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Intern  Joined: 10 Nov 2015
Posts: 4
Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers  [#permalink]

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2
D: I and II only.
Easy to realize that P must be 1 because maximum 99+99 = 198 < 200, P cannot be 2 or greater.
XY + YX = 10X + Y + 10Y + X = 11(X+Y) = 1Q5
=> 1Q5 divisible by 11
=> Q = 1+5 = 6, PQ5 = 165
=> X+Y = 15 => (X,Y) e {(9,6),(8,7),(7,8),(6,9)}. But notice that X,Y,P,Q is different from each other so we cannot take 6 twice.
Conclusion : X e (7,8). Choice D.
Current Student Joined: 13 Apr 2015
Posts: 6
Location: India
GPA: 3.7
Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers  [#permalink]

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1
X Y + Y X = PQ5

This implies that, X + Y should have a units digit as 5.
there are 2 options:
1) One of the numbers is 1 and the other is 4. But this condition is ruled out as it is mentioned in the question that X, Y, P, Q are different non-zero digits, hence P cannot be equal to 0
2) The only other option is when one of the numbers is 7 and the other is 8. So when we add X Y + Y X, we get the desired 3 digit number with the units digit as 5.

So the correct answer is D.
Manager  Joined: 20 Sep 2008
Posts: 68
Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers  [#permalink]

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x,y single digits > 0 because only sum of two digit numbers can give a three digit number.

Also, x + y <= 18

Unit digit = 5
Tens digit = Q
Hundreds digit = P

I) If x = 7. For the unit digit of (x + y) to end on 5, the value of y = 15 - 7 = 8. So, the two numbers are 78 and 87. Sum is 165. P=1, Q = 6. Values are all different, which is a solution.
II) If x = 8. For the unit digit of (x + y) to end on 5, the value of y = 15 - 8 = 7. So, the two numbers are 87 and 78. Sum is 165. P=1, Q = 6. Values are all different, which is a solution.
III) If x = 9. For the unit digit of (x + y) to end on 5, the value of y = 15 - 9 = 6. so, the two numbers are 69 and 96. Sum is 165. All values are different, which is a solution.
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GMAT 1: 630 Q44 V32 GMAT 2: 680 Q47 V35 Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers  [#permalink]

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It should be E. All the 3 values are possible for X.

X Y
+ Y X
_____
P Q 5

Possibilities for adding units digits:
Let, Y + X = N5 (where N can be any number from 0 to 9)
Now, since, Y and X both can only range between 1-9, Y+X will can only range between 2-18
So, Y + X can only yield 5 or 15

Possibilities for adding tens digits:
Given that X + Y at tens place yield a 2-digit number (P cant be 0). If Y + X from units yield only a 5 then X + Y from tens will never yield a two-digit number.
Therefore Y + X = 15

Possible values of (X,Y):
(7,8) or (8,7) or (9,6) or (6,9)

Substituting values of and summing XY + YX gives 165.

Since, all the numbers X, Y, P, Q have to be different, X cannot be equal to 6 (as Q=6), therefore X can take values- 7, 8 or 9.

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Math Expert V
Joined: 02 Sep 2009
Posts: 58421
Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers  [#permalink]

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Bunuel wrote:

Jamboree and GMAT Club Contest Starts

QUESTION #1:

X Y
+Y X
________

The sum of the two digit numbers above is a three digit number PQ5, where each letter X, Y, P, and Q represents a different non zero digit. Which of the following can be the value of X?

I) 7
II) 8
III) 9

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I , II and III

Check conditions below:

For the following two weekends we'll be publishing 4 FRESH math questions and 4 FRESH verbal questions per weekend.

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, respective forum moderators will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize. He/She can opt for one of the following as a Grand Prize. It will be a choice for the winner:
-- GMAT Online Comprehensive (If the student wants an online GMAT preparation course)
-- GMAT Classroom Program (Only if he/she has a Jamboree center nearby and is willing to join the classroom program)

Bookmark this post to come back to this discussion for the question links - there will be 2 on Saturday and 2 on Sunday!

There is only one Grand prize and student can choose out of the above mentioned too options as per the conditions mentioned in blue font.
All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time.

NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!

JAMBOBREE OFFICIAL SOLUTION:

Since XY and YX are 2 digit numbers, there sum has to be less than 200. Thus, P = 1. Also sum of the digits X and Y should end with 5. Hence X + Y = 5 or 15 or 25… so that 5 is the units digit and tens digit is carried over. However, X + Y is less than 18 (as they are digits) and greater than 10 (else sum won’t be a 3-digit number). Thus, X + Y = 15, which gives
X Y
+ Y X
-------
P Q 5
=> 1 6 5

Alternatively: when a two digit number is added to another two digit number formed by reversing the digits of the first two digit number,the result is a multiple of 11. So in this case as PQ5 has to be a multiple of 11. So, using the divisibility rule of 11, P =1 and Q = 6.

Now, X+Y = 15,so all possible values of x and Y are
Case 1 X = 6 Y = 9
Case 2 X = 7 Y = 8
Case 3 X = 8 Y =7
Case 4 X =9 Y = 6

But as it is given that each letter X, Y, P, and Q represents a different non zero digit AND we know that Q = 6 so case 1 and case 4 are not possible

So the only possible values of x are 7 and 8

Hence the correct answer is option D.
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Intern  Joined: 13 Mar 2011
Posts: 21
Jamboree and GMAT Club Contest: The sum of the two digit numbers  [#permalink]

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MelodyofMew wrote:
D: I and II only.
Easy to realize that P must be 1 because maximum 99+99 = 198 < 200, P cannot be 2 or greater.
XY + YX = 10X + Y + 10Y + X = 11(X+Y) = 1Q5
=> 1Q5 divisible by 11
=> Q = 1+5 = 6, PQ5 = 165
=> X+Y = 15 => (X,Y) e {(9,6),(8,7),(7,8),(6,9)}. But notice that X,Y,P,Q is different from each other so we cannot take 6 twice.
Conclusion : X e (7,8). Choice D.

I like the solution above. However, for me, it isn't understandable why Q = 1 + 5 = 6 ???
So, In my humble opinion, much easier will be to solve this problem like below :

the system of thoughts will be :

Easy to realize that P must be 1 because maximum 99+99 = 198 < 200, P cannot be 2 or greater. (1) and X and Y are integer (1.1)

XY + YX = 10X + Y + 10Y + X = 11(X+Y) = 1Q5 (2) and the unit digit is 5 (2.1)

because of (1) and (1.1) => 100 < 11(X+Y) < 200 => 9 < X+Y < 19 (2.2)

because of (2.1) and (2.2) and (1.1) => X + Y = 15 (3) - in other words 15 will be the single integer from 9 to 19 that gives unit digit of 5

because of (2) and (3) 11(X+Y) = 11*15 = 165 (4)

X+Y = 15 => (X,Y) e {(9,6),(8,7),(7,8),(6,9)}. (5)

But because of (4) => X e (7,8). Choice D
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Please, press the +1 KUDOS button , if you find this post helpful Intern  B
Joined: 10 Aug 2015
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GMAT 1: 700 Q48 V38 GPA: 3.5
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Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers  [#permalink]

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Bunuel wrote:

Jamboree and GMAT Club Contest Starts

QUESTION #1:

X Y
+Y X
________

The sum of the two digit numbers above is a three digit number PQ5, where each letter X, Y, P, and Q represents a different non zero digit. Which of the following can be the value of X?

I) 7
II) 8
III) 9

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I , II and III

Every one is missing a small fact that XY and YX are 2 digit numbers one with reversed digit. Now their sum will always be divisible by 11. As both are 2 digit numbers then P=1 so only possibility where 1Q5 is divisible by 11 is when Q=6.

So Essentially we have XY
+YX
_____
165

Now you can check numbers that give X+Y =15 to get answer.
VP  P
Joined: 07 Dec 2014
Posts: 1224
Jamboree and GMAT Club Contest: The sum of the two digit numbers  [#permalink]

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X Y
+Y X
________

The sum of the two digit numbers above is a three digit number PQ5, where each letter X, Y, P, and Q represents a different non zero digit. Which of the following can be the value of X?

I) 7
II) 8
III) 9

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I , II and III

XY+YX is a multiple of 11 and 5
100<XY+YX<187
XY+YX=11*15=165
XY=78 and YX=87; or
XY=87 and YX=78
D I and II only
Intern  B
Joined: 17 Jan 2018
Posts: 44
Schools: ISB '20 (A)
Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers  [#permalink]

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Given -

10X + Y + 10Y + X = 100P + 10Q + 5

11(X+Y) = 5(20P +2Q + 1)

Now, common sense says that P needs to be 1. (Max 99 + 99 is still 198 only)
So,
11(X + Y) = 5(21 + 2Q)

Now, we can see that 21 + 2Q should be a multiple of 11, to satisfy this equation.
We also know that Q can be at max 9, (So, 21 + 18 = 39), so a multiple of 11, less than 39 is 33.

Hence, 21 + 2Q = 33..... Shows that Q = 6

Hence, X + Y = 5*3 = 15

Also, P and Q are taken, 1 and 6 respectively.

The only way to make (X,Y) as 15 are (9,6) and (8,7).
As, 6 is taken by Q, we are left with just (8,7)

So, X can be either 7 or 8.

Ans) D

Bunuel wrote:

Jamboree and GMAT Club Contest Starts

QUESTION #1:

X Y
+Y X
________

The sum of the two digit numbers above is a three digit number PQ5, where each letter X, Y, P, and Q represents a different non zero digit. Which of the following can be the value of X?

I) 7
II) 8
III) 9

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I , II and III

Check conditions below:

For the following two weekends we'll be publishing 4 FRESH math questions and 4 FRESH verbal questions per weekend.

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, respective forum moderators will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize. He/She can opt for one of the following as a Grand Prize. It will be a choice for the winner:
-- GMAT Online Comprehensive (If the student wants an online GMAT preparation course)
-- GMAT Classroom Program (Only if he/she has a Jamboree center nearby and is willing to join the classroom program)

Bookmark this post to come back to this discussion for the question links - there will be 2 on Saturday and 2 on Sunday!

There is only one Grand prize and student can choose out of the above mentioned too options as per the conditions mentioned in blue font.
All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time.

NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!

Non-Human User Joined: 09 Sep 2013
Posts: 13408
Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers  [#permalink]

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_________________ Re: Jamboree and GMAT Club Contest: The sum of the two digit numbers   [#permalink] 25 Aug 2019, 18:26

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