, so you can't really say the above is true, unless you know for a fact that y > 0

(1)
6 copper coins and 3 brass coins are removed, which leaves 14 copper coins

SUFFICIENT

(2)
four of the first six are copper, remainder could be any combination of copper and bronze coins
INSUFFICIENT

Should be (A)

OA pls?