Class 6

Students can Download Chapter 6 Integers Ex 6.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 6 Integers Ex 6.2

Question 1.
Using the number line write the integer which is:
Solution:
a) 3 more than 5

Hence, 8

b) 5 more than -5

c) 6 less than 2

Hence -4

d) 3 less than -2

Hence -5

Question 2.
Use number line and add the following integers:
a) 9 + (-6)

Hence ,3

b) 5 + (-11)

Hence, -6

c) (-1) + (-7)

Hence-8

d) (-5) + 10

Hence, 5

e) (-1) + (-2) + (-3)

Hence, -6

f) (-2) + 8 + (-4)

Hence, 2

Question 3.
Add without using number line:
a) 11 + (-7)
11 + (-7) = 4

b) (-13) + (+18)
(-13) + (+18) = 5

c) (-10) + (+19)
(-10) + (+19) = 9

d) (-250) + (+150)
(-250) + (+150) = -100

e) (-380) + (-270)
(-380) + (-270) = -650

f) (-217) + (-100)
(-217) + (-100) = -317

Question 4.
Find the sum of
a) 137 and-354
137 + (- 354) = -217

b) -52 and 52
-52 + 52 = 0

c) -312, 39 and 192
-312 + 39 + 192 = -312 + 231 = -81

d) -50, -200 and 300
-50 + (-200) + 300 = -250 + 300 = 50

Question 5.
Find the sum:
a) (-7) + (-9) + 4 + 16
(-7) + (-9) + 4 + 16 = -16 + 20 = 4

b) (37) + (-2) + (-65) + (-8)
37 + (-2) + (-65) + (-8) = 37 + (-75) = -38

Students can Download Chapter 6 Integers Ex 6.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 6 Integers Ex 6.1

Question 1.
Write opposites of the following
Solution:
a) Increase in Weight
Decrease in weight

b) 30 km north
30km south

c) 326 BC
326 A.D

d) Loss of Rs 700
Gain of Rs.700

e) 100m above sea level
100m below sea level

Question 2.
Represent the following numbers as integers with appropriate signs
Solution:
a) An aeroplane is flying at a height two thousand metre above the ground
+2000

b) A submarine is moving at a depth, eight hundred metre below the sea level.
– 800

c) A deposit of rupees two hundred
+ 200

d) Withdrawal of rupees seven hundred.
– 700

Question 3.
Represent the following numbers on a number line:
Solution:
a) +5

b) -10

c) +8

d) -1

e) -6

Question 4.
Adjacent figure is a vertical number line, representing integers. Observe it and locate the following points:
Solution:
a) If point D is +8, then which point is -8?
-F

b) Is point G a negative integer or a positive integer?
Negative integer (-6)

c) Write integer for points B and E.
Point B represents 4 and point E represents -10

d) Which point marked on this number line has the least value?
E has the least value as it represents -10

e) Arrange all the list of temperatures of five places in India on a particular day of the year.
D > O > B > A > O > H > G > F > E.

Question 5.
Following is the list of temperatures of five places in India on a particular day of the year.

a) Write the temperatures of these places in the form of integers in the blank column.
b) Following is the number line representing the temperature in degree Celsius.

Plot the name of the city against its temperature.
c) Which is the coolest place?
d) Write the names of the places where temperatures are above 10°C.
Solution :

c) Siachin
d) Delhi, Ahmedabad

Question 6.
In each of the following pairs, Which number is to the right of the other on the number line?
Solution:
a) 2, 9
9(9 > 2)

b) -3, -8
-3 (-3 > -8)

c) 0, -1
0( 0 > -1)

d) -11, 10
10(10 > -11)

e) -6, 6
6(6 > -6)

f) 1, -100
1(1 > -100)

Question 7.
Write all the integers between the given pairs ( Write them in the increasing order)
Solution:
a) 0 and -7
-6, -5, -4, -3, -2, -1

b) -4 and 4
-3, -2, -1, 0, 1, 2, 3

c) -8 and -15
-14, -13, -12, -11, -10, -9

d) -30 and -23 .
-29, -28, -27, -26, -25, -24

Question 8.
a)Write four negative integers greater than -20
Solution:
-19, -18, -17, -16

b) Write four negative integers less than -10
Solution:
– 11, – 12, – 13, – 14

Question 9.
For the following statements, Write True(T) or False(F). If the statement is false, correct the statement
Solution:
a) -8 is to the right of -10 on a number line
True (-8 > -10)

b) -100 is to the right of -50 on a number line.
False (-50 > -100)
-100 is to the left of -50 on a number line.

c) smallest negative integer is -1
False as the greater negative integer is -1

d) – 26 is greater than – 25
False as -26 is smallest than -25

Question 10.
Draw a number line and answer the following:
Solution:
a) Which number will we reach if we move 4 numbers to the right of -2.

We will reach at 2 if we move 4 numbers to the right of -2

b) Which number will we reach if we move 5 numbers to the left of 1.

We will reach at -4, if we move 5 numbers to the left of 1.

c) If we are at -8 on the number line, in which direction should we move to reach -1?

Clearly, -1 is to the right of -8 Therefore, we should move towards the right direction.

d) If we are at -6 on the number line, in Which direction should we move to reach -1?

Clearly, -1 is to the right of -6. Therefore, we should move towards the right direction.

Students can Download Hindi Lesson 25 वनमहोत्सव Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 6 Hindi helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Hindi Chapter 25 वनमहोत्सव

वनमहोत्सव Questions and Answers, Summary, Notes

अभ्यास

1. इनके उत्तर लिखो :

क. स्कूल में कौन – सा उत्स्व मनाया जा रहा था?
उत्तर:
स्कूल में वनमहोत्सव मनाया जा रहा था

ख. विश्व पर्यावरण दिवस किस तरीख को मनाया जाता है?
उत्तर:
जून विश्व पर्यावरण दिवस मनाया जाता है।

ग. लता और गीता ने क्या कहा ?
उत्तर:
लता और गीता ने कहा कि – हम उनके चारों तरफ तारों की बाड़ लगायेंगी।

घ. विश्व पर्यावरण दिवस क्यों मनाया जाता है?
उत्तर:
पर्यावरण के प्रति जागति लाने के लिए यह दिवस मनाया जाता है।

ङ सबसे पहले किनसे पेड़ लगवाया गया ?
उत्तर:
सबसे पहले मुख्याध्यापक से पेड़ लगवाया गया

च. पेड़ लगाने से क्या – क्या लाभ है
उत्तर:
पेड़ लगाने से बहुत सारें लाभ होते हैं जैसे – पेडों से हमें प्राणवायु मिलती है कि व होती है शुद्ध हवा मिलती हैं पेडो से छाया मिलती है मकान बनाने के लिए लकडी मिलती हैं

छ. राजीव ने गुरु जी से क्या कहा ?
उत्तर:
राजीव ने गुरु जी से कहा कि -” मैं हर साल अपना जन्म दिन एक पेड़ लगाकर मनाऊँगा'”

2. कोष्टक में सही या गलत लिखो :

क. हमें पंड़ नहीं काटना चाहिए – सही
ख. पेड़ों से हानि होती है – गलत
ग. पेडों से छाया मिलती है – सही
घ. पेडों से अधिक वर्षा होती है – सही

3. चित्रों को देखो और पेडों के नाम लिखो:

वनमहोत्सव Summary In Hindi

सारांश :
प्रस्तुत गद्य भाग नाटक के रुप में है। इस गद्य भाग का सारांश यह है कि – गुरु जी छात्रों से कहते हैं कि – आज स्कूल में वनमहोत्सव मनाएँगें वे यह भी समझाते हैं कि वनमहोत्सव क्यों मनाना चाहिए? उससे क्या लाभ? आदि प्रश्नों का समाधान भी दिया।

गुरु जी कहने लगे कि वर्ण के दिनों में पौधे लगाने से वे सुखते नहीं, बल्की जल्दी बढ़ते हुए पेड़ बन जाएँगें ज्यादा पेड़ लगाने से अच्छा वर्ण होती है’ पौधों को लगाने के बाद हम रोज उसकी चारों तरफ बाड़ बनाकर उन पौधों को पानी डालना चाहिए वनमहोत्सव कार्यक्रम का आरंभ – मुख्याध्यानक पौधा लगाने के साथ हुआ इसके बाद सभी

अध्यापक गण तथा छात्र पौधों कां लगाए और उन पौधों को पानी भी दिया गुरुजी, पेड़ लगाने से और लाभ के बारे में बताने लगे कि – पेड़ों से हमें प्राणवायु मिलती हैं पेड़ों से वर्ण होती हैं हमें शुद्ध हवा मिलती हैं पेड़ों से छाया मिलती है। इन्हीं पेडों से मकान बनाने के लिए लकड़ी भी मिलती हैं इन बातों को सुन कर राजीव ने कहा कि – वह हर साल अपना जन्म दिन एक पौधा लगाकर मनाएगा गुरु जी सलाह दी कि – “सभी ऐसा ही करेंगे तो यह धरती हरी – भरी हो जाएगी

शब्दार्थ:
चन = जंगल, महोत्सव = बड़ा उत्सव, बाड़ = घेरा, ओर = तरफ, प्राणवायु = अक्सिजन

वनमहोत्सव Summary In English

Teacher: Children, today in our school we celebrate a function to encourage a forestation
Ramesh: Teacher, why should we celebrate forestation?
Teacher: See Ramesh, in the rainy season, if we plant a tree it won’t dry. So, now we celebrating this forestation
Asha: Sir, what is the benefit by doing this plantation?
Teacher: Asha, By trees, we get more rain.
John: Sir, what shall we do?
Teacher: Right question. We protect trees and plants first. We should make place around them and pour water every day.
Rahim & Ramesh: Sir, definitely, we plant the tree.
Seetha & Geetha: Sir, we make the arrangements.
Rakesh: Sir, I will water them every day.
Teacher: Yes, Alright, I am very happy to hear this. Come let us go and do the first plantation from our Headmaster. All: Yes sir, we all plant a one-one tree and water them.
Basanti: Sir, what is the benefit from doing plantation
Teacher: Basanti, very good question. By growing more trees, we get good air, means more oxygen, more trees mean more raining, From trees we get good air and shadow and wood for building construction.
All: Sir, we will celebrate forestation every year.
Rajeev: Sir, I will plant a tree on my birthday every year.
Teacher: Children, you all do like this, our earth will become greenery.

वनमहोत्सव Summary In Kannada

Students can Download Chapter 4 Basic Geometrical Ideas Ex 4.6 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.6

Question 1.
From the figure, identify:
Solution:
a) the centre of circle
O

e) a sector
O.P

f) a segment
Q

g) a sector
AOB (Shaded region)

h) a segment
DE (Shaded region)

Question 2.
a) Is every diameter of a circle also a chord?
Solution:
Yes, the diameter is the longest possible chord of the circle

b) Is every chord of a circle also a diameter?
Solution:
No.

Question 3.
Draw any circle and mark
Solution:
a) its centre
O

b) a radius
\(\overrightarrow{\mathrm{OA}}\)

c) a diameter
\(\overrightarrow{\mathrm{AB}}\)

d) a sector
COA

e) a segment
DE

f) a point in its interior
O

g) a point in its exterior
F

h) an arc

Question 4.
Say true or false
Solution:
a) Two diameters of a circle win necessarily intersect
True, They will along interest each other at the centre of the circle

b) The centre of a circle is always in its interior.
True

Students can Download Chapter 7 Fractions Ex 7.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 7 Fractions Ex 7.3

Question 1.
Write the fractions. Are all the these fractions equivalent?

Solution:
a) In the given circles, 1 out of 2, 2 out of 4, 3 out of 6, and 4 out of 8 equal part of shaded respectively. Therefore these circles represent.
Also, all these fractions are equivalent.
b) In the given rectangles, 4 out of 12, 3 out of 9, 2 out of 6, 1 out of 3, and 6 out of 15 equal parts (i.e, circles) are shaded respectively Therefore these rectangles represent, Not, not all of these fractions are equivalent.

Question 2.
Write the fractions and pair up the equivalent fractions from each row.

Solution:
a) Here, 1 parts is shaded out of 2 equal parts ( i,e., rectangle) Hence this, Figure represents a fraction \(\frac{1}{2}\)
b) Here, 4 parts is shaded out of 6 equal parts ( i,e., rectangle) Hence this, Figure represents a fraction \(\frac{4}{6}\) = \(\frac{2}{3}\)
c) Here, 3 parts is shaded out of9 equal parts ( i,e., rectangle) Hence this, Figure represents a fraction \(\frac{3}{9}\) = \(\frac{1}{3}\)
d) Here, 2 parts is shaded out of 8 equal parts ( i,e., rectangle) Hence this, Figure represents a fraction \(\frac{2}{8}\) = \(\frac{1}{4}\)
e) Here,3 parts is shaded out of4 equal parts ( i,e., rectangle) Hence this, Figure represents a fraction \(\frac{3}{4}\)
i) Here,6 parts are shaded out of 18 equal parts (i.e, triangles) Hence, this figure represents a fraction \(\frac{6}{18}\) = \(\frac{1}{3}\)
ii) Here, 4 parts are shaded out of 8 equal parts (i,e, rectangle) Hence, this figure represents a fraction \(\frac{4}{8}\) = \(\frac{1}{2}\)
iii) Here, 12 parts are shaded out o f 16 equal parts (i.e , squares) Hence, this figure \(\frac{12}{16}\) = \(\frac{3}{6}\)
iv) Here, 8 parts are shaded out of 12 equal parts (i,e . rectangle) Hence, this figure represents a fraction \(\frac{8}{12}\) = \(\frac{2}{3}\)
v) Here,4 parts are shaded out of 16equal parts (i,e , triangles) Hence, this figure represents a fraction \(\frac{4}{16}\) = \(\frac{1}{4}\)
Now these figures can be, matched correctly as (a)(ii), (b)(iv), (c)(i), (d)(v), (e)(iii).

Question 3.
Replace and mnsq² in each of the following by the correct number:

Solution:

Hence, & mnsq² Can be replaced by 28.

Hence, & mnsq² Can be replaced by 16.

Hence, & mnsq² Can be replaced by 12.

Hence, & mnsq² Can be replaced by 20.

Hence, & mnsq² Can be replaced by 3.

Question 4.
Find the equivalent fraction of \(\frac{3}{5}\) having
a) denominator 20
b) numerator 9
c) denominator 30
d) numerator 27
Solution:
a)


Hence, the required fraction is \(\frac{12}{20}\)

b)


Hence, the required fraction is \(\frac{18}{30}\)

d)


Hence, the required fraction is \(\frac{27}{45}\)

Question 5.
Find the equivalent fraction of \(\frac{36}{48}\) with
a) numerator 9
b) denominator 4
Solution:
a)


Hence, the required fraction is \(\frac{3}{4}\)

Question 6.
Check whether the given fractions are equivalent:

Solution:
a)


Clearly, both the fractions are equivalent.

b)


Clearly, both the fractions are not equivalent.

c)


clearly, both the fractions are not equivalent

Question 7.
Reduce the following fractions to simplest form:

Solution:

Question 8.
Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used tip 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/ his pencils?
Solution:
Fraction used by Ramesh = \(\frac{10}{20}=\frac{1}{2}\)
Fraction used by sheelu = \(\frac{25}{50}=\frac{1}{2}\)
Fraction used by Jamaal = \(\frac{40}{80}=\frac{1}{2}\)
Yes, all of them used equal fraction of pencils, i.e., \(\frac{1}{2}\)

Question 9.
Match the equivalent fractions and write two more for each.

Solution:
i)

Two more fractions are \(\frac{25}{40}, \frac{30}{48}\)

ii)

Two more fractions are \(\frac{18}{20}, \frac{27}{30}\)

iii)

Two more fractions are \(\frac{20}{30}, \frac{200}{300}\)

iv)

Two more fractions are \(\frac{20}{40}, \frac{30}{60}\)

v)

Two more fractions are \(\frac{20}{40}, \frac{40}{100}\)
Now these can be matched as
(i) – (d),
(ii) – (e),
(iii) – (a),
(iv) – (c),
(v) – (b)

Students can Download Chapter 7 Fractions Ex 7.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 7 Fractions Ex 7.2

Question 1.
Draw number lines and locate the points on them:

Solution:

Question 2.
Express the following as mixed fraction:

Solution:

Question 3.
Express the following as improper fractions:

Solution:

Students can Download Chapter 7 Fractions Ex 7.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 7 Fractions Ex 7.1

Question 1.
Write the fraction representing the shaded portion.

i) The given figure represents 2 shaded parts out of 4 equal parts. Hence \(\frac{2}{4}\)
ii) The given figure represents 8 shaded parts out of 9 equal parts. Hence \(\frac{8}{9}\)
iii) The given figure represents 4 shaded parts out of 8 equal parts. Hence \(\frac{4}{8}\)
iv) The given figure represents 1 shaded parts out of 4 equal parts. Hence \(\frac{1}{4}\)
v) The given figure represents 3 shaded parts out of 7equalparts. Hence \(\frac{3}{7}\)
vi) The given figure represents 3 shaded parts out of 12 equal parts. Hence \(\frac{3}{12}\)
vii) The given figure represents 10 shaded parts out of 10 equal parts. Hence \(\frac{10}{10}\)
viii) The given figure represents 4 shaded parts out of 9 equal parts. Hence \(\frac{4}{9}\)
xi) The given figure represents 4 shaded parts out of 8 equal parts. Hence \(\frac{4}{8}\)
x) The given figure represents 1 shaded parts out of 2 equal parts. Hence \(\frac{1}{2}\)

Question 2.
Colour the part according to the given fraction.

Solution:

Question 3.
Identify the error, if any

Solution :
The given figures do not represent the fractions as here each shape is not divided in equal parts.

Question 4.
What fraction of a day is 8 hours?
Solution:
There are 24 hour in a day. Therefore 8, hours of a day represent \(\frac{8}{24}\)

Question 5.
What fraction of an hour is 40 minutes?
Solution:
There are 60 minutes in a hour. Therefore 40, minutes of an hour represent \(\frac{40}{60}\)

Question 6.
Arya, Abhimanyu, and vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sand wiches so that each person will have an equal share of each sandwich.
a) How can Ary a divide his sandwiches so that each person has an equal share?
b) What part of a sand wich will each boy receive?
Solution:
a) Arya will divide each sandwich in three equal parts then,he will give one part of each sandwich to each one of them.
b) each boy will receive \(\frac{1}{3}\) part each sandwich.

Question 7.
Kanchan dyes dresses, she had to day 30 dresses. she has so far finished 20 dresses. What fraction of dresses has she finished?
Solution:
Dress dyed so far = 20
Total dress = 30

Question 8.
Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Solution:
Natural numbers from 2 to 12 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, & 12
Prime numbers among these are 2, 3, 5, & 11
Therefore, out of 11 numbers, 5 are prime numbers. It represents a fraction \(\frac{5}{11}\)

Question 9.
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Solution:
Natural numbers from 102 to 113 are 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113.
Among these numbers, the prime numbers are 103, 107, 109 and 113
Therefore, out of 12 numbers, 4 are prime numbers, It represents a fraction \(\frac{4}{12}\)

Question 10.
What fraction of these circles have X’s in them?

Solution:
There are 4 circle, out of 8, having X’s in them,
Therefore it represents a fraction \(\frac{4}{8}\)

Question 11.
Kristin received a CD player for her birthday. She bought 3 CD s and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?
Solution:
Total CDs Kristin had on her birthday 3 + 5 = 8
out of 8 CDs, She bought 3 CDs and also got 5 CDs as gifts
Therefore, she bought and received CDs as gifts in a fraction of \(\frac{3}{8}\) and \(\frac{5}{8}\) respectively

Students can Download Hindi Lesson 21 चतुर बंदर Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 6 Hindi helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Hindi Chapter 21 चतुर बंदर

चतुर बंदर Questions and Answers, Summary, Notes

अभ्यास

1. उत्तर लिखो :

क. किन – किन में गहरी दोस्ती थी ?
उत्तर:
गहरी दोस्ती बंदर तथा मगरमच्छ में थी

ख. नदी की सैर कौन कराता था ?
उत्तर:
नदी की सैर मगरमच्छ कराता था

ग. बंदर मगरमच्छ को क्या देता था ?
उत्तर:
बंदर मगरमच्छ को जामुन देता था

घ. मगरमच्छ की पत्नी ने क्या माँगा ?
उत्तर:
मगरमच्छ की पत्नी ने बंदर का कलेजा माँगा

च. मगरमच्छ बंदर से क्या बोला ?
उत्तर:
मगरमच्छ बंदर से बोला कि -” आज तुम्हारी भाभी ने बढिया पकवान बनाए हैं और तुम घर बुलाया है।

छ. चतुर बंदर ने मगरमच्छ से क्या कहा ?
उत्तर:
चतुर बंदर मगरमच्छ से कहा कि – :”अरे मित्र ! मेरा कलेजा तो पेड़ की कोटर में ही रखा हैं चलो भाभी के लिए कलेजा ले आएँ ”

2. निक्तस्थानों में सही शब्द भरो :

क. बंदर पेड़ पर रहता था (पर / में)
ख. नदी में मगरमच्छ तैरता था (में / पर)
ग. बंदर मगरमच्छ को जामुन देता (देता / देती)
घ. बंदर मीठे जामुन खाता था। (केला / जामून)
ङ. उसका कलेजा कितना मीठा होगा? (उसका / उसकी)

3. श्ब्दों को पूरा करो

1. म ग रमच्छ
2. जामु न
3. रो ज
4. भा भी
5. पे ड़
6. क ले जा
{पेड़, रोज़, जामुन, मगरमच्छ, मित्र, कलेजा }

4. नमूने की तरह नीचे दिए गए शब्दों की सूची में एक वर्गेतर श्ब्द है, उस पर गोला लगाओ :

5. नीचे दिये गये जीवियों में कौन पेड पर चढ सकता है? निशान लगाओ :

चतुर बंदर Summary In Hindi

सारांश:
एक नदी के किनारे लगे जामुन के पेड़ पर एक बंदर रहता था उसी नदी में एक मगरमच्छ भी रहता था इन दोनों में गहरी दोस्ती थी मगरमच्छ, बंदर को अपनी पीठ पर बिठाकर नदी की सेर कराता था बदले में बंदर भी मगर को मीठे – मीठे जामुन तोड़कर देता था एक दिन मगरमच्छ कुछ जामुन अपनी पत्नी के लिए ले गया पत्नी जामुन खाकर खुश हुई उसने कहा – ” आपका मित्र बंदर, रोज़ मीठे जामुन खाता है, तो उसका कलेजा कितना मीठा होगां मैं तो उसका कलेजा ही खाऊँगी”.

पत्नी की ज़िद पर मगरमच्छ पेड़ के पास पहुँचा और बंदर से बोला ” आज तुम्हारी भाभी ने बढ़िया – बढ़िया पकवान बनाए हैं और तुम्हे घर बुलाया है ” मगरमच्छ की बात सुनकर बंदर झट से उसकी पीठ पर बैठ गया आधी नदी पार हो जाने पर उसने बंदर को सच्चाई बता दी – सच्चाई सुनकर बंदर ने चतुराई से काम लिया और बोला – ” अरे मित्र ! मेरा कलेजा तो पेड़ की कोटर में ही रखा है ” यह सुनकर मगर उसे वापस पेड़ के पास ले गया।

पेड़ पर पहुँचते ही बंदर मगरमच्छ से बोला – ” अरे मूर्ख ! इतना भी नहीं जानता कि कलेजा तो अपने भीतर ही होता हैं पत्नी की बातों में आकर अपने मित्र को मारने चला था आज से हमारी मित्रता टूट गई.” मगरमच्छ अपना मैं लटकाए घर चला गया

शब्दार्थ :
गहरी = घनिष्ट, जिद = हठ, बढ़िया = अच्छा, भाभी = बड़े भाई की पत्नी, कलेजा = हृदय, पकवान = मिष्टान, झट = जल्दी, चतुर = होशियार, कोटर =पेड़ का खोखला भाग, मित्र = दोस्त, भीतर = अंदर मुँह लटकाना = निराश होना

चतुर बंदर Summary In English

Once upon a time, there was two friends monkey and a Crocodile. They were best friends. Monkey lived on a tree near river bank and Crocodile in the river. Monkey gives sweet fruits to the Crocodile from the trees. Crocodile takes the ride to the monkey on his back.

Both are very happy. Crocodile takes some fruits to his house and gives to his wife. Once wife asks with the Crocodile that, the fruits given by the monkey are very sweet. So that monkey are very sweet. So that monkey’s heart may be sweeter than this. I want monkey’s heart.

Crocodile went to the river bank and said to the monkey that his wife wants to give a party to him. Monkey happily sit on the back of the Crocodile. On the midway, the crocodile disclosed the matter that his wife wants to cat monkey’s heart. Monkey told that he kept his heart in the tree nest.

So let’s go back and bring the heart. They returned back. When they reached the riverbank, the monkey jumped on the tree and told the crocodile that you fool crocodile by your’s wife wish you are ready to kill me. Are you a friend? Crocodile return back with sad.

चतुर बंदर Summary In Kannada

Students can Download Chapter 3 Playing with Numbers Ex 3.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 3 Playing with Numbers Ex 3.5

Question 1.
Which of the following statements are true?
a) If a number is divisible by 3, it must be divisible by 9.
b) If a number is divisible by 9, it must be divisible by 3.
c) A number is divisible by 18, if it is divisible by both 3 and 6.
d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
e) If two numbers are co-primes, at least one of them must be prime.
f) All numbers which are divisible by 4 must also be divisible by 8.
g) All numbers which are divisible by 8 must also be divisible by 4.
h) If a number exactly divides two numbers separately, it must exactly divide their sum.
i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Solution:
(a) False
6 is divisible by 3, but not by 9.

(b) True, as 9 = 3 × 3
Therefore, if a number is divisible by 9, then it will also be divisible by 3.

(c) False
30 is divisible by 3 and 6 both, but it is not divisible by 18.

(d) True, as 9 × 10 = 90
Therefore, if a number is divisible by 9 and 10 both, then it will also be divisible by 90.

(e) False
15 and 32 are co-primes and also composite.

(f) False
12 is divisible by 4, but not by 8.

(g) True, as 8 = 2 × 4
Therefore, if a number is divisible by 8, then it will also be divisible by 2 and 4.

(h) True
2 divides 4 and 8 as well as 12. (4 + 8 = 12)

(i) False
2 divides 12, but does not divide 7 and 5.

Question 2.
Here are two different factor trees for 60. Write the missing numbers.
Solution:

Question 3.
Which factors are not included in the prime factorisation of a composite number ?
Solution:
1 and the number itself.

Question 4.
Write the greatest 4-digit number and express it in terms of its prime factors.
Solution:
Greatest four-digit number = 9999
9999 = 3 × 3 × 11 × 101

Question 5.
Write the smallest 5-digit number and express it in the form of its prime factors.
Solution:

10,000
= 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

Question 6.
Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.
Solution:
1729 = 7 × 13 × 19
13 – 7 = 6, 19 – 13 = 6

The difference of two consecutive prime factors is 6.

Question 7.
The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Solution:
2 × 3 × 4 = 24, which is divisible by 6
9 × 10 × 11 = 990, which is divisible by 6
20 × 21 × 22 = 9240, which is divisible by 6

Question 8.
The sum of two consecutive odd numbers is divisible by 4. Verily this statement with the help of some examples.
Solution:
3 + 5 = 8, which is divisible by 4
15 + 17 = 32, which is divisible by 4
19 + 21 = 40, which is divisible by 4

Question 9.
In which of the following expressions, prime factorisation has been done?
(a) 24 = 2 × 3 × 4
(b) 56 = 7 × 2 × 2 × 2
(c) 70 = 2 × 5 × 7
(d) 54 = 2 × 3 × 9
Solution:
a) 24 = 2 × 3 × 4
Since 4 is composite, prime factorisation has not been done.

(b) 56 = 7 × 2 × 2 × 2
Since ail the factors are prime, prime factorisation has been done.

(c) 70 = 2 × 5 × 7
Since all the factors are prime, prime factorisation has been done.

(d) 54 = 2 × 3 × 9
Since 9 is composite, prime factorisation has not been done.

Question 10.
Determine if 25110 is divisible by 45.
|Hint: 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9|.
Solution:
45 = 5 × 9
Factors of 5 = 1, 5
Factors of 9 = 1, 3, 9
Therefore, 5 and 9 are co-prime numbers. Since the last digit of 25110 is 0, it is divisible by 5.
Sum of the digits of 25110 = 2 + 5 + 1 + 1 + 0 = 9
As the sum of the digits of 25110 is divisible by 9. therefore, 25110 is divisible by 9.
Since the number is divisible by 5 and 9 both, it is divisible by 45.

Question 11.
18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 × 6 = 24? If not, give an example to justify your answer.
Solution:
No. It is not necessary because 12 and 36 are divisible by 4 and 6 both, but are not divisible by 24.

Question 12.
I am the smallest number, having four different prime factors. Can you find me?
Solution:
Since it is the smallest number of such type, it will be the product of 4 smallest prime numbers.
2 × 3 × 5 × 7 = 210

Students can Download Chapter 3 Playing with Numbers Ex 3.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 3 Playing with Numbers Ex 3.4

Question 1.
Find the common factors of:
a) 20 and 28
b) 15 and 25
c) 35 and 50
d) 56 and 120
Solution:

Question 2.
Find the common factors of
a) 4, 8 and 12
b) 5, 15 and 25
Solution:

Question 3.
Find first three common multiples of:
a) 6 and 8
b) 12 and 18
Solution:
a) 6 and 8:-

3 common multiples = 24, 48, 72

b) 12 and 18:-
Multiples of 12 = 12, 24, 36, 48, 60, 72
Multiples of 18= 18, 36, 54, 72
3 Common multiples = 36, 72, 108

Question 4.
Write all the number less than 10(1 which are common multiples of 3 and 4
Solution:
Multiples of 3 = 3, 6, 0, 12, 15, 18, 21, 24, 27, 30
Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
Common multiples = 12, 24, 36, 48, 60, 72, 84, 96

Question 5.
Which of the following numbers are co-prime?
a) 18 and 35
b) 15 and 37
c) 30 and 415
d) 17 and 68
e) 216 and 215
f) 81 and 16
Solution:
a) Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 35 = 1, 5, 7, 35
Common factors = 1
Therefore, the given two numbers are co-prime

b) Factors of 15 = 1, 3, 5, 15
Factors of 37 = 1, 37
Common factors = 1
The given two numbers are co-prime

c) Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
Factors of 415 = 1, 5, 83, 415
Common factors = 1, 5
As these numbers have a common factors other than 1, the given two numbers arc not co-prime

d) Factors of 17 = 1, 17
Factors of 68 = 1, 2, 4, 17, 34, 68
Common factors = 1, 17
As these numbers have a common factors other than 1, the given two numbers are not co-prime

e) Factors of 216 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 17, 36, 54, 72, 108, 216
Factors of 215 = 1, 5, 43, 215
Common factors = 1
The given two numbers are prime.

f) Factors of 81= 1, 3, 9, 27, 81
Factors of 16 = 1, 2, 4, 8, 16
Common factors = 1
Therefore the given two numbers are co-prime.

Question 6.
A number is divisible by both 5 and 12. By which other number will that number be always divisible?
Solution:
Factors of 5 = 1, 5
Factors of 12 = 1, 2, 3, 4, 5, 6, 12
As the common factors of these number is 1, the given two numbers are co-prime and the number will also be divisible by their product , i,e 60 and the factors of 60.
i.e, 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

Question 7.
A number is divisible by 12. By what other numbers will that number be divisible?
Solution:
Since, the number is divisible by 12, it will also be divisible 1,2,3,4 and 6 are numbers other than 12 by which this number is also divisible.