Q.Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Step 1 — clear denominators (LCM = 6)

Step 2 — expand and collect

Step 3 — isolate $x$

Dividing by the positive $5$ keeps the inequality sign unchanged. …

Step 1 — Define the variables

Let $x=$ number of experienced persons and $y=$ number of freshers.

Step 2 — Write the output

Each experienced person does $10$ units and each fresher $5$ units, so total daily output is $10x+5y$.

Step 3 — Translate the constraints

"Maintain at least 50 units" means the output must be $50$ or more:

The number of people cannot be negative, so:

$x\ge 0,\qquad y\ge 0$ …

Step 1 — Express all sides via the shortest side $s$

• Shortest = $s$
• Longest = $2s$
• Third = longest $-4 = 2s - 4$

Step 2 — Write the perimeter inequality

'Perimeter at least 61 cm' means

Step 3 — Solve the inequality

The smallest permissible value is $s = 13$ cm (sides then 13, 26, 22 — a valid triangle since $13+22>26$).

Why the other options are wrong: (A) 7, (B) 9, (C) 11 all give a perimeter below 61 cm, violating the condition. …

Step 1 — Remove the fractions

Multiply every term of $\dfrac{5x}{3} \le \dfrac{x}{6} - 5$ by $6$ (positive, so the direction is unchanged):

Step 2 — Isolate $x$

$10x - x \le -30 \Rightarrow 9x \le -30 \Rightarrow x \le -\tfrac{30}{9} = -\tfrac{10}{3}.$

Step 3 — Write the solution set

Because the sign is $\le$, $-\tfrac{10}{3}$ belongs to the set: $\left(-\infty, -\tfrac{10}{3}\right]$. …

Step 1 — Interpret "at least"

The mixture must contain AT LEAST the stated units, so each nutrient total $\ge$ its minimum.

Step 2 — Form one constraint per nutrient

With $x$ units of Food-1 and $y$ units of Food-2:

Step 3 — Add non-negativity

Quantities cannot be negative: $x \ge 0,\ y \ge 0$.

This is exactly option (C). …

Concept and Intuition

The triangle formed by $x+y=10$ and the axes has vertices $(0,0),(10,0),(0,10)$. "Inside" the triangle (excluding the boundary) with $x,y$ natural numbers means $x\ge1$, $y\ge1$, and strictly $x+y<10$.

Step-by-Step Solution

1. Condition: $x\ge1$, $y\ge1$, $x+y<10 \Rightarrow x+y\le9$ (since $x,y$ are integers).
2. For each fixed $x$ from 1 to 8 (must leave room for $y\ge1$, so $x\le8$), $y$ can range from $1$ to $9-x$, giving $9-x$ choices.
3. Total count $= \sum_{x=1}^{8}(9-x) = 8+7+6+5+4+3+2+1 = 36$.

Common Mistakes …

Concept and Intuition

For any affine function $L(x,y)=\alpha x+\beta y+\gamma$, its range over a triangle (a convex region) is exactly the interval between its smallest and largest vertex value — every interior point is a convex combination of vertices. So a candidate inequality is "necessarily satisfied" throughout the triangle exactly when the expression has the same sign at all three vertices.

Step-by-Step Solution

1. Vertices: $(1,3),\,(5,0),\,(-1,2)$.
2. Test $3x+2y$: at $(1,3)$: $9$; at $(5,0)$: $15$; at $(-1,2)$: $1$. All positive, minimum value $1>0$ — so $3x+2y>0$ holds for the whole triangle, including every interior point.
3. Test $2x-3y-12$: at $(1,3)$: $-19$; $(5,0)$: $-2$; $(-1,2)$: $-20$. All negative, so "$>0$" (option C) is false everywhere. …

Concept and Intuition

To make the largest number as big as possible while the total sum is fixed, you must make the other numbers as small as possible — but they're bounded below by the smallest number itself (since it IS the smallest). So the extremal configuration has the three "middle" numbers all equal to the smallest value.

Step-by-Step Solution

1. Sum of 5 natural numbers with mean 40: total $=200$.
2. Let the smallest be $s$; the largest is then $s+10$.
3. The remaining three numbers must each be $\ge s$ (since $s$ is the minimum of all five) and $\le s+10$.
4. To maximize $s$ (and hence the largest, $s+10$), minimize the sum of the other three — set them all equal to $s$: total becomes $s+(s+10)+3s=5s+10$. …

Concept and Intuition

An equation like $|L(x,y)| = k$ (where $L$ is linear, here $L = x+y$) represents two parallel lines, $L = k$ and $L = -k$. A point lies strictly between these two lines exactly when its value of $L$ lies strictly between $-k$ and $k$, i.e., $|L| < k$. This is the natural extension of 'between two numbers on a line' to 'between two parallel lines in the plane,' since $L(x,y)$ is constant along each of the two lines and varies monotonically as you cross from one to the other.

Step-by-Step Solution

1. The equation $|x+y| = 4$ represents two parallel lines: $x + y = 4$ and $x + y = -4$.
2. A point lies in the strip between these two lines exactly when its value of $x+y$ is strictly between $-4$ and $4$:

3. Substitute the point $(a,a)$, so $x = a, y = a$:

4. Divide through by 2: …