Linear Equations

linear equations class 8 rs aggarwal

Step by Step solutions of Exercise- CASE STUDY BASED QUESTIONS of RS Aggarwal ICSE Class-8 Maths chapter 15- Linear Equations by Goyal Brothers Prakashan is provided.

Table of Contents

Case Study based Questions

I. Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond.


Q1: If x represents the total number of deer in the herd, which of the following equations holds true?

Step 1:Total deer = \(x\)
Half are grazing: \(\frac{x}{2}\)
Remaining deer = \(x – \frac{x}{2} = \frac{x}{2}\)
Step 2:Three-fourths of remaining are playing:
\(\frac{3}{4} \cdot \frac{x}{2} = \frac{3x}{8}\)
Step 3:The rest are drinking water = 9 deer
Step 4:Sum of all groups = total deer \[ \frac{x}{2} + \frac{3x}{8} + 9 = x \\ \frac{x}{2} + \frac{3x}{8} = x – 9 \]Answer: \(\frac{x}{2} + \frac{3x}{8} = x – 9\) (Not listed in options)


Q2: The total number of deer in the herd is:

Step 1:From Q1, the equation for total deer \(x\) is: \[ \frac{x}{2} + \frac{3x}{8} = x – 9 \]Step 2:Simplify the left-hand side: \[ \frac{4x}{8} + \frac{3x}{8} = x – 9 \\ \frac{7x}{8} = x – 9 \]Step 3:Subtract \(\frac{7x}{8}\) from both sides: \[ 0 = x – \frac{7x}{8} – 9 \\ 0 = \frac{x}{8} – 9 \]Step 4: \[ \frac{x}{8} = 9 \\ x = 9 \times 8 = 72 \]Answer: b. 72 deer


Q3: If there is one attendant for every four grazing, deer, how many attendants are there?

Step 1:Total number of deer = 72 (from Q2)
Number of grazing deer = half of total deer: \[ \text{Grazing deer} = \frac{72}{2} = 36 \]Step 2:One attendant for every 4 grazing deer: \[ \text{Number of attendants} = \frac{36}{4} = 9 \]Answer: b. 9 attendants


Q4: The ratio between the number of deer grazing those playing and those drinking water from the pond is:

Step 1:Total number of deer = 72 (from Q2)
Number of grazing deer = half of total deer: \[ \text{Grazing deer} = \frac{72}{2} = 36 \]Step 2:Remaining deer = 72 – 36 = 36
Number of deer playing = three-fourths of remaining deer: \[ \text{Playing deer} = \frac{3}{4} \cdot 36 = 27 \]Step 3:Number of deer drinking water = rest of remaining deer: \[ \text{Drinking deer} = 36 – 27 = 9 \]Step 4:Ratio of grazing : playing : drinking = 36 : 27 : 9
Divide each by 9: \[ 36:27:9 = 4:3:1 \]Answer: b. 4 : 3 : 1



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