Question

Let a and b be natural numbers. If $a^2+ab+a=14$ and $b^2+ab+b=28$, then (2a+b) equals

• A. 7
• B. 10
• C. 9
• D. 8

Answer 1

Answer: (D)

8

Answer 2

The correct answer is D: 8
Let's solve this step by step:
Given equations:
$a^2 + ab + a = 14$ ...(i)
$b^2 + ab + b = 28$ ...(ii)
Subtracting equation (i) from equation (ii):
$(b^2 - a^2) + (ab - ab) + (b - a) = 28 - 14$
(b+a)(b-a)+0+(b-a)=14
(b-a)(b+a+1)=14
Since a and b are natural numbers, and 14 is a positive integer, we need to find pairs of factors of 14 that differ by 1:
$14=1\times14$
$14 = 2\times7$
So, (b-a)(b+a+1) can be either (1)(14) or (2)(7).
Case 1: $(b-a)(b+a+1)=1\times14$
This gives us two equations:
b-a=1 ...(iii)
b+a+1=14 ...(iv)
Adding equations (iii) and (iv):
2b+1=15
2b=14
b=7
Substitute the value of b into equation (iii):
7-a=1
a=6
So, in this case, $2a+b=2\times6 + 7 = 12 + 7 = 19$.
Case 2: $(b - a)(b + a + 1) = 2\times7$
This gives us two equations:
b-a=2 ...(v)
b+a+1=7 ...(vi)
Adding equations (v) and (vi):
2b+1=9
2b=8
b=4
Substitute the value of b into equation (v):
4-a=2
a=2
So, in this case, $2a+b=2\times2 + 4 = 4 + 4 = 8$.
Comparing the two cases, we see that the value of (2a+b) is indeed 8.
Hence, the correct answer is (2a+b)=8.