Find the lowest of three nummbers as described :
The cube of the first number exceeds their product by 2, the cube of the second number is smaller than their product by 3 and the cube of the third number exceeds their product by 3.

A. 3^(1/3)
B. 9^(1/3)
C. 2
D. 2^(1/3)
E. Any of these

Dear Mike,
I really appreciate your concern. As a matter of fact, I posted this question coz I was myself confused about its phrasing.I have reproduced it here exactly the same way as it was stated in the source text book.
Anyways, I think the "their product" in the question is refered to the product of all the 3 i.e., a,b &c as -

Find the lowest of three numbers (a, b, & c) as described :
The cube of the first number (i.e., a) exceeds the product of a, b & c by 2, the cube of the second number(i.e., b) is smaller than their product by 3 and the cube of the third number(i.e., c) exceeds their product by 3.

I hope this might help us reach the solution now & accordingly I would b able to edit the error in the question. Although I was struck in this way also but would appericiate your help.
Thanks
Pari....

Each installment is of 2000 $ (two equal installments adding up to 4000)

On the 1st installment he will get 1 year interest: 20% of 2000 = 400
On the 2nd installment he will get no interest as it would be the final installment

So the total paid for the Loan (L) is:
2400 + 2000 = (1.2)ˆ2 * L

Solving for L you get :
L = 3055,55

So, the total interest paid was 4000 - 3055,55 = approximately 944 $
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look at this equation as PT....put d=2...then equation becomes a^2+b^2=c^2...
and if you look at the triplets that can satisfy this equation are 3-4-5, 6-8-10,9-40-41, 10-60-61..so on
What we get from this is 2 is never between or maximum or any value between a,b,c...so we left with only option A and E...
Well A is exactly what the triplets show...hence the answer......hope it helps.
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This is a question of lcm where frequencies (or cycle time period, whatever you prefer to solve by. Frequency = 1/T) of each word is to be worked out and then after finding lcm, we can calculate when would all three words be on again.
Now for each word there is a cycle of on and off.

Ram = 3 (on) + 47/6 (off) = 65/6

And = 3(on) + 4/3 (off) = 13/3

Shyam = 3(on) +17/3(off) = 26/3

Clearly after 2 cycles of Ram, 10 cycles of 'And' and 5 cycles of Shyam, we get all words off at 130/3 seconds and the next second they all would start their cycle again.

so the answer is 130/3 D

DJ

For the sum of digits to be equal to 3 only three combinations are possible

1. 1,1,1 and 10 0s (zeroes)
2. 1,2 and 11 (zeroes)
3. 3 and 12 (zeroes)

So the answer shall be arranging a thirteen digit number with cases 1,2 and 3

1. For this case total arrangements = (3(for first digit) * 12! (for rest of the digits))/3!*10!(because 3 1s and 10 zeroes are similar)
this equals 66

2. Similarly for 2nd case arrangements = (2(again for first digit)*12!(for rest of the digits))/11!(for similar 11 zeroes)
this equals to 24

3. For this case only one arrangement is possible, hence 1

total is 66+24+1 = 91
So answer is C

DJ

rearrange the equation as x^2 +1=4x
square both sides
x^4 +1 +2x^2=16x^2
x^4+1=14x^2
divide by x^2 on both sides.
x^2 + 1/x^2 = 14
square again and rearrange

x^4 + 1/x^4 =196-2=194

Hence E

DJ

x^2 - 4x + 1 = 0 --> dived by x: x-4+\frac{1}{x}=0 --> x+\frac{1}{x}=4 --> square it: x^2+2+\frac{1}{x^2}=16 --> x^2+\frac{1}{x^2}=14 --> square again: x^4+2+\frac{1}{x^4}=196 --> x^4+\frac{1}{x^4}=194.

Answer: E.
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In the following figure find angle ACB. Given C is the point of contact of two circles and A, B are the points of contact of the common tangent PP1 to the two circles.
(Find figure in the attched file)
a. 60
b. 90
c. 120
d. 89
e. Cannot be determined

Source: T.I.M.E- CAT'11 Series

Dear Pariearth

This is a very hard Geometry problem, much harder than you will see on the GMAT. I'm pretty good with Geometry, and it took me several minutes to puzzle out this particular problem.

In the diagram above, D and E are centers of the circle. Of course, AD = CD, so ACD is an isosceles triangle, and similarly, CE = BE, so BCE is an isosceles triangle as well. We will need Mr. Euclid's remarkable theorem, the Isosceles Triangle Theorem.

Let angle DAC = p, and let the complement of p be q ----> p + q = 90, and 2p + 2q = 180.

In triangle ACD, by the Isosceles Triangle Theorem, angle DAC = angle ACD = p, and because the three angles of triangle ACD must sum to 180, we know angle D = 180 - 2p = 2q. The angles of triangle ACD are p & p & (2q).

Now, look at angle E. Segments AD and BE are parallel (they are both perpendicular to line AB, and perpendicular to the same thing means they're parallel). Because AD // BE, angles D & E are same side interior angles, which must be supplementary. Since angle D = (2q), angle E = (2p).

We know angle (BCE) = angle (CBE), again by the ITT. Because the three angles of triangle BCE must sum to 180, we can deduce that angle (BCE) = angle (CBE) = q. The angles of triangle BCE are q & q & (2p).

Now, look at the three angles at point C. The three angles form a straight line, so their sum must be 180.

(angle ACD) + (angle ACB) + (angle BCE) = 180

p + (angle ACB) + q = 180

(angle ACB) + (p + q) = 180

(angle ACB) + 90 = 180

(angle ACB) = 90

Answer = B

Does all this make sense?

Mike
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Mike McGarry
Magoosh Test Prep