Math, asked by gixaha2322, 8 months ago

A person has Rs. 35,000 that he has divided into three investments. Part of the
money is invested in a savings account with annual rate of interest of 6%, part in
7% annual yield bonds, and the remainder in a business. In 2006, when he lost 6%
of the money that he invested in that business, his net income from all three
investments was Rs. 660. If he invested Rs. 3,000 more in the business than in
savings account, how much was invested in each? Use matrix method.

Answers

Answered by Anonymous
1

Given:

Amount of money = Rs. 35000

Investments = 3

Rate of interest = 6%, 7% and -6%

Net income from three investmens = 660

More amount invested = 3000

To Find:

Total amount invested

Solution:

Let the amount invested in three installments is a, b and c.

Let the amount invested be = 1000 Rs.

Thus,

First = 1000a and 6%

Second = 1000b and 7%

Third = 1000c and ( 6%) ( loss)

ATQ<

1000a + 1000b + 1000c = 35000

= a + b + c  = 35

Now,

1000a × 6/100  + 1000b × 7/100  + 1000c × (-6/100) = 660

= 6a + 7b - 6c  = 66

= 6(a - c) + 7y = 66 --- 1

Calculating value of (a - c)

1000c = 1000a + 3000 ( If 3000 more was invested)

= a - c = - 3 --- 2

Substituting in 1.

= 6(-3) + 7b = 66

= - 18 + 7b = 66

= 84 = 7b

b = 12

Finding other values -

a + b + c  = 35  

= a + b + c + 3  = 35 ( from 2)

= 2a + b = 32

= 2a + 12 = 32

= a = 10 and c = 13

Thus,

1000a = 1000 × 10 = 10,000

1000b = 1000 × 12 = 12,000

1000c = 1000 × 13, 000

Answer: The amount invested is 10,000 in savings account, 12,000 in bonds and 13000 in business.

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