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AP State Syllabus AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic Ex 6.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic Ex 6.4

Question 1.
Express each of the following percents as fractions in the simplest form,
a) 15%
b) 35%
c) 50%
d) 75%
Answer:
a) Given percent is 15%
= \(\frac{15}{100}\) = \(\frac{3}{20}\) (fraction form) [∵ To express a percentage as a fraction first drop the symbol % and then divide it by 100.]

b) Given percent is 35%
= \(\frac{35}{100}\) = \(\frac{7}{20}\) (fraction form)

c) Given percent is 50%
= \(\frac{50}{100}\) = \(\frac{1}{2}\) (fraction form)

d) Given percent is 75%
= \(\frac{75}{100}\) = \(\frac{3}{4}\) (fraction form)

Question 2.
Express each of the following fractions as percents.
a) \(\frac{15}{2}\)
b) 8\(\frac{1}{4}\)
c) 5\(\frac{3}{4}\)
d) 3\(\frac{1}{3}\)
Answer:
a) Given number \(\frac{15}{2}\) is in fraction form.
To convert the fraction into percent we have to multiply the fraction by 100%
\(\frac{15}{2}\) = \(\frac{15}{2}\) × 100 = 15 × 50 = 750% (percent form)

b) Given number 8\(\frac{1}{4}\) is in fraction form.
To convert the fraction into percent we have to multiply the fraction by 100%
8\(\frac{1}{4}\) = \(\frac{33}{4}\) = \(\frac{33}{4}\) × 100 = 33 × 25 = 825%,(percent form).

c) Given number 5\(\frac{3}{4}\) is in fraction form.
To convert the fraction into percent we have to multiply the fraction by 100 and assign the percentage symbol %
5\(\frac{3}{4}\) = \(\frac{23}{4}\) × 100 = 23 × 25 = 575% (percent form)

d) Given number is 3\(\frac{1}{3}\) in fraction form.
To convert the fraction into percent we have to multiply the fraction by 100%
3\(\frac{1}{3}\) = \(\frac{10}{3}\) × 100 = \(\frac{1000}{3}\) = 333\(\frac{1}{3}\) % (percent form)

Question 3.
Express each of the following ratios as percents.
(a) 3 : 5 (b) 5 : 8 (c) 2.5 : 55 (d) 4 : 36
Answer:
a) 3 : 5
Given ratio is 3 : 5.
3 : 5 = \(\frac{3}{5}\) (fraction form)
To convert the fraction into percent we have to multiply the fraction by 100%
\(\frac{3}{5}\) = \(\frac{3}{5}\) × 100 = 60% (percent form)

b) 5 : 8
Given ratio is 5 : 8
5 : 8 = \(\frac{5}{8}\) (fraction form)
To convert the fraction into percent we have to multiply the fraction by 100%
\(\frac{3}{5}\) = \(\frac{3}{5}\) × 100 = 62 \(\frac{1}{2}\) % (percent form)

c) 2.5 : 55
Given ratio is 2.5 : 55
2.5 : 55 = \(\frac{2.5}{55}\)
To convert the fraction into percent we have to multiply the fraction by 100%

d) 4 : 3 6
Given ratio is 4 : 36
4 : 36 = \(\frac{4}{36}\) (fraction form)
To convert the fraction into percent we have to multiply the fraction by 100%
\(\frac{4}{36}\) = \(\frac{4}{36}\) × 100 = \(\frac{100}{9}\) = 11 \(\frac{1}{9}\) % (percent form)

Question 4.
Express each of the following percents as ratios in the simplest form.
(a) 12% (b) 25% (c) 45% (d) 84%
Answer:
a) Given percent is 12%
= \(\frac{12}{100}\) = \(\frac{3}{25}\) (fraction form)
= 3 : 25 (ratio form)

b) Given percent is 25%
= \(\frac{25}{100}\) = \(\frac{1}{4}\) (fraction form)
= 1 : 4 (ratio form)

c) Given percent is 45%
= \(\frac{45}{100}\) = \(\frac{9}{20}\) (fraction form)
= 9 : 20 (ratio form)

d) Given percent is 84%
= \(\frac{84}{100}\) = \(\frac{21}{25}\) (fraction form)
= 21 : 25 (ratio form)

Question 5.
Express each of the following percents as decimals,
(a) 1% (b) 6% (c) 19% (d) 67%
Answer:
a) Given percent is 1%
= 1% = \(\frac{1}{100}\) (fraction form)
= 0.01 (decimal form)

b) Given percent is 6%
= 6% = \(\frac{6}{100}\) = \(\frac{3}{50}\) (fraction form)
= 0.06 (decimal form)

c) Given percent is 19%
= 19% = \(\frac{19}{100}\) (fraction form)
= 0.19 (decimal form)

d) Given percent is 67%
= 67% = \(\frac{67}{100}\)
= 0.67 (decimal form)

Question 6.
Express each of the following decimals as percent.
(a) 0.04 (b) 0.52 (c) 0.125 (d) 0.0006
Answer:
a) Given decimal number is 0.04.
Convert this into fraction

b) Given decimal number is 0.52.
Convert this into fraction

c) Given decimal number is 0.125.
Convert this into fraction

d) Given decimal number is 0.0006.
Convert this into fraction

Question 7.
Find the number, which is 12\(\frac{1}{2}\)% of 75.
Answer:
Given 12\(\frac{1}{2}\)% of 75
We know that x% of y = \(\frac{x}{100}\) × y

(OR)

Question 8.
Pavani secured 85% marks in mathematics paper. If maximum marks in the paper are 80, find the marks secured by her in that paper.
Answer:
Given maximum marks of the paper = 80
Percentage of marks secured by Pavani = 85%
Marks secured by Pavani

∴ Marks secured by Pavani = 68

Question 9.
Siva spends 78% of his monthly income. If he saves Rs. 7,700/- per month, what is his monthly income?
Answer:
Let Siva’s monthly income = Rs. x
Percentage of income spent = 78%
Percentage of income saved = 100% – percentage of spent income
= 100% – 78% = 22%
Saved income = 22% of monthly income = 7700
= 22% of x = 7700
⇒ \(\frac{1}{2}\) × x = 7700
⇒ x = \(\frac{7700×100}{2}\)
⇒ x = Rs. 35000
∴ Siva’s monthly income = Rs. 35000

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic Ex 6.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic Ex 6.3

Question 1.
If the cost of 3 apples is Rs. 60/-, then find the cost of 7 apples.
Answer:
Cost of 3 apples = Rs. 60
Cost of 1 apple = \(\frac{60}{3}\) = Rs. 20
∴ Cost of 7 apples = 7 × Rs. 20 = Rs. 140

Question 2.
Uma bought 8 books for a total of ? 120. How much would she pay for just 5 books?
Answer:
Cost of 8 books = Rs. 120
Cost of 1 book = \(\frac{120}{8}\) = Rs. 15
∴ Cost of 5 books = 5 × 15 = Rs. 75

Question 3.
The cost of 5 fans is Rs. 11,000/-. Find the number of funs that can be purchased for Rs. 4,400/-.
Answer:
Cost of 5 fans = Rs. 11000
Cost of 1 fan = \(\frac{11000}{5}\) = Rs. 2200
Number of fans can be purchased for Rs. 4400 = 4400 ÷ cost of 1 fan = \(\frac{4400}{2200}\) = 2

Question 4.
A car is moving at a constant speed covers a distance of 180 km in 3 hours. Find the time taken by the car to cover a distance of 420 km at the same speed.
Answer:
Distance covered = 180 km
Time taken = 3 Hrs.
Distance to be covered = 420 km
Time required = ? = x (say)
Since the car is moving at a constant speed, we have
180 : 3 is as to 420 : x
180 : 3 :: 420 : x
Its a proportion.
So, product of extremes = Product of means
180 × x = 3 × 420

∴ Time required = 7 Hrs.
(OR)
Distance covered by a car in 3 hours = 180 km
Distance covered by a car in 1 hour = 180 ÷ 3
Time taken to cover 60 km distance = \(\frac{180}{3}\) = 60 km
Time taken to cover 420 km distance = 420 ÷ 60 = \(\frac{420}{60}\) = 7 hours

Question 5.
A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Answer:
108 : 594 :: x : 1650
So, 108 × 1650 = x × 594
∴ x = \(\frac{108 \times 1650}{\hline 594}\) = 300 litres
(OR)
Distance covered by truck with 108 litres diesel = 594 km
Distance covered by truck with 1 litre of diesel = \(\frac{594}{108}\) km
Diesel required to cover 1650 km distance = 1650 ÷ \(\frac{594}{108}\) = 1650 × \(\frac{108}{594}\)
Diesel required to cover 1650 km distance = 300 litres.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic Ex 6.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic Ex 6.2

Question 1.
Check whether the following are in proportion? or not?
(a) 10, 12, 15, 18
(b) 11, 16, 16, 21
(c) 8, 13, 17, 19
(d) 30, 24, 20, 16
Answer:
a) 10, 12, 15, 18
Given numbers are 10, 12, 15, 18.
If a, b, c, d are in proportion i.e., a : b :: c : d,
then product of extremes = product of means (a.d = b.c)

a : b :: c : d
10 × 18 = 12 × 15
180 = 180
If the product of extremes is equal to the product of means, then the given numbers are in proportion.
So, 10, 12, 15 and 18 are in proportion,

b) 11, 16, 16, 21
Given numbers are 11, 16, 16, 21.
If a, b, c, d are in proportion i.e., a : b :: c : d, then product of extremes = product of means (a.d = b.c)

11 × 21 = 16 × 16
231 ≠ 256
Here the product of extremes is not equal to the product of means, then the given numbers are not in proportion.
So, 11,16, 16 and 21 are not in proportion.

c) 8, 13, 17, 19
Given numbers are 8, 13, 17,19.
If a, b, c, d are in proportion i.e., a : b :: c : d,
then the product of extremes = product of means (a.d = b.c)

8 × 19 = 13 × 17
152 ≠ 221
Here the product of extremes is not equal to the product of means, then the given numbers are not in proportion.
So, 8,13, 17 and 19 are not in proportion.

d) 30, 24, 20, 16
Given numbers are 30, 24, 20, 16.
If a, b, c, d are in proportion i.e., a : b :: c : d,
then the product of extremes = product of means (a.d = b.c)
30 : 24 :: 20 : 16

480 = 480
Here the product of extremes is equal to the product of means, then the given numbers are in proportion.
So, 30, 24, 20 and 16 are in proportion.

Question 2.
Write true or false for each of the following.
a) 4 : 2 :: 14 : 7
b) 21 : 7 :: 15 : 5
c) 13 : 12 :: 12 : 13
d) 5 : 6 :: 7 : 8
Answer:
a) 4 : 2 :: 14 : 7
Given proportion is 4 : 2 :: 14 : 7
If a, b, c, d are in proportion, then a.d = b.c

4 × 7 = 2 × 14
28 = 28
So, the given proportion is true.

b) 21 : 7 :: 15 : 5
Given proportion is 21 : 7 : : 15 : 5
If a, b, c, d are in proportion, then a.d = b.c

21 × 5 = 7 × 15
105 = 105
So, the given proportion is true.

c) 13 : 12 :: 12 : 13
Given proportion is 13 : 12 :: 12 : 13
If a, b, c, d are in proportion, then a.d = b.c

13 × 13 = 12 × 12
169 ≠ 144
So, the given proportion is false.

d) 5 : 6 :: 7 : 8
Given proportion is 5 : 6 :: 7 : 8
If a, b, c, d are in proportion, then a.d = b.c

5 × 8 = 6 × 7
40 ≠ 42
So, the given proportion is false.

Question 3.
Check whether the following form a proportion? Write middle terms and extremes where the ratios form a proportion.
a) 15 cm, 1 m and Rs. 45, Rs. 300
b) 20 ml, 2 l and Rs. 100, Rs. 10,000
Answer:
a) 15 cm, 1 m and Rs. 45, Rs. 300
Given values are 15 cm, 1 m and Rs. 45, Rs. 300
Ratio of lengths = 15 cm : 1 m (convert metres into cm)
= 15 cm : 100 cm = 3 : 20 (write in simplest form)
Ratio of amounts = Rs. 45 : Rs. 300 = 45 : 300 = 3 : 20
The two ratios are equal.
So, 3 : 20 :: 3 : 20 They are in proportion.
Middle terms are 20, 3 and extremes are 3, 20.

b) 20 ml, 2 l and Rs. 100, Rs. 10,000
Given values are 20 ml, 2 l and Rs. 100, Rs. 10,000
Ratio of quantities = 20 ml : 2 l (convert litres into ml)
= 20 : 2000 (write in simplest form)
= 1 : 100
Ratio of amounts = Rs. 100 : Rs. 10,000 (write in simplest form)
= 1 : 100
The two ratios are equal.
So, 1 : 100 = 1 : 100
∴ They are in proportion.
Middle terms are 100, 1 and extremes are 1, 100.

Question 4.
Find the missing numbers in the following proportions.
a) 8 : 12 :: : 48
b) 15 : :: 105 : 98
c) 34 : 102 :: 27 :
Answer:
a) 8 : 12 :: : 48
Given proportion is 8 : 12 :: : 48
Let the missing number = x
If the given numbers are in proportion, then
Product of extremes = Product of means

8 × 48 = 12 × x (or) 12x = 8 × 48

∴ Missing number in the proportion = 32

b) 15 : :: 105 : 98
Given proportion is 15 : : 105 : 98
Let the missing number = a
If the given numbers are in proportion, then Product of extremes = product of means

15 × 98 = 105 × a (or) 105 a = 15 × 98

∴ a = 14
Missing number in the proportion

c) 34 : 102 :: 27 :
Given proportion is 34 : 102 :: 27 :
Let the missing number = y
If the given numbers are in proportion,
then the product of extremes = product of means

Missing number in the proportion = 81

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic Ex 6.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic Ex 6.1

Question 1.
Express the following in the terms of ratios.
i) The length of a rectangle is 5 times to its breadth.
ii) For preparing coffee, 2 cups of water require to 1 cup of milk.
Answer:
i) Let the breadth of rectangle = x = 1 part
length of rectangle = 5x = 5 parts
Ratio = length : breadth = 5x : x = \(\frac{5x}{x}\) = \(\frac{5}{1}\) = 5 : 1

ii) To prepare coffee,
Required cups of water = 2
Required cups of milk = 1
Ratio = water : milk = 2 : 1

Question 2.
Express the following in the simplest form.
i) 24 : 9
ii) 144 : 12
iii) 961 : 31
iv) 1575 : 1190
Answer:
i) 24 : 9
Given ratio is 24 : 9
To write the given ratio in the simplest form.
First find the HCF of two terms.

HCF of 24 and 9 is 3.
Then, divide each term by their HCF.
Simplest form of the ratio = 24 ÷ 3 : 9 ÷ 3
Required ratio = 8 : 3

ii) 144 : 12
Given ratio is 144 : 12
To write the given ratio in the simplest form.
First find the HCF of two terms.

HCF of 144 and 12 is 12.
Then, divide each term by their HCF.
Simplest form of the ratio = 144 ÷ 12 : 12 ÷ 12
Required ratio = 12 : 1

iii) 961 : 31
Given ratio is 961 : 31
To write the given ratio in the simplest form.
First find the HCF of two terms.

HCF of 961 and 31 is 31.
Then, divide each term by their HCF.
Simplest form of the ratio = 961 ÷ 31 : 31 ÷ 31
Required ratio = 31 : 1

iv) 1575 : 1190
Given ratio is 1575 : 1190
To write the given ratio in the simplest form.
First find the HCF of two terms.

HCF of 1575 and 1190 is 35.
Then, divide each term by their HCF.
Simplest form of the ratio = 1575 ÷ 35 : 1190 ÷ 35
∴ Required ratio = 45 : 34

Question 3.
Write the antecedents and consequents of the following ratios.
(i) 36 : 73
(ii) 65 : 84
(iii) 58 : 97
(iv) 69 : 137
Answer:

Question 4.
Find the ratios of the following in their simplest form.
i) 25 minutes to 55 minutes
ii) 45 seconds to 30 minutes
iii) 4 m 20 cm to 8 m 40 cm
iv) 5 litres to 0.75 litres
v) 4 weeks to 4 days
vi) 5 dozen to 2 scores (1 score = 20 items)
Answer:
i) 25 minutes to 55 minutes
Given 25 minutes to 55 minutes = 25 : 55

HCF of 25 and 55 is 5 = 25 ÷ 5 : 55 ÷ 5 (Divide each term by 5)
∴ Required ratio = 5:11

ii) 45 seconds to 30 minutes
Given 45 seconds to 30 minutes.
To write the given ratio in the simplest form.
First convert the two terms into same units.
1 minute = 60 seconds
30 minutes = 30 × 60 = 1800 seconds
Then, find the HCF of two terms and divide them by their HCF.

HCF of 45 and 1800 is 45.
= 45 ÷ 45 : 1800 ÷ 45 (divide each term by 45)
∴ Required ratio = 1 : 40

iii) 4 m 20 cm to 8 m 40 cm
Given 4 m 20 cm to 8 m 40 cm
To write the given ratio in the simplest form.
First we convert them into cm.
1 m = 100 cm
4 m = 400 cm
8 m = 800 cm
4 m 20 cm = 400 + 20 = 420 cm
8 m 40 cm = 800 + 40 = 840 cm
Then, find the HCF of 420 and 840 and divide them by HCF.

HCF of 420 and 840 is 420.
= 420 ÷ 420 : 840 ÷ 420 (divide each term by 420)
Required ratio = 1 : 2.

iv) 5 litres to 0.75 litres
Given 5 litres to 0.75 litres
To write the given ratio in the simplest form, first we convert litres into millilitres.
1 litre = 1000 ml
5 litres = 5000 ml
0.75 litres = 750 ml
Then, find the HCF of 5000; 750 and divide them by HCF.

HCF of 5000 and 750 is 250,
= 5000 ÷ 250 : 750 ÷ 250 (divide each term by 250)
Required ratio = 20 : 3

v) 4 weeks to 4 days
Given 4 weeks to 4 days.
To write the given ratio in the simplest form, first we convert them into same units (days).
1 week = 7 days
4 weeks = 4 × 7 = 28 days
Then, find the HCF of 28 and 4, then divide them by HCF.

HCF of 28 and 4 is 4 = 28 ÷ 4 : 4 ÷ 4 (divide each term by 4)
Required ratio = 7 : 1

vi) 5 dozen to 2 scores (1 score = 20 items)
Given 5 dozen to 2 scores.
To write the given ratio in the simplest form, first we convert them into same units.
1 dozen = 12 items
5 dozens = 5 × 12 = 60 items
1 score = 20 items
2 scores = 2 × 20 = 40 items
Then, find the HCF of 60 and 40, divide them by HCF.

HCF of 60 and 40 is 20 = 60 ÷ 20 : 40 ÷ 20 (divide each term by 20)
Required ratio = 3 : 2

Question 5.
Rahim works in a software company and earns Rs. 75,000/- per month. He saves Rs. 28,000/- per month from his earnings. Find the ratio of
i) His savings to his income
ii) His income to his expenditure
iii) His savings to his expenditure
Answer:
Given Rahim’s monthly income = Rs. 75000
monthly savings = Rs. 28000
Monthly expenditure = income – savings
= 75000 – 28000 = 47,000/-
i) Ratio of savings to income = 28000 : 75000
To convert the ratio into the simplest form divide each term by their HCF is 1000
= 28000 ÷ 1000 : 75000 ÷ 1000 (divide by 1000)
= 28 : 75
Required ratio of savings to income = 28 : 75

ii) Ratio of income to expenditure = 75000 : 47000
To convert the ratio into the simplest form divide each term by their HCF is 1000 = 75000 ÷ 1000 : 47000 ÷ 1000 (divide by 1000)
= 75 : 47
Required ratio of income to expenditure = 75 : 47

iii) Ratio of savings to expenditure = 28000 : 47000
To convert the ratio into the simplest form divide each term by their HCF is 1000 = 28000 ÷ 1000 : 47000 ÷ 1000 (divide by 1000)
= 28 : 47
Required ratio of savings to expenditure = 28 : 47

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Unit Exercise Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Unit Exercise

Question 1.
The sum of two fractions is 5\(\frac{3}{9}\). If one fraction is 2\(\frac{3}{4}\), then find the other fraction.
Answer:
Given one fraction = 2\(\frac{3}{4}\) = \(\frac{11}{4}\)
Sum of two fractions = 5\(\frac{3}{9}\) = \(\frac{48}{9}\)
Now, \(\frac{11}{4}\) + second fraction = \(\frac{48}{9}\)
Second fraction = \(\frac{48}{9}\) – \(\frac{11}{4}\)
LCM of 9 and 4 is 36

Question 2.
A rectangle sheet of paper is of length 12\(\frac{1}{2}\) and breadth 10\(\frac{2}{3}\). Find its perimeter.
Answer:
Length of rectangular sheet = l = 12\(\frac{1}{2}\) = \(\frac{25}{2}\) m
Breadth of rectangular sheet = b = 10\(\frac{2}{3}\) = \(\frac{32}{3}\) m
Perimeter of rectangular sheet = 2(1 + b)

Question 3.
Simplify: \(\left(3 \frac{1}{6}-1 \frac{1}{3}\right)+\left(4 \frac{1}{6}-2 \frac{1}{3}\right)\)
Answer:
\(\left(3 \frac{1}{6}-1 \frac{1}{3}\right)+\left(4 \frac{1}{6}-2 \frac{1}{3}\right)\)
First convert the given mixed fractions into improper fractions.
\(\left(\frac{19}{6}-\frac{4}{3}\right)\) + \(\left(\frac{25}{6}-\frac{7}{3}\right)\)
Then subtract the fractions which are in brackets by making them like fractions.

Now add the fractions = \(\frac{11+11}{6}\) = \(\frac{22}{6}\) = 3\(\frac{4}{6}\)

Question 4.
By what number should 3\(\frac{1}{16}\) be multiplied to get 9\(\frac{3}{16}\)?
Answer:
Given fraction = 3\(\frac{1}{16}\) = \(\frac{49}{16}\)
Product = 9\(\frac{3}{16}\) = \(\frac{147}{16}\)
Fraction × fraction to be multiplied = \(\frac{147}{16}\)
\(\frac{49}{16}\) × other number = \(\frac{147}{16}\)
other number = \(\frac{147}{16}\) ÷ \(\frac{49}{16}\) = \(\frac{147}{16}\) × \(\frac{16}{49}\)
Other number = 3
∴ Number to be multiplied = 3.

Question 5.
The length of the staircase is 5\(\frac{1}{2}\) m. If one step is set at \(\frac{1}{4}\) m. then, how many steps will be there in the staircase?
Answer:
Given length of staircase = 5\(\frac{1}{2}\) m = \(\frac{11}{2}\) m
Length of each step = \(\frac{1}{4}\) m
Length of each step × number of steps = length of staircase
\(\frac{1}{4}\) × Number of steps = \(\frac{11}{2}\)
Number of steps = \(\frac{11}{2}\) ÷ \(\frac{1}{4}\) = \(\frac{11}{2}\) × \(\frac{4}{1}\) = 11 × 2 = 22
∴ Number of steps in the staircase = 22

Question 6.
Simplify: 23.5 – 27 + 35.4 – 17
Answer:
Given 23.5 – 27 + 35.4 – 17
First make them like decimals.
= 23.5 – 27.0 + 35.4 – 17.0
Add the positive numbers
= (+23.5 + 35.4) + (-27.0 – 17.0)
= 58.9 + (-44.0)
= 58.9 – 44.0
= 14.9

Question 7.
Sailaja bought 3.350kg of potatoes, 2.250kg of tomatoes and some onions. If the weight of the total items are 10.250 kg. Then, find the weight of onions.
Answer:
Weight of potatoes = 3.350 kg
Weight of tomatoes = 2.250 kg
Weight of total items = 10.250 kg
Weight of onions = ?
Weight of (potatoes + tomatoes + onions) = 10.250 kg
(3.350 + 2.250) + onions weight = 10.250
5.600 + onions weight = 10.250
Onions weight = 10.250 – 5.600
∴ Weight of onions = 4.650 kg.

Question 8.
What should be subtracted from 7.1 to get 0.713?
Answer:
Given number = 7.1
Difference = 0.713
Number to be subtracted = Number – difference
= 7.1 – 0.713
= 7.100 – 0.713
∴ Number to be subtracted = 6.387.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Ex 5.5 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Ex 5.5

Question 1.
Add the following:
i) 5.702, 5.2, 6.04 and 2.30
ii) 40.004; 44.444; 40.404 and 4.444
Answer:
i) 5.702, 5.2, 6.04 and 2.30
Given decimals fractions are 5.702; 5.2; 6.04 and 2.30
Convert the given decimals into like decimals.
5.702; 5.200; 6.040; 2.300
Then write the decimals in column with the decimal points directly below each other.
So, that tenth’s come under tenths, hundredths come under hundredths and so on, then add.

ii) 40.004; 44.444; 40.404 and 4.444
Given decimals are 40.004; 44.444; 40.404 and 4.444
They are like decimal fractions.
So, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on, then add.

Question 2.
Do the following:
i) 426.326 – 284.482
ii) 5 – 3.009
iii) 2.107 – 0.31
Answer:
i) 426.326 – 284.482
Given decimals are 426.326 and 284.482.
They are like fractions.
So, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on, then subtract.

ii) 5 – 3.009
Given decimals are 5 and 3.009 Make them as like fractions.
5.000 – 3.009
So, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on. Then subtract.

iii) 2.107 – 0.31
Given decimals are 2.107 and 0.31, make them as like fractions.
2.107 – 0.310
So, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on. Then subtract.

Question 3.
Akshara bought 3 m 40 cm cloth for her shirt and 1 m 10 cm cloth for skirt. Find the total cloth bought by her.
Answer:
Length of cloth bought for shirt = 3.40
Length of cloth bought for skirt = 1.10
Total length of cloth bought by Akshara = 3.40 + 1.10 = 4.50 cm

Question 4.
Write in decimals using the units written in brackets.
i) 90 rupees 75 paisa (Rs.)
ii) 49 m 20 cm (m)
iii) 12 kg 450 g (kg)
iv) 50 l 500 ml (l)
Answer:
i) 90 rupees 75 paisa = Rs. 90.75
ii) 49 m 20 cm = 49.20 m
iii) 12 kg 450 g = 12.450 kg
iv) 50 l 500 ml = 50.500 l

Question 5.
Convert into decimals and add.
i) 58 kg 100 g; 60 kg 350 g
ii) 80 m 15 cm; 72 m 30 cm
Answer:
i) 58 kg 100 g; 60 kg 350 g
Given values are 58 kg 100 g and 60 kg 350 g
First convert the given values into like decimal fractions.
58.100 kg and 60.350 kg
Then, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on.
Then add:

ii) 80 m 15 cm; 72 m 30 cm
Given values are 80 m 15 cm and 72 m 30 cm.
First convert the given values into decimal fractions.
80.15 m and 72.30 m
Then, write the decimals in column with the decimal points directly below each other.
So, that tenths come under tenths, hundredths come under hundredths and so on. Then add:

Question 6.
Convert into decimals and subtract.
i) 14 kg 720 g from 16 kg 744 g
ii) 1 l 12 ml from 2 l 20 ml
Answer:
i) 14 kg 720 g from 16 kg 744 g
Given values are 14 kg 720 g and 16 kg 744 g.
First convert the given values into decimal fractions.
14.720 kg and 16.744 kg
Then, write the decimals in column with the decimal points directly below each other. So, that tenths come under tenths, hundredths come under hundredths and so on.
Then subtract:

ii) 1 l 12 ml from 2 l 20 ml
Given values are 1 l 12 ml and 2 l 20 ml
First convert the given values into decimal fractions.
1.012 l and 2.020 l
Then write the decimals in column with the decimal points directly below each other. So, that tenths come under tenths, hundredths come under hundredths and so on. Then subtract:

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Ex 5.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Ex 5.4

Question 1.
Which of the following are unlike decimal fractions ?
i) 5.03, 6.185
ii) 42.7, 7.42
iii) 16.003, 5.301
iv)15.81, 1.36
Answer:
i) 5.03, 6.185
5.3 has 2 decimal places
6.185 has 3 decimal places
Their decimal places are not equal.
So, these decimal fraction are unlike decimal fractions.

ii) 42.7, 7.42
42.7 has one decimal place
7.42 has two decimal places
Their decimal places are not equal.
So, these are unlike decimal fractions.

iii) 16.003, 5.301
16.3 has three decimal places
5.301 has three decimal places
Their decimal places are equal.
So, these are like decimal fractions.

iv) 15.81, 1.36
15.81 has two decimal places
1.36 has two decimal places
Their decimal places are equal.
So, these are like decimal fractions.

Question 2.
Change the following into like decimal fractions.
i) 0.802, 54.32, 873.274
ii) 4.78, 9.193, 11.3
iii) 16.003, 16.9, 16.19
Answer:
i) 0.802, 54.32, 873.274
Given decimal fractions are 0.802, 54.32, 873.274
The greatest number of decimal places is 3.
So, we convert all of them to equivalent decimals with 3 decimal places by placing sufficient zeroes.
0.802 = 0.802
54.32 = 54.320
873.274 = 873.274
Thus, 0.802, 54.32, 873.274 when converted to like decimals becomes 0.802, 54.320, 873.274.

ii) 4.78, 9.193, 11.3
Given decimal fractions are 4.78, 9.193, 11.3.
So, we convert all of them to equivalent decimals with 3 decimal places.
4.78 = 4.780
9.193 = 9.193
11.3 = 11.300
Thus, 4.78, 9.193, 11.3 when converted to like decimals becomes 4.780, 9.193, 11.300.

iii) 16.003, 16.9, 16.19
Given decimal fractions are 16.003, 16.9, 16.19
So, we convert all of them to equivalent decimals with three decimal places by providing sufficient zeroes.
16.3 = 16.003
16.9 = 16.900
16.19 = 16.190
Thus, 16.003, 16.9, 16.19 when converted to like decimals becomes 16.003, 16.900, 16.190.

Question 3.
Write the following in ascending order.
a) 7.26, 7.62, 7.2
b) 0.464, 0.644, 0.446, 0.664
c) 30.000, 30.060, 30.30
Answer:
a) 7.26, 7.62, 7.2
Given decimal fractions are 7.26, 7.62, 7.20
First compare the whole numbers.
Here whole numbers are equal. So, compare tenth’s place digits of the decimal fractions.
In 7.26 and 7.20 Tenths places are same.
But 7.62 Tenth place is more than the remaining numbers.
So,7.62 > 7.26 and 7.62 > 7.20
To compare 7.26 and 7.20 we should compare hundredth’s places. Then 7.26 > 7.20
∴ 7.62 > 7.26 > 7.20
(or) 7.20 < 7.26 < 7.62
Ascending order: 7.20; 7.26; 7.62

b) 0.464, 0.644, 0.446, 0.664
Given decimal fractions are 0.464, 0.644, 0.446, 0.664
Then, compare whole numbers.
Here whole numbers are same. Then, compare tenth’s place digits of the decimal fractions.
0.464, 0.446 are less than 0.644 and 0.664.
First make or check the given fraction are all like fractions or not?
Then, compare hundredth’s place digits
∴ 0.464 > 0.446, 0.664 > 0.644
and 0.664 > 0.644 > 0.464 > 0.446
(or) 0.446 < 0.464 < 0.644 < 0.664
Ascending order: 0.446; 0.464; 0.644; 0.664

c) 30.000, 30.060, 30.30
Given decimal fractions are 30.000, 30.060, 30.30
First make or check the given fractions are all like fractions or not?
So, 30.000; 30.060; 30.300 are like fractions.
Then, compare the whole numbers. They are same. So, compare, Tenth’s places of the decimal fractions.
30.300 > 30.060 > 30.000
(or) 30.000 < 30.060 < 30.300
Ascending order: 30.000; 30.060; 30.300

Question 4.
Arrange these numbers in descending order:
16.96; 16.42; 16.3; 16.03; 16.1; 16.99; 16.01
Answer:
Given decimal fractions are
16.96; 16.42; 16.3; 16.03; 16.1; 16.99; 16.01
First make or check the given fractions are all like fractions or not?
So, 16.96; 16.42; 16.30; 16.03; 16.10; 16.99; 16.01 are like fractions.
Then, compare whole numbers. They are same.
So, compare Tenth’s place digits of the decimal fractions.
16.96 < 16.99; 16.42 > 16.30 > 16.10 > 16.03 > 16.01
Then compare hundredth’s place digits.
So, 16.99 > 16.96 > 16.42 > 16.30 > 16.10 > 16.03 > 16.01 (or)
16.01 < 16.03 < 16.10 < 16.30 < 16.42 < 16.96 < 16.99
Descending order: 16.99; 16.96; 16.42; 16.30; 16.10; 16.03; 16.01.

Question 5.
Fill in the blanks by using appropriate symbols >, < or =.
i) 0.005 …… 0.0005
ii) 4.353 …… 4.2
iii) 58.3 …… 58.30
Answer:
i) 0.005 …… 0.0005
First make like decimal fractions
0.0050 …… 0.0005
By comparing Thousandth’s place digits.
So, 0.0050 > 0.0005

ii) 4.353 …… 4.2
First make like decimal fractions.
4.353 …… 4.200
Compare Tenth’s place digits, if whole numbers are same.
So, 4.353 > 4.200

iii) 58.3 …… 58.30
First make like decimal fractions.
58.30 …… 58.30
In these whole numbers and decimal place digits are same.
So, 58.30 = 58.30

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Ex 5.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Ex 5.3

Question 1.
Find the reciprocal of each of the following fractions.
i) \(\frac{5}{9}\)
ii) \(\frac{12}{7}\)
iii) 2\(\frac{1}{5}\)
iv) \(\frac{1}{8}\)
v) \(\frac{13}{11}\)
vi) \(\frac{8}{3}\)
Answer:

(Reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\)).

Question 2.
Simplify:
i) 15 ÷ \(\frac{3}{4}\)
ii) 6 ÷ 1\(\frac{4}{7}\)
iii) 3 ÷ 2\(\frac{1}{3}\)
iv) \(\frac{4}{9}\) ÷ 15
v) 4\(\frac{3}{7}\) ÷ 14
Answer:
i) 15 ÷ \(\frac{3}{4}\)
While dividing a whole number by a fraction, multiply the whole number with the reciprocal of the fraction.
Thus,

ii) 6 ÷ 1\(\frac{4}{7}\)
While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction and then multiply the whole number with the recipro¬cal of the improper fraction.
Thus,

iii) 3 ÷ 2\(\frac{1}{3}\)
While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction and then multiply the whole number with the reciprocal of the improper fraction.
Thus,

iv) \(\frac{4}{9}\) ÷ 15
While dividing a proper fraction by a whole number, multiply the proper fraction with the reciprocal of a whole number.
Thus,

v) 4\(\frac{3}{7}\) ÷ 14
While dividing mixed fraction by a whole number first convert the mixed fraction into improper fraction and then multiply the improper fraction with the reciprocal of a whole number.
Thus,

Question 3.
Find:
i) \(\frac{4}{9}\) ÷ \(\frac{2}{3}\)
ii) \(\frac{4}{11}\) ÷ \(\frac{8}{11}\)
iii) 2\(\frac{1}{3}\) ÷ \(\frac{3}{5}\)
iv) 5\(\frac{4}{7}\) ÷ 1\(\frac{3}{10}\)
Answer:
i) \(\frac{4}{9}\) ÷ \(\frac{2}{3}\)
While dividing a proper fraction by a proper fraction, multiply the proper fraction with the reciprocal of the fraction.
Thus,

ii) \(\frac{4}{11}\) ÷ \(\frac{8}{11}\)
While dividing a proper fraction by a proper fraction, multiply the proper fraction with the reciprocal of the fraction.
Thus,

iii) 2\(\frac{1}{3}\) ÷ \(\frac{3}{5}\)
While dividing a mixed fraction by a proper fraction, first convert the mixed fraction into improper fraction, then multiply it with the reciprocal of the proper fraction.
Thus,

iv) 5\(\frac{4}{7}\) ÷ 1\(\frac{3}{10}\)
While dividing a mixed fraction by a mixed fraction first convert the two mixed fraction into improper fraction, then multiply the improper fraction with the reciprocal of the second improper fraction.
Thus,

Question 4.
The product of two numbers is 25\(\frac{5}{6}\). If one of the number is 6\(\frac{2}{3}\). Find the other.
Answer:

Question 5.
By what number should 9\(\frac{3}{4}\) be multiplied to get 5\(\frac{2}{3}\)?
Answer:
Given one number = 9\(\frac{3}{4}\) = \(\frac{39}{4}\)
Product = 5\(\frac{2}{3}\) = \(\frac{17}{3}\)
\(\frac{39}{4}\) × Required Number = \(\frac{17}{3}\)
Required Number = \(\frac{17}{3}\) ÷ \(\frac{39}{4}\)
= \(\frac{17}{3}\) × \(\frac{4}{39}\)
= \(\frac{17 \times 4}{3 \times 39}\)
= \(\frac{68}{117}\)
Number to be multiplied = \(\frac{68}{117}\)

Question 6.
A bucket contains 34\(\frac{1}{2}\) litres of water. How many times do you get 1\(\frac{1}{2}\) litres of water?
Answer:
Capacity of Bucket = 34\(\frac{1}{2}\) litres = \(\frac{69}{2}\) litres
Capacity required = 1\(\frac{1}{2}\) = \(\frac{2 \times 1+1}{2}\) =\(\frac{3}{2}\)
Number of times we get = \(\frac{69}{2}\) ÷ \(\frac{3}{2}\) = \(\frac{69}{2}\) × \(\frac{2}{3}\) = 23 times.

Question 7.
The cost of 3\(\frac{3}{4}\) kg. of sugar is Rs. 121\(\frac{1}{2}\). Find its cost per 1 kg.
Answer:
Cost of 3\(\frac{3}{4}\) kg of sugar = Rs. 121\(\frac{1}{2}\)
Cost of 1kg of sugar

Cost of 1kg of sugar = Rs. 32.40.

Question 8.
The length of a rectangular field is 12\(\frac{1}{4}\) m. and its area is 65\(\frac{1}{3}\)m². Find its breadth.
Answer:
Length of a rectangular field = 12\(\frac{1}{4}\) = \(\frac{49}{4}\) m
Breadth of a rectangular field = ?
Area of a rectangular field = 65\(\frac{1}{3}\) = \(\frac{196}{3}\) sq.m
length × breadth = Area of a rectangular

∴ Breadth of a rectangular field = 5\(\frac{1}{3}\) m.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals Ex 5.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals Ex 5.2

Question 1.
Find the product of the following.
i) 3 × \(\frac{5}{12}\)
ii) \(\frac{15}{8}\) × 12
iii) 1\(\frac{3}{4}\) × \(\frac{12}{21}\)
iv) \(\frac{4}{5}\) × \(\frac{12}{7}\)
Answer:

Question 2.
Which is greater?
i) \(\frac{1}{2}\) of \(\frac{6}{7}\) or \(\frac{2}{3}\) of \(\frac{3}{7}\)
ii) \(\frac{2}{7}\) of \(\frac{3}{4}\) or \(\frac{3}{5}\) of \(\frac{5}{8}\)
Answer:
i) \(\frac{1}{2}\) of \(\frac{6}{7}\) or \(\frac{2}{3}\) of \(\frac{3}{7}\)

\(\frac{3}{7}\) or \(\frac{2}{7}\)
So, \(\frac{3}{7}\) > \(\frac{2}{7}\)
∴ \(\frac{1}{2}\) of \(\frac{6}{7}\) > \(\frac{2}{3}\) of \(\frac{3}{7}\)

ii) \(\frac{2}{7}\) of \(\frac{3}{4}\) or \(\frac{3}{5}\) of \(\frac{5}{8}\)

Convert them into like fractions
LCM of denominators = 2 × 2 × 2 × 7 = 56
\(\frac{3 \times 4}{14 \times 4}\) or \(\frac{3 \times 7}{8 \times 7}\)
\(\frac{12}{56}\) or \(\frac{21}{56}\)
So, \(\frac{12}{56}\) < \(\frac{21}{56}\)
∴ \(\frac{2}{7}\) of \(\frac{3}{4}\) < \(\frac{3}{5}\) of \(\frac{5}{8}\)

Question 3.
Find:
i) \(\frac{7}{11}\) of 330
ii) \(\frac{5}{9}\) of 108
iii) \(\frac{2}{7}\) of 16
iv) \(\frac{1}{7}\) of \(\frac{3}{10}\)
Answer:
i) \(\frac{7}{11}\) of 330
= \(\frac{7}{11}\) × 330
= \(\frac{7}{11}\) × 11 × 30
= 7 × 30 = 210

ii) \(\frac{5}{9}\) of 108
= \(\frac{5}{9}\) × 108
= \(\frac{5}{9}\) × 9 × 12
= 5 × 12 = 60

iii) \(\frac{2}{7}\) of 16
= \(\frac{2}{7}\) × 16
= \(\frac{2×16}{7}\)
= \(\frac{32}{7}\) or 4\(\frac{4}{7}\)

iv) \(\frac{1}{7}\) of \(\frac{3}{10}\)
= \(\frac{1}{7}\) × \(\frac{3}{10}\)
= \(\frac{1 \times 3}{7 \times 10}\)
= \(\frac{3}{70}\)

Question 4.
If the cost of a notebook is Rs. 10\(\frac{3}{4}\). Then, find the cost of 36 books.
Answer:
Cost of each notebook = Rs. 10\(\frac{3}{4}\) (or) \(\frac{43}{4}\)
Cost of 36 notebooks = 36 × 10\(\frac{3}{4}\) = 36 × \(\frac{43}{4}\) = \(\frac{9 \times 4 \times 43}{4}\) = 9 × 43
Cost of 36 notebooks = Rs. 387

Question 5.
A motor bike runs 52\(\frac{1}{2}\) km using 1 litre of petrol. How much distance will it cover for 2\(\frac{3}{4}\) litres of petrol ?
Answer:
Distance covered by the motor bike 1 litre petrol = 52\(\frac{1}{2}\) km (or) \(\frac{105}{2}\) km
Distance covered by the motor bike for 2\(\frac{3}{4}\) litres of petrol = 2\(\frac{3}{4}\) of 52\(\frac{1}{2}\)
= \(\frac{11}{4}\) × \(\frac{105}{2}\)
= \(\frac{11 \times 105}{4 \times 2}\)
∴ Distance covered by the motor bike for 2\(\frac{3}{4}\) litres of petrol = \(\frac{1155}{8}\) = 144\(\frac{3}{8}\) km.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers Unit Exercise Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Unit Exercise

Question 1.
Write the integers for the following situations.
i) A kite is flying at a height of 225 m in the sky. ( )
ii) A whale is at a depth of 1250 m in the ocean. ( )
iii) The temperature in Sahara desert is 12°C below freezing temperature. ( )
iv) Ravi withdrawn Rs. 3800 from ATM using his debit card. ( )
Answer:
i) 225 m
ii) – 1250 m
iii) – 12°C
iv) -3800

Question 2.
Justify the following statements with an example.
i) A positive integer is always greater than a negative integer.
ii) All positive integers are natural numbers.
iii) Zero is greater than a negative integer.
iv) There exist infinite integers in the number system.
v) All whole numbers are integers.
Answer:
i) A positive integer is always greater than a negative integer:
Consider
a) 3, -5 are two integers
3 > -5
b) – 10, 1 are two integers
1 > -10
∴ A positive integer is always greater than a negative integer.

ii) All positive integers are natural numbers:
Positive integers = 1, 2, 3, …..
Natural numbers = 1, 2, 3, …..
So, all positive integers are natural numbers.

iii) Zero is greater than a negative integer:
On a number line, for a given pair of numbers the number on R.H.S is always greater than the number on L.H.S.
All negative integers lie on the L.H.S. of zero, on a number line.
As such all negative numbers are less than zero or zero is greater than all negative integers.

iv) There exist infinite integers in the number system:
If we write integer on a number line, as the line extends on both sides endlessly so as the integers.
(Or)
For every integer there exists another integer which is 1 more than the given integer. Hence the integers are infinite.

v) All whole numbers are integers:
Whole numbers : 0, 1, 2, 3,……
Integers: ……, -3, -2, -1, 0, 1, 2, 3,…..
So, whole number are part of integers.
∴ All whole numbers are integers.

Question 3.
Represent
i) 3+4
ii) 8+(-3)
iii) -7-2
iv) 6-(5)
v)-5-(-4)
on number line.
Answer:
i) 3 + 4 = +7.

On the number line we first move 3 steps to the right of 0 to reach +3.
Then, we move 4 steps to the right of +3 and reach + 7.

ii) 8 + (-3) = +5.

On the number line we first move 8 steps to the right of 0 to reach +8.
Then, we move 3 steps to the left of +8 and reach +5.

iii) -7 -2 = -9

On the number line we first move 7 steps to the left of 0 to reach -7.
Then we move 2 steps to the left of -7 and reach -9.

iv) 6 – (5) = + 1

On the number line we first move 6 steps to the right of 0 to reach +6.
Then, we move 5 steps to the left of +6 and reach +1.

v) -5 – (-4) = -1

-5 – (-4) = -5 + 4 = -1 (∵ -(-a) = a)
On the number line, we first move 5 steps to the left of 0 to reach -5.
Then, we move 4 steps to the right of -5 and reach -1.

Question 4.
Write all the integers lying between the given numbers.
i) 7 and 12
ii) -5 and -1
iii) -3 and 3
iv) -6 and 0
Answer:
i) 7 and 12
Integers lying between 7 and 12 are 8, 9, 10, ll.

ii) -5 and -1
Integers lying between -5 and -1 are -2, -3 and -4.

iii) -3 and 3
Integers lying between -3 and 3 are -2, -1, 0, 1, 2.

iv) -6 and 0
Integers lying between -6 and 0 are -5, -4, -3, -2, -1.

Question 5.
Arrange the following integers in ascending order and descending order.
-1000, 10, -1, -100, 0, 1000, 1, -10
Answer:
Given numbers -1000, 10, -1, -100, 0, 1000, 1, -10
Ascending order: – 1000, -100, -10, -1, 0, 1, 10, 1000
Descending order: 1000, 10, 1, 0, -1, -10, -100, -1000

Question 6.
Write a real life situation for each of the following integers.
i) -200 m
ii) +42°C
iii) -4800(cr)
iv) -3.0 kg
Answer:
i) -200 m
In the Singareni coal mines workers will go 200 m below the ground level.
ii) +42°C
In summer average temperature of May month is 42°C.
iii) Rs. 4800 (crores)
Central Government sanctioned Rs. 4800(crores) for education in the Annual Budget.
iv) – 3.0 kg
In a Primary School the ground balance of rice = -3 kg

Question 7.
Find:
i) (-603) + (603)
ii) (-5281) +(1825)
iii) (-32) + (-2) + (-20) + (-6)
Answer:
i) (-603) + (603)
Sum of a number and its additive number is 0.
Additive inverse of – 603 is 603.
So, -603 + 603 = 0

ii) (-5281) + (1825)
= – 5281 + 1825 = – 3456

iii) (-32) + (-2) + (-20) + (-6)
= -32 – 2 – 20 – 6 = -60

Question 8.
Find:
i)(-2) – (+1)
ii) (-270) – (-270)
iii) (1000) – (-1000)
Answer:
i) -2 – (+1)
= -2 -1 = -3

ii) – 270 – (-270)
= -270 + 270 [∵ -(-a) = a]
= 0 [-a + a = 0]

iii) 1000 – (-1000)
= 1000 + 1000 [∵ -(-a) = a]
= 2000

Question 9.
In a quiz competition, where negative score for wrong answer is taken,Team A scored +10, -10, 0, -10, 10, -10 and Team B scored 10,10, -10,0,0,10 in 6 rounds successively .Which team wins the competition? How?
Answer:
Score of Team A = (+10), (-10), 0, (-10), 10, (-10)
Total score of Team A = (+10) + (-10) + (0) + (-10) + (10) + (-10)
= +10 – 10 + 0 – 10 + 10 – 10
= +20 – 30 = -10
Score of Team B = 10, 10, -10, 0, 0, 10
Total Score of Team B = (+10) + (+10) + (-10) + 0 + 0 + (+10)
= +10 + 10 – 10 + 10
= + 30 – 10
= + 20
-10 < + 20
Team A < Team B
So, Team B is the winner. Because, Team B got more score than Team A.

Question 10.
An apartment has 10 floors and two cellars for car parking under the basement. A lift is now, at the ground floor. Ravi goes 5 floors up and then 3 floors up, 2 floors down and then 6 floors down and come to lower cellar for taking his car. Count how many floors does Ravi travel all together? Represent the result on a vertical number line.
Answer:

First move:
Ground to 5th floor up = 5
(Can reach 5th floor)
Second move:
5th floor to 3 floors up = 3
(Can reach 8)
Third move:
8th floor to 2 floors down = 2
(Can reach 6th floor)
Fourth move:
6th floor to 6 floors down = 6
(Can reach ground floor)
Final move:
Ground to lower cellar = 2
(Can reach cellar -2)
No. of floors Ravi travelled = 18

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers Ex 4.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Ex 4.4

Question 1.
Find:
i) 40 – (22)
ii) 84 – (98)
iii) (-16) + (-17)
iv) (-20) – (13)
v) (38) – (-6)
vi) (-17) – (-36)
Answer:
i) 40 – (22) = +40 – 22
= +(18 + 22)-22
= 18 +(22-22)
= 18 + 0
∴ 40-(22) = + 18

ii) 84 – (98) = 84 – 98
= + 84 – 84 – 14
= (84 – 84) – 14
= 0 – 14
∴ 84 – (98) = – 14

iii) (-16) + (-17) = -16 – 17 = – 33
∴ (-16) + (-17) = -33

iv) (-20) – (13) = – 20 – 13
= -33
∴ (-20) – (13) = – 33

v) 38 – (-6) = 38 + 6
We know – (-a) = a = + 44
∴ 38 – (-6) = + 44

vi) (-17) – (-36) = -17 + 36
We know – (- a) = a
= -17 + 17 + 19
= (-17+17) + 19
= + 19
∴ (-17) – (-36) = + 19

Question 2.
Fill in the boxes with >, < or = sign:
(i) (-4) + (-5) (-5) – (-4)
(ii) (-16) – (-23) (-6) + (-12)
(iii) 44 – (-10) 47 +(-3)
Answer:
(i) (-4) + (-5) (-5)- (-4)
-4-5  -5 + 4
-9 -1

(ii) (-16) – (-23) (-6) + (-12)
-16 + 23 -6-12  [∵ -(-a) = a]
-16 + 16 + 7 -18
+7 -18

(iii) 44 – (-10) 47 + (-3)
44+10 47-3.
54 44 + 3 – 3
54 44

Question 3.
Fill in the blanks:
i) (-13) + ———– = 0
ii) (-16) + 16 = ———–
iii) (-5) + ———– = -14
iv) ———– + (2 – 16) = – 22
Answer:
i) (-13) + ———– = 0
We know, additive inverse of a is -a (or) – a is a.
Additive inverse of -13 is 13. So, -13 + 13 = 0

ii) (-16) + 16 = ———–
We know, sum of a number and its additive inverse is 0.
i.e., (-a) + a = 0
So, (-16) + (16) = 0

iii) (-5) + ———– = -14
-5 + (-9) = -14

iv) ———– + (2 – 16) = – 22
———– + (-14) = -22
– 8 + (-14) = – 22

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers Ex 4.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers Ex 4.3

Question 1.
Add the following integers using number line.
i) 7 + (-6)
ii) (-8) + (-2)
iii) (-6) + (-5) + (+2)
Answer:
i) 7 + (-6)

On the number line, we first move 7 steps to the right of 0 to reach +7.
Then, we move 6 steps to the left of +7 and reach + 1.
So, 7 + (-6) = 1

ii) (-8) + (-2)

On the number line, we first move 8 steps to the left of 0 to reach -8.
Then, we move 2 steps to the left of -8 and reach -10.
So,(-8) + (-2) = -10

iii) (-6) + (-5) + (+2)

On the number line, we first move 6 steps to the left of 0 to reach -6.
We move 5 steps to the left of -6 and reach -11.
Then, we move 2 steps to the right of -11 and reach -9.
So, (-6) + (-5) + (+2) = -9

Question 2.
Add without using number line.
(i) 10 + (-3)
(ii) 10 + (+16)
(iii) (-8) + (+8)
Answer:
i) 10 + (-3)
10 + (-3) = 7 + 3 + (-3) = 7 + (3 + (-3))
= 7 + 3 – 3 = 7 + 0
∴ 10 + (-3) = +7

ii) -10 + (+16)
-10 + (+16)
= -10 + 10 + 6
= (-10 + 10) + 6
= 0 + 6
∴ -10 + (+16) = + 6

iii) (-8) + (+ 8)
-8 + (+8) = -8 + 8 = 0
∴ -8 + (+8) = 0

Question 3.
Find the sum of i) 120 and -274 ii) -68 and 28
Answer:
i) 120 and-274
Method -1: Sum = + 120 + (-274)
= + 120 + (- 120 – 154)
= + 120 – 120 – 154
∴ 120 + (-274) = -154

Method – II:
As the given numbers have opposite sign, we subtract one from other

As 274 is having negative (-) sign, the answer is – 154.

ii) -68 and 28
Sum = – 68 + (28)
= – 40 – 28 + 28
∴ – 68 + 28 = – 40

Question 4.
Simplify:
i) (-6) + (-10) + 5 + 17
ii) 30 + (-30) + (-60) + (-18)
Answer:
i) (-6) + (-10) + 5 + 17
– 6 – 10 + 5 + 17 = -16 + 22
= -16 + 16 + 6
= + 6
∴ (-6) + (-10) + 5 + 17 = + 6

ii) 30 + (-30) + (-60) + (-18)
= 30 + (-30 -60 -18)
= 30 + (-108)
= 30 – 108 = – 78
∴ 30 + (-30) + (-60) + (-18) = -78