CAT 2022

Slot 1

QA

Question & Solution

1. Concept Used

• Topic: Algebra — Factoring and Systems of Equations
• Formula: $$ a(a + b + 1) = 14 \quad \text{and} \quad b(a + b + 1) = 28 $$

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2. Calculation

We start by factoring both given equations cleverly.

From the first equation: (a^2 + ab + a = 14), we factor out (a): $$a(a + b + 1) = 14 \quad \cdots (1)$$

From the second equation: (b^2 + ab + b = 28), we factor out (b): $$b(a + b + 1) = 28 \quad \cdots (2)$$

Now, we divide equation (2) by equation (1): $$\frac{b(a + b + 1)}{a(a + b + 1)} = \frac{28}{14}$$

The common factor ((a + b + 1)) cancels out (since it is non-zero for natural numbers), giving: $$\frac{b}{a} = 2 \implies b = 2a$$

Substituting (b = 2a) into equation (1): $$a(a + 2a + 1) = 14$$ $$a(3a + 1) = 14$$ $$3a^2 + a - 14 = 0$$

We solve this quadratic by splitting the middle term: $$3a^2 - 6a + 7a - 14 = 0$$ $$3a(a - 2) + 7(a - 2) = 0$$ $$(3a + 7)(a - 2) = 0$$

This gives (a = 2) or (a = -\frac{7}{3}).

Since (a) must be a natural number, we take (a = 2).

Therefore, (b = 2a = 2 \times 2 = 4).

Verification:

• (a^2 + ab + a = 4 + 8 + 2 = 14) ✓
• (b^2 + ab + b = 16 + 8 + 4 = 28) ✓

Finally: $$2a + b = 2(2) + 4 = 4 + 4 = 8$$

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3. Solution

Answer = Option A ✅

The final calculated value of ((2a + b)) is 8.