Sandhya invested a certain amount in a bank. When the rate of interest

1.	Sandhya invested a certain amount in a bank. When the rate of interest changed from \( 10 \% \) to \( 121 / 2 \% \) her annual income increased by \( ₹ 1250 \). How much did Sandhya invest?
Answer» Let sandhya invest Rs. x. ∴ When r = 10% total amount = x + 10x/100 = x + x/10 = 11x/10 When r = 12\(\frac12\)% = \(\frac{25}2\)% Total amount = x + \(\cfrac{\frac{25}2x}{100}\) = x + \(\frac x8=\frac{9x}8\) According to given condition, we have \(\frac{9x}8-\frac{11x}{10}=1250\) ⇒ \(\frac{45x-44x}{40}=1250\) ⇒ x = 1250 x 40 = 50000 Hence, sandhya invest Rs.50,000.

Sandhya invested a certain amount in a bank. When the rate of interest changed from \( 10 \% \) to \( 121 / 2 \% \) her annual income increased by \( ₹ 1250 \). How much did Sandhya invest?

Answer»

Let sandhya invest Rs. x.

∴ When r = 10%

total amount = x + 10x/100 = x + x/10 = 11x/10

When r = 12\(\frac12\)% = \(\frac{25}2\)%

Total amount = x + \(\cfrac{\frac{25}2x}{100}\) = x + \(\frac x8=\frac{9x}8\)

According to given condition, we have

\(\frac{9x}8-\frac{11x}{10}=1250\)

⇒ \(\frac{45x-44x}{40}=1250\)

⇒ x = 1250 x 40

= 50000

Hence, sandhya invest Rs.50,000.

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Sandhya invested a certain amount in a bank. When the rate of interest changed from \( 10 \% \) to \( 121 / 2 \% \) her annual income increased by \( ₹ 1250 \). How much did Sandhya invest?

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