Veritas Prep 10 Year Anniversary Promo Question #1 : GMAT Problem Solving (PS)
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# Veritas Prep 10 Year Anniversary Promo Question #1

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Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:00
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Question Stats:

81% (02:25) correct 19% (02:08) wrong based on 94 sessions

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Veritas Prep 10 Year Anniversary Promo Question #2

One quant and one verbal question will be posted each day starting on Monday Sept 17th at 10 AM PST/1 PM EST and the first person to correctly answer the question and show how they arrived at the answer will win a free Veritas Prep GMAT course (\$1,650 value). Winners will be selected and notified by a GMAT Club moderator. For more questions and details please check here: veritas-prep-10-year-anniversary-giveaway-138806.html

To participate, please make sure you provide the correct answer (A,B,C,D,E) and explanation that clearly shows how you arrived at it.
Winners will be announced the following day at 10 AM Pacific/1 PM Eastern Time.

If A, X, Y, and Z are unique nonzero digits in the equation:

XY
*YX
____
AYZ

And Y > X, the 2-digit number XY = ?

A) 13
B) 21
C) 23
D) 24
E) 25
[Reveal] Spoiler: OA

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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:01
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Winner:

harshvinayak

Official Explanation:

With abstract calculation problems, start with what you know. Here, you know a few things:

-you’re multiplying two 2-digit numbers and in doing so you’ll end up with a 3-digit number
-the units digit of the solution is different from the units digit of either number in the problem (which means you cannot use 1)
-the two numbers you’re multiplying are in the form that the digits are the same, just flipped

This means that A and B are both eliminated (you cannot have a units digit of 1, or the units digit in the answer would be either X or Y). And then try choice D: 42*24 is greater than 1000, so its answer will have more than three digits. The same, then, is true of choice E (52*25 is bigger than 42*24), so choice C is the only plausible answer.
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:08

The reasoning is as follows:-
B) can't be answer because Y<X
D) can't be answer because it XY x YX will give 4 digit number. Thus wrong
E) can't be answer because it XY x YX will give 4 digit number. Thus wrong
A) can't be answer because it XY x YX will give 3 digit number but the middle digit will not be equal to Y. Thus wrong

C) is the correct answer because Y>x & will give 3 digit number & middle digit is Y
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Last edited by fameatop on 17 Sep 2012, 09:13, edited 2 times in total.
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:09
fameatop wrote:

The reasoning is as follows:-

FYI: to be fair to everyone, we won't be able to take into consideration incomplete posts (answer only without an explanation) and will count complete answers/replies by the the last edit time/date stamp on the post (once you edit it or post again, there will be an update time stamp on it).

Thank you for participating and good luck!
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:10
The question could be solved easily using brute force approaching through the options:

A) 13 * 31 would give units digit as 3 which cannot be Z
B) 21 * 12 would give units digit as 2 which cannot be z
C) 23 * 32 would give 736 and satisfies Y> X, hence XY = 23 !
D and E will give answers not in the form of AYZ
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:12
Answer is C by back solving

Eliminate B from the get go then plug in numbers from C (if it wasn't C work through other numbers).

023
032
736
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:19
Since Y>X, answer choice B is eliminated. Then I test the answer choices.

A) (13)(31)= 403

C) (23)(32)= 736

D) (24)(42)= 1008

E) (25)(52)=1300

Since the answer choice is 3-digit number, D and E are eliminated. That leaves A and C. The question states that the second digit is Y thus C is the correct answer.
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:26
:OA is C.
Y>X--so eliminate B,because 1 is not greater than 2.

A,D,E doesn't result in pattern AYZ when XY is multiplied with YX.So,we are left with C.
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:31
Back solving we find that only 23 satisfies the the conditions .
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:50
I wasn't able to understand the equation initially...looked more like an expression.

A. 13*31 = 403
B. Doesn't satisfy Condition Y>X
C. 23*32=736
D. 24*42=1008
E. 25*52=1300

D & E are not AYZ format.
C is only AYZ format. Hence Ans is 'C'
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:53

Why do we have to solve problem backward or do lenghty calculations for each number? Just need to read and understand question and possible answers carefully and we can eliminate incorrect choices one by one.

The solution to this problem is ---

A, X, Y, and Z are unique and are nonzero

XY
*YX
____
AYZ

As per question Y >X and hence ans choice B is eliminated

Next in remaining choices X = 1 or X =2.

However,if X =1 then unit digit of product (Z) would not be unique hence X <>1 . Hence X =2 And choice A also eliminiated

Now we have 3 possible values of Y= 3 ,4 or 5. However if Y=5, unit digit of multiplication would be 0 as X=2. But we know all numbers are non zero. Therefore Ans choice E also eliminated.

We are left with C and D. ie 24 or 23. But 24*42 is 4 digit and hence choice D is also eliminated.

we are left with one choice that is C. which is the correct answer.
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 09:57
A) 13 13 *31= 403 the last digit of 403 is equal to Y, but we are said that it must be equal to Z,not Y , answ choice A is out
B) 21 is out ,since we are told that Y>X
C) 23 23*32=736 bingo the second digit of 736 is equal to Y and other digits are in line with the required condition
D) 24 24*42=1008 is out, since 1008 is a 4 -digit number, but we need only 3-digit number
E) 25 no need to calculate, since we already know that 24*42 is a big number . accordingly, 25*52 is a 4-digit number
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 12:03
Quote:
If A, X, Y, and Z are unique nonzero digits in the equation:

XY
*YX
____
AYZ

And Y > X, the 2-digit number XY = ?

A) 13
B) 21
C) 23
D) 24
E) 25

A digit is a number from a set of numbers from 0 to 9.

B) eliminated at first glance; the condition is given as Y>X but 12 makes it just the opposite ( i.e. X>Y)
A) by multiplying 13 and 31 ( i.e. X = 1 and Y = 3), the multiplication does not give Y as 3 in AYZ. Hence, eliminated
C) 23*32 gives Y as 3 in AYZ (736); this satisfies all the conditions
D) E) Both give result that is greater than 999 i.e. it makes the result of 4 digits. However, A should be a digit and X & Y are single digit numbers as given in the condition that XY is a 2-digit number. Hence both options eliminated

This left with C.
I wonder if there is a better method to solve such kind of problem. That would be helpful.
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 15:55

because the question says Y>X, so it eliminates answer choice B.
start putting the values and looking at the value. value for answer choice A violates the desired format (13*31 = 403) (XY * YX = **Y). values for answer choice D and answer choice E are in 4 digit so out of scope.

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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 18:13
Hi, i also think of the answer as C, but i approached different method to solve: here is my explanation:
XY and YX as two digit numbers can be written as following:
XY=10X + Y
YX=10Y + X
and
AYZ=100A + 10Y + Z
So XY * YX = AYZ can be (10X + Y)*(10Y + X) = 100A + 10Y + Z
which is also 100XY + 10(X^2 + Y^2) + XY = 100A + 10Y + Z
From:
XY
*YX
AYZ
we can know that XY or Z is one digit number otherwise the last digit of X^2 + Y^2 cann't be equal to Y, and considering Y>X we have to check numbers which match the following cases:
1) XY=Z
2) last digit of X^2+Y^2 is Y
X Y
1 9
1 8
till
1 2
but none of them matches the second requirement
next set is

2 3
2 4
the products of remaining are two digit numbers.
let's check the first case
2^2 + 3^2= 13 which is the last digit is 3 ( Y itself)
2^2 + 4^2 = 20 no
that is the anwer is 23 which results 23*32=736
thank you for attention
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 19:06
Really nice question. I see many have already answered by back solving, which I assume would be the best way to go about it on the test. However, I'm hoping that you are looking for slightly different way to go about it, so here is my attempt:

A,X,Y,Z are unique and non-zero 1-9

Xy*yx =ayz since the product is 3 digits, x and y must both be less than 5, and the product of x and y must be less than 10 which gives 1,2,3,4 as the possible values for x and y

moving on since the left most digit of the product (a) is not equal to z, it is understood that that this must be because of a carry over from the operation xx+yy. It is also understood that a must be greater than z.

-> xx+yy> 10

x*y=z

x*x+y*y= 10(x*y - a)+y

a>z>y>x, or a>x*y>y>x

The only way you will get x*x + y*y, as being greater than 10 while each is less than 5, and the product of x and y is less than 10 is if one is the combos os 2,3 or 2,4. As seen, y is the greater value in either of the combos.

However x*x+y*y should have the units digit as y, and only 2,3 -> 2*2+3*3 give the units digit as y (3 in this case).

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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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17 Sep 2012, 19:14
Really nice question. I see many have already answered by back solving, which I assume would be the best way to go about it on the test. However, I'm hoping that you are looking for slightly different way to go about it, so here is my attempt:

A,X,Y,Z are unique and non-zero 1-9

Xy*yx =ayz since the product is 3 digits, x and y must both be less than 5, and the product of x and y must be less than 10 which gives 1,2,3,4 as the possible values for x and y

moving on since the left most digit of the product (a) is not equal to z, it is understood that that this must be because of a carry over from the operation xx+yy. It is also understood that a must be greater than z.

-> xx+yy> 10

x*y=z

x*x+y*y= 10(x*y - a)+y

a>z>y>x, or a>x*y>y>x

The only way you will get x*x + y*y, as being greater than 10 while each is less than 5, and the product of x and y is less than 10 is if one is the combos os 2,3 or 2,4. As seen, y is the greater value in either of the combos.

However x*x+y*y should have the units digit as y, and only 2,3 -> 2*2+3*3 give the units digit as y (3 in this case).

Posted from my mobile device

Posted from my mobile device
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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18 Sep 2012, 07:34
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We have a winner!

Winner:

harshvinayak

Official Explanation:

With abstract calculation problems, start with what you know. Here, you know a few things:

-you’re multiplying two 2-digit numbers and in doing so you’ll end up with a 3-digit number
-the units digit of the solution is different from the units digit of either number in the problem (which means you cannot use 1)
-the two numbers you’re multiplying are in the form that the digits are the same, just flipped

This means that A and B are both eliminated (you cannot have a units digit of 1, or the units digit in the answer would be either X or Y). And then try choice D: 42*24 is greater than 1000, so its answer will have more than three digits. The same, then, is true of choice E (52*25 is bigger than 42*24), so choice C is the only plausible answer.
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Re: Veritas Prep 10 Year Anniversary Promo Question #1 [#permalink]

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11 Sep 2015, 01:46
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Re: Veritas Prep 10 Year Anniversary Promo Question #1   [#permalink] 11 Sep 2015, 01:46
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# Veritas Prep 10 Year Anniversary Promo Question #1

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