(a) Solve 3x + 8 > 2, when (i) x is an integer. (ii) x — CBSE Class 12 Applied Mathematics 2024 PYQ

Question

3x + 8 > 2

(i)

x

is an integer.

(ii)

x

is a natural number.

(iii)

x

is a whole number.

(OR)

(b) A man goes 12 km downstream and comes back to the starting point by swimming non-stop in 3 hours. If the speed of the stream is 3 km/h, find the speed with which the man can swim in still water.

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★est

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$3x+8>2$ reduces to $x>-2$ , and the listed solution set changes with the number system;
solving the 3-hour round-trip equation gives still-water speed $9$ km/h.

x

Part (a): Solve $3x+8>2$

$3x+8>2\Rightarrow 3x>-6\Rightarrow x>-2$ .
(i) $x$ an integer: solution set $=\{-1,0,1,2,3,\dots\}$ .
(ii) $x$ a natural number: solution set $=\{1,2,3,\dots\}$ .
(iii) $x$ a whole number: solution set $=\{0,1,2,3,\dots\}$ .

Part (b): Speed in still water

5. Let the man's speed in still water be $x$ km/h; stream speed $=3$ km/h.

6. Downstream speed $=(x+3)$ km/h, upstream speed $=(x-3)$ km/h; each leg is 12 km and total time is 3 h:

$\dfrac{12}{x+3}+\dfrac{12}{x-3}=3.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 465/S/WXYZ/41 markMCQ
$\dfrac{|x+1|}{x+1} > 0, x \in R$
›Reveal solutionSolution
$\dfrac{|x+1|}{x+1}>0$ holds precisely where $x+1>0$ , i.e. $x>-1$ , excluding $x=-1$ where it is undefined.

$t \ne 0$

Let $t = x+1$ ; the expression is $\dfrac{|t|}{t}$ .

If $t>0$ , $|t| = t$ , so $\dfrac{|t|}{t} = 1 > 0$ . ✓

If $t<0$ , $|t| = -t$ , so $\dfrac{|t|}{t} = -1 < 0$ . ✗ …
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CBSE 2025Set 465/S/WXYZ/41 markMCQ
$|3x - 2| \leq \dfrac{1}{2}$
›Reveal solutionSolution
The Assertion's solution set $[\tfrac12,\tfrac56]$ is correct, but the Reason misstates the modulus rule (it describes $\ge$ , not $\le$ ), so (C).

$|x-a|\le r \iff a-r \le x \le a+r$

Solve the Assertion: $|3x-2|\le \tfrac12 \iff -\tfrac12 \le 3x-2 \le \tfrac12$ .

Add 2 throughout: $\tfrac32 \le 3x \le \tfrac52$ .

Divide by 3: $\tfrac12 \le x \le \tfrac56$ , i.e. solution set $\left[\tfrac12,\tfrac56\right]$ — Assertion is true. …
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CBSE 2024Set 465/RQPS/41 markMCQ
$x > y$
›Reveal solutionSolution
Dividing both sides of $x>y$ by the negative number $z$ reverses the inequality, so $\dfrac{x}{z}<\dfrac{y}{z}$ .

$a>b$

Given $x>y$ and $z<0$ . …
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CBSE 2023Set 465/EF1GH/41 markMCQ
$\dfrac{|x+1|}{x+1} > 0$
(a) $x \in [-1, \infty)$
(b) $x \in (-1, \infty)$
(c) $x \in (-\infty, -1)$
(d) $x \in (-\infty, -1]$
›Reveal solutionSolution
$\dfrac{|x+1|}{x+1}>0$ forces $x+1>0$ , i.e. $x\in(-1,\infty)$ .

$t\neq 0$

Let $t=x+1$ . The expression $\dfrac{|t|}{t}>0$ holds only when $t>0$ .

So $x+1>0 \Rightarrow x>-1$ . …
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Q.(a) Solve $3x + 8 > 2$ , when

Concept understanding — Linear and Modulus Inequalities

Q.(a) Solve 3x+8>23x + 8 > 23x+8>2, when

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Q.(a) Solve $3x + 8 > 2$ , when

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