Category: Class 7

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 3

Question 1.
Find each of the following products.
(i) \(\frac{5}{6} \times \frac{7}{11}\)
(ii) \(6 \times \frac{1}{5}\)
(iii) \(2 \frac{1}{3} \times 3 \frac{1}{5}\)
Solution:

Question 2.
Multiply and reduce to lowest form.
(i) \(\frac{2}{3} \times 5 \frac{1}{5}\)
(ii) \(\frac{2}{7} \times \frac{1}{3}\)
(iii) \(\frac{9}{3} \times \frac{5}{5}\)
Solution:

Question 3.
Which one is greater’?
(i) \(\frac{2}{5} \text { of } \frac{4}{7} \text { or } \frac{3}{4} \text { of } \frac{1}{2}\)
(ii) \(\frac{1}{2} \text { of } \frac{4}{7} \text { or } \frac{2}{3} \text { of } \frac{3}{7}\)
Solution:
(i) \(\frac{2}{5} \text { of } \frac{4}{7} \text { or } \frac{3}{4} \text { of } \frac{1}{2}\)

(ii) \(\frac{1}{2} \text { of } \frac{4}{7} \text { or } \frac{2}{3} \text { of } \frac{3}{7}\)

Question 4.
Rehana works 2\(\frac{1}{2}\) hours each day on her embroidery. She completes the work in 7 days.
How many hours did she take to complete her work?
Solution:
Number of hours Rehana worked each day = 2\(\frac{1}{2}\) hrs
Number of days Rehana worked on embroidery = 7 days
Total time worked = 2½ x 7 = \(\frac{5}{2}\) x 7 = \(\frac{5 \times 7}{2}=\frac{35}{2}\) = = 17\(\frac{1}{2}\) hrs.

Question 5.
A truck runs 8 km using 1 litre of petrol. How much distance will it cover using 10\(\frac{2}{3}\) litres of petrol?
Solution:
bistance covered wIth 1 litre of petrol 8 km
∴ Distance covered for 10\(\frac{2}{3}\) litres petrol = 10\(\frac{2}{3}\) x 8
= \(\frac{32}{3} \times 8=\frac{32 \times 8}{3}=\frac{256}{3}=85 \frac{1}{3}\)km.

Question 6.
Raja walks 1\(\frac{1}{2}\) meters in 1 second. How much distance will he walk in 15 minutes?
Solution:
Distance covered by Raja in 1 second = 1\(\frac{1}{2}\) m
Distance covered by Raja in 15 minutes 15 x 60 x 1\(\frac{1}{2}\)
= 900 x \(\frac{3}{2}\) = \(\frac{900\times 3}{21}\) =1350 m

Question 7.
Provide the number in the box to make the statement true.

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

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

Question 1.
Multiply the following. Write the answer as a mixed fraction.
(i) \(\frac { 3 }{ 6 }\) x 10
(ii) \(\frac { 1 }{ 3 }\) x 4
(iii) \(\frac { 6 }{ 7 }\) x 2
(iv) \(\frac { 2 }{ 9 }\) x 5
(v) 15 x \(\frac { 2 }{ 5 }\)
Solution:
(i) \(\frac { 3 }{ 6 }\) x 10 = \(\frac { 30 }{ 6 }\) = 5 i.e., integer part 5; fraclon part = 0
(ii) \(\frac { 1 }{ 3 }\) x 4 = \(\frac { 4 }{ 3 }\) = 1\(\frac { 1 }{ 3 }\)
(iii) \(\frac { 6 }{ 7 }\) x 2 = \(\frac { 12 }{ 7 }\) = 1\(\frac {5 }{ 7 }\)
(iv) \(\frac { 2 }{ 9 }\) x 5 = \(\frac { 10 }{ 9 }\) = 1\(\frac { 1 }{ 9 }\)
(v) 15 x \(\frac { 2 }{ 5 }\) = \(\frac { 30 }{ 5 }\) = 6 i.e., integer part = 6; fraction part = o

Question 2.
Shade: (i) \(\frac { 1 }{ 2 }\) of the circles in box (a)
(ii) \(\frac { 2 }{ 3 }\) of the triangles in box (b)
(iii) \(\frac { 3 }{ 5 }\)of the rectangles in box (c)
(iv) \(\frac { 3 }{ 4 }\) of the circles in box (d)


Solution:

Question 3.
Find (i) \(\frac { 1 }{ 2 }\) of 12 (ii) \(\frac { 2 }{ 5 }\) of 15
Solution:
(i) \(\frac { 1 }{ 2 }\) of 12 = \(\frac { 1 }{ 2 }\) x 12 = \(\frac { 12 }{ 3 }\) = 4

(ii) \(\frac { 2 }{ 5 }\) of 15 = \(\frac { 2 }{ 5 }\) x 15 = \(\frac { 30 }{ 5 }\) = 6

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)

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 2

Question 1.
Represent the following additions on a number line.
(i) 5 + 7
(ii) 5 + 2
(iii) 5 + ( – 2)
Solution:

Question 2.
Compute the following.
(i) 7 + 4
(ii) 8+( – 3)
(iii) 11 + 3
(iv) 14 + ( – 6)
(v) 9 + ( – 7)
(vi) 14+( – 10)
(vii) 13 + ( – 15)
(viii) 4 + ( – 4)
(ix) 10 +( – 2)
(x) 100 +( – 80)
(xi) 225 +( – 145)
Solution:
(i) 7 + 4 = 11
(ii) 8 +( – 3) = 5
(iii) 11 + 3 = 14
(iv) 14 + ( – 6) = 8
(v) 9 + ( – 7) = 2
(vi) 14 +( – 10) = 4
(vii) 13 + ( – 15) = -2
(viii) 4 + ( – 4) = 0
(ix) 10 +( – 2) = 8
(x) 100 +( – 80) = 20
(xi) 225 +( – 145) = 80

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

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

Question 1.
Some integers are marked on the number line. Which is the biggest and which is the
smallest?

Solution:
The numbers marked = -3 and 2
biggest number = 2
smallest number = -3

Question 2.
Write the integers between the pairs of integers given below. Also, choose the biggest and
smallest intergers from them.
(i) – 5, – 10
(ii) 3, – 2
(iii) – 8, 5
Solution:
i) – 5, – 10
Integers between – 5, – 10 are – 6, – 7, – 8, – 9
smallest Integer – 9
biggest integer = – 6

ii) 3, – 2
Integers between 3 and – 2 are – 1,0, 1, 2
smallest integer = – 1
biggest integer 2

iii) – 8, 5
Integers between – Sand 5 are 4, 3, 2, 1, 0, – 1, – 2, – 3, – 4, – 5, – 6, – 7
smallest integer = – 7
biggest integer = 4

Question 3.
Write the following integers in ascending order (smallest to biggest).
(i) – 5, 2, 1, – 8
(ii) – 4, – 3, – 5, 2
(iii) – 10, – 15, – 7
Solution:
(i) – 8. – 5, 1, 2
(ii) – 5, – 4, – 3, 2
(iii) – 15, – 10, – 7

Question 4.
Write the following integers in descending order (biggest to smallest).
(i) – 2, – 3, – 5
(ii) – 8, – 2, – 1
(iii) 5, 8, – 2
Solution:
(i) – 2, – 3, – 5
(ii) – 1, – 2, – 8
(iii) 8, 5, – 2

Question 5.
Represent 6, – 4, 0 and 4 on a number line.
Solution:

Question 6.
Fill the missing integers on the number line given below

Solution:

Question 7.
The temperatures of 5 cities in india on a particular day are shown on the number line below.

(i) Write the temperatures of the cities marked on it?
(ii) Which city has the highest temperature?
(iii) Which city has the lowest temperature?
(iv) Which cities have temperature less than 0°C?
(v) Name the cities with temperature more than 0°C?
Solution:
i) The temperatures of the cities shown on the nL.lber line an
Kasauli
Manali = – 7°C
Nainital = – 3°C
Ooty = 15°C
l3anglore = 20°C

ii) The city with highest temperature is Banglore.
iii) The city with lowest temperature is Kasauli.
iv) The cities with temperatures less than 0°C are Nainital, Manali and Kasauli.
v) The cities with temperatures more than 0°C are Ooty and Banglore.