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HKD1710
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Back solving is the best on in my opinion .

The best answer is B, 6.
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4+8+2 = 14

1 gets carried over to the next number so it is 3 + 1 + x + x = 1x where x is the units digit. another way to write the equation is 3 + 1 + x +x =10 + x

7+5+9 = 21 that means we had a carry over from the previous sum to get 22.

Try picking 5 it becomes 3 + 5 + 5 +1 = 14. Where x = 5 so this is not true.

Try picking 6 it becomes 3 + 6 + 6 + 1 = 16 where x = 6 so it is true.

3 + x + x + 1 = 10 + x

4 + 2x = 10 + x

x = 6

Answer choice B
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HKD1710

In the addition problem above, the number * must be?

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9


Attachment:
question.JPG
Unit's digit is 4 + 8 + 2 = 14 ( Carry Over 1 to ten's digit )

Ten's Digit is 1 ( Carry Over ) + 3 + x + x = x

So, we have Ten's digit as 4 + x + x = x ( Digit )

Now, Check using the options.....

(A) 4 + 5 + 5 = 14 ( Not Possible)

(B) 4 + 6 + 6 = 16 ( Possible )

(C) 4 + 7 + 7 = 18 ( Not Possible)

(D) 4 + 8 + 8 = 20 ( Not Possible)


Thus, Answer must be (B)­
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I still couldn't understand why 5 was eliminated.
Kindly explain
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since the result in the right was bigger than 10 you carry 1. now 3+1 supply you a 4
4+5+5 will be 4 in the units and not 5
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Usually, in such special cases, when you have just one digit to fill, it's best to try values given in the options.
Start putting values from option A (didn't get the resutlt) -> option B (got the result, STOP).
Answer B.
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Hi Bunuel, please could you explain this question - I am not clear on what has been asked?
HKD1710

In the addition problem above, the number * must be?

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9


Attachment:
question.JPG
­
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HKD1710

In the addition problem above, the number * must be?

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9


Attachment:
question.JPG
8+4+2=14, 4 remains and 1 carries over
3+1+x+x=x+10 (as x remains and 1 carries over)
4+2x=x+10
x=6

Can be verified by plugging it back: 734+568+962=2264
Hence, B.­
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HKD1710

In the addition problem above, the number * must be?

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9

STRATEGY: Upon reading any GMAT Problem Solving question, we should always ask, Can I use the answer choices to my advantage?
In this case, we can easily replace each * with each answer choice until we find the one that creates the correct addition.
Now let's give ourselves up to 20 seconds to identify a faster approach.
In this case, we can also assign variables to the missing digits, but I can already see that testing the values will be super fast.
So, let's go with that...


When we replace each * with 5, the addition doesn't work.
When we replace each * with 6, the addition works!



Answer: B
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We're clearly carrying a 1 into the tens column. So in the tens column we're adding 1 + 3 + star + star = 4 + 2*star, which is clearly even. But in the bottom row we see that sum must end in star, so star must be even, and now we know the answer can only be 6 or 8, and testing just one answer lets us figure out which is right (if we test 8, and see it's wrong, we don't need to test 6, because it has to be right).
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HKD1710

In the addition problem above, the number * must be?

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9


We don't need the left two columns and can arrive at the correct answer just from the right two.
We carry a 1 from the right column.
1+3+x+x yields something that ends in x.

(A) 1+3+5+5 = 14 Doesn't end in 5. Eliminate
(B) 1+3+6+6 = 16 Ends in 6. Keep it.
(C) 1+3+7+7 = 18 Doesn't end in 7. Eliminate.
(D) 1+3+8+8 = 20 Doesn't end in 8. Eliminate.
(E) 1+3+9+9 = 22 Doesn't end in 9. Eliminate.

Answer choice B.
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Simplifying, we get

=>(3+1)+x+x=10+x => x=6

Hence B

Another approach, verifying options one by one only B.6 and C.7 fits however if you look closely through the addition of unit digits, you will eliminate C.7 and we are left with B.6
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