The sum and difference of the L.C.M and H.C.F. of two numbers are $592$ and $518$ respectively. If the sum of the numbers be $296$, find the numbers.

### Concept & Strategy This multi-step problem requires solving simple simultaneous equations to find the L.C.M. and H.C.F. values first. Once the H.C.F. is known, you can use the strict mathematical rule that any number is a direct multiple of its H.C.F. to quickly eliminate incorrect options, bypassing complex quadratic equations. ### Step-by-Step Solution **Step 1: Find the numerical values of L.C.M. ($L$) and H.C.F. ($H$).** * We are given two simultaneous equations: $L + H = 592$ ...(Equation 1) $L - H = 518$ ...(Equation 2) * Add both equations together to eliminate $H$: $2L = 1110$ $L = 555$ * Substitute $L$ back into Equation 1 to find $H$: $555 + H = 592$ $H = 592 - 555$ $H = 37$ **Step 2: Form an equation for the numbers (Standard Method).** * Let the two numbers be $x$ and $y$. We are given $x + y = 296$, so $y = 296 - x$. * We use the core property: Product of numbers = Product of H.C.F. and L.C.M. $x \times (296 - x) = 555 \times 37$ $296x - x^2 = 20535$ $x^2 - 296x + 20535 = 0$ * Solving this quadratic yields roots $111$ and $185$. However, this is too slow for an exam. Use the shortcut below instead. ### Exam Strategy & Shortcut Solving the quadratic $x^2 - 296x + 20535 = 0$ is a massive time sink. Instead, use the **Options and H.C.F. Rule**: Once you calculate the H.C.F. is $37$, you know **both numbers MUST be perfectly divisible by $37$**. Check the multiple-choice options for divisibility by $37$: * Option B ($120$ and $176$): $120$ is not divisible by $37$. (Eliminate) * Option C ($137$ and $159$): $137$ is not divisible by $37$. (Eliminate) * Option D ($148$ and $148$): $148$ is $37 \times 4$. Sum is $296$. However, if both numbers are identical, their H.C.F. and L.C.M. are both $148$. Their sum would be $296$, not $592$. (Eliminate) * Option A ($111$ and $185$): $111 / 37 = 3$. $185 / 37 = 5$. Both are perfectly divisible. This confirms Option A is correct without needing the quadratic equation. ### Common Pitfall Attempting to solve the large quadratic equation using the standard $abc$ formula without recognizing that the roots must be multiples of the H.C.F. will waste precious minutes. Always leverage the H.C.F. to check divisibility and eliminate options first. ### Final Answer **Therefore, the correct answer is $111$ and $185$.**