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

The Correct Option is D

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.

Answer 2

\(a^2 + ab + a = 14\)    — (1)

\(b^2 + ab + b = 28\)    ------(2)

Eq 2 – Eq 1

\(b^2  - a^2 + b – a  = 14\)

\((b-a)(b+a) + (b-a) = 14\)

\((b-a)(b+a+1) = 14\)

Now, \(14 = 1×14\ or \ 2×7\)

But \(a < 4\) since \(a^2 < 14\)

And \(b < 5\) since \(b^2 + b < 28\)

So,\( a+b+1 < 4+5+1 = 10\)

So, \(a+b+1 = 7\) and \(b-a = 2\)

Solving, \(b = 4\) and \(a = 2\)

So, \(2a+b = 2(2) + 4 = 8\)