Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
It gets pretty easy when both the statements are used.
If we take the numbers as 1, 2 ,3 ,4 ,5 ,6. ( Dont know if 0 can be a possibility )
10ac+bc= 100d+10e+f
or , c ( 10a + b) = 100d + 10e + f { putting the values of b and f )
or , c ( 10a + 4) = 100d + 10e + 2 {LHS will have 2 in the units place.
In order to match this the value of c
has to be 3 , because 3 * 4 = 12 }
or, 3 (10a + 4) = 100d + 10e + 2
or, 30a + 12 = 100d + 10e + 2
or, 30a + 10 = 100d + 10e
or, 3a + 1 = 10d + e {Now put a = 5 , d = 1 , and e = 6 }
This can be solved. So a = 5
However, my doubt is can't we come to the same conclusion without knowing the values of b and f ?
If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?
(1) a+b+c+d+e+f < 22
(2) b=4 and f=2
So I got soooo bored of writing applications that I had to find something to do with my brain. That is how I am back here. btw.. hope everyone's prep is going on well.
ok here is my attempt.
I satrted with statement B (looks like it has more info than A)
so if (10a+4)c = 100d+10e+2
=> c has to be equal to 3 (you multiply a number ending in 4 with a number ending in 2 the multiple has to be 3.
this quick step tells me b = 4 ; c = 3 & f =2.
but a, d,e could be any numbers
no statement A: this condition gives many possibilities.
combining
a+d+e < 13 (22-4-3-2).....(i)
also 30a+12 = 100d+10e+2
=> 3a +1 = 10d +e.....(ii)
case 1: d= 1
then a+e< 12
only possible value of a that will give each number as a distinct integer is 4
If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?
(1) a+b+c+d+e+f < 22 (2) b=4 and f=2
So I got soooo bored of writing applications that I had to find something to do with my brain. That is how I am back here. btw.. hope everyone's prep is going on well.
ok here is my attempt.
I satrted with statement B (looks like it has more info than A) so if (10a+4)c = 100d+10e+2 => c has to be equal to 3 (you multiply a number ending in 4 with a number ending in 2 the multiple has to be 3. this quick step tells me b = 4 ; c = 3 & f =2.
but a, d,e could be any numbers
no statement A: this condition gives many possibilities.
combining a+d+e < 13 (22-4-3-2).....(i) also 30a+12 = 100d+10e+2 => 3a +1 = 10d +e.....(ii) case 1: d= 1 then a+e< 12 only possible value of a that will give each number as a distinct integer is 4
case 2: d=2 then a+e<11 no possible values
case3: d=3 a+e<10 no possible values.
Thus C.
Anyone else tried this? Any other ideas?
Excellent approach, but were you too quick to dismiss A?
if a+b+c+...+f< 22, then (a,b,c,d,e,f) must be a permutation of the first six positive integers.
If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?
(1) a+b+c+d+e+f < 22 (2) b=4 and f=2
So I got soooo bored of writing applications that I had to find something to do with my brain. That is how I am back here. btw.. hope everyone's prep is going on well.
ok here is my attempt.
I satrted with statement B (looks like it has more info than A) so if (10a+4)c = 100d+10e+2 => c has to be equal to 3 (you multiply a number ending in 4 with a number ending in 2 the multiple has to be 3. this quick step tells me b = 4 ; c = 3 & f =2.
but a, d,e could be any numbers
no statement A: this condition gives many possibilities.
combining a+d+e < 13 (22-4-3-2).....(i) also 30a+12 = 100d+10e+2 => 3a +1 = 10d +e.....(ii) case 1: d= 1 then a+e< 12 only possible value of a that will give each number as a distinct integer is 4
case 2: d=2 then a+e<11 no possible values
case3: d=3 a+e<10 no possible values.
Thus C.
Anyone else tried this? Any other ideas?
Excellent approach, but were you too quick to dismiss A?
if a+b+c+...+f< 22, then (a,b,c,d,e,f) must be a permutation of the first six positive integers.
Intuitively, I know that (A) is sufficient, just can't seem to come up with the right combination in under two minutes. Gosh, the GMAT seems to be getting harder all the time
Keep in mind that this is NOT a GMAT question, so don't estimate the difficulty of the exam by the questions I post. GMATPrep is by far the best indicator. These questions are just for fun and meant to help us to look for creative solutions.
If a+b+c+d+e+f<22 and each is a distinct positive integer, it follows that (a,b,c,d,e,f) is a permutation of (1,2,3,4,5,6).
We are told that 10ac+bc=100d+10e+f
Let's write it as:
.ab
x.c
___
def
Focussing on the units digits, it's clear that none of them can be 1 or 5.
Could b and c be 2 and 3 or vice versa? If that case, f would be 6, the two digit number de would simply be ac, which is not possible since the six numbers have to be distinct. This makes it clear that bc>9, since we NEED a carry-over. Only 3 and 4 will do! I have to start a class! Can somebody continue?
Keep in mind that this is NOT a GMAT question, so don't estimate the difficulty of the exam by the questions I post. GMATPrep is by far the best indicator. These questions are just for fun and meant to help us to look for creative solutions.
If a+b+c+d+e+f<22 and each is a distinct positive integer, it follows that (a,b,c,d,e,f) is a permutation of (1,2,3,4,5,6).
We are told that 10ac+bc=100d+10e+f
Let's write it as:
.ab x.c ___ def
Focussing on the units digits, it's clear that none of them can be 1 or 5. Could b and c be 2 and 3 or vice versa? If that case, f would be 6, the two digit number de would simply be ac, which is not possible since the six numbers have to be distinct. This makes it clear that bc>9, since we NEED a carry-over. Only 3 and 4 will do! I have to start a class! Can somebody continue?
so if b and c are 3 and 4 (or vice versa) f would be 2
Let's see if (b,c)=(3,4)
.a3
X 4
----
de2
The two digit number de is 4*a+1- a would have to be 5 or 6, but in each case d would be 2- not permissible
How about (b,c)=(4,3)?
.a4
X.3
----
de2
This time, the two digit number de is 3*a+1- a must be 5, so de is 16