Following is the question asked in a recent aptitude exam:

Given that : $$ a+b+ab=10\\ b+c+bc=20 \\ c+a+ac=30$$ What is the value of $a+b+c+abc$ ?

I can solve it by finding the individual values by: $$ a = \frac{10-b}{b+1} ,\\ c = \frac{20-b}{b+1}$$ Putting these in the third equation: $$ \frac{10-b}{b+1}+\frac{20-b}{b+1} + \frac{(10-b)\times(20-b)}{(b+1)^2}=30\\ (30-2b)\times(b+1)+(b-20)\times(b-10)=30\times(b+1)\\{30\times(b+1)}-2b(b+1)+b^2-30b+200={30\times(b+1)}\\-2b^2-2b+b^2-30b+200=0\\-b^2-32b+200=0\\b^2+32b-200=0$$ Which can be solved to: $$b= \frac{-32+-\sqrt{32^2-4\times(-200)}}{2} \approx \frac{-32+-42.71}{2} = 5.35 \text{ or} -37.35 $$

And we can get the values for $a$ and $c$ as well: $$a \approx 9.157 \text{ or} -1.297\\ c\approx 2.30 \text{ or} -1.577$$ Here, the for the first two equations, $b=-37.35$, $a=-1.297$ and $c=-1.577$ works. Whereas, the third equation is satisfied by $a=9.157$ and $c=2.30$. This doesn't seem correct. Since only a single value of $b$ works for the first two equations, therefore the other value of $b$ is rejected. This also rejects the derived values of $a$ and $c$. So, the positive values of $a$ and $c$ should not be used. But the former values don't satisfy the third equation, but the latter values do. What is the problem here?

Moreover, is there any shorter way of solving this problem?

  • 1
    $\begingroup$ You have an error in the second line after "Putting these in the third equation". The right-hand side should be $30 (b + 1)^2$, not $30 (b+1)$. $\endgroup$ Sep 3, 2018 at 14:41

1 Answer 1


Just add one

Add one to each equation and factorize to get : $$ (a+1)(b+1) = 11 \\ (b+1)(c+1) = 21 \\ (a+1)(c+1) = 31 \\ $$

Now, set $x ,y,z = a+1,b+1,c+1$ respectively.This gives $xy = 11, yz = 21,zx = 31$, so $x^2y^2z^2 = 11 \times 21\times 31$ by multiplying these.

Now divide this equation by the other equations suitably to get the values of $x,y,z$, and use the fact that $a+b+c + abc = x+y+z - 3 + (x-1)(y-1)(z-1)$ to get the answer.


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