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Difficulty:
25%
(medium)
Question Stats:
75% (01:25) correct
25%
(01:04)
wrong
based on 314
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In the multiplication problem below, each capital letter represents a positive digit, and the tens digit of the product is zero. What is the value of digit B? 
Arithmetic_2.jpg [ 11.68 KiB | Viewed 7619 times ]
A. 2
B. 4
C. 7
D. 8
E. 9
Kudos for a correct solution.
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Attachment:
Arithmetic_2.jpg [ 11.68 KiB | Viewed 7619 times ]
A. 2
B. 4
C. 7
D. 8
E. 9
Kudos for a correct solution.
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Here we go:
9*9 = 63
This will generate 6 to be carried forward
and B + 6 = 10 (out of which 1 will be carried forward leaving 0)
B has to 4
Option B is correct
9*9 = 63
This will generate 6 to be carried forward
and B + 6 = 10 (out of which 1 will be carried forward leaving 0)
B has to 4
Option B is correct
Here we go:
9*9 = 81
This will generate 8 to be carried forward
and B*9 + 8 = X0
B has to 8
Option D is correct
9*9 = 81
This will generate 8 to be carried forward
and B*9 + 8 = X0
B has to 8
Option D is correct
hi..
ans D 8...
initially it becomes a bit confusing as we have almost 16 letters for 10 digits...
but then since the question stems asks us to concentrate only at B, the method is as follows...
HB9
x 9
S0W
it means w=1 as 9*9=81 and 9*B+8=a0..
B=8 as only this value gives us a 0 as units digit for eq
_________________
ans D 8...
initially it becomes a bit confusing as we have almost 16 letters for 10 digits...
but then since the question stems asks us to concentrate only at B, the method is as follows...
HB9
x 9
S0W
it means w=1 as 9*9=81 and 9*B+8=a0..
B=8 as only this value gives us a 0 as units digit for eq
_________________
DesiGmat
hi desigmat u have wrongly written 9*9 as 63 it is 81
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Schools: Erasmus '17
Posts: 5
My solution revolves around the trick of testing the anwser choices, however I would love to see an algebraic approach for this problem.
The information provided is that 9*9 = 81, the result of this is that the 8 is carried one place to the left. The problem asks you to seek for a solution that adds 2 to the tens digit place.
By testing the answer choices, what in my opinion is a tedious process, we can figure out the answer.
A) 29*9 = (9*9) + (9*20) = 81 + 180 = 261 - incorrect
B) 49*9 = (9*9) + (9*40) = 81 + 360 = 441 - incorrect
C) 79*9 = (9*9) + (9*70) = 81 + 630 = 711 - incorrect
D) 89*9 = (9*9) + (9*80) = 81 + 720 = 801 - correct
E) 99*9 = (9*9) + (9*90) = 81 + 810 = 891 - incorrect (for completeness)
Answer: D
Again, I would love to see an algebraic approach less tedious than this one.
The information provided is that 9*9 = 81, the result of this is that the 8 is carried one place to the left. The problem asks you to seek for a solution that adds 2 to the tens digit place.
By testing the answer choices, what in my opinion is a tedious process, we can figure out the answer.
A) 29*9 = (9*9) + (9*20) = 81 + 180 = 261 - incorrect
B) 49*9 = (9*9) + (9*40) = 81 + 360 = 441 - incorrect
C) 79*9 = (9*9) + (9*70) = 81 + 630 = 711 - incorrect
D) 89*9 = (9*9) + (9*80) = 81 + 720 = 801 - correct
E) 99*9 = (9*9) + (9*90) = 81 + 810 = 891 - incorrect (for completeness)
Answer: D
Again, I would love to see an algebraic approach less tedious than this one.
ModularArithmetic
hi i dont think you have to get into such calculations..
you have rightly said 9*9=81..
and next step= 9*b+8=a0... this means a0 -8 =9*B...only b as 8 gives 2 inunits digit
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PareshGmat
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
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Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,531
Answer = D = 8
9*9 = 81, which gives w = 1 & creates a carry of 8
Setting up the equation (For tens digit)
9*B + 8 = 0
To give a "0" at the end, we require 9*B to give "2" in the units place
Only 9*8 = 72 satisfies the condition
9*9 = 81, which gives w = 1 & creates a carry of 8
Setting up the equation (For tens digit)
9*B + 8 = 0
To give a "0" at the end, we require 9*B to give "2" in the units place
Only 9*8 = 72 satisfies the condition
Bunuel
This question works toward helping you develop the "Math instinct" we talk about.
You get B*9 + 8 = ..0
So you know that B*9 should end with 2. Immediately 8*9 = 72 will come to your mind. Along with it, what should also come to your mind is that in the multiplication table of 9, every product ends with a different digit.
9*1 = 9
9*2 = 18
9*3 = 27
9*4 = 36
9*5 = 45
9*6 = 54
9*7 = 63
9*8 = 72
9*9 = 81
9*10 = 90
This happens for tables of 1, 3, 7 and 9 - i.e. all odd numbers except 5. So if this were a DS question, you would know that this is sufficient information without running through the table of 9.
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