CA Pakistan PRC 2 Quantitative Analysis for Business ke linear equations aur word problems wale topic ke tamam questions ke step-by-step solutions share kar dein?

Question

Ahsan Malik · Accepted Answer

Yeh math/algebra ka topic hai jo CA Pakistan ke **PRC 2 (Quantitative Analysis for Business)** syllabus ka hissa hai aur PRC 1 (Accounting) ke scope se baahar hai. Lekin aapki help ke liye, is page par maujood tamam (all 15) questions ke complete step-by-step solutions niche diye gaye hain:

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### **Exercises (1-8): Solve the linear equations**

#### **1. $3x - 12 + 4x = 6x + 6$**
* $x$ waali terms jama karein: $7x - 12 = 6x + 6$
* Shifting karein: $7x - 6x = 6 + 12$
* **Answer:** $x = 18$

#### **2. $3(2y - 4) + 3y = 5(y - 4)$**
* Brackets open karein: $6y - 12 + 3y = 5y - 20$
* Simplify karein: $9y - 12 = 5y - 20$
* Shifting: $9y - 5y = -20 + 12 \implies 4y = -8$
* **Answer:** $y = -2$

#### **3. $3x - 2(2x - 5) = 2(x + 3) - 8$**
* Brackets open: $3x - 4x + 10 = 2x + 6 - 8$
* Simplify: $-x + 10 = 2x - 2$
* Shifting: $10 + 2 = 2x + x \implies 12 = 3x$
* **Answer:** $x = 4$

#### **4. $3x + \frac{1}{5} = 2 - x$**
* Shifting: $3x + x = 2 - \frac{1}{5}$
* $4x = \frac{9}{5} \implies x = \frac{9}{20}$
* **Answer:** $x = 0.45$

#### **5. $\frac{x}{4} + \frac{x}{6} = \frac{x}{2} + \frac{3}{4}$**
* Poori equation ko LCM $12$ se multiply karein:
  $12\left(\frac{x}{4}\right) + 12\left(\frac{x}{6}\right) = 12\left(\frac{x}{2}\right) + 12\left(\frac{3}{4}\right)$
  $3x + 2x = 6x + 9 \implies 5x = 6x + 9$
* **Answer:** $x = -9$

#### **6. $\frac{y+4}{2y-3} = \frac{2}{5}$**
* Cross-multiply karein: $5(y + 4) = 2(2y - 3)$
* $5y + 20 = 4y - 6$
* $5y - 4y = -6 - 20$
* **Answer:** $y = -26$

#### **7. $x - \left(2x - \frac{3x-4}{7}\right) = \frac{4x-27}{3} - 3$**
* Bracket ko simplify karein: $2x - \frac{3x-4}{7} = \frac{14x - 3x + 4}{7} = \frac{11x+4}{7}$
* Equation: $x - \frac{11x+4}{7} = \frac{4x-27-9}{3}$
* Left side solve karein: $\frac{7x - 11x - 4}{7} = \frac{-4x-4}{7}$
* Ab cross-multiply karein: $\frac{-4x-4}{7} = \frac{4x-36}{3}$
  $3(-4x - 4) = 7(4x - 36) \implies -12x - 12 = 28x - 252$
  $240 = 40x$
* **Answer:** $x = 6$

#### **8. $0.3x + 0.4 = 0.28x + 1.16$**
* Shifting: $0.3x - 0.28x = 1.16 - 0.4$
* $0.02x = 0.76 \implies x = \frac{0.76}{0.02}$
* **Answer:** $x = 38$

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### **9. Select the ordered pairs which are solution points**
*(Points equation mein put karke check karein ke LHS = RHS aata hai ya nahi)*

* **(a) $3x - y = -8$**
  * $(-1, 10) \rightarrow 3(-1) - 10 = -13$ (No)
  * $(3, 10) \rightarrow 3(3) - 10 = -1$ (No)
  * **$(-2, 2)$** $\rightarrow 3(-2) - 2 = -8$ (Yes)
  * **$(-1, 5)$** $\rightarrow 3(-1) - 5 = -8$ (Yes)
* **(b) $2x - y = -1$**
  * **$(0, 1)$** $\rightarrow 2(0) - 1 = -1$ (Yes)
  * **$(2, 5)$** $\rightarrow 2(2) - 5 = -1$ (Yes)
  * $(0, 0) \rightarrow 2(0) - 0 = 0$ (No)
  * **$(-2, -3)$** $\rightarrow 2(-2) - (-3) = -1$ (Yes)
* **(c) $2x + y = 4$**
  * $(2, 3) \rightarrow 2(2) + 3 = 7$ (No)
  * $(1, 0.5) \rightarrow 2(1) + 0.5 = 2.5$ (No)
  * **$(0, 4)$** $\rightarrow 2(0) + 4 = 4$ (Yes)
  * $(-2, 2) \rightarrow 2(-2) + 2 = -2$ (No)

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### **10. Find three ordered pairs $(x, y)$**

* **(a) $y - x = 3 \implies y = x + 3$**
  * Agar $x = 0 \implies y = 3 \rightarrow$ **$(0, 3)$**
  * Agar $x = 1 \implies y = 4 \rightarrow$ **$(1, 4)$**
  * Agar $x = 2 \implies y = 5 \rightarrow$ **$(2, 5)$**
* **(b) $3x + 2y = -6 \implies 2y = -6 - 3x$**
  * Agar $x = 0 \implies 2y = -6 \implies y = -3 \rightarrow$ **$(0, -3)$**
  * Agar $x = -2 \implies 2y = 0 \implies y = 0 \rightarrow$ **$(-2, 0)$**
  * Agar $x = 2 \implies 2y = -12 \implies y = -6 \rightarrow$ **$(2, -6)$**

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### **Word Problems (11-15)**

#### **11. Sum of four consecutive odd numbers is 56. Greatest number?**
* Numbers assume karein: $n, n+2, n+4, n+6$
* Sum equation: $n + n+2 + n+4 + n+6 = 56 \implies 4n + 12 = 56$
* $4n = 44 \implies n = 11$
* Greatest number: $n + 6 = 11 + 6 = 17$
* **Answer:** **17**

#### **12. Number is multiplied by 2, and 8 is added, result is 50. Find the number.**
* Equation: $2x + 8 = 50 \implies 2x = 42$
* **Answer:** $x = 21$

#### **13. Saima's sister is 5. Her age is 2 more than $\frac{1}{4}$ of Saima's age ($S$).**
* Equation: $5 = \frac{1}{4}S + 2 \implies 3 = \frac{1}{4}S$
* **Answer:** Saima's age is **12 years**

#### **14. One number ($x$) is 7 times another ($y$). If 14 is added to sum, result is 38.**
* Conditions: $x = 7y$ aur $(x + y) + 14 = 38 \implies x + y = 24$
* Substitute: $7y + y = 24 \implies 8y = 24 \implies y = 3$
* $x = 7(3) = 21$
* **Answer:** Numbers are **21 and 3**

#### **15. Ahsan is 3 times as old as Fahad. In 3 years, he will be twice as old as Fahad.**
* Ahsan ($a$), Fahad ($f$): $a = 3f$
* In 3 years: $a + 3 = 2(f + 3) \implies 3f + 3 = 2f + 6$
* $3f - 2f = 6 - 3 \implies f = 3$
* Ahsan's age: $a = 3(3) = 9$
* **Answer:** Ahsan is **9 years old** aur Fahad **3 years old** hai.

[[SUGGEST_GROUP:MATHEMATICS]]