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A person borrowed a certain sum at the rate of 20% per annum [#permalink]

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03 Jan 2013, 03:06

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A person borrowed a certain sum at the rate of 20% per annum compound interest. He paid a total of $4000 in two annual equal installments. What interst did he pay approximately ?

A. $955 B. $944 C. $933 D. $922 E. $911

Last edited by Bunuel on 03 Jan 2013, 03:08, edited 1 time in total.

Find the lowest of three nummbers as described : [#permalink]

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03 Jan 2013, 03:10

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

Re: Find the lowest of three nummbers as described : [#permalink]

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03 Jan 2013, 12:00

Expert's post

pariearth wrote:

Find the lowest of three numbers 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 pariearth, First of all, your phrasing in the problem is deeply ambiguous. You need to understand --- the phrasing in GMAT PS problems is clear and precise. When you say "their product", you are using a pronoun with no antecedent --- something that would always be incorrect on GMAT SC. It's important to understand: the grammar & syntax on all parts of the GMAT are always up to the standards of GMAT SC questions.

I think what you mean is the following: Let a, b, and c be three numbers. The cube of a exceeds the product of b and c by 2; the cube of b is smaller than the product of a and c by 3; and the cube of c exceeds the product of a and b by 3. This is a much more precise wording.

Another problem is: one phrased this way --- assuming this was the problem you had in mind --- this is near impossible to solve with simple algebra. This is WAY too hard for GMAT math. I solve the system of equations on Wolfram Alpha, which found three different sets of real value solutions as well as five different values of complex value solutions. All the real value answers involve negative numbers, so in each case, the "lowest value" is a negative. Your answer of 3^(1/3) doesn't appear anywhere among the answers.

So I believe even when we clarify what the question is asking, the core question & answers are flawed.

Re: Find the lowest of three nummbers as described : [#permalink]

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04 Jan 2013, 00:02

mikemcgarry wrote:

pariearth wrote:

Find the lowest of three numbers 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 pariearth, First of all, your phrasing in the problem is deeply ambiguous. You need to understand --- the phrasing in GMAT PS problems is clear and precise. When you say "their product", you are using a pronoun with no antecedent --- something that would always be incorrect on GMAT SC. It's important to understand: the grammar & syntax on all parts of the GMAT are always up to the standards of GMAT SC questions.

I think what you mean is the following: Let a, b, and c be three numbers. The cube of a exceeds the product of b and c by 2; the cube of b is smaller than the product of a and c by 3; and the cube of c exceeds the product of a and b by 3. This is a much more precise wording.

Another problem is: one phrased this way --- assuming this was the problem you had in mind --- this is near impossible to solve with simple algebra. This is WAY too hard for GMAT math. I solve the system of equations on Wolfram Alpha, which found three different sets of real value solutions as well as five different values of complex value solutions. All the real value answers involve negative numbers, so in each case, the "lowest value" is a negative. Your answer of 3^(1/3) doesn't appear anywhere among the answers.

So I believe even when we clarify what the question is asking, the core question & answers are flawed.

Does all this make sense?

Mike

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....

Re: Find the lowest of three nummbers as described : [#permalink]

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04 Jan 2013, 12:34

Expert's post

pariearth wrote:

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....

Dear Pari, Well, I don't know what source you are using, but I can absolutely guarantee --- this question is much more advanced mathematically than anything you will see on the GMAT. If the source you are using purports to be a GMAT source, then I would say this question gives evidence that it is quite poorly informed about what the test actually is likely to ask.

I'm pretty good with algebra, and I see no algebraic solution possible that would not involve having to factor a general cubic, which is leagues beyond anything the GMAT would ask.

I solved this using Wolfram Alpha as well, and found the following

Solution Set #1 a = 2 b = 3^(1/3) c = 3^(2/3)

Solution Set #2 a = 1/[2^(1/3)] b = -[3^(2/3)]/[2^(1/3)] c = (3/2)^(1/3)

There are also five sets of complex roots not on the real number line.

The value b will always be the lowest value of the three. For the set of positive roots, Set #1, the lowest value, b, is indeed 3^(1/3), but in Set #2, b has a negative value. I don't know if the source meant to specify that the three numbers should be positive and forgot to do so (poorly written sources typically forget to specify such things.)

Overall, I would say: if this question is representative in this source, then immediately stop using this source, and find a source that actually is designed to prepare you for what is really on the GMAT.

Re: A person borrowed a certain sum at the rate of 20% per annum [#permalink]

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04 Jan 2013, 14:39

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 $

If a, b, c and d are natural numbers such that [#permalink]

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14 Jan 2013, 06:15

If a, b, c and d are natural numbers such that \(a^d+b^d=c^d\), then which of the following is always true?

a.) d is never more than the minimum of a, b and c b.) d is never less than the maximum of a, b and c c.) d lies between the minimum of a, b and c and the maximum of a, b and c d.) d is more than any of a, b and c e.) None of these

Re: If a, b, c and d are natural numbers such that [#permalink]

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14 Jan 2013, 07:06

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. _________________

The signboard outside the department store "Ram and Shyam" [#permalink]

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14 Jan 2013, 22:04

The signboard outside the department store "Ram and Shyam" lights up as described below: When the switch is turned on, all the three words light up and remain lighted for 3 seconds. After that, the first word is switched off for \(7\frac{5}{6}\) seconds, the second word is switched off for \(1\frac{1}{3}\)seconds and the third word is switched off for \(5\frac{2}{3}\) seconds. Then each word is again switched on for 3 seconds and switched off for the time duration mentioned. This process continues, repeatedly. after how many seconds of switching on the signboard will the entire board be switched on for the second time for 3 seconds?

A.) \(40\frac{1}{3}\) B.) \(41\frac{2}{3}\) C.) \(42\frac{2}{3}\) D.) \(43\frac{1}{3}\) E.) \(40\frac{2}{3}\)

Re: The signboard outside the department store "Ram and Shyam" [#permalink]

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14 Jan 2013, 23:49

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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.

Re: N is a natural number having the sum of its digits as 3. [#permalink]

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15 Jan 2013, 03:00

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

Re: If [m]x^2 - 4x + 1 = 0[/m], find the value of [m]x^4 + [frac [#permalink]

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15 Jan 2013, 03:45

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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

Re: N is a natural number having the sum of its digits as 3. [#permalink]

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15 Jan 2013, 03:47

akashganga wrote:

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

hey how did you come to a conclusion that it is a combination problem?

Re: N is a natural number having the sum of its digits as 3. [#permalink]

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15 Jan 2013, 03:58

chiccufrazer1 wrote:

akashganga wrote:

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

hey how did you come to a conclusion that it is a combination problem?

Posted from my mobile device

I think the more you practice, the easier it gets to tell what kind of a problem it is.

But also the question says how many values are possible. And when we look at options, if they would have been less than 10 I would have done each and every case manually, but since they are large in number, there is no other way of doing it in short period of time. So it is a Permutation and Combination problem.

In the following figure find angle ACB. Given C is the poin [#permalink]

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15 Jan 2013, 04:23

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

Re: N is a natural number having the sum of its digits as 3. [#permalink]

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15 Jan 2013, 04:41

wow that now makes sense..we are really not supposed to solve for large numbers manually but with a simple trick like combination..i was really confused..thanks

Re: In the following figure find angle ACB. Given C is the poin [#permalink]

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15 Jan 2013, 11:29

Expert's post

pariearth wrote:

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

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.

Attachment:

tangent circles tangent to line.JPG [ 26.62 KiB | Viewed 1226 times ]

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 _________________

Mike McGarry Magoosh Test Prep

gmatclubot

Re: In the following figure find angle ACB. Given C is the poin
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15 Jan 2013, 11:29

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