If x is a two digit number (so that x = b a, with b and a as digits), what is the units' digit a of x?
(1)The number 3 x is a three digit number whose units' digit is a.
(2)The digit a is less than 7.
digit a of x
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- sanju09
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- albatross86
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There is a conceptual way to do this problem which is difficult to explain, but it is simply by observing the possibilities of a that satisfy this condition in your head. That should be easy for some, but anyway here's an algebraic explanation.
The number x can be expressed as 10b + a
1. 3*(10b + a)
= 30b + 3a
The units digit is therefore "3a" but truncated to its own units digit
You can confirm this with an example: 24*3 = 72 ... 4*3 = 12... so units digit of 24*3 = 2 which is the truncated units digit of 12, i.e. 12 - 10
Another example 98*3 = 294 ... 8*3 = 24 so units digit of 98*3 = 4 which is truncated units digit of 24 i.e. 24 - 20
Sometimes you won''t need to truncate, like 22 * 3 = 66, where the units digit is simply 2*3 = 6
So, either 3a = a => a = 0
OR 3a - 10 = a => a = 5
OR 3a - 20 = a => a = 10 ... but a is a digit so it cannot be greater than 9.
So a could be 0 or 5.
INSUFFICIENT
2. a < 7
Could be anything from 0 to 6
INSUFFICIENT
Both 1 and 2:
Could be 0 or 5.
INSUFFICIENT
Pick E.
The number x can be expressed as 10b + a
1. 3*(10b + a)
= 30b + 3a
The units digit is therefore "3a" but truncated to its own units digit
You can confirm this with an example: 24*3 = 72 ... 4*3 = 12... so units digit of 24*3 = 2 which is the truncated units digit of 12, i.e. 12 - 10
Another example 98*3 = 294 ... 8*3 = 24 so units digit of 98*3 = 4 which is truncated units digit of 24 i.e. 24 - 20
Sometimes you won''t need to truncate, like 22 * 3 = 66, where the units digit is simply 2*3 = 6
So, either 3a = a => a = 0
OR 3a - 10 = a => a = 5
OR 3a - 20 = a => a = 10 ... but a is a digit so it cannot be greater than 9.
So a could be 0 or 5.
INSUFFICIENT
2. a < 7
Could be anything from 0 to 6
INSUFFICIENT
Both 1 and 2:
Could be 0 or 5.
INSUFFICIENT
Pick E.
~Abhay
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So my approach here is to pick numbers and disprove ....
stmt 1 - say x = 31, 41 it works but x could also be 72, 82 ....... hence stmt 1 does not suffice
stmt 2 - does not suffice as shown above.
Combined not enough
Hence E
stmt 1 - say x = 31, 41 it works but x could also be 72, 82 ....... hence stmt 1 does not suffice
stmt 2 - does not suffice as shown above.
Combined not enough
Hence E
- albatross86
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Hi, could you please explain how those examples you've given for statement 1 work?mj78ind wrote:So my approach here is to pick numbers and disprove ....
stmt 1 - say x = 31, 41 it works but x could also be 72, 82 ....... hence stmt 1 does not suffice
stmt 2 - does not suffice as shown above.
Combined not enough
Hence E
As far as I can see, only numbers with 0 or 5 in the units digit would work.
~Abhay
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
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- selango
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I am not sure whether my approach is correct.
Let x=10b+a
From stmt1,
3x is a 3 digits number(let pqr)
100p+10q+r=3x=30b+3a
100p+10q+r=30b+3a
r=a
-->100p+10q+a=30b+3a
2a=100p+10q-30b
a=50p+5q-15b
a/5=10p+q-3b
The equation will be satisified only if a is divisible by 5.Since a is a digit it can be 0 or 5 only.
a=0 or 5
Insufficient.
From stmt2,
a can be any no from 0 to 6.Insufficient.
Combining,
Again a can be 0 or 5.
Insufficient.
Hence E
Let x=10b+a
From stmt1,
3x is a 3 digits number(let pqr)
100p+10q+r=3x=30b+3a
100p+10q+r=30b+3a
r=a
-->100p+10q+a=30b+3a
2a=100p+10q-30b
a=50p+5q-15b
a/5=10p+q-3b
The equation will be satisified only if a is divisible by 5.Since a is a digit it can be 0 or 5 only.
a=0 or 5
Insufficient.
From stmt2,
a can be any no from 0 to 6.Insufficient.
Combining,
Again a can be 0 or 5.
Insufficient.
Hence E
--Anand--
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Unit digits range from 0 -9sanju09 wrote:If x is a two digit number (so that x = b a, with b and a as digits), what is the units' digit a of x?
(1)The number 3 x is a three digit number whose units' digit is a.
(2)The digit a is less than 7.
St 1:
Multiply the unit digit(a) by 3 and see which one exactly matches with" a" in its unit's place.
a * 3= a( expected)
0 *3 = 0 (YES)
1*3= 3(NO)
2*3=6 (NO)
3*3=9 (NO)
4*3=2 (NO)
5*3=5 YES
6*3=8 (NO)
7*3 =1(NO)
8*3 =4 (NO)
9*3 =7 (NO)
a can be 0 or 5.. Insifficient
ST 2: again no much of any use.
combining..Not really of any use
Sp pick E
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@abhayalbatross86 wrote:Hi, could you please explain how those examples you've given for statement 1 work?mj78ind wrote:So my approach here is to pick numbers and disprove ....
stmt 1 - say x = 31, 41 it works but x could also be 72, 82 ....... hence stmt 1 does not suffice
stmt 2 - does not suffice as shown above.
Combined not enough
Hence E
As far as I can see, only numbers with 0 or 5 in the units digit would work.
My bad I was trying to make the hundred's digit of 3x same as units of x. And when I reread I saw my error.
Thanks