AP Board Class 7 Maths Chapter 7 Ratio and Proportion InText Questions Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 7 Ratio and Proportion InText Questions Book Answers |

## Andhra Pradesh Board Class 7th Maths Chapter 7 Ratio and Proportion InText Questions Textbooks Solutions PDF

Andhra Pradesh State Board STD 7th Maths Chapter 7 Ratio and Proportion InText Questions Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 7th Maths Chapter 7 Ratio and Proportion InText Questions Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 7th Maths Chapter 7 Ratio and Proportion InText Questions solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 7th Maths Chapter 7 Ratio and Proportion InText Questions Textbooks. These Andhra Pradesh State Board Class 7th Maths Chapter 7 Ratio and Proportion InText Questions Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.## Andhra Pradesh State Board Class 7th Maths Chapter 7 Ratio and Proportion InText Questions Books Solutions

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## AP Board Class 7th Maths Chapter 7 Ratio and Proportion InText Questions Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 7th Maths Chapter 7 Ratio and Proportion InText Questions Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:There are several situations in our daily life where we use ratio and proportion. Let us look at the following pictures and answers to the given questions:

Question 1.

Can we say the ratio of speeds of Cheetah to Man?

Answer:

Speed of Cheetah = 120 kmph

Speed of a man = 20 kmph

∴ Ratio of speed of Cheetah to a man

= 120 : 20

= 6 : 1

Question 2.

What will be the ratio of heights of HematoAmir?

Answer:

Height of Hema = 150 cm

Height of Amir 75 cm

∴ Ratio of heights 150 : 75

= 2 : 1

Question 3.

Answer:

Cost of 2 Hands of banana = ₹ 80

∴ Cost of 3 Hands of banana

= 32 × 80 = ₹ 120

Question 4.

Do you know which aspect ratio makes constructions more attractive and beautiful?

Answer:

Golden ratio is more beautiful. It is the ratio of a line segment cut in to two pieces of different lengths such that the ratio of whole segment to that of the longer segment is equal to that of longer segment to the shorter segment.

Check your Progress [Page No. 4]

Question 1.

Write the compound ratio of the following given ratios.

(i) 3 : 5 and 4 : 3

Answer:

Given ratios are 3 : 5 and 4 : 3.

If a : b and c : d are ratios, then their compound ratio is product of antecedents : product of extremes.

that is a × c : b : d

3 × 4 : 5 × 3

12 : 15

∴ Compound ratio = 4 : 5

(ii) 8 : 3 and 6 : 5

Answer:

Given ratios are 8 : 3 and 6 : 5.

If a : b and c : d are ratios, then their compound ratio is product of antecedents : product of extremes, that is

a × c : b : d

8 × 6 : 3 × 5

∴ Compound ratio = 16 : 5

(iii) 2: 1 and 8: 7

Answer:

Given Ratios are 2 : 1 and 8 : 7.

If a : b and c: d are ratios, then their compound ratio is product of antecedents : product of extremes that is a × c : b : d

2 × 8 : 1 × 7

16 : 7

∴ Compound ratio= 16 : 7

Question 2.

Fill in the boxes with correct answers.

Answer:

[Page No. 6]

Discuss and answer the following questions related to real-life situations:

Question 1.

If the cost of 3 ball pens is ₹ 15, then what is the cost of 6 such pens?

Answer:

3 : 15 : : 6 : cost

Cost x 3 = 15 × 6

∴ Cost = 15×63 = ₹ 30

Question 2.

In a school, for implementing ‘Mid day Meal’ scheme for class VII of 40 students they need 6 kgs of rice for one day.

(i) How much rice is needed for 6 days?

Answer:

Rice required for 40 students for one day = 6 kgs

Rice required for 6 days = 6 × 6 = 36 kg

(ii) How much rice is needed for 10 days?

Answer:

Rice required for 10 days = 10 × 6 = 60 kg

(iii) How much rice is needed for 5 days?

Answer:

Rice required for 5 days = 5 × 6 = 30 kg

Let’s Explore [Page No. 6]

Question 1.

Given below the number of workers working for construction of a house and their wages in total.

Answer:

Let’s Think [Page No. 7]

Question 1.

If the cost of 4 note books is 80. What would be the cost of7 note books? The above problem can be solved by using unitary method. Can you solve? How? Think.

Answer:

Given the cost of 4 note books = ₹ 80

Cost of 1 note book = 80 ÷ 4 = ₹ 20

Cost of 7 note books = ₹ 20 × 7

= ₹ 140

Check Progress [Page No 7]

Question 1.

Fill in the blanks, if the given quantities are in direct proportion.

Answer:

Let’s’ Do Activity [Page No. 8]

Take a clock and fix Its minute hand at a particular number on clock (if it is 12 It will be easy). Then find and note the angles made by minute hand in every 15 minutes interval of time:

Check whether they are in direct proportion or not?

Answer:

15 : 30 and 90 : 180

Here 1 :2 = 1 :2

If ratios are equal, then they are in direct proportion.

So, time passed and angle are in direàt proportion.

Let’s Think [Page No. 8]

Question 1.

What is the angle made by minutes hand in a minute?

Answer:

We know angle made by minutes hand in 15 m = 90°

Let angle made by minutes hand in 1 m = x°

Then 15 : 1 = 90°: x°

If the ratios are equal,

The product of means = The product of extremes

⇒ 15 × x = 1 × 90°

⇒ 15𝑥15=90∘15

⇒ x = 6°

∴ Angle made by minutes hand in 1 minute = 6°

Question 2.

What is the angle made by hours hand in one minute?

Answer:

We know angle made by hours hand in 1 hour (60 minutes) = 30°

Let angle made by hours hand 1 minute = x°

Then 60 : 1 = 30°: x°

If the ratios are equal,

The product of means = The product of the extremes

⇒ 60 × x = 1 × 30

⇒ 60𝑥60=3060=12

⇒ x = 12 (or) 0.5°

∴ Angle made, by hours hand in 1 minute = 12° (or) 0.5°

Check your Progress [Page No. 9]

Question 1.

Fill in the blanks if the given quantities are in inverse proportion.

Answer:

Check your Progress [Page No. 13]

Question 1.

Analyse how the three quantities given below are related and find ‘x’.

Answer:

From the above,

- Ration and the number of days are in inversely proportional.
- Ration and number of men are in directly proportional.
- Number of men is depends on both Ration and number of days.

So we have to take compound ratio of 108 : 70 and 25 : 15

i. e., 108 × 25 : 70 × 15

∴ 18 : x = 108 × 25 : 70 × 15

If the ratios are equal.

The Product of means = The Product of extremes

⇒ 108 × 25 × x = 18 × 70 × 15

∴ x = 7

Let’s Think [Page No. 18]

Question 1.

A man buys chocolates 10 for 10 rupees and sells them at 10 for ₹ 12. Does this result profit or loss ? What percent?

Answer:

Given cost price of 10 chocolates = ₹ 10

Selling price of 10 chocolates = ₹ 12

SP > CP, then man gets profit.

Profit = S.P – C.P

= 12 – 10 = ₹ 2

We know profit percent

Profit percentage = 20%

Question 2.

If a shopkeeper bought sofa sets and increases their prices by 50% and sell at 50% less, is it a loss or gain?

Answer:

Let the cost price of sofa set = ₹ 100

= ₹ 5o

Total printed price = ₹ 100 + ₹ 50

= ₹ 150

Loss percent = 50%

Loss = ₹ 75

Selling price Printed Price – Loss

= 150 – 75 = ₹ 75

C.P= ₹ 100; S.P = ₹ 75

and C.P > S.P

So, the shopkeeper will get loss of 25%.

Check Your Progress [Page No. 20]

Answer:

Check Your Progress [Page No. 24]

Answer:

Let’s Think [Page No. 24]

Question 1.

At what rate per annum will the principle doubles in 10 years?

Answer:

Method 1:

Let the principle = 100

Interest rate = x% ; Time = 10 years

Amount = Principle + Interest

= Double the principle

= 100 + 10x = 200

⇒ 100 + 10x – 100 = 200 – 100

⇒ 10x = 100

⇒ 10𝑥10=10010

⇒ x = 10%

∴ Rate of interest = 10%

Method 2:

Let the principle =?

Rate of Interest = R%

T = 10 years

Given that A = 2P = P + P [∵ A = P + I]

∴ I = P

∴ P×T×R100 = P

⇒ P×10×R100 = P

R = 10%

∴ The principle become double in 10 years at 10% rate of interest.

Question 2.

At what rate per annum will the principle be 4 times in 15 years?

Answer:

Method 1:

Let the principle = ₹ 100

Interest rate = x %

Tirne = 15 years

Interest = PTR100

Amount = Principle + Interest

= 4 times the principle

= 100 + 15x = 400

⇒ 100 + 15x – 100 = 400 – 100

⇒ 15x = 300

⇒ 15𝑥15=30015 = 20%

∴ x = 20%

Method 2:

Let the principle = P

Time = 15 years

A = 4timesP = 4P

⇒ A = P + 3P

But A = P + I

So I = 3P

We know that I = PTR 100

⇒ 3P = 𝑃×15×𝑅100

⇒ R = 100×315 = 20%

∴ The principle become 4 times in 15 years at 20% rate of interest.

Examples:

Question 1.

Find the compound ratio of 8 : 7 and 9 : 13.

Answer:

Given ratio = 8 : 7 :: 9 : 13

Compound ratio = 8 × 9 : 7 × 13 = 72 : 91

(∵ Compound ratio product of antecedent product consequents).

Question 2.

Two friends Prabhu and Suresh started a business with ₹ 1,00,000 each. After 3 months Suresh left the business. At the end of the year, there was a profit of ₹ 20,000. Calculate the profits shared by Prabhu and Suresh?

Answer:

Here Prabhu and Suresh started a business with ₹ 1,00,000.

Prabhu continued till the end of the year.

Suresh continued only for 3 months.

The ratio of the contributions

= 100000: 100000 = 1 : 1

Ratio of their period of business

= 12 : 3 = 4 : 1

So, their profits sho’uld be divided on the basis of compound ratio

Compound ratio = 1 × 4 : 1 × 1

Profit = ₹ 20,000

Total parts= 4 + 1 = 5

Suresh’s profit = 20000 × 15

= ₹ 4000

Prabhus profit = 20000 – 4000

= ₹ 16000

Question 3.

Rani started a beauty parlour with an amount ₹ 75,000. After 4 months, Vani also joined with Rani with an amount ₹ 50,000. After one year, they got a profit of ₹ 52,000. Calculate the profits shared by Rani and Vani.

Answer:

Rani’s investment = ₹ 75,000

Rani’s period in business = 1 year = 12 months

Vani’s investment = ₹ 56,000

Vani’s period in business = 8 months

The ratio of investments of Rani to Vani = 75,000 : 50,000 = 3:2

The ratio of periods of business of Rani to Vani =12:8 = 3:2

The profit should be distributed on the, basis of compound ratio.

Compound ratio = 3 ×. 3 : 2 × 2

= 9 : 4

Profit = ₹ 52,000

Total parts = 9 + 4 = 13

Rani’s profit = 52,000 × 913

= ₹ 36000

Vani’s profit = 52,000 – 36,000 = ₹ 16,000

Question 4.

If 3 : 4 and 9 : x are in direct proportion, then what is the value of x?

Answer:

If 3 : 4 and 9 : x are in direct proportion, then their ratio is constant.

∴ 34=9𝑥

⇒ 3 × x = 4 × 9

⇒ x = 4×93 = 12

Question 5.

If the cost of 4 note books is ₹ 80. What would be the cost of 7 note books?

Answer:

We know that as number of note books increases, the cost also increases such that the ratio of number of note books and the ratio of their costs will remain the same. That means here number of note books and the cost are in direct proportion.

Let the cost of 7 note books be ‘x’.

Then, 4 : 80 = 7 : x

If the ratios are equal, the product of means = The product of the extremes

4 × x = 80 × 7

⇒ x = 80×74 = ₹ 140

Thus, the cost of 7 note book is equal to ₹ 140.

Question 6.

The scale of a map is given as 1 : 30000. If two cities are 20 cm apart on the map, then o-j, . find the actual distance between them.

Answer:

Let the actual distance be ‘x’ cm. Since the distance on the map is directly proportional to the actual distance.

1 : 30000 = 20 : x

If the ratios are equal, the product of means = The product of the extremes

∴ 1 × x = 30,000 × 20

⇒ x = 6,00,000 cm = 6 km. [1,00,000 cm = 1000 m = 1 km.]

Thus, two cities which are 20cm apart on the map are actually 6 km away from each other.

Question 7.

If 4, 7 and 2, x are inverse proportion, then what is the value of x?

Answer:

4, 7 and 2, x are in inverse proportion . Therefore, 4 × 7 = 2 × x

⇒ x = 4×72 = 14

Question 8.

If 18 workers can build a wall in 12 days, how many days will eight workers take to build the same wall?

Answer:

If the number of workers decreases, the time taken to build the wall increases in the same proportion.

So, number of workers and the number of days to complete work are in inverse proportion.

Let the number of days to complete the work be ‘x’.

Number of workers | Number of days |

18 | 12 |

8 | x |

By taking inverse proportion,

18 : 8 = x : 12

Then,

∴ 18 × 12 = 8 × x

⇒ 8 × x = 18 × 12

⇒ x = 18×128 = 27days

∴ Eight workers can complete the will in 27 days.

Observe this:

In inverse Proportion, product is always constant.

∴ 18 × 12 = 8 × x

⇒ 8 × x = 18 × 12

⇒ x = 18×128 = 27days

Question 9.

4 Pumps are required to fill a tank in 1 hr 30 min. How long will it take if only 3 pumps of the same type are used?

Answer:

Let the time be ‘x’.

1 hr 30min = 60 + 30 = 90 minutes ,

If the number of pumps are decrease, the time taken to fill the tank increases.

So, the number of pumps and time taken to fill the tank are in inverse proportion.

Number of pipes | Time taken to fill (min) |

4 | 90 |

3 | x |

By taking inverse proportion,

4 : 3 = x : 90

Thus,

∴ 4 × 90 = 3 × x

⇒ 3 × x = 4 × 90

⇒ x = 4×903 = 120 min

∴ 3 pumps will fill the tank in 120 min or 2 hrs.

Observe this:

In inverse proportion, product is always constant

∴ 4 × 90 = 3 × x

⇒ 3 × x = 4 × 90

⇒ x = 4×903 = 120 min

Question 10.

If 30 persons use 40 kg. of sugar in 10 days, find in how many days 80 persons will use 320 kg. of sugar? How can we solve?

Answer:

Here, the 3 quantities are persons, weight and number of days.

From the above comparison mustbe in between unknown form to know form.

- Here days and persons are in inverse we denote it with Tmark.
- Here days and sugar are in direct proportion we denote it with 4r mark.

Here number of days is depends on both persons and weight of sugar. So we have to take compound ratio of 80 : 30 and 40 : 320.

∴ 10 : x = 80 × 40 : 30 × 320 = 3200 : 9600

Since the ratios are equal, the product of extremes is equal to product of means.

10 : x = 3200 : 9600 ⇒ 10 × 9600 = x × 3200

⇒ x × 3200 = 10 × 9600 ⇒ x = 10×96003200 = 30

Question 11.

8 painters can paint a wall of 160m long in 5 days. How many painters are required to paint 240kn. wall in 10 days?

Answer:

Here we have three quantities – number of painters, length of wall and number of days.

Number of painters, is directly proportional to length of the wall.

Number of painters is inversely proportional to the number of days.

Since the number of painters depends both on length of wall and number of days. We will take the compound ratio of 160 : 240 and 10 : 5 1

∴ 8 : x = 160 × 10 : 240 × 5

Since the ratios are equal, product of means is equal to product of extremes

8 : x = 160 × 10 : 240 × 5

⇒ x × 160 × 10 = 8 × 240 × 5

⇒ x = 8×240×5160×10 = 6

Hence the required number of painters = 6

Question 12.

195 men working 10 hours a day can finish a job in 20 days. How many men are employed to finish the job in 15 days if they work 13 hours a day?

Answer:

Here we have three quantities – number of workers, number of hours and number of days.

Here the number of workers inversely proportional to number of hours per day.

Here the number of workers inversely proportional to number of days.

Since number of workers depends on both number of hours and number of days, we will take compound ratio of 13 : 10 and 15 : 20.

Since the ratios are equal, the product of means is equal to product of extremes.

195 : x = 13 × 15 : 20 × 10

⇒ x × 13 × 15 = 195 × 20 × 10

⇒ x = 195×20×1013×15 = 200

Hence the required number of workers = 200

Question 13.

Express the following percentages as fraction, decimal and In ratio.

(i) 45%

Answer:

45% = 45100=920(Fraction)

= 0.45 (decimal form)

= 9:20 (ratio)

(ii) 62%

Answer:

62% = 62100=3150 = (Fraction)

= 0.62 (decimal form)

= 31:50 (ratio)

Question 14.

Find 24% of 150 and àlso the remaining of that number.

Answer:

24% of 150 = 24100 × 150 = 36

The remaining of that number

= 150 – 36 = 114

Question 15.

Raghu bought pens for 400 and he sold them for ₹ 480 what Is his profit or loss percent?

Answer:

Jyothi solved It this way:

Pen C.P of = ₹ 400, S.P = ₹ 480

S.P > C.P.

So, Raghu gets a profit P = 480 – 400 = ₹ 80

Profit Percentage = 80400 × 100 = 20%

Anwar solved it this way :

Raghu’s Profit = S.P. – C.P = 480 – 400 = ₹ 80

The ratio of profit and cost or the fraction is 80400

∴ Profit Percentage = 80400 × 100 = 20%

Suresh solved it using proportion :

When C.P is ₹ 400, the profit is ₹ 80 C.P. is ₹ 100, let the profit be ₹ x.

Here C.P and profit are in direct proportion.

x : 80 = 100 : 400 ⇒ 𝑥80=100400

⇒ x × 400 = 100 × 80

⇒ x = 100×80400 = 20%

Profit = 20 per 100

Profit Percentage = 20%

Question 16.

Ramana bought a cycle for ₹ 1200 and sold it to his friend Rehman for ₹ 900, then what is Ramanas profit or loss percentage?

Answer:

Here Ramana’s Cost price = ₹ 1200

Selling price = ₹ 900

S.P < C.P, then Ramana got loss

Loss = C.P – S.P = 1200 – 900 = ₹ 300

Loss percentage = 3001200 × 100 = 25%

Question 17.

If John buys a ear for ₹ 1,50,000 and gains 10% on selling it. Then find the selling price.

Answer:

Cost Price = ₹ 1,50,000

Gain % = 10%

Profit = 10% of ₹ 1,50,000

= 10100 × 150000 = ₹ 15000

Selling Price = C.P. + Profit

= ₹ 1,50,000 + ₹ 15,000

= ₹ 1,65,000

This can be solved using proportion:

Gain 10% means

If CP is ₹ 100,the gain is ₹ 10

Thus S.P = 100 + 10 = 110

Now, here C.P = ₹ 1,50,000

Let S.P = x

CP and SP are directly proportional.

110100=𝑥1,50,000

𝑥1,50,000=110100

x = 1,50,000×110100 = ₹ 1,65,000

Question 18.

Kiran sold a refrigerator for ₹ 16800 at a gain of 12%. Then what is the cost price of it?

Answer:

Roopa did the problem uslnj unitary method:

S.P = ₹ 16800 ; Gain% = 12%

If CP is ₹ 100, then Profit is ₹ 12 and thus S.P = ₹ 112

So when S.P is ₹ 112, then C.P is ₹ 100

When S.P is One Rùpee, C.P is 110112 .

Here S.P is ₹ 16800

So, C.P = 110112 × 16800 = ₹ 15000

Sneha solve did the prot1em using proportion as follows:

Gain % = 12% ;S.P = ₹ 16800

If C.P is ₹ 100 then profit is ₹ 12, thus SP = ₹ 12

here S.P = ₹ 16800

C.P be x

C.P and S.P are directly proportional.

∴ 𝑥16800=100112

⇒ x = 100×16800112 = ₹ 15000

Question 19.

The cost of an article goes down every year by 10% of its previous value. Find its original cost, if its cost after 2 years is ₹ 32400.

Answer:

Let the cost at the beginning of 1st year be ₹ 100. At the beginning of,2nd year i.e at the end of 1st year it will be decreased by 10% means cost will be ₹ 90.

At the end of 2nd year i.e at the beginning of 3rd year it will be reduced by 10%

i. e. ₹ 90 is reduced by 10%.

90 – 9 = ₹ 81

If the cost of object is ₹ 100 at the beginning, then after 2 years it’s cost will be ₹ 81.

Let the cost of the object ₹ ‘x’ at the beginning, after 2 years it’s cost is ₹ 32,400 .

Thus ratio of original costs = ratio of costs after 2 years

⇒ x : 100 = 32400 : 81

⇒ 𝑥100=3240081

⇒ x = 32400×10081 = ₹ 40000

Question 20.

Find discount if,

(i) marked price is ₹ 450, selling price = ₹ 415

Answer:

Discount = marked price – selling price ⇒ 450 – 415 = ₹ 35.

(ii) marked price = ₹ 810, selling price = ₹ 765.

Answer:

Discount = marked price – selling price ⇒ 810 – 765 = ₹ 45.

Question 21.

If discount is ₹ 40, marked price is ₹ 400, then find discount percentage.

Answer:

Discount = ₹ 40, marked price = ₹ 400,

then discount percentage = 40400 × 100 = 10

Question 22.

A shopkeeper marks his goods 20% above the cost price and allows a discount of 10% on them. What percent does he gain?

Answer:

Let the cost price be ₹ 100.

Then the marked price = 100 + 20 = ₹ 120

Discount = 10%, so discount = x 120 = 12% /

SP = Marked price – Discount = ₹ 120 – ₹ 12 = ₹ 108

Gain = 8100 × 100 = 8%

The shopkeeper gains 8% after discount.

Question 23.

Calculate simple interest and the total amount, if

(i) principal = ₹ 5000, time = 2 years, rate = 10%

Answer:

Simple interest

I = P×T×R100=5000×2×10100 = ₹ 1000

Total amount = principal + interest = 5000 + 1000 = ₹ 6000

(ii) principal = ₹ 25000, time = 3 years, rate = 12%

Answer:

Simple interest = P×T×R100=25000×3×12100 = ₹ 9000

Total amount = principal + interest = 25000 + 9000 = ₹ 34000

Question 24.

If Raheem borrowed a sum of ₹ 25000 at a rate 10% per annum, what is the simple interest and total amount he has to pay for 3 years?

Answer:

Rajesh did like this :

Principal = ₹ 25000 ; Time = 3 years ; Rate of interest = 10%

Simple interest = I = P×T×R100=25000×3×10100 = ₹ 7500

Total amount = ₹ 25000 + ₹ 7500 = ₹ 32500

Sangeetha did like this :

For 1 year we have to pay 10%,

for 3 years we have to pay 3 × 10 = 30% as interest.

Interest = 30100 × 25000 = ₹ 7500

Total amount = ₹ 25000 + ₹ 7500 = ₹ 32500

Question 25.

What sum will yield an interest of ₹ 6000 at 9% per annum in 3 years 4 months?

Answer:

S.I = ₹ 6000 ; R = 9%

Question 26.

At what rate per annum will ₹ 70000 yield an interest of ₹ 14000 in 212 years?

Answer:

Principal = ₹ 70000 ; Time = 212years = 52 years ; Simple interest = ₹ 14000

## AP Board Textbook Solutions PDF for Class 7th Maths

- AP Board Class 7
- AP Board Class 7 Maths
- AP Board Class 7 Maths Chapter 1 Integers Review Exercise
- AP Board Class 7 Maths Chapter 1 Integers Ex 1.1
- AP Board Class 7 Maths Chapter 1 Integers Ex 1.2
- AP Board Class 7 Maths Chapter 1 Integers Ex 1.3
- AP Board Class 7 Maths Chapter 1 Integers Ex 1.4
- AP Board Class 7 Maths Chapter 1 Integers Unit Exercise
- AP Board Class 7 Maths Chapter 1 Integers InText Questions
- AP Board Class 7 Maths Chapter 2 Fractions, Decimals and Rational Numbers Review Exercise
- AP Board Class 7 Maths Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.1
- AP Board Class 7 Maths Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.2
- AP Board Class 7 Maths Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.3
- AP Board Class 7 Maths Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.4
- AP Board Class 7 Maths Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.5
- AP Board Class 7 Maths Chapter 2 Fractions, Decimals and Rational Numbers Unit Exercise
- AP Board Class 7 Maths Chapter 2 Fractions, Decimals and Rational Numbers InText Questions
- AP Board Class 7 Maths Chapter 3 Simple Equations Ex 3.1
- AP Board Class 7 Maths Chapter 3 Simple Equations Ex 3.2
- AP Board Class 7 Maths Chapter 3 Simple Equations Ex 3.3
- AP Board Class 7 Maths Chapter 3 Simple Equations Ex 3.4
- AP Board Class 7 Maths Chapter 3 Simple Equations Unit Exercise
- AP Board Class 7 Maths Chapter 3 Simple Equations InText Questions
- AP Board Class 7 Maths Chapter 4 Lines and Angles Review Exercise
- AP Board Class 7 Maths Chapter 4 Lines and Angles Ex 4.1
- AP Board Class 7 Maths Chapter 4 Lines and Angles Ex 4.2
- AP Board Class 7 Maths Chapter 4 Lines and Angles Ex 4.3
- AP Board Class 7 Maths Chapter 4 Lines and Angles Ex 4.4
- AP Board Class 7 Maths Chapter 4 Lines and Angles Unit Exercise
- AP Board Class 7 Maths Chapter 4 Lines and Angles InText Questions
- AP Board Class 7 Maths Chapter 5 Triangles Review Exercise
- AP Board Class 7 Maths Chapter 5 Triangles Ex 5.1
- AP Board Class 7 Maths Chapter 5 Triangles Ex 5.2
- AP Board Class 7 Maths Chapter 5 Triangles Ex 5.3
- AP Board Class 7 Maths Chapter 5 Triangles Ex 5.4
- AP Board Class 7 Maths Chapter 5 Triangles Ex 5.5
- AP Board Class 7 Maths Chapter 5 Triangles Ex 5.6
- AP Board Class 7 Maths Chapter 5 Triangles Unit Exercise
- AP Board Class 7 Maths Chapter 5 Triangles InText Questions
- AP Board Class 7 Maths Chapter 6 Data Handling Ex 6.1
- AP Board Class 7 Maths Chapter 6 Data Handling Ex 6.2
- AP Board Class 7 Maths Chapter 6 Data Handling Ex 6.3
- AP Board Class 7 Maths Chapter 6 Data Handling Ex 6.4
- AP Board Class 7 Maths Chapter 6 Data Handling Unit Exercise
- AP Board Class 7 Maths Chapter 6 Data Handling InText Questions
- AP Board Class 7 Maths Chapter 7 Ratio and Proportion Review Exercise
- AP Board Class 7 Maths Chapter 7 Ratio and Proportion Ex 7.1
- AP Board Class 7 Maths Chapter 7 Ratio and Proportion Ex 7.2
- AP Board Class 7 Maths Chapter 7 Ratio and Proportion Ex 7.3
- AP Board Class 7 Maths Chapter 7 Ratio and Proportion Ex 7.4
- AP Board Class 7 Maths Chapter 7 Ratio and Proportion Ex 7.5
- AP Board Class 7 Maths Chapter 7 Ratio and Proportion Ex 7.6
- AP Board Class 7 Maths Chapter 7 Ratio and Proportion Ex 7.7
- AP Board Class 7 Maths Chapter 7 Ratio and Proportion Unit Exercise
- AP Board Class 7 Maths Chapter 7 Ratio and Proportion InText Questions
- AP Board Class 7 Maths Chapter 8 Exponents and Powers Ex 8.1
- AP Board Class 7 Maths Chapter 8 Exponents and Powers Ex 8.2
- AP Board Class 7 Maths Chapter 8 Exponents and Powers Ex 8.3
- AP Board Class 7 Maths Chapter 8 Exponents and Powers Unit Exercise
- AP Board Class 7 Maths Chapter 8 Exponents and Powers InText Questions
- AP Board Class 7 Maths Chapter 9 Algebraic Expressions Review Exercise
- AP Board Class 7 Maths Chapter 9 Algebraic Expressions Ex 9.1
- AP Board Class 7 Maths Chapter 9 Algebraic Expressions Ex 9.2
- AP Board Class 7 Maths Chapter 9 Algebraic Expressions Ex 9.3
- AP Board Class 7 Maths Chapter 9 Algebraic Expressions Ex 9.4
- AP Board Class 7 Maths Chapter 9 Algebraic Expressions Unit Exercise
- AP Board Class 7 Maths Chapter 9 Algebraic Expressions InText Questions
- AP Board Class 7 Maths Chapter 10 Construction of Triangles Review Exercise
- AP Board Class 7 Maths Chapter 10 Construction of Triangles Ex 10.1
- AP Board Class 7 Maths Chapter 10 Construction of Triangles Ex 10.2
- AP Board Class 7 Maths Chapter 10 Construction of Triangles Ex 10.3
- AP Board Class 7 Maths Chapter 10 Construction of Triangles Unit Exercise
- AP Board Class 7 Maths Chapter 10 Construction of Triangles InText Questions
- AP Board Class 7 Maths Chapter 11 Area of Plane Figures Review Exercise
- AP Board Class 7 Maths Chapter 11 Area of Plane Figures Ex 11.1
- AP Board Class 7 Maths Chapter 11 Area of Plane Figures Ex 11.2
- AP Board Class 7 Maths Chapter 11 Area of Plane Figures Ex 11.3
- AP Board Class 7 Maths Chapter 11 Area of Plane Figures Ex 11.4
- AP Board Class 7 Maths Chapter 11 Area of Plane Figures Unit Exercise
- AP Board Class 7 Maths Chapter 11 Area of Plane Figures InText Questions
- AP Board Class 7 Maths Chapter 12 Symmetry Review Exercise
- AP Board Class 7 Maths Chapter 12 Symmetry Ex 12.1
- AP Board Class 7 Maths Chapter 12 Symmetry Ex 12.2
- AP Board Class 7 Maths Chapter 12 Symmetry Ex 12.3
- AP Board Class 7 Maths Chapter 12 Symmetry Ex 12.4
- AP Board Class 7 Maths Chapter 12 Symmetry Unit Exercise
- AP Board Class 7 Maths Chapter 12 Symmetry InText Questions
- AP Board Class 7 Maths Chapter 1 పూర్ణ సంఖ్యలు Review Exercise
- AP Board Class 7 Maths Chapter 1 పూర్ణ సంఖ్యలు Ex 1.1
- AP Board Class 7 Maths Chapter 1 పూర్ణ సంఖ్యలు Ex 1.2
- AP Board Class 7 Maths Chapter 1 పూర్ణ సంఖ్యలు Ex 1.3
- AP Board Class 7 Maths Chapter 1 పూర్ణ సంఖ్యలు Ex 1.4
- AP Board Class 7 Maths Chapter 1 పూర్ణ సంఖ్యలు Unit Exercise
- AP Board Class 7 Maths Chapter 1 పూర్ణ సంఖ్యలు InText Questions
- AP Board Class 7 Maths Chapter 2 భిన్నాలు మరియు దశాంశాలు Review Exercise
- AP Board Class 7 Maths Chapter 2 భిన్నాలు మరియు దశాంశాలు Ex 2.1
- AP Board Class 7 Maths Chapter 2 భిన్నాలు మరియు దశాంశాలు Ex 2.2
- AP Board Class 7 Maths Chapter 2 భిన్నాలు మరియు దశాంశాలు Ex 2.3
- AP Board Class 7 Maths Chapter 2 భిన్నాలు మరియు దశాంశాలు Ex 2.4
- AP Board Class 7 Maths Chapter 2 భిన్నాలు మరియు దశాంశాలు Ex 2.5
- AP Board Class 7 Maths Chapter 2 భిన్నాలు మరియు దశాంశాలు Unit Exercise
- AP Board Class 7 Maths Chapter 2 భిన్నాలు మరియు దశాంశాలు InText Questions
- AP Board Class 7 Maths Chapter 3 సామాన్య సమీకరణాలు Ex 3.1
- AP Board Class 7 Maths Chapter 3 సామాన్య సమీకరణాలు Ex 3.2
- AP Board Class 7 Maths Chapter 3 సామాన్య సమీకరణాలు Ex 3.3
- AP Board Class 7 Maths Chapter 3 సామాన్య సమీకరణాలు Ex 3.4
- AP Board Class 7 Maths Chapter 3 సామాన్య సమీకరణాలు Unit Exercise
- AP Board Class 7 Maths Chapter 3 సామాన్య సమీకరణాలు InText Questions
- AP Board Class 7 Maths Chapter 4 రేఖలు మరియు కోణాలు Review Exercise
- AP Board Class 7 Maths Chapter 4 రేఖలు మరియు కోణాలు Ex 4.1
- AP Board Class 7 Maths Chapter 4 రేఖలు మరియు కోణాలు Ex 4.2
- AP Board Class 7 Maths Chapter 4 రేఖలు మరియు కోణాలు Ex 4.3
- AP Board Class 7 Maths Chapter 4 రేఖలు మరియు కోణాలు Ex 4.4
- AP Board Class 7 Maths Chapter 4 రేఖలు మరియు కోణాలు Unit Exercise
- AP Board Class 7 Maths Chapter 4 రేఖలు మరియు కోణాలు InText Questions
- AP Board Class 7 Maths Chapter 5 త్రిభుజాలు Review Exercise
- AP Board Class 7 Maths Chapter 5 త్రిభుజాలు Ex 5.1
- AP Board Class 7 Maths Chapter 5 త్రిభుజాలు Ex 5.2
- AP Board Class 7 Maths Chapter 5 త్రిభుజాలు Ex 5.3
- AP Board Class 7 Maths Chapter 5 త్రిభుజాలు Ex 5.4
- AP Board Class 7 Maths Chapter 5 త్రిభుజాలు Ex 5.5
- AP Board Class 7 Maths Chapter 5 త్రిభుజాలు Ex 5.6
- AP Board Class 7 Maths Chapter 5 త్రిభుజాలు Unit Exercise
- AP Board Class 7 Maths Chapter 5 త్రిభుజాలు InText Questions
- AP Board Class 7 Maths Chapter 6 దత్తాంశ నిర్వహణ Ex 6.1
- AP Board Class 7 Maths Chapter 6 దత్తాంశ నిర్వహణ Ex 6.2
- AP Board Class 7 Maths Chapter 6 దత్తాంశ నిర్వహణ Ex 6.3
- AP Board Class 7 Maths Chapter 6 దత్తాంశ నిర్వహణ Ex 6.4
- AP Board Class 7 Maths Chapter 6 దత్తాంశ నిర్వహణ Unit Exercise
- AP Board Class 7 Maths Chapter 6 దత్తాంశ నిర్వహణ InText Questions
- AP Board Class 7 Maths Chapter 7 నిష్పత్తి మరియు అనుపాతం Review Exercise
- AP Board Class 7 Maths Chapter 7 నిష్పత్తి మరియు అనుపాతం Ex 7.1
- AP Board Class 7 Maths Chapter 7 నిష్పత్తి మరియు అనుపాతం Ex 7.2
- AP Board Class 7 Maths Chapter 7 నిష్పత్తి మరియు అనుపాతం Ex 7.3
- AP Board Class 7 Maths Chapter 7 నిష్పత్తి మరియు అనుపాతం Ex 7.4
- AP Board Class 7 Maths Chapter 7 నిష్పత్తి మరియు అనుపాతం Ex 7.5
- AP Board Class 7 Maths Chapter 7 నిష్పత్తి మరియు అనుపాతం Ex 7.6
- AP Board Class 7 Maths Chapter 7 నిష్పత్తి మరియు అనుపాతం Ex 7.7
- AP Board Class 7 Maths Chapter 7 నిష్పత్తి మరియు అనుపాతం Unit Exercise
- AP Board Class 7 Maths Chapter 7 నిష్పత్తి మరియు అనుపాతం InText Questions
- AP Board Class 7 Maths Chapter 8 ఘాతాంకాలు మరియు ఘాతాలు Ex 8.1
- AP Board Class 7 Maths Chapter 8 ఘాతాంకాలు మరియు ఘాతాలు Ex 8.2
- AP Board Class 7 Maths Chapter 8 ఘాతాంకాలు మరియు ఘాతాలు Ex 8.3
- AP Board Class 7 Maths Chapter 8 ఘాతాంకాలు మరియు ఘాతాలు Unit Exercise
- AP Board Class 7 Maths Chapter 8 ఘాతాంకాలు మరియు ఘాతాలు InText Questions
- AP Board Class 7 Maths Chapter 9 బీజీయ సమాసాలు Review Exercise
- AP Board Class 7 Maths Chapter 9 బీజీయ సమాసాలు Ex 9.1
- AP Board Class 7 Maths Chapter 9 బీజీయ సమాసాలు Ex 9.2
- AP Board Class 7 Maths Chapter 9 బీజీయ సమాసాలు Ex 9.3
- AP Board Class 7 Maths Chapter 9 బీజీయ సమాసాలు Ex 9.4
- AP Board Class 7 Maths Chapter 9 బీజీయ సమాసాలు Unit Exercise
- AP Board Class 7 Maths Chapter 9 బీజీయ సమాసాలు InText Questions
- AP Board Class 7 Maths Chapter 10 త్రిభుజాల నిర్మాణం Review Exercise
- AP Board Class 7 Maths Chapter 10 త్రిభుజాల నిర్మాణం Ex 10.1
- AP Board Class 7 Maths Chapter 10 త్రిభుజాల నిర్మాణం Ex 10.2
- AP Board Class 7 Maths Chapter 10 త్రిభుజాల నిర్మాణం Ex 10.3
- AP Board Class 7 Maths Chapter 10 త్రిభుజాల నిర్మాణం Unit Exercise
- AP Board Class 7 Maths Chapter 10 త్రిభుజాల నిర్మాణం InText Questions
- AP Board Class 7 Maths Chapter 11 సమతల పటాల వైశాల్యాలు Review Exercise
- AP Board Class 7 Maths Chapter 11 సమతల పటాల వైశాల్యాలు Ex 11.1
- AP Board Class 7 Maths Chapter 11 సమతల పటాల వైశాల్యాలు Ex 11.2
- AP Board Class 7 Maths Chapter 11 సమతల పటాల వైశాల్యాలు Ex 11.3
- AP Board Class 7 Maths Chapter 11 సమతల పటాల వైశాల్యాలు Ex 11.4
- AP Board Class 7 Maths Chapter 11 సమతల పటాల వైశాల్యాలు Unit Exercise
- AP Board Class 7 Maths Chapter 11 సమతల పటాల వైశాల్యాలు InText Questions
- AP Board Class 7 Maths Chapter 12 సౌష్ఠవము Review Exercise
- AP Board Class 7 Maths Chapter 12 సౌష్ఠవము Ex 12.1
- AP Board Class 7 Maths Chapter 12 సౌష్ఠవము Ex 12.2
- AP Board Class 7 Maths Chapter 12 సౌష్ఠవము Ex 12.3
- AP Board Class 7 Maths Chapter 12 సౌష్ఠవము Ex 12.4
- AP Board Class 7 Maths Chapter 12 సౌష్ఠవము Unit Exercise
- AP Board Class 7 Maths Chapter 12 సౌష్ఠవము InText Questions

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