AP State Syllabus AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 6 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 6

Question 1.

How long will it take for a sum of ₹ 12600 invested at 9% per annum beocme to ₹ 15624?

Solution:

HereA = ₹ 15,624 R = 9% T = ?

P = ₹ 12,600

∴ I = A – P = 15,624 – 12,600 = ₹ 3024

Also I = \(\frac{P R T}{100}\)

∴ 3024 = \(\frac{12,600 \times 9 \times \mathrm{T}}{100}\)

Question 2.

At what rate a sum doubles itself in 8 year 4 months?

Solution:

Given that A = double the sum

Let the principle be P

then A = 2P

T =8 years 4 months = 8y \(\frac { 4 }{ 12 }\) y = 8\(\frac { 1 }{ 3 }\) years = \(\frac { 25 }{ 3 }\)

R = R% say

We know that

I = \(\frac{\text { PTR }}{100}\)

Here I = A – P

= 2P – P = P

∴ P = \(\frac{P \cdot \frac{25}{3} \times R}{100}\)

∴ R = 100 x \(\frac{3}{25}\) = 12%

Question 3.

A child friendly bank announces a savings scheme for school children. They will give kiddy banks to children. Children have to keep their savings in it and the bank collects all the money once in a year. To encourage children savings, they give 6% interest if the amount exceeds by ₹ 10000. and other wise 5%. Find the interest received by a school if they deposit is ₹ 9000 for one year.

Solution:

Money deposited = ₹ 9000

Interest applicable 5% on ₹ 9000

= 5 × \(\frac{9000}{100}\) = ₹ 450

Question 4.

A sum of money invested at 8% per annum for simple interest amounts to ₹ 12122 in 2 years. What will it amounts to in 2 year 8 months at 9% rate of interest?

Solution:

First part

Pirnciple = P say

R=8%

T = 2 years

A = ₹ 12,122

Second part

P = ₹10,450

R = 9%

T = 2 years 8 months, A = ?

Question 5.

In 4 years, ₹ 6500 amounts to ₹ 8840 at a certain rate of interest. In what time will ₹ 1600 amounts to 1816 at the same rate?

Solution:

First part

T = 4 years

P = ₹ 6500

A = ₹ 8840

R = R% = ?

I = A – P = 8840 – 6500 = ₹ 2340

But I = \(\frac{P T R}{100}\)

∴ 2340 = \(\frac{6500 \times 4 \times R}{100}\)

∴ R = \(\frac{2340 \times 10 \theta}{6500 \times 4}=\frac{36}{4}\) = 9%

Second part

P = ₹ 1600

A = ₹ 1816

T = ?

R = 9%

I = A – P = 1816 – 1600 = ₹ 216

But I = \(\frac{P T R}{100}\)

∴ 2340 = \(\frac{1600 \times T \times 9}{100}\)

∴ R = \(\frac{216 \times 100 \theta}{1600 \times 9}=\frac{3}{2}\) = 1 \(\frac { 1 }{ 2 }\) years.