The sum of positive integers x and y is 77. What is value of xy?

Actually we don't get fractions.

The sum of positive integers x and y is 77. What is value of xy?

Given that \(x+y=77\) find the value of \(xy\).

(1) x = y + 1 --> \((y+1)+y=77\) --> \(y=38\) and \(x=39\) --> \(xy=39*38\). Sufficient.

(2) x and y have the same tens digit. In order the sum to be 77 the tens digit of of x and y must be 3, thus \(x=38\) and \(y=39\) or vise-versa, in either case \(xy=39*38\). Sufficient.

Answer: D.

Hope this helps.
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32+35 is not equal to 77

There are only 2 numbers that add up to 77 and have the same 10 digits. They are 38 and 39. Therefore either statement is sufficient - D

St1:
x = y+1
so y+1+y = 77 --> can solve for y, then can solve for x and finally xy. Sufficient.

st2:
x and y have the same tens digit. We can rule out the tens digit 1,2,4,5-9 because that would require the other integer to take either a bigger or smaller value. The only value that works is 38,39. Since it's multiplication, we don't care if x took 38 or x took 39. xy will be the same. Sufficient.

Ans D

Here is the data sufficiency problem from GMATPrep;

The sum of positive integers x and y is 77, what is the value of xy?

(1) x=y+1
(2) x and y have the same tens digit.

Hi ,

I got the answer as D
The sum of positive integers x and y is 77, what is the value of xy?

(1) x=y+1
(2) x and y have the same tens digit.

Statement A : X=Y+1 so using eqn X+Y=77
2Y+1=77
Y= 38
X= 39
thus we can find the value of XY
Hence A is sufficient
Statement 2 : Says X & Y have the same 10's digit => X= 10 * A + B & Y = 10 * A + C

For X + Y = 77 & the above constraint we can only opt for a number between 30 - 39 thus the choice of 38 & 39
Hence B is sufficient

Thus Answer D

Welcome to GMAT Club. Hope below solution clears your doubts.

The sum of positive integers x and y is 77, what is the value of xy?

Given: x+y=77. Question: xy=?

(1) x=y+1 --> together with x+y=77 we have two distinct linear equations, hence we can solve the for variables and obtain the value of xy. Sufficient.

(2) x and y have the same tens digit --> x and y cannot have the tens digit of 2 or 4 (as 29+29<77 and 40+40>77) --> the units digit is 3 --> now, if x=y then x=y=77/2=38.5 --> as both are integers then x and y are 38 and 39 or vise versa (neither of them can be less than 38 as in this case the sum will be less than 77: 37+39=76). Therefore xy=38*39. Sufficient.

Answer: D.

The problem with your solution is that (30,37), (31,36), (32,35), (33,34) add up to 67 not 77.

Hope it's clear.
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sorry for the inconvenience:) it may happen (mistakes in simple math), thanks anyway.

No need to apologize at all: errare humanum est. We all do similar mistakes from time to time.

You are most welcome to post any question you like.
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The question says that X and Y are two positive integers but solving statement 1 you will get X and Y as fractions. If the statement refutes the question data then what is the answer?

Questions says 10's digit same, but isn't it assumption that it should be 3? Like this we can assume anything and solve the question.

No. The tens digit of x and y cannot be any digit but 3: if it's less than 3, then x+y<77 and if it's greater than 3 then x+y>77.

Does this make sense?
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In the statement 2, the restriction is given on the tenth digit. However, there was no restriction on unit digit. Hence, x and y can be 31, 32, 33, 34, 35, 36, 37, 38 and 39. If x = 35, y = 32, then X*Y = 1120. Again, if x= 33, y = 34, then X*Y = 1122.


I can't understand the explanation. Please elaborate the second statement in more details.


Thanks.


Regards
Jahid

The numbers you consider give the sum of 67 not 77: 35+32=67 and 33+34=67. The only numbers with the tens digit of 3 which give the sum of 77 are 38 and 39.

Hope it's clear.
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The algebraic way for the second statement is as follows.

\(10a+b+10a+c=77 \implies 20a+b+c=77 \implies b+c=77-20a\)

The limitation here is that \(b \cap c <10 \implies a=3 \implies b+c=17\)

Since \(b \cap c <10 \implies (b=9, c=8) \cup (b=8, c=9)\)
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