gmatnmein2010 wrote:For how many two digit positive numbers will tripling the tens digit give us a two digit number that is triple the original.
A) None
B) 1
C) 2
D) 3
E) 4

Say ab is the original number the second number is (3a)b

Now 3(10a+b) = 3a*10 +b
30a + 3b = 30a +b
2b=0
b =0
a can be 1,2,3 ( numbers can be 10,20 and 30)

So there are 3 numbers
D

gmatnmein2010 wrote:For how many two digit positive numbers will tripling the tens digit give us a two digit number that is triple the original.
A) None
B) 1
C) 2
D) 3
E) 4

_______
Let the two digit positive number be xy.x-tens place,y-units place. [Suppose 63,x=6,y=3]
Thus the no. can be represented as 10x+y.
Now,tripling the tens digit means x should be replaced 3x.
Thus the no. now becomes 30x + y.

Now,according to the question,
30x+y=3(10x + y)
=>30x + y = 30x + 3y
=>y = 0,y should be 0.
Therefore,the tens place can be taken by:-
1(10 tripling gives 30)
2(20 tripling gives 60)
3(30 tripling gives 90)

For 40 and above,the no. chnages to a 3 digit number.[40 x 3=120(3 digit no.)]
Hence,the answer is D.

a can be 1,2,3 ( numbers can be 10,20 and 30)


1(10 tripling gives 30)
2(20 tripling gives 60)
3(30 tripling gives 90)

Ajit and Harsh you both have posted correct solution and used plug-in method for getting values of a. but if the problem is related with some bigger number may be more than 4 digits, then you will have to solve it algebraically.
for solving you will have to put the condition that tripling the no. should be less than 100(as in this problem it is a 2 digit no. only).

so 3(10a+b) <100
also b=0 as explained in your posts.
hence 30a < 100.
--> a<10/3=3.33..
also a is integer so a=1,2,3.
So, the nos. are 10, 20 ,30.
Hence,the answer is D.