AUnVerifiedTaxPayer

Sorry, the answer choices got mixed up as well, here's the original problem.

If 3x−y=12, what is the value of 8x/2y?

A) 2^12
B) 4^4
C) 8^2
D) The value cannot be determined from the information given.

here's my solution. 

One way is to write

8x/2y

so that the numerator and denominator are writen with the same base. Since 2 and 8 are powers of 2, pluging in 2^3 for 8 in the numerator of 8x/2y gives

(2^3)x/2y

which can be written

2^3*x/2y (confusing)

Since the numerator and denominator of have a common base, this expression can be written as 2(3x−y). In the problem, it says that 3x−y=12, so can substitute 12 for the exponent, 3x−y, which means that

8x/2y=2^12

Heres a solution that I found. 

Multiply each side of the given equation by ax−2 (so you can get rid of the fraction). When you multiply each side by ax−2, you should have:

24x^2+25x−47=(−8x−3)(ax−2)−53

You should then multiply (−8x−3) and (ax−2) using FOIL.

24x^2+25x−47=−8ax^2−3ax+16x+6−53

Then, reduce on the right side of the equation

24x^2+25x−47=−8ax^2−3ax+16x−47

Since the coefficients of the x^2 term have to be equal on both sides of the equation, −8a=24, or a=−3.

The final answer is B.

I'm sorry for the exponents and the fractions, I'm just new to latex and don't understand it.

Sorry, I had a typo. its A circular table is pushed into a corner of the room, where two walls meet at a right angle. A point P on the edge of the table (as shown below) has a distance of 8 from one wall, and a distance of 9 from the other wall. Find the radius of the table.

Alright, here are the solutions that I got. 

1. It is given that the two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, hundreds digit and units digit must be either 2 or 3 i.e. 2_2, 2_3, 3_2 or 3_3 which means that there are four possible numbers - 242, 263, 362 and 393.

Now, it is also given that - the difference between its reverse and it is 99. So 263 and 362 satisfy this condition. Hence, the sum of the three digits is 11 in each case.

2. It is given that the man has 40 litres container of milk. Also, he will drink 1 litre on the first day and refill the container with water, will drink 2 litres on the second day and refill the container, will drink 3 litres on the third day and refill the container, and so on till 40th day. Thus at the end of 40 days, he must have drunk (1 + 2 + 3 + 4 + ..... +38 + 39 + 40) = 820 litres of liquid.

Out of those 820 litres, 40 litres is the milk which he had initially. Hence, he must have drunk 780 litres of water.

3. As there are 11 balls along one side, it means that there are 11 layers of balls. The top most layer has 1 ball. The second layer has 3 (1+2) balls. The third layer has 6 (1+2+3) balls. The fourth layer has 10 (1+2+3+4) balls. The fifth layer has 15 (1+2+3+4+5) balls. Similarly, there are 21, 28, 36, 45, 55 and 66 balls in the remaining layers.

Hence, the total number of balls is = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 = 286 balls

4. Let's start with JKL = 9 * LQ. Note that L appear on both the side. Also, after multiplying LQ by 9 the answer should have L at the unit's place. The possible values of LQ are 19, 28, 37, 46, 55, 64, 73, 82 and 91; out of which only 64, 73 and 82 satisfies the condition. (As all alphabets should represent different digits)

Now, consider PQR = 6 * LQ. Out of three short-listed values, only 73 satisfies the equation.

Also, ZYX = 3 * LQ is satisfied by 73.

Hence, Z=2, Y=1, X=9, P=4, Q=3, R=8, J=6, K=5, L=7

219/3 = 438/6 = 657/9 = 73

5. The old man temporarily added his dog to the 17, making a total of 18 dogs.

First son got 1/2 of it = 9

Second son got 1/3 of it = 6

Third son got 1/9 of it = 2 for a total of 17.

He then steals his dog back and goes away......

I will give you that you got the three right! Nice job, and the method for number 3 was my strategy too. Great job NotThatSmart!

Yeah, C and P is really hard, but it teaches you a lot