Math, asked by sanjeebsarangi, 1 month ago

A man had 35000. He lent *15000 at 4% p.a. and 10000 at 3% p.a. simple interest. At what rate must he lend the remaining money so that the total income is 5% on 35000 annually? ​

Answers

Answered by BeAuTyBLusH
5

Answer:

Solution−

Total Principal Amount, p = Rs 35000

Rate of interest, r = 5 %

Time, t = 1 year

We know,

Income, I on a certain sum of money Rs P invested at the rate of r % per annum for t years is

\boxed{ \tt{ \: I = \frac{p \times r \times t}{100} \: \: }}

I=

100

p×r×t

So, on substituting the values of p, r and t, we get

\rm :\longmapsto\:I = \dfrac{35000 \times 5 \times 1}{100}:⟼I=

100

35000×5×1

\bf\implies \:\boxed{ \tt{ \: I = 1750 \: \: }}⟹

I=1750

Case :- 1

Amount Lent, p = Rs 15000

Rate of interest, r = 4 %

Time, t = 1 year

We know,

Income, I on a certain sum of money Rs P invested at the rate of r % per annum for t years is

\boxed{ \tt{ \: I = \frac{p \times r \times t}{100} \: \: }}

I=

100

p×r×t

So, on substituting the values, we get

\rm :\longmapsto\:I_1 = \dfrac{15000 \times 4 \times 1}{100}:⟼I

1

=

100

15000×4×1

So, on substituting the values, we get

\rm :\longmapsto\:I_1 = \dfrac{15000 \times 4 \times 1}{100}:⟼I

1

=

100

15000×4×1

\bf\implies \:\boxed{ \tt{ \: I_1 = 600 \: \: }}⟹

I

1

=600

Case :- 2

Amount Lent, p = Rs 10000

Rate of interest, r = 3 %

Time, t = 1 year

We know,

Income, I on a certain sum of money Rs P invested at the rate of r % per annum for t years is

\boxed{ \tt{ \: I = \frac{p \times r \times t}{100} \: \: }}

I=

100

p×r×t

So, on substituting the values, we get

\rm :\longmapsto\:I_2 = \dfrac{10000 \times 3 \times 1}{100}:⟼I

2

=

100

10000×3×1

\bf\implies \:\boxed{ \tt{ \: I_2 = 300 \: \: }}⟹

I

2

=300

Case :- 3

Now,

Remaining amount = 35000 - (15000+10000) = 10000

Anount lent, p = Rs 10000

Rate of interest, r = r %

Time, t = 1 year

We know,

Income, I on a certain sum of money Rs P invested at the rate of r % per annum for t years is

\boxed{ \tt{ \: I = \frac{p \times r \times t}{100} \: \: }}

I=

100

p×r×t

\rm :\longmapsto\:I_3 = \dfrac{10000 \times r \times 1}{100}:⟼I

3

=

100

10000×r×1

\bf\implies \:\boxed{ \tt{ \: I_3 = 100r \: \: }}⟹

I

3

=100r

According to statement

\rm :\longmapsto\:I_1 + I_2 + I_3 = I:⟼I

1

+I

2

+I

3

=I

\rm :\longmapsto\:600 + 300 + 100r = 1750:⟼600+300+100r=1750

\rm :\longmapsto\:900 + 100r = 1750:⟼900+100r=1750

\rm :\longmapsto\:100r = 1750 - 900:⟼100r=1750−900

\rm :\longmapsto\:100r = 850:⟼100r=850

\bf\implies \:\boxed{ \tt{ \: r \: = \: 8.5 \: \% \: \: }}⟹

r=8.5%

Answered by ItzparthXx99
2

Answer:

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