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Manager
Joined: 28 May 2021
Posts: 135
Location: Uzbekistan
Concentration: Strategy, General Management
Schools: Carlson (A$$$) Kenan-Flagler '25 (A$$$$) Goizueta '25 (A$$$) Olin '25 (A$$$$) Fuqua '25 (M$$$$) Marshall '25 (A$$$$)
GMAT 1: 740 Q50 V40 
GMAT 2: 770 Q50 V46 (Online)
GPA: 4
a, b, c and d are different digits. if a + b + c = 7
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02 Jun 2021, 02:36
02 Jun 2021, 02:36
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a, b, c and d are different digits. If a + b + c = 7, \((a+b)^2\) = d and \(abc\neq{0}\), find \(\frac{c^2-c}{a+b}\)
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
a, b, c and d are different digits. if a + b + c = 7
[#permalink]
02 Jun 2021, 03:54
02 Jun 2021, 03:54
Expert Reply
sherxon wrote:
a, b, c and d are different digits. If a + b + c = 7, \((a+b)^2\) = d and \(abc\neq{0}\), find \(\frac{c^2-c}{a+b}\)
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
So each of a, b, c and d are values from 0 to 9.
But \(abc\neq 0\), meaning none of them are 0.
Given that \((a+b)^2=d\): d is a square, so it can be 0, 1, 4 or 9.
But the minimum value of a and b is 1 and 2, so d has to be (1+2)^2=9.
Any increase in value of a and b will make d as 2-digit number.
Thus a and b are 1 and 2 in any order. => c=7-(1+2)=4.
Thus \(\frac{c^2-c}{a+b}\)= \(\frac{4^2-4}{1+2}\)=\(\frac{12}{3}\)=4
D
_________________
Manager
Joined: 28 May 2021
Posts: 135
Location: Uzbekistan
Concentration: Strategy, General Management
Schools: Carlson (A$$$) Kenan-Flagler '25 (A$$$$) Goizueta '25 (A$$$) Olin '25 (A$$$$) Fuqua '25 (M$$$$) Marshall '25 (A$$$$)
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q50 V46 (Online)
GPA: 4
Re: a, b, c and d are different digits. if a + b + c = 7
[#permalink]
02 Jun 2021, 04:06
02 Jun 2021, 04:06
1
Kudos
chetan2u wrote:
sherxon wrote:
a, b, c and d are different digits. If a + b + c = 7, \((a+b)^2\) = d and \(abc\neq{0}\), find \(\frac{c^2-c}{a+b}\)
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
I am sure the question is missing something.
You can have hundreds of answers.
a+b+c=7...let a=-1, b=-2, so c=10 and d=(-1-2)^2=9
c^2-c/(a+b)=90/(-3)=-30
a+b+c=7...let a=1, b=2, so c=4 and d=(1+2)^2=9
c^2-c/(a+b)=12/(3)=4.
There is almost no role of d except telling that (a+b)^2 cannot be c.
I think everything is clear
Definition of digit from Merriam-Webster:
1a: any of the Arabic numerals 1 to 9 and usually the symbol 0
Re: a, b, c and d are different digits. if a + b + c = 7
[#permalink]
02 Jun 2021, 04:11
02 Jun 2021, 04:11
1
Kudos
Expert Reply
sherxon wrote:
chetan2u wrote:
sherxon wrote:
a, b, c and d are different digits. If a + b + c = 7, \((a+b)^2\) = d and \(abc\neq{0}\), find \(\frac{c^2-c}{a+b}\)
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
I am sure the question is missing something.
You can have hundreds of answers.
a+b+c=7...let a=-1, b=-2, so c=10 and d=(-1-2)^2=9
c^2-c/(a+b)=90/(-3)=-30
a+b+c=7...let a=1, b=2, so c=4 and d=(1+2)^2=9
c^2-c/(a+b)=12/(3)=4.
There is almost no role of d except telling that (a+b)^2 cannot be c.
I think everything is clear
Definition of digit from Merriam-Webster:
1a: any of the Arabic numerals 1 to 9 and usually the symbol 0
Hi
You are correct. I for some reasons read digit as numbers.
Let me edit my answer.
_________________
Re: a, b, c and d are different digits. if a + b + c = 7
[#permalink]
02 Jun 2021, 04:33
02 Jun 2021, 04:33
Is answer 4 that is D ?
Manager
Joined: 28 May 2021
Posts: 135
Location: Uzbekistan
Concentration: Strategy, General Management
Schools: Carlson (A$$$) Kenan-Flagler '25 (A$$$$) Goizueta '25 (A$$$) Olin '25 (A$$$$) Fuqua '25 (M$$$$) Marshall '25 (A$$$$)
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q50 V46 (Online)
GPA: 4
Re: a, b, c and d are different digits. if a + b + c = 7
[#permalink]
02 Jun 2021, 04:38
02 Jun 2021, 04:38
maheshhchaudhari wrote:
Is answer 4 that is D ?
Your answer is correct, but it would be useful for others if you could post how you arrived at your answer ))
Re: a, b, c and d are different digits. if a + b + c = 7
[#permalink]
02 Jun 2021, 04:42
02 Jun 2021, 04:42
1
Kudos
Within two minutes, I rush to assume a = 1, b = 2, c = 4 as d is just to confirm whether a and b are correct.
Substitute values in C^2-C/a+b give 4.
Later on I verified that (1+2)^2 = 9 = d.
Substitute values in C^2-C/a+b give 4.
Later on I verified that (1+2)^2 = 9 = d.
Re: a, b, c and d are different digits. if a + b + c = 7
[#permalink]
04 Jun 2021, 23:29
04 Jun 2021, 23:29
chetan2u wrote:
sherxon wrote:
a, b, c and d are different digits. If a + b + c = 7, \((a+b)^2\) = d and \(abc\neq{0}\), find \(\frac{c^2-c}{a+b}\)
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
So each of a, b, c and d are values from 0 to 9.
But \(abc\neq 0\), meaning none of them are 0.
Given that \((a+b)^2=d\): d is a square, so it can be 0, 1, 4 or 9.
But the minimum value of a and b is 1 and 2, so d has to be (1+2)^2=9.
Any increase in value of a and b will make d as 2-digit number.
Thus a and b are 1 and 2 in any order. => c=7-(1+2)=4.
Thus \(\frac{c^2-c}{a+b}\)= \(\frac{4^2-4}{1+2}\)=\(\frac{12}{3}\)=4
D
Wonderful explanation. thank you very much
Re: a, b, c and d are different digits. if a + b + c = 7
[#permalink]
04 Jun 2021, 23:53
04 Jun 2021, 23:53
[quote="maheshhchaudhari"] keeping a=4 b=2 c=1 answer coming 0
Posted from my mobile device
Posted from my mobile device
gmatclubot
Re: a, b, c and d are different digits. if a + b + c = 7 [#permalink]
04 Jun 2021, 23:53
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