The sum of four numbers $A$, $B$, $C$ and $D$ is 110. $C$ is twice $A$, sum of $A$ and $D$ is equal to that of $B$ and $C$ and $A : B$ is $3 : 5$. The numbers are

### Concept & System of Linear Equations Define all given relationships in terms of a single variable by chaining the ratios and given conditions, then substitute into the sum equation. ### Step-by-Step Solution 1. **Given:** $A+B+C+D = 110$. Ratio $A : B = 3 : 5$. Let $A = 3x$ and $B = 5x$. 2. Given $C = 2A$. Therefore, $C = 2(3x) = 6x$. 3. Given $A + D = B + C$. Substitute the known expressions: $3x + D = 5x + 6x$. 4. Solve for $D$: $D = 11x - 3x = 8x$. 5. Now, use the sum equation: $3x + 5x + 6x + 8x = 110$. 6. $22x = 110 \implies x = 5$. 7. Calculate the numbers: $A = 3(5) = 15$, $B = 5(5) = 25$, $C = 6(5) = 30$, $D = 8(5) = 40$. ### Exam Strategy & Shortcut Use option elimination. Check option (b): $15+25+30+40 = 110$. Check the conditions: is $C$ twice $A$? Yes ($30 = 2 \times 15$). Is $A+D = B+C$? Yes ($15+40 = 55$ and $25+30 = 55$). Option (b) strictly satisfies all conditions within seconds. ### Common Pitfall Trying to solve for variables purely through elimination instead of defining them in terms of a common proportional variable $x$. ### Final Answer Therefore, the correct answer is **15, 25, 30, 40**.