Class 7

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 4

Question 1.
In a school X, 48 students appeared for 10th class exam out of which 36 students passed. In another school Y, 30 students appeared and 24 students passed. If the District Educational Officer wants to give an award on the basis of pass percentage. To which school will he give the award?
Solution:
From the first school:
Number of students appeared for the exam = 48
Number of students passed = 36
as a percentage = \(\frac { 36 }{ 48 }\) × 100 = 75
Pass percentage = 75%

From the second school:
Number of students appeared = 30
Number of students passed = 24
Pass percentage = \(\frac { 24 }{ 30 }\) × 100 = 80%
As 80% > 75%
D.E.O should give award to the 2nd school.

Question 2.
Last year the cost of 1000 articles was ₹ 5000. This year it goes down to ₹ 4000. What is the percentage of decrease in price?
Solution:
Last year
Cost of 1000 articles = ₹ 5000
∴ Cost of one article = \(\frac { 5000 }{ 1000 }\) = ₹ 5
This year
Cost of 1000 articles = ₹ 4000
∴ Cost of one article = \(\frac { 4000 }{ 1000 }\) = ₹ 4
Decrease in price = ₹ 5 – ₹ 4 = ₹ 1
Decrease In percentage = \(\frac { 1 }{ 5 }\) × 100% = 20%

Question 3.
Sri Jyothi has a basket full of bananas, oranges and mangoes. If 50% are bananas, 15% are oranges, then what percent are mangoes?
Solution:
Percent of bananas = 50%
Percent of oranges = 15%
∴ Percent of mangoes = 100% – (50% + 15%)
= 100% – 65% = 35%

Question 4.
64% + 20% + …..? = 100%
Solution:
(64% + 20%) +…………… = 100%
∴ ?=100% – 84% = 16%

Question 5.
On a rainy day, out of 150 students in a school 25 were absent. Find the percentage of students absent from the school? What percentage of students are present?
Solution:
Total students = 150
Number of students absent = 25
∴ Number of students present = 150 – 25 = 125

Question 6.
Out of 12000 voters in a constituency, 60% voted. Find hte number of people voted in the
constituency?
Solution:
Percentage of people who voted = 60% of I 2,000
∴ Number of people voted = \(\frac{60 \times 12000}{100}\) = 7,200

Question 7.
A local cricket team played 20 matches in one season. If it won 25% of them and lost rest. How many matches did it loose?
Solution:
Total number of matches played = 20
Matches won = 25%
∴ Matches lost = (100 –  25)% of total played
= 75% of 20
\(\frac{75 \times 20}{100}\) = 15

Question 8.
In every gram of gold, a goldsmith mixes 0.25 grams of silver and 0.05 grams of copper. What is the percentage of gold, silver and copper in every gram of gold?
Solution:
In 1 gm
The quantity of silver = 0.25 gm
The quantity of copper = 0.05 gm
∴ The quantity of gold = 1 gm – (0.25 . 0.05)
= 1 – 0.3 = 0.7gm
∴ Percentage of gold = \(\frac { 0.7 }{ 1 }\) × 100 = 70%
∴ Percentage of silver = \(\frac { 0.25 }{ 1 }\) × 100 = 25%
∴ Percentage of copper = \(\frac { 0.05 }{ 2 }\) × 100 = 5%

Question 9.
40% of a number is 800 then find the number?
Solution:
Let the number be = x
40% of x = 800
(i.e.,) \(\frac{40 \times x}{100}\) = 800
x = \(\frac{800 \times 100}{40}\)
x = 2000
∴ The required number = 2000

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Ex 5 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 5

Question 1.
Draw the following pairs of angles as adjacent angles. Check whether they fonn linear pair.

Solution:
Linear pair of angles

Linear pair of angles

Do not form linear pair.

Question 2.
Niharika took two angles- 1300 and 500 and tried to check whether they form a linear pair. She made the following picture.
Solution:
Yes. These two angles do not form a pair of linear angles.
Because she made angle more than 180°.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4

Question 1.
Which of the following are adjacent angles?

Solution:
(i)
i) a and b are adjacent angles.

ii) c and d are adjacent angles.

iii) e, f are not adjacent angles.

Question 2.
Name all pairs of adjacent angles in the figure. How many pairs of adjacent angles are formed? Why these angles are called adjacent angles?
Solution:
(∠AOL). ∠AOC). (∠AOC. ∠BOC)
(∠BOC. ∠BOD), (∠BOD. ∠AOD) are the four pairs of adjacent angles.
These are adjacent since they have a common vertex, common arm and on either sides of the common arm.

Question 3.
Can two adjacent angles be supplementary? Draw figure.
Solution:
Yes. Two adjacent angles can be supplementary.

Question 4.
Can two adjacent angles be complementary? Draw figure.
Solution:

Yes. Two adjacent angles can be complementary.

Question 5.
Give four examples of adj acent angles in daily life.
Example: Angles between the spokes at the centre ofa cycle wheel.
(i) ____________________________
(ii) ____________________________
(iii) ____________________________
(iv) ____________________________
Solution:
i) Angles between (Door and Wall). (Door and frame)
ii) Angles between the 3 hands of a clock at a point of time.
iii) Adjacent angles in the Rangoli.
iv) Angles between spokes of the window design.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 3

Question 1.
Which of the following pairs of angles are supplementary?

Solution:
i) Sum of the angles = 110° + 70° =180°
Hence they are supplementary.

ii) Sum of the angles = 90° + 90° =180°
hence they are supplementary.

iii) Sum of the angles = 50°+ 140° =190° ≠ 180°
Hence they are not supplementary.

Question 2.
Find the supplementary angles of the given angles.
(i) 105° (ii) 95° (iii) 150° (iv) 20°
Solution:
(i) 105°
Supplementary angle of 105° = 180° – 105° = 75°
(ii) 95°
Supplementary angle of 95° = 180° – 95° = 85
(iii) 150°
Supplementary angle of 150° = 180° – 150° = 30°
(iv) 20°
Supplementary angle of 20° = 20° = 180° – 20° = 60°

Question 3.
Two acute angles cannot form a pair of supplementary angles. Justify.
Solution:
As the measure of an acute angle is less than 90°, sum of two acute angles < 90° + 90° = 180°
Hence two acute angles cant brin a pair of supplementary angles.

Question 4.
Two angles are equal and supplementary to each other. Find them.
Solution:
Let the angle be x°
then its supplementary 180° – x°
By problem, x° =180° – x°
x° + x° = 180°
x° = \(\frac{180^{\circ}}{2}\)
x° = 90°

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles Exercise 2

Question 1.
Which of the following pairs of angles are complementary?

Solution:
In fig (i) 110° – 30° = 140° ≠ 90. Hence they are not complementary.
In fig (ii) 90° + 90° = 180 ≠ 90°. Hence they are not complementary.
In fig (iii) 80° + 10° = 90° = 90. Hence they are complementary.

Question 2.
Find the complements of the given angles. Find the complementary angles of the following.
(i) 25°
(ii) 40°
(iii) 89°
(iv) 55°
Solution:
i) Complementary angle of 25° = 90° – 25° = 65°
ii) Complementary angle of 40° = 90° – 40° = 50°
ill) Complementary angle of 89° = 90° – 89° = 1°
iv) Complementary angle of 55° = 90° – 55 = 35°

Question 3.
Two angles are complement to each other and are also equal. Find them.
Solution:
Let the angle be x°
Then Its complement = (900_ x°)
By problem, x° = 90° – x°
x° + x°= 90°
2x = 90°
x = \(\frac{90^{\circ}}{2}\)
x = 45°

Question 4.
Manasa says, “Each angle in any pair of complementary angles is always acute”. Do
Solution:
Yes. Each angle ¡n any pair of complementary angles is always acute.
As sum of two angles is 90°, each of the angle must be less than 90°.

AP State Syllabus AP Board 7th Class Science Important Questions Chapter 1 Food Components

AP State Syllabus 7th Class Science Important Questions 1st Lesson Food Components

7th Class Science 1st Lesson Food Components Important Questions and Answers

Question 1.
What type of food is required to keep us healthy?
Answer:

1. Generally every food item contains all the components of food.
2. But some components may be more while some may be less.
3. We require different quantities of carbohydrates, proteins and fats according to age and need of individuals.
4. Growing children and adolescents need more protein-containing food like milk, meat, pulses etc.
5. We also need minute quantities of some other components called vitamins and minerals to keep us healthy.

Question 2.
What are roughages? In what way are they useful to us?
Answer:

1. There are some components of food that are necessary for our body called roughages or dietary fibres.
2. Vegetables like a ribbed gourd, bunch beans, lady’s finger or some boiled sweet potato etc. are called roughages.
3. Roughages are a kind of carbohydrate that our body fails to digest.
4. They help in free bowel movement in the digestive tract and prevent constipation.

Question 3.
Mention some sources of roughages.
Answer:
Sources of Roughages:

1. Bran, shredded wheat, cereals, fruits and vegetables, sweet and plain potato, peas and berries, pumpkins, palak, apples, banana, papaya and many kinds of beans are the sources of roughages.
2. We must take care to include sufficient fibre foods in our daily diet.

Question 4.
Why should we eat fruits with peels? What should be done before eating them.
Answer:

1. Generally we have a habit of eating some fruits without peels.
2. We eat banana without peel but fruits like apples, grapes, sapota are eaten along with peels.
3. Most of the vegetable are also used along with peels, sometimes we make some special dishes like chutneys etc., with peels.
4. So don’t peel or discard outer layers of fruits or vegetables.
5. They are rich in nutrients.
   
6. Peel contains fibre which helps in digestion.
7. But now a days farmers use many pesticides in the fields.
8. They are very dangerous for our health so we must wash fruits and vegetables with salt water thoroughly.
9. Then only it becomes safe to eat them along with peels.

Question 5.
Do fruits and vegetables contain water in them? Give some examples of such fruits.
Answer:

1. Water is also an essential component needed by our body.
2. We should drink sufficient water for our body.
3. We get water from fruits and vegetables also.
   
4. Most fruits and vegetables contain water.
5. Cut these fruits and vegetables. We find water in them.
6. Most vegetables like potatoes, beans, kheera, tomatoes, gourds and fruits like apples, papaya and melons etc., contain water.

Question 6.
Why does our body need water? Explain with an example.
Answer:

1. Take a piece of sponge and try to move it in a pipe.
2. It moves with some difficulty. Remove the sponge from the pipe, dip it in water and try to move it again in the pipe.
   
3. It moves freely or smoothly.
4. Water is food and it also helps the food to move easily in the digestive tract.
5. Water helps in many other processes in our body as well.
6. Hence, we must drink plenty of water.

Question 7.
How would you make your diet a balanced one?
Answer:

1. Taking green salads and vegetables everyday.
2. Taking foods like cereals, pulses, milk etc., adequately.
3. Taking a bit of fat (Oil, Ghee, Butter etc.)
4. Don’t forget to supplement your daily diet with green salads and vegetables.

Question 8.
Write the history of food and nutrition.
Answer:
History of food and nutrition:

1. Until about 170 years ago there was little scientific knowledge in the West about nutrition.
2. The founder of modern science of nutrition was Frenchman named Lavoisier (1743 to 1793) whose contribution paved new ways to nutrition research.
   
3. In the year 1752 James Lind’s discovered “Scurvy” which could be cured or prevented by eating fresh fruits and vegetables.
4. It was known that diseases could be cured by eating certain kinds of foods.
5. In 19th century it was known that the body obtains food from three substances namely proteins, fats and carbohydrates.

Question 9.
How do you test the presence of starch in the food item given to you.
Answer:
Test for Starch: N – line Preparation of dilute iodine solution

1. Take a test tube or a cup and add few drops of Iodine solution to it.
2. Then dilute it with water till it becomes light yellow / brown in colour.
   
3. Take a sample of food item in the test tube.
4. Add a few drops of dilute Iodine solution on the sample we have collected.
5. Observe the change in colour.
6. If the substance turns dark-blue or black, it contains starch.

Question 10.
Describe how do you test the presence of fats in the food item given to you.
Answer:
Test for fats:

1. Take a small quantity of each sample.
   
2. Rub it gently on a piece of paper.
3. If the paper turns translucent the substance contains fats.

Question 11.
What test do you conduct to detect the presence of proteins in the food item given to you?
Answer:
A) Preparation of solutions:

1. Preparation of 2% copper sulphate solution: To make 2% copper sulphate solution dissolve 2 gms of copper sulphate in 100 ml. of water.
2. To make 10% of sodium hydroxide solution: Dissolve 10 gms of sodium hydroxide in 100 ml. of water.

B) Test for proteins:

1. If the substance to test is a solid, grind it into powder or paste.
2. Add a little of it in the test tube and add 10 drops of water to the powder and stir well.
3. Add 10 drops of this solution in a clean test tube. Add 2 drops of copper sulphate solution and 10 drops of sodium hydroxide solution to the test tube and shake well.
4. Change of colour to violet or purple confirms presence of protein.

Question 12.
What do the above tests confirm?
Answer:

1. The above tests show the presence of components of food which are usually present in larger amounts as compared to others,
2. All types of food that we eat contain all the above-mentioned food components.
3. The quantity of each component varies from type to type.

Question 13.
Collect some food packets like chips, coffee, milk, juice … etc., and put a tick mark if you find the listed food components present in food items.
Answer:
Table: Food items and components

Question 14.
What are the components found in biscuits?
Answer:
In biscuits the following components are present.

1. carbohydrates
2. proteins
3. fat
4. sugars
5. saturated fatty acids
6. mono unsaturated fatty acids
7. poly unsaturated fatty acids
8. trans-fatty acids
9. cholesterol
10. calcium
11. iron
12. iodine
13. vitamins.

Question 15.
What components are most common in your list?
Answer:
The common components in the list are

1. carbohydrates
2. proteins
3. fats
4. sugars
5. minerals and
6. vitamins.

Question 16.
Do you find any vitamins and minerals in the biscuits? What are they?
Answer:

1. I find vitamins and minerals in the biscuits.
2. They are
   a) Vitamin – D, Vitamins B, B₆ and B₁₂
3. a) Calcium, b) Iron, c) Iodine are the minerals present.

Question 17.
Where do you write salt and sugar? Why?
Answer:

1. Salt and sugar are written separately.
2. These two components are not present in all the food items as common.

Question 18.
Are there any food items with similar components?
Answer:
Many ready made food items packed will have similar components.

Question 19.
What are the essential components of food?
Answer:

1. Our food consists of carbohydrates, proteins, fats, vitamins and minerals.
2. Besides these, water and fibres are also present.

Question 20.
Different food items are given in the table below. Find out the type of components in them and fill the information on the basis of your observations.
Answer:
Table: Testing of food items for carbohydrates, proteins, fats

Question 21.
Which foods show the presence of starch?
Answer:

1. Rice (78.2%)
2. Potato (22.6%)
3. Milk (5%)
4. Curd (3%)
5. Egg (0%)

Question 22.
What nutrients are present in milk?
Answer:
In buffalo’s milk

1. Minerals are present (0.8%)
2. Calcium (210 mg per 100 g) is present
3. Phosphorus (130 mg per 100 g) is present
4. Iron (0.2 mg per 100 g) is present.

Question 23.
Which components of food could you identify in potatoes?
Answer:
The following components are identified in potatoes.
a) (1.6 gm per 100 gm) proteins.
b) (0.1 gm per 100 gm) fat
c) (0.6 gm per 100 gm) minerals
d) (0.4 gm per 100 gm) fibre
e) (22.6 gm per 100 gm) carbohydrates
f) (10 mg per 100 gm) calcium
g) (40 mg per 100 gm) phosphorus
h) (0.48 mg per 100 gm) iron

Question 24.
Which food item contain more fat?
Answer:
Buffalo’s milk contains more fat (6.5 gm / 100 gm)

Question 25.
Which food items contain more protein?
Answer:
a) Rice contains proteins of 6.8 gm / 100 gm
b) Potato contains proteins of 1.6 gm / 100 gm.
c) Milk contains proteins of 4.3 gm / 100 gm
d) Curd contains proteins of 3.1 gm / 100 gm.
e) Egg contains proteins of 13.3 gm / 100 gm.
So egg and milk contain more proteins compared to other food items.

Question 26.
Why should we eat food?
Answer:
We should eat food because food supplies the energy we need to do many tasks in our day to day activities.

Question 27.
Do we need energy while sleeping?
Answer:

1. While we are sleeping, we breathe.
2. The circulation of blood in our body goes on.
3. So we need energy while we are sleeping.

Question 28.
Suppose you do not get food for lunch how do you feel?
Answer:

1. If I do not get food for lunch, I shall be very much tired and exhausted with lack of supply of energy at the end of the day.
2. I shall feel very weak. The hunger will be in an alarming condition.
3. I shall be ready to eat anything that is available.

Question 29.
If you do not get food for many days what will happen to you?
Answer:

1. If I do not get food for many days, I shall become very weak and the resistive power of my body decreases very much.
2. The energy resources in my body will also be exhausted.
3. There will be a danger to my life.

Question 30.
Mention some food items which keep you healthy.
Answer:
Dry fruits like dates, plums, raisins, cashew nuts, pistachios, etc., also keep us healthy.

Question 31.
Is balanced diet cheap? Explain.
Answer:

1. Scientists have found out that a balanced diet need not necessarily be costly.
2. Everyone can afford it, even the poor.
3. If a person eats dal, rice, rotis, green vegetables, little oil and jaggery all the food requirements of the body are fulfilled.
4. Just balancing our diet with different kinds of foods is not enough.
5. It should be cooked in a proper way.

Question 32.
List the food items eaten by you yesterday from breakfast to dinner.
a) Does your diet contain all necessary components of food in it.
Answer:
Following is the list of the food items eaten by me yesterday.

a) Yes, my food contains all necessary components of food.

Question 33.
Look at the food ‘THAU’ with many food items and list out the food items and food components in it. You need not eat all items as shown in the “THAU” rather you should ensure that your food contains all food components everyday in adequate quantity. For example, a diet containing food items having more of carbohydrates and protein along with a little fat, vitamins and minerals makes a balanced diet.

Answer:

Question 34.
How and why the nutrients in the food are lost?
Answer:
You know many nutrients are lost by over cooking, re-heating many times, washing the vegetables after cutting them into small pieces.

Question 35.
Write which foods are to be eaten moderately, adequately, plenty and sparingly.
Answer:

1. Foods like cereals, pulses, milk etc. should be taken adequately.
2. Fruits, leafy vegetables and other vegetables should be used in plenty.
3. Cooking oils and animal foods should be used moderately.
4. Vanaspathi, Ghee, Butter, Cheese must be used sparingly.

Question 36.
Why should we avoid Junk foods?
Answer:

1. If we are eating only pizzas and sandwiches daily.
   
2. Our body is being deprived of other food substances.
3. Junk food causes damages to our digestive system.
4. It is better to avoid eating junk food.

Question 37.
On what factors do the food habits of people depend?
Answer:

1. Food habits of the people depend upon climatic conditions and cultural practices of the particular place.
2. We eat rice in large quantities but people living in north India eat chapathies as daily food.
3. Because wheat is grown widely in that region.
4. The way of cooking and eating food also reflects the cultural practices of people.

Question 38.
Kiran wants to know about nutrients in the milk. He added 2 drops of Copper Sulphate solution and 10 drops of Sodium Hydroxide solution to the Milk. What colour he may have observed in it? Which nutrient does he find in the milk?
Answer:

1. Take 10 drops of milk in a clean test tube.
2. Add 2 drops of Copper Sulphate solution and 10 drops of Sodium Hydroxide solution to the test tube and shake well.
3. We observe the change of colour to violet or purple.
4. The colour confirms presence of proteins in milk.

Question 39.
Who need the highest proportion of Proteins in their daily diet-a 15 year old boy, a 55 year old female writer and 35 year old male bank officer? Explain why.
Answer:

1. The boy with an age of 15, requires high quantity of proteins in his diet.
2. Because proteins are very essential for his physical growth.
3. The boy is in adolescence, hence he has rapid physical growth. So, he needs high quantity of proteins in his diet.

Question 40.
We are suffering from various gastro intestinal diseases. What precautions do you take to avoid these diseases?
Answer:

1. We should take healthy nutritious diet.
2. We should consume high fibre content food like leafy vegetables. They have roughages. They are very helpful in preventing constipation.
3. Avoid junk foods completely.
4. We should cultivate healthy food habits.

Question 41.
You did an experiment in your classroom to test the proteins. In that
i) What apparatus did you use in that experiment?
ii) What are the solutions you prepared?
iii) What change did you observe in the colour?
iv) What precautions did you take in this experiment?
Answer:
a) Test tubes, Dropper.
b) 2% Copper Sulphate solution and 10% Sodium Hydroxide solution.
c) Violet or purple colour.
d) Take care of measuring chemicals while preparing the solutions.

Question 42.
What are the major nutrients found in our food?
Answer:
There are three major nutrients present in our body. They are: 1) Carbohydrates, 2) Proteins and 3) Fats.

Question 43.
Define the term balanced diet. Which food items are to be eaten moderately, ade¬quately, plenty and sparingly to make our diet a balanced one?
Answer:

1. The diet that is with all the required nutrients in adequate quantities is called balanced diet.
2. Energy is required according to the age and nature of work.
3. Carbohydrates provide energy and are present in cereals, pulses, milk etc. They should be taken adequately.
4. Vitamins, minerals available in vegetables, fruits and leafy vegetables should be used in plenty.
5. Fats present in cooking oils and animal foods should be used moderately.
6. Vanaspati, ghee, butter and cheese must be used sparingly.

Question 44.
Madhavi eats only biryani and chicken daily. Do you think it is a balanced diet? Why? If not, write a few suggestions to make her diet a balanced diet.
Answer:

1. Madhavi’s diet is not a balanced diet.
2. We should take a balanced diet containing all the nutrients.
3. Madhavi can get only carbohydrates, proteins and fats in large quantities through her meal.
4. This diet leads to obesity in future.
5. Madhavi should take fruits, vegetables, in order to get vitamins and minerals as well as roughages.

Question 45.
How do you test the presence of starch in the food item given to you? (OR) Write the procedure of test for carbohydrates conducted in your classroom.
Answer:
Test for Starch: N – line Preparation of dilute iodine solution

1. Take a test tube or a cup and add few drops of Iodine solution to it.
2. Then dilute it with water till it becomes light yellow/brown in colour.
   
3. Take a sample of food item in the test tube.
4. Add a few drops of dilute Iodine solution on the sample we have collected.
5. Observe the change in colour.
6. If the substance turns dark-blue or black, it contains starch.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 1

Question 1.
Solve the following.
(i) 2 + \(\frac { 3 }{ 4 }\)
(ii) \(\frac{7}{9}+\frac{1}{3}\)
(iii) 1 – \(\frac{4}{7}\)
(iv) \(2 \frac{2}{3}+\frac{1}{2}\)
(v) \(\frac{5}{8}-\frac{1}{6}\)
(vi) \(2 \frac{2}{3}+3 \frac{1}{2}\)
Solution:

Question 2.
Arrange the following in ascending order.
(i) \(\frac{5}{8}, \frac{5}{6}, \frac{1}{2}\)
(ii) \(\frac{2}{5}, \frac{1}{3}, \frac{3}{10}\)
Solution:

Question 3.
Check whether in this square the sum of the numbers in each row and in each column and along the diagonals is the same.

Solution:

Question 4.
A rectangular sheet of paper is 5\(\frac{2}{3}\) cm long and 3\(\frac{1}{5}\) cm wide. Find its perimeter.
Solution:
Length of the rectangular sheet = 5\(\frac{2}{3}\) cm
Breadth/width of the rectangular sheet = 5\(\frac{2}{3}\) cm
Perimeter = 2 x (length + breadth)

Question 5.
The recipe requires 3\(\frac{1}{4}\) cups of flour. Radha has 1\( \frac{3}{8}\) cups of flour. How many more cups of flour does she need?
Solution:
Flour required for the recipe = 3\(\frac{1}{4}\) cups
Flour with Radha = 1\(\frac{3}{8}\) cups
More cups of flour required = \(3 \frac{1}{4}-1 \frac{3}{8}\)
= \(\frac{3 \times 4+1}{4}-\frac{1 \times 8+3}{8}\)

Question 6.
Abdul is preparing for his final exam. He has completed \(\frac{5}{12}\) part of his course content. Find out how much course content is left?
Solution:
Take content as 1 (i.e., full) Course completed = \(\frac{5}{12}\)
Course yet to be completed = 1 – \(\frac{5}{12}\)
= \(\frac{12 \times 1-5}{12}\)
\(\frac{12-5}{12}=\frac{7}{12}\)

Question 7.
Find the perimeters of(i) ΔABE (ii) the rectangle BCDE in this figure. Which figure has greater perimetre and by how much?

Solution:
i) Perimeter of ΔABE = AB + BE + AE

ii) Perimeter of BCDE = 2(BE + BC)

As \(\frac{116}{15}<\frac{153}{15}\), we conclude that the perimetre of ΔABE > Perimeter of BCDE

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 7 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 7

Question 1.
In a class test containing 15 questions, 4 marks are given for every correct answer and (-2) marks are given for every incorrect answer. (i) I3harathi attempts all questions but only 9 answers are correct. What is her total score’? (ii) One of her friends Hema answers only 5 questions correct. What will be her total score?
Solution:
i) Bharathi’s score = 4 × Number of correct answers + (-2) × Number of wrong answers
=4 × 9 + (-2) × 6 = 36 + (-12) = 24
ii) Hema’s score = 4 × (5) + (-2) × 10 = 20 + (-20) = 0

Question 2.
A cement company earns a profit of ₹ 9 per bag of white cement sold and a loss of ₹ 5 per bag of grey cement sold.
(i) The company sells 7000 bags of white cement and 6000 bags of grey cement in a month. What is its profit or loss?
(ii) What is the number of white cement bags it must sell to have neither profit nor loss, ifthe number of grey bags sold is 5400.
Solution:
i) Profit on 7000 white cement bags = 9 × 7000 = ₹ 63,000
Loss on 6000 grey cement bags = 5 × 6000 = ₹ 30,000
Net profit or loss = 63,000 – 30,000 = ₹ 33,000 profit
ii) When there is no loss then
Profit on white cement bags = loss on grey cement bags
Loss on grey cement bags = 5 × 5400 = ₹ 27,000
∴ Number of white cement bags = \(\frac{27,000}{9}\) = 3,000

Question 3.
The temperature at 12 noon was 10°C above zero. ¡fit decreases at the rate of 2°C per hour until midnight. at what time would the temperature be 8°C below zero? What would be the temperature at midnight’?
Solution:
The temperature at 12 noon = 10°C
Given temperature = 8°C below zero = – 8°C
Difference between the temperature = 10 – (- 8°) = 18°C
∴ Time lasted = 18 + 2 = 9 hours
i.e., at 9 o’ clock the temperature would be – 8C
At mid-night the temperature 10 + 12(-2) = 10 + (- 24) = – 14°C

Question 4.
In a class test (+3) marks are given for every correct answer and (-2) marks are given for every incorrect answer and no marks for not attempting any question.
(i) Radhika scored 20 marks. if she has got 12 correct answers, how many questions has she attempted incorrectly?
(ii) Mohini scores (-5) marks in this test, though she has got 7 correct answers.
How many questions has she attempted incorrectly?
Solution:
i) Let the no. of incorrect answers = x
So score = 3 × No. of correct answers + (-2) No. of incorrect answers
Radhika score = 20 marks
⇒ 20 = 3 × 12 + (-2) x ⇒ 20 = 36 – 2x ⇒ 2x = 36 – 20 ⇒ x = 16/2 = 8
∴ No. of incorrect questions = 8

ii) Mohini scores = -5
Marks for correct answers = 7 × 3 = 21
Marks for incorrect answers = -5-21 = -26
∴ Number of incorrect answers = -26 ÷ 2 = -13

Question 5.
An elevator descends into a mine shaft at the rate of 6 meters per minute. If the decent starts from 10 rn above the ground level, how long will it take to reach -350 m.
Solution:
Total length to be descend = -350 – 10 = -360 m
Rate of the speed of the elevator 6m per minute
Total time to descend = \(\frac{360}{6}\) = 60 minutes = 1 hour

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 6 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 6

Question 1.
Fill the following blanks.
(i) -25 ÷ ……………… = 25
(ii) ………………. ÷ 1 = -49
(iii) 50 ÷ 0 =……………….
(iv) 0 ÷ 1 = …………………
Solution:
(i) -25 ÷ -1 = 25
(ii) -49 ÷ 1 = -49
(iii) 50 ÷ 0 = 50 cannot divide by 0
(iv) 0 ÷ 1 = 0

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 5 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 5

Question 1.
Verify the following.
(i) 18 × [7 + (-3)] = [18 × 7] + [18 x (-3)]
(ii) (-21) × [(-4) + (-6)] = [(-21) × (-4) ] + [(-21) × (-6)]
Solution:
(i) 18 × [7 + (-3)] = [18 x 7] + [18 x (-3)]
LHS: 18 × (7 + (-3)] = 18 × (4) = 72
RHS:[18 × 7] + [18 × (-3)] = 126 + (-54) = 72
∴ LHS = RHS

(ii) (-21) × [(-4) + (-6)] = [(-21) × (-4) ] + [(-21) x (-6)]
LHS:(-21) × [(-4) + (-6)] = (-21) × (-10) = 210
RHS: [(-21) × (-4)] + [(-21) × (-6)] = (84) + (126)= 210
∴ LHS=RHS

Question 2.
(1) For any integer a, what is(-1) × a equal to9
(ii) Determine the integer whose product with (-1) is 5
Solution:
i) For any integer a
(- 1) × a = -a
ii) …………. × (-1) = 5
∴ (-5) × (-1) = 5

Question 3.
Find the product, using suitable properties.
(i) 26 × (-48)+(-48) × (-36)
(ii) 8 × 53 × (-125)
(iii) 15 × (-25) × (4) × (-10)
(iv) (-41) × 102
(v) 625 × (-35) +(625) × 65
(vi) 7 × (50 – 2)
(vii) (-17) × (-29)
(viii) (-57) × (-19) + 57
Solution:
(i) 26 × (-48) + (-48) × (-36)
= -48 × [26 + (-36)]
= -48 × (-10)
= 480

(ii) 8 × 53 × (- 125)
424 × (- 125)
= -53000

(iii) 15 × (-25) × (-4) × (-10)
= (-375) × (40)
= -15000

(iv) (-41) × 102
= -4182

(v) 625 × (-35) + (-625) × 65
= 625 × [(-35) + (-65)]
=625 × (-100)
= – 62500

(vi) 7 × (50 – 2)
= 7 × 48
= 336

(vii) (-17) × (-29)
= 493

(viii) (-57) × (-19) + 57
=1083 + 57
= 1140

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 4

Question 1.
Fill in the blanks.
(i) ( – 100) × ( -6) =
(ii) ( – 3) × …………. = 3
(iii) 100 × ( – 6) = ………….
(iv) ( – 20) × (- 10) = ………….
(v) 15 × (-3) = ………….
Solution:
(i) ( – 100) × ( -6) =
(ii) ( – 3) × …………. = 3
(iii) 100 × ( – 6) = ………….
(iv) ( – 20) × (- 10) = ………….
(v) 15 × (-3) = ………….

Question 2.
Find each of the following products.
(i) 3 × ( – 1)
(ii) ( – 1) × 225
(iii) ( – 2 ) × ( – 30)
(iv) ( – 316) × ( – 1)
(v) (-15) × 0 × (-18)
(vi) (-12) × (-11) × (10)
(vii) 9 × ( – 3) × ( – 6)
(viii) ( – 18) × ( – 5) × ( – 4)
(ix) ( – 1) × ( – 2) × ( – 3) × 4
(x) ( – 3) × ( – 6) × ( – 2) × ( – 1)
Solution:
(i) 3 × ( – 1) = -3
(ii) ( – 1) × 225 = -225
(iii) ( – 2 ) × ( – 30) = 630
(iv) ( – 316) × ( – 1) = 316
(v) (-15) × 0 × (-18) = 0
(vi) (-12) × (-11) × (10) = 1320
(vii) 9 × ( – 3) × ( – 6) = 162
(viii) ( – 18) × ( – 5) × ( – 4) = -360
(ix) ( – 1) × ( – 2) × ( – 3) × 4 = -24
(x) ( – 3) × ( – 6) × ( – 2) × ( – 1) = 36

Question 3.
A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?
Solution:
At present, the room temperature 40°C
The total decrease in temperature after 10 hours at the rate of 5°C a hour = 10 × 5 = 50°
∴ Temperature after 10 hours = 40° – 50° = – 10°C.

Question 4.
In a class test containing 10 questions, ‘3’ marks are awarded for every correct answer and ( – 1) mark is for every incorrect answer and ‘0’ for questions not attempted.
(i) Gopi gets 5 correct and 5 incorrect answers. What is his score?
(ii) Reshma gets 7 correct answers and 3 incorrect answers. What is her score?
(iii) Rashmi gets 3 correct and 4 incorrect answers out of seven questions she attempts. What is her score?
Solution:
i) GopI’sscoe =5 × 3 + 5( – 1)= 15 + ( – 5) = 10
ii) Reshmasscore = 7 × 3 + 3 × ( – 1) = 21 +( – 3) = 18
iii) Rashmis score = 3 × 3 + 4( – 1) = 9 + ( – 4) = 5

Question 5.
A merchant on selling rice earns a profit of ₹ 10 per bag of basmati rice sold and a loss of ₹ 5 per bag of 1 non-basmati rice.

(i) He sells 3,000 bags of basmati rice and 5,000 bags ofnon-basmati rice in a month. What is his profit or loss in a month?
(ii) What is the number of basinati rice bags he must sell to have neither profit nor loss, if the number of bags of non-basmati rice sold is 6,400.
Solution:
i) Profit on 3,000 bags of basmati rice = 3,000 × 10 = ₹ 30,000
Loss on 5,000 bags of non-basmati rice = 5,000 × 5 = ₹ 25,000
Net profit or loss = 30,000 – 25,000 = ₹ 5000 profit
Another Method : Net profit or loss = 3,000 × 10 + 5,000(-5)
= 30,000 – 25,000 = ₹ 5,000

ii) As there is no loss or gain
Profit on basmati rice= Loss on non-basmati rice
But loss on nonbasrnati rice = 6,400 × 5 = ₹ 32,000
∴ Number of bags of basmati rice sold = \(\frac{32,000}{10}\) = 3200

Question 6.
Replace the blank with an integer to make it a true statement.
(i) (- 3) × ______________ = 27
(ii) 5 × ______________ = – 35
(iii) _______ × ( – 8) = – 56
(iv) _______ × ( – 12) = 132
Solution:
(i) (- 3) × -9 = 27
(ii) 5 × -7 = – 35
(iii) 7 × ( – 8) = – 56
(iv) 11 × ( – 12) = 132

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 3

Question 1.
Represent the following subtractions on the number line.
(i) 7 – 2
(ii) 8 – ( – 7)
(iii) 3 – 7
(iv) 1 5 – 14
(v) 5 – ( – 8)
(vi) ( – 2) – ( – 1)
Solution:
(i) 7 – 2 = 5

(ii) 8 – ( – 7) = 15

(iii) 3 – 7 = -4

(iv) 1 5 – 14 = 1

(v) 5 – ( – 8) = 13

(vi) ( – 2) – ( – 1) = -2 + 1 = -1

Question 2.
Solve the following.
(i) 17 – ( – 14)
(ii) 13 – ( – 8)
(iii) 19 – ( – 5)
(iv) 15 – 28
(v) 25 – 33
(vi) 80 – ( – 50)
(vii) 150 – 75
(viii) 32 – ( – 18)
Solution:
(i) 17 – ( – 14) = 31
(ii) 13 – ( – 8) = 21
(iii) 19 – ( – 5) = 24
(iv) 15 – 28 = -13
(v) 25 – 33 = -8
(vi) 80 – ( – 50) = 130
(vii) 150 – 75 = 75
(viii) 32 – ( – 18) = 50

Question 3.
Express ‘ – 6’ as the difference between a negative integer and a whole number.
Solution:
– 6=( – 4) – (2) ( – 5) – (1)=( – 2) – (4)