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

Question 13

Step 1: Assume numbers of bananas.

Let the number of bananas in Lot A = a.

Let the number of bananas in Lot B = b.

Step 2: Write cost for Lot A and B at first selling rates.

• Lot A: Rs 2 for 3 bananas → price of 1 banana = Rs \(\tfrac{2}{3}\).
• So cost of a bananas = \(\tfrac{2}{3}a\).
• Lot B: Re 1 each → cost of b bananas = \(1 \times b = b\).

Total cost = 400 → equation (1):

\(\tfrac{2}{3}a + b = 400\)

Step 3: Write cost at changed rates.

• Lot A: Re 1 each → cost = \(a\).
• Lot B: Rs 4 for 5 bananas → price of 1 banana = Rs \(\tfrac{4}{5}\).
• So cost of b bananas = \(\tfrac{4}{5}b\).

Total cost = 460 → equation (2):

\(a + \tfrac{4}{5}b = 460\)

Step 4: Remove fractions for easier solving.

• Multiply equation (1) by 3: \(2a + 3b = 1200\).
• Multiply equation (2) by 5: \(5a + 4b = 2300\).

Step 5: Solve the two equations.

We have:

\(2a + 3b = 1200\) … (i)

\(5a + 4b = 2300\) … (ii)

Multiply (i) by 5: \(10a + 15b = 6000\).

Multiply (ii) by 2: \(10a + 8b = 4600\).

Subtract: \((10a + 15b) - (10a + 8b) = 6000 - 4600\).

So, \(7b = 1400 → b = 200\).

Put \(b = 200\) in (i):

\(2a + 3(200) = 1200 → 2a + 600 = 1200\).

So, \(2a = 600 → a = 300\).

Step 6: Find total bananas.

Total bananas = \(a + b = 300 + 200 = 500\).

Answer: 500 bananas (Lot A: 300, Lot B: 200).