The question asks for the maximum possible value of the variable $a$, given an equation and constraints on the other variables.
We are given the equation:
$a + b + c = 8$
Additionally, we have the constraints that $b$ and $c$ must be positive integers. This means:
To find the maximum value of $a$, we need to rearrange the equation:
$a = 8 - (b + c)$
This shows that maximizing $a$ is equivalent to minimizing the sum $(b + c)$.
Given the constraints $b \ge 1$ and $c \ge 1$, the smallest possible integer value for $b$ is 1, and the smallest possible integer value for $c$ is 1.
Therefore, the minimum value of the sum $(b + c)$ is:
$(b + c)_{\text{min}} = 1 + 1 = 2$
Now, substitute the minimum value of $(b + c)$ back into the equation for $a$ to find its maximum value:
$a_{\text{max}} = 8 - (b + c)_{\text{min}}$
$a_{\text{max}} = 8 - 2$
$a_{\text{max}} = 6$
The maximum value 'a' can take is 6. This corresponds to Option A.
Simplify: $3((\frac{5}{3})x^2 - 28x + 15) - 5(x^2 + 6x - 15)$
The numerator of a fraction is 5 less than its denominator. If 2 is subtracted from the numerator and 2 is added to the denominator, the fraction becomes $\frac{2}{5}$. Find the original fraction.
A group of 630 children is seated in rows for a group photo session. Each row contains three less children than the row in front of it. Which one of the following number of rows is not possible?
The letters L, M, N, 0, P, Q, R, S and T in their order are substituted by nine integers 1 to 9 but not in that order. 4 is assigned to P. The difference between P and T is 5. The difference between N and T is 3.
What is the integer assigned to N?
Four persons, Alok, Bhupesh, Chander and Dinesh have a total of Rs. 100 among themselves. Alok and Bhupesh between them have as much money as Chander and Dinesh between them, but Alok has more money than Bhupesh; and Chander has only half the money that Dinesh has. Alok has in fact Rs. 5 more than Dinesh has.
Who has the maximum amount of money?
If x=3/2, then the value of 27x3-54x2+36x-11 is
If a+b+c = 6 and ab+bc+ca = 11, then the value of bc(b+c) + ca(c+a) +ab(a+b) +3abc is