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Mathematics — Standard (041) — CBSE Class 10 board question

Q1. [5] § 3.1 Introduction § 3.3.2 Elimination Method
Vijay invested certain amounts of money in two schemes $A$ and $B$, which offer interest at the rate of 8% per annum and 9% per annum, respectively. He received ₹1,860 as the total annual interest. However, had he interchanged the amounts of investments in the two schemes, he would have received ₹20 more as annual interest. How much money did he invest in each scheme?
Previously asked in CBSE board exam
2025 30/1/1 Q32
Generated by claude-sonnet-4-6 · 2026-06-15 10:32 · grounding rag
Model Answer

Let amount invested in Scheme A = ₹x and in Scheme B = ₹y.

Equation 1 (original investment):
$$\frac{8x}{100} + \frac{9y}{100} = 1860$$
$$\Rightarrow 8x + 9y = 186000 \tag{1}$$

Equation 2 (interchanged investment gives ₹20 more):
$$\frac{9x}{100} + \frac{8y}{100} = 1880$$
$$\Rightarrow 9x + 8y = 188000 \tag{2}$$

Elimination Method:

Multiply (1) by 9 and (2) by 8:
$$72x + 81y = 1674000 \tag{3}$$
$$72x + 64y = 1504000 \tag{4}$$

Subtract (4) from (3):
$$17y = 170000 \Rightarrow y = 10000$$

Substitute y = 10000 in (1):
$$8x + 90000 = 186000 \Rightarrow 8x = 96000 \Rightarrow x = 12000$$

∴ Vijay invested ₹12,000 in Scheme A and ₹10,000 in Scheme B.

Verification: 8% of 12000 + 9% of 10000 = 960 + 900 = ₹1860 ✓

Source: Pair of Linear Equations in Two Variables, Section 3.3.2 (Elimination Method)

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Explanation
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