KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Students can Download Chapter 2 Fractions and Decimals Ex 2.5, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Question 1.
Which is greater ?
i) 0.5 or 0.05
Solution:
0.5 > 0.05 (∵ 0.5 = 0.50)

ii) 0.7 or 0.5
Solution:
0.7 > 0.5

iii) 7 or 0.7
Solution:
7 > 0.7 (∵ 7 = 7.0)

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

iv) 1.37 or 1.49
Solution:
1.49 > 1.37

v) 2.03 or 2.30
Solution:
2.30 > 2.03

vi) 0.8 or 0.88
Solution:
0.88 > 0.8 (∵ 0.8 = 80)

1 1/2 as a decimal is equal to 1.5.

Question 2.
Express as rupees using decimals :
i) 7 paise
Solution:
7 paise = Rs. 0.07

ii) 7 rupees 7 paise
Solution:
7 rupees 7 paise Rs. 7.07

iii) 77 rupees 77paise
Solution:
77 rupees 77 paise = ₹ 77.77

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

iv) 50 paise
Solution:
50 paise = 0.50

v) 235 paise
Solution:
235 paise = 2.35

Question 3.
i) Express 5cm in metre and kilometre
Solution:
5 cm = 0.05m = 0.00005km.

ii) Express 35mm in cm, m and km
Solution:
35mm = 3.5cm = 0.000035km

Question 4.
Express in kg :
i) 200g
Solution:
200g = 0.200kg = 0.2 kg

ii) 3470g
Solution:
3470g = 3.470kg

iii) 4 kg 8 g
Solution:
4kg 8g = 4.008 kg

Question 5.
Write the following decimal numbers in the expanded form :
i) 20.03
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 1

ii) 2.03
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 21

iii) 200.03
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 22

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

iv) 2.034
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 23

Question 6.
Write the place value of 2 in the following decimal numbers :

i) 2.56
Solution:
2.56 place value of 2 in the decimal number 2.56 = 2 ones place)

ii) 21.37
Solution:
21.37 place value of 2 in the decimal number 21.37 = 2 × 10 = 20. (2 is in tens place)

iii) 10.25
Solution:
10.25 = palce value of 2 in the decimal
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 24

iv) 9.42
Solution:
9.42 = place value of 2 in the decimal number
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 25
(2 is in hundredths place)

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

v) 63.352
Solution:
63.352 = place value of 2 in the decimal number
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 26

Question 7.
Dinesh went from place A to place B and from 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelles more and by how much ?
Solution:
Distance travelled from A to B = 7.5kms
Distance travelled from B to C = 12.7kms
Total distance travelled from A to B & B to C = 20.2 kms.
Ayub
Distance travelled from A to D = 9.3 kms
Distance travelled from D to C = 11.8 kms
Total distance travelled from A to D & D to C = 21.1 kms
∴ Ayub travelled more distance than Dinesh by = 21.1 – 20.2 = 0.9 kms.

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Question 8.
Shyama bought 5 kg 300g apples and 3 kg 250 g mangoes. Sarala bought 4 kg of 800 g oranges and 4 kg 150g bananas. Who bought more fruits?
Solution:
Shyama:
Weight of Apples bought = 5.300kgs
Weight of Mangoes bought = 3.250kgs
Total weight of fruits = 8.550kgs

Sarala:
Weight of oranges bought = 4.800kgs
Weight of Bananas bought = 4.150kgs
Total weight of fruits = 8.950kgs
8.950 >8.550
∴ Sarala bought more fruits than Shyama.

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Question 9.
How much less is 28 km than 42.6 km?
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5 30
∴ 28 km is less than 42.6 km by 14.6 kms.

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

Students can Download Chapter 8 Comparing Quantities Ex 8.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

Percentage difference calculator tool makes the calculation faster, and it displays the difference in the percentage in a fraction of seconds.

Question 1.
Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.
a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
Solution:
C. P = Rs 250
S. P = Rs. 325
S.P > C.P
∴ It is profit.
Total profit = S.P – C.P
= 325 – 250 = Rs 75
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 1

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.
Solution:
C. P = Rs. 12,000
S.P = Rs. 13,500
S.P > C.P
It is profit
Total profit = S. P – C. P
= 13,500 – 12,000
= Rs.1,500/-
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 2

c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
Solution:
C.P = Rs. 2,500
S. P = Rs. 3,000
S.P > C.P
It is profit
Total profit = S. P – C. P
= 3000 – 2500
= Rs. 500
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 3

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

d) A skirt bought for 250 and sold at ₹ 150.
C.P = Rs. 250/-
S.P = Rs. 150/-
S.P < C.P
∴ It is loss
Total loss = C.P – S.P
= 250 – 150 = Rs. 100
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 4

Percentage Difference Calculator is very simple app to calculate percentage difference between 2 values.

Question 2.
Convert each part of the ratio to percentage :
a) 3 : 1
Solution:
Total parts = 3 + 1 = 4
Percentage of First part of the ratio =
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 5
Percentage of 2nd part of the ration =
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 6

b) 2 : 3 : 5
Solution:
Total parts = 2 + 3 + 5 = 10
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 100
% of second part of the ratio =
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 101
% of third part of the ratio =
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 102

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

c) 1 : 4
Total parts = 1 + 4 = 5
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 8

d) 1 : 2 : 5
Total parts. = 1 + 2 + 5 = 8
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 9
% of second part of the ratio =
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 103

Question 3.
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:
Decreased population = 25,000 – 24,500
= 500
Percentage of decrease =
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 11

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

Question 4.
Arun bought a car for 3, 50, 000. The next year, the price went upto ₹ 3,70,000. What was the percentage of price increase ?
Solution:
Price of a car = Rs. 3,50,000
Increased price of a car = Rs. 3,70,000
Total increase = Rs. 20,000
Percentage of price increase =
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 12

Question 5.
I buy a T. V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it ?
Solution:
C.P of a T.V = Rs. 10,000
Profit = 20%
20% of Rs. 10,000
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 13
∴ Amount I get = C. P + profit
= 10,000 + 2,000
= Rs. 12,000

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

Question 6.
Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she bought it ?
Solution:
S. P = C. P – Loss
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 15
∴ She bought that washing machine for Rs. 16,875.

CGPA into Percentage … cumulative grading percentage of current grade.

Question 7.
i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the percentage of carbon in chalk.
Solution:
Total parts = 10 + 3 + 12 = 25
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 16
% of carbon in chalk =
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 17

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

ii) If in a stick of chalk. carbon is 3g, what is the chalk stick?
Solution:
Let the weight of chalk stick be ‘M’
Then 12% of M = 3
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 18
∴ The weight of the chalk stick is 25 grams.

Question 8.
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for ?
Solution:
C. P of a book = Rs. 275
loss = 15% of C. P
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 19
S.P = C.P – Loss
= 275 – 41.25
= Rs. 233.751
∴ She sells it for Rs. 233.75.

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

Question 9.
Find the amount to be paid at the end of 3 years in each case:
a) Principal = ₹ 1,200 at 12 % p.a
P = Rs. 1200/-
R = 12%
T = 3 years
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 70
A = P + I = 1200 + 432 = Rs. 1,632
∴ The amount to paid at the end of 3 years is Rs. 1,632

b) Principal = ₹ 7,500 at 5% p.a
P = Rs. 7,500
R = 5%
T = 3 years
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 71
A = P + I
= 7500 + 1125
= Rs. 8,625
∴ The amount to be paid at the end of 3 years is Rs. 8,625.

Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years ?
Solution:
P = Rs. 56,000/-
R = ?
T = 2 years
I = Rs. 280
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 72
∴ The rate of interest is 0.25% per annum.

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3

Question 11.
If Meena gives an interest of ? 45 for one year at 9% rate p.a. What is the sum she has borrowed ?
Solution:
P = ?
R = 9%
T = 1 years
I = Rs. 45
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 73
∴ She has borrowed the sum of Rs. 500

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Students can Download Chapter 8 Comparing Quantities Ex 8.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

The percentage difference calculator is here to help you compare two numbers.

Question 1.
Convert the given fractional numbers to percents.
a) \(\frac{1}{8}\)
Solution:
Percent means per hundred. So multi-plied by 100.
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 1

b) \(\frac{5}{4}\)
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 2

c) \(\frac{3}{40}\)
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 3

d) \(\frac{2}{7}\)
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 4

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 2.
Convert the given decimal fractions to per cents.
a) 0.65
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 5

b) 2.1
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 6

c) 0.02
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 7

d) 12.35
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 8

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

The percent decrease calculator calculates the amount of decrease, in which we would say, “x percent decrease”.

Question 3.
Estimate what part of the figures is coloured and hence find the per cent which is coloured.
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 9
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 10
∴ In the first figure 25% is coloured.
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 11
∴ in the second figur e 50% Is coloured,.
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 111
∴ In the third figure 37.5 % is coloured.

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 4.
Find :
a) 15% of 250
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 12

b) 1% of 1 hour
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 13

c) 20% of ₹ 2500
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 14

d) 75% of ₹ 1 kg
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 15

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 5.
Find the whole quantity if
a) 5% of it is 600.
Solution:
Let the whole quantity be ‘m’
∴ 5% of m = 600
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 16
∴ The whole quantity is 12,000.

b) 12% of its 1080.
Solution:
Let the whole quantity be ‘m’
∴ 12% of m = Rs. 1080
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 17
∴ The whole amount be Rs. 9.000

c) 40 % of it is 500 km
Solution:
Let the whole quantity be ‘m’
∴ 40% of m = 500 km
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 18
∴ The whole amount is 1250 kms

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

d) 70% of it is 14 minutes
Let the whole quantity be ‘m’
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 19
∴ The whole amount is 20 minutes

e) 8% of it is 40 litres.
Let the whole quantity be ‘m’
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 20
∴ The whole amount be 500 liters

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 6.
Convert given per cents to decimal fractions and also to fractions in simplest forms:
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 21

Question 7.
In a city, 30% are females, 40% are males and the remaining are children. What percent are children?
Solution:
females = 30%
males = 40%
children = 100 – (30 + 40)
remaining are children = 100 – 70 = 30%
∴ Percentage of children = 30%

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Percentage difference calculator tool makes the calculation faster, and it displays the difference in the percentage in a fraction of seconds.

Question 8.
Out of 15,000 votes in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Solution:
Total no. of voters in a constituency } = 15,000
Percentage of voted = 60%
∴ Percentage of voters who did not vote
= 100 – 60 = 40%
Actual number of voters who did not vote} = 40% of 15,000
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 80
6000 voters who did not vote.

I hope you guys are now clear about how to convert CGPA to percentage in the SPPU mark sheet.

Question 9.
Meeta saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?
Solution:
Let Meeta’s salary be ‘M’
10% of M is = Rs. 400
∴ M =?
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 81
∴ Her salary be Rs. 4,000/-

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Let the total matches won by them be M 25% of 20 = M
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 82
∴ They win 5 matches.

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Students can Download Chapter 2 Fractions and Decimals Ex 2.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Multiplying and Dividing Rational Expressions Calculator.

Question 1.
Solve:
i)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 1
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 2

ii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 3
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 4

iii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 5
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 6
L.C.M = 5 × 7 × 1 × 1 = 35
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 7
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 8

iv)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 9
Solution:

Online Subtracting Fractions Calculator subtracts the fractions 9/11 and 4/5 i.e. 1/55

The given fractions are 9/11 and 4/5

Firstly the L.C.M should be done for the denominators of the two fractions 9/11 and 4/5

9/114/5

The LCM of 11 and 5 (denominators of the fractions) is 55

Given numbers has no common factors except 1. So, there LCM is their product i.e 55

 

= 9 x 5 – 4 x 11/55

= 45 – 44/55

= 1/55

Result: 1/55

v)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 12
Solution:
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 13
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 14

vi)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 15
⇒ convert mixed fractions to improper fraction
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 16

vii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 17
convert mixed fractions to improper fraction
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 18

Question 2.
Arrange the following in descending order:
i)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 19
= We need to arrange these in descending order,
To find which number is greater or smaller, we make their denominators equal.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 20

ii)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 21
⇒ We make their denominators equal, to find the descending order.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 22

Free quadratic equation completing the square calculator – Solve quadratic equations using completing the square step-by-step.

Question 3.
In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square ?
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 23
Solution:
For Row,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 24
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 25
For diagonals,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 26
Since Sum of air rows, columns and diagonals are equal.

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Question 4.
A rectangular sheet of paper is 12\(\frac{1}{2}\) cm long and 10\(\frac{2}{3}\) cm wide. Find its perimeter.
Solution:
Length of rectangular sheet of paper = 12\(\frac{1}{2}\) cm
(breadth) width of rectangular sheet paper = 10\(\frac{2}{3}\) cm
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 27
Perimeter of rectangle = 2 (length + breadth)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 28
converting the above fraction to mixed fraction,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 29

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Question 5.
Find the perimeters of
(i) ∆ ABE
(ii) the rectangle BCDE in this figure. Whose perimeter is greater ?
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 30
Solution:
(i) ∆ ABE
perimeter of ∆ ABE
perimeter = AB + AE + BE
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 31
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 311

ii) The rectangle BCDE in this figure
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 32
Perimeter of rectangle BCDE,
As it is a rectangle, opposite sides are equal
BC = DE CD = BE
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 33
perimeter of rectangle = 2(l + b)
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 34
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 35
Also,
We have to find which perimeter is greater
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 36
To find which fraction is greater, we make its denominator equal.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 37
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 38
Perimeter of ∆ ABE > Perimeter of BCDE
(Thus, Perimeter of ∆ ABE is greater)

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Question 6.
Salil wants to put a picture in a frame. The picture is 7\(\frac{3}{10}\) cm wide. How much should the picture be trimmed ?
Solution:
There are two things here – picture, and frame
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 39
so, picture has to be trimmed
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 40
\(=\frac{3}{10}\)
∴ Picture has to trimmed by = \(=\frac{3}{10}\) cm

Question 7.
Ritu ate \(\frac{3}{5}\) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ? Who had the larger share ? By how much ?
Solution:
Total part = 1
Part eaten by Ritu = \(\frac{3}{5}\)
Part eaten by Somu = Total part – part eaten by Ritu
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 41
Now,
We have to tell who ate the larger share so, we have to compare,
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 42
∴ Ritu ate the larger share.
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 43
Ritu ate large share by \(\frac{1}{5}\)

KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1

Question 8.
Michael finished colouring in \(\frac{7}{12}\) hour. Vaibhav finished colouring the same picture in \(\frac{3}{4}\) hour. Who worked longer ? By what fraction was it longer ?
Solution:
Michael finished work in = \(\frac{7}{12}\) hour
Vaibhav finished work in = \(\frac{3}{4}\)
We need to find who worked longer.
i.e., we need to find greater of = \(\frac{7}{12}\) & \(\frac{3}{4}\)
We make their denominators equal
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 44
∴ Vaibhav worked longer.
We also need to find by how much
KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1 45
Vaibhav worked longer by \(\frac{1}{6}\) hours.

KSEEB Solutions for Class 7 Science Chapter 1 Nutrition in Plants

Students can Download Chapter 1 Nutrition in Plants Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Science, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Science Chapter 1 Nutrition in Plants

Class 7 Science Nutrition in Plants NCERT Textbook Questions and Answers

Question 1.
Why do organisms need to take food?
Answer:
Food is needed by all organisms for many purposes :

  1. The main function of food is to help in growth.
  2. Food provides energy for movements such as running, walking or raising our arm.
  3. Food is also needed for replacement and repairing damaged parts of body.
  4. Food gives us resistance to fight against diseases and protects us from infections.

Question 2.
Distinguish between a parasite and a saprotroph.
Answer:

Parasite Saprotrophs
(a) Parasite derives nutrients from living organisms. (a) They deserve nutrients from dead and decaying organisms
(b) The organism on which it feeds is called the host. (b) They do not feed on a living organism.
(c) Example: Cuscuta and orchids. (c) Example: Mushrooms and fungi.

Question 3.
How would you test the presence of starch in leaves?
Answer:
The presence of starch in leaves is tested in the following ways:

  • Take two potted plants of the same kind.
  • Keep one in the dark and the other in the sunlight for some time.
  • Take one leaf from each of the plants.
  • Put a few drops of iodine solution on each of the leaves.
  • The leaf kept in the sunlight will turn blue-black due to the presence of starch.
  • The leaf kept in the dark will not turn blue-black because of the absence of starch.

Question 4.
Give a brief description of the process of synthesis of food in green plants.
Answer:
Photosynthesis is defined as the process in which the chlorophyll-containing plant cells synthesize food in the form of carbohydrates, using carbon dioxide and water in the presence of solar energy.

Sources of raw materials required for photosynthesis are :

  1. Water is taken in, from the roots of the plant and is transported to the leaves.
  2. carbon dioxide from the air enters the leaves through the pores called stomata and diffuses to the cells containing chlorophyll.
  3. Solar energy is used to break water into hydrogen and oxygen. This hydrogen is combined with carbon dioxide to form food for the plants, which is ultimately used by the animals as well.

Thus, photosynthesis can be represented by the following equation.
KSEEB Solutions for Class 7 Science Chapter 1 Nutrition in Plants 1

Question 5.
Show with the help of a sketch that the plants are the ultimate source of food.
Answer:
All the living being depends on plants whether directly or indirectly. For example, the plant-eater animals depend directly on plants but carnivore depends indirectly on plants. The following sketch shows some examples of plant dependency.
KSEEB Solutions for Class 7 Science Chapter 1 Nutrition in Plants 2

Question 6.
Fill in the blanks:
(a) Green plants are called autotrophs since they synthesise their own food.
(b) The food synthesised by the plants is stored as starch.
(c) In photosynthesis solar energy is captured by the pigment called Chlorophyll.
(d) During photosynthesis plants take in carbon dioxide and oxygen release.

Question 7.
Name the following:
(i) A parasitic plant with yellow, slender and tubular stem.
Answer:
Cuscuta

(ii) A plant that has both autotrophic and heterotrophic mode of nutrition.
Answer:
Pitcher plant, Venus Flytrap (Insectivorous Plants)

(iii) The pores through which leaves exchange gases.
Answer:
Stomata

Question 8.
Tick the correct answer:
(a) Amarbel is an example of:
(i) autotroph
(ii) parasite ✓
(iii) saprotroph
(iv) host
Answer:
(ii) parasite

(b) The plant which traps and feeds on insects is:
(i) Cuscuta
(ii) china rose
(iii) pitcher plant
(iv) rose
Answer:
(iii) pitcher plant

Question 9.
Match the items given in Column I with those in
KSEEB Solutions for Class 7 Science Chapter 1 Nutrition in Plants 3
Answers:
1 – d,
2 – a,
3 – e,
4 – b,
5 – c

Question 10.
Mark ‘T’ if the statement is true and ‘F’ if it is false:
(i) Carbon dioxide is released during photosynthesis. (T/F) False
(ii) Plants which synthesise their food themselves are, called saprotrophs. (T/F) False
(iii) The product of photosynthesis is not a protein. (T/F) True
(iv) Solar energy is converted into chemical energy during photosynthesis. (T/F) True

Question 11.
Choose the correct option from the following:
Which part of the plant gets carbon dioxide from the air for photosynthesis,
(i) root hair
(ii) stomata
(iii) leaf veins
(iv) sepals
Answer:
(ii) stomata

Question 12.
Choose the correct option from the following:
Plants take carbon dioxide from the atmosphere mainly through their:
(i) roots
(ii) stem
(iii) flowers
(iv) leaves
Answer:
(iv) leaves

Question 13.
Why do farmers grow many fruits and vegetable crops inside large greenhouses? What are the advantages to the formers?
Answer:
The greenhouse protects the plants inside it from the climatic conditions outside and gives plants a suitable temperature for growth and development. It also protects from birds and animals. The greenhouse is made up of glass. Glass is a bad conductor of heat and traps the heat inside it in cold places. Thus, it provides a suitable temperature for the plants to grow. This is the advantage of tire farmers.

Categories Class 7 Science MCQ ·

Class 7 Science Nutrition in Plants Additional Important Questions and Answers

Question 1.
What are the nutrients?
Answer:
Carbohydrates, proteins, fats, vitamins, and minerals are called the components of food, These components are called nutrients.

Question 2.
What is nutrition?
Answer:
Nutrition is the mode of taking food by an organism and its utilization by the body.

Question 3.
What are autotrophs?
Answer:
All plants are called autotrophs plants to prepare food for themselves in the process of photosynthesis. Auto means self and trophies mean nourishment.

Question 4.
What are heterotrophs?
Answer:
All animals depend upon plants for their food are called heterotrophs. heteros mean other.

Question 5.
Write a neat diagram of a cell and label the parts.
Answer:
KSEEB Solutions for Class 7 Science Chapter 1 Nutrition in Plants 4

Question 6.
What are Carbohydrates?
Answer:
Carbohydrates are produced in the process of photosynthesis. This is one of the components of food consists of Carbon, hydrogen and oxygen.

Question 7.
Is photosynthesis takes place in red, brown and other colours of leaves ?
Answer:
Yes, photosynthesis takes place in all colours of leaves. All colour leaves have chlorophyll in them but the other colour pigment dominate the green colour chlorophyll, hence it looks red or brown colour.

Question 8.
Write a neat diagram of the section of the leaf and label the parts.
Answer:
KSEEB Solutions for Class 7 Science Chapter 1 Nutrition in Plants 5

Question 9.
What are insectivorous plants?
Answer:
Some plants eat insects for their food. These plants are insectivorous plants.
eg: pitcher plant.

Question 10.
Write a neat diagram of the pitcher plant. Label the parts and describe them.
Answer:
KSEEB Solutions for Class 7 Science Chapter 1 Nutrition in Plants 6
Observe the above figure. The pitcher or pot like or Jug like structure is the modified part of leaf. The apex of the leaf forms a lid which can open and colse the mouth of pitcher. Inside the pitcher there are hair which are directed downwards. When an insect lands in
the pitcher, the lid closes and trapped insects gets entangled into the hair. The lid closes and the insect is trapped. The insect is digested by the digestive Juices secreted in the pitcher and its nutrients are absorbed

Question 11.
What are saprotrophs?
Answer:
The Organisms with the saprotrophic mode of nutrition are called saprotrophes.

Question 12.
What is meant by Saprotrophic nutrition?
Answer:
The Organisms which take nutrients from dead and decaying matter is called as saprotrophic nutrition.

Question 13.
What are parasites?
Answer:
The organism which live on other plants and absorb nutrients from them are called para-sites. Cuscuta is a good example for parosites. It grows on some trees and absorb nutrients from the host tree.

Question 14.
What is symbiosis?
Answer:
Some organisms live together and share both shelter and nutrients. This relationship is called symbiosis.

Question 15.
Explain the symbiosis in Lichens.
Answer:
In lichens, the chlorophyll-containing part-ner Alga and Fungus live together. The fungus provides shelter, water and minerals to the Alga and in return the Alga prepares food and provides to the Fungus.

Question 16.
What is the role of Rhizobium in plants’ life?
Answer:
The plants need nitrogen for their growth. In the atmosphere, plenty of nitrogen is there but plants cannot use it directly. They need nitrogen in a soluble form. The Rhizobium a kind of bacteria which lives in the roots of a gram, peas, moong, beans and other legumes. It can take atmospheric nitrogen and convert it into a usable form. Rhizobium provides nitrogen to plants, in return the plants provide food and shelter to the Rhizobium. Thus they have a symbiotic relationship. This is of great significance to the formers.

Question 17.
Write a diagram showing photosynthesis and label the parts.
Answer:
KSEEB Solutions for Class 7 Science Chapter 1 Nutrition in Plants 7

Question 18.
Where and how Fungic will grow?
Answer:
The Fungi spores are present in the air when these spores land on moist and warm things grow. eg: Wet bread, pickles, deather goods, and other articles left in hot and humid weather for more days, Fungi grow on them and spoil the things.

II. Fill in the blanks :

  1. Leaves are the food factories of plants.
  2. Humans and animals are directly or indirectly dependent on plants.
  3. Nutrition is the mode of taking food by an organism and its utilisation by the body.
  4. The bodies of living organisms are made of tiny units called cells.
  5. The cell is enclosed by the thin outer boundary called a cell membrane.
  6. The nucleus is surrounded by a jellylike substance called cytoplasm.
  7. The tiny pores present on the surface of leaves are called stomata.
  8. The leaves have green pigment known as Chlorophyll.
  9. Sun is the ultimate source of energy for all living Organisms.
  10. Carbohydrates are synthesized in the process of photosynthesis.
  11. Mushroom is growing on dead and decayed material.
  12. Rhizobium lives in leguminous plants
  13. Light is very important to plants, so as to absorb maximum sunlight they grow in many patterns.
  14. The plant on which it climbs is called the host.
  15. Insectivorous plants eat insects to fulfill their needs for Nitrogen.
  16. The raw material for photosynthesis is Carbon dioxide and water.

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