NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 3: Pair of Linear Equations in Two Variables - Exercise 3.4

Question 6

Step 1: Assume the speeds.

Let the speed of the rickshaw be \(r\) km/h.

Let the speed of the bus be \(b\) km/h.

Step 2: Recall the formula for time.

\(\text{Time} = \dfrac{\text{Distance}}{\text{Speed}}\)

Step 3: Write the first condition (Case 1).

Distance by rickshaw = 2 km, so time = \(\dfrac{2}{r}\).

Distance by bus = 12 km, so time = \(\dfrac{12}{b}\).

Total time = 30 minutes = \(\dfrac{1}{2}\) hour.

So, \(\dfrac{2}{r} + \dfrac{12}{b} = \dfrac{1}{2}\). … (1)

Step 4: Write the second condition (Case 2).

Distance by rickshaw = 4 km, so time = \(\dfrac{4}{r}\).

Distance by bus = 10 km, so time = \(\dfrac{10}{b}\).

Total time = 30 min + 9 min = 39 min = \(\dfrac{13}{20}\) hour.

So, \(\dfrac{4}{r} + \dfrac{10}{b} = \dfrac{13}{20}\). … (2)

Step 5: Make the equations easier.

Put \(u = \dfrac{1}{r}\) and \(v = \dfrac{1}{b}\).

Equation (1): \(2u + 12v = \dfrac{1}{2}\).

Equation (2): \(4u + 10v = \dfrac{13}{20}\).

Step 6: Solve the equations.

From (1): \(2u + 12v = \dfrac{1}{2}\) → Divide by 2 → \(u + 6v = \dfrac{1}{4}\). … (3)

From (2): \(4u + 10v = \dfrac{13}{20}\) → Divide by 2 → \(2u + 5v = \dfrac{13}{40}\). … (4)

Multiply (3) by 2: \(2u + 12v = \dfrac{1}{2}\). … (5)

Now subtract (4) from (5):

\((2u + 12v) - (2u + 5v) = \dfrac{1}{2} - \dfrac{13}{40}\)

\(7v = \dfrac{20}{40} - \dfrac{13}{40} = \dfrac{7}{40}\)

So, \(v = \dfrac{1}{40}\).

Step 7: Find bus speed.

Since \(v = \dfrac{1}{b}\), we get \(b = 40\) km/h.

Step 8: Find rickshaw speed.

Put \(v = \dfrac{1}{40}\) in (3):

\(u + 6 \times \dfrac{1}{40} = \dfrac{1}{4}\)

\(u + \dfrac{6}{40} = \dfrac{1}{4}\)

\(u + \dfrac{3}{20} = \dfrac{1}{4}\)

\(u = \dfrac{1}{4} - \dfrac{3}{20} = \dfrac{5}{20} - \dfrac{3}{20} = \dfrac{2}{20} = \dfrac{1}{10}\)

So, \(r = 10\) km/h.

Final Answer: Rickshaw speed = 10 km/h, Bus speed = 40 km/h.