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Nidhi received simple interest off Rs. 1200 when invested Rs x at 6% p.a. and Rs y at 5% p.a. for 1 year. Had she invested Rs. x at 3% p.a. and Rs. y at 8% for that year, she would have received simple interest of Rs 1,260. Find the values Rs. x an y.
Nidhi received simple interest off Rs. 1200 when invested Rs x at 6% p.a. and Rs y at 5% p.a. for 1 year. Had she invested Rs. x at 3% p.a. and Rs. y at 8% for that year, she would have received simple interest of Rs 1,260. Find the values Rs. x an y.
This Question has 1 answers.
Let the amounts invested by Nidhi be x and y in Rs.
The formula for simple interest is:
$ SI = \frac{P \times R \times T}{100} $
where P is the principal, R is the rate, and T is the time in years.
$ \text{Given conditions:} $
$ \frac{x \times 6 \times 1}{100} + \frac{y \times 5 \times 1}{100} = 1200 $
$ \Rightarrow \frac{6x}{100} + \frac{5y}{100} = 1200 $
$ \Rightarrow 6x + 5y = 120000 $ \quad \cdots (1)
$ \text{Similarly, for the second case:} $
$ \frac{x \times 3 \times 1}{100} + \frac{y \times 8 \times 1}{100} = 1260 $
$ \Rightarrow \frac{3x}{100} + \frac{8y}{100} = 1260 $
$ \Rightarrow 3x + 8y = 126000 $ \quad \cdots (2)
$ \text{Solving equations (1) and (2):} $
$ \text{Multiply (1) by 3 and (2) by 6:} $
$ 18x + 15y = 360000 $
$ 18x + 48y = 756000 $
$ \text{Subtracting:} $
$ (18x + 48y) - (18x + 15y)$
The formula for simple interest is:
$ SI = \frac{P \times R \times T}{100} $
where P is the principal, R is the rate, and T is the time in years.
$ \text{Given conditions:} $
$ \frac{x \times 6 \times 1}{100} + \frac{y \times 5 \times 1}{100} = 1200 $
$ \Rightarrow \frac{6x}{100} + \frac{5y}{100} = 1200 $
$ \Rightarrow 6x + 5y = 120000 $ \quad \cdots (1)
$ \text{Similarly, for the second case:} $
$ \frac{x \times 3 \times 1}{100} + \frac{y \times 8 \times 1}{100} = 1260 $
$ \Rightarrow \frac{3x}{100} + \frac{8y}{100} = 1260 $
$ \Rightarrow 3x + 8y = 126000 $ \quad \cdots (2)
$ \text{Solving equations (1) and (2):} $
$ \text{Multiply (1) by 3 and (2) by 6:} $
$ 18x + 15y = 360000 $
$ 18x + 48y = 756000 $
$ \text{Subtracting:} $
$ (18x + 48y) - (18x + 15y)$
$ = 756000 - 360000 $
$ 33y = 396000 $
$ y = \frac{396000}{33} = 12000 $
$ \text{Substituting } y = 12000 \text{ in (1):} $
$ 6x + 5(12000) = 120000 $
$ 6x + 60000 = 120000 $
$ 6x = 60000 $
$ x = \frac{60000}{6} = 10000 $
$ \text{Final Answer:} $
$ x = 10000, \quad y = 12000 $
$ 33y = 396000 $
$ y = \frac{396000}{33} = 12000 $
$ \text{Substituting } y = 12000 \text{ in (1):} $
$ 6x + 5(12000) = 120000 $
$ 6x + 60000 = 120000 $
$ 6x = 60000 $
$ x = \frac{60000}{6} = 10000 $
$ \text{Final Answer:} $
$ x = 10000, \quad y = 12000 $
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