NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-2/

BoardCBSE
TextbookNCERT
ClassClass 6
SubjectMaths
ChapterChapter 2
Chapter NameWhole Numbers
Exercise Ex 2.1
Number of Questions Solved8
CategoryNCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1

Question 1.
Write the next three natural numbers after 10999.
Solution :
The next three natural numbers after 10999 are 11000, 11001 and 11002.

Question 2.
Write the three whole numbers occurring just before 10001.
Solution :
The three whole numbers occurring just before 10001 are 10000, 9999 and 9998.

Question 3.
Which is the smallest whole number?
Solution :
0 is the smallest whole number.

Question 4.
How many whole numbers are there between 32 and 53?
Solution :
There are 20 whole numbers between 32 and 53.
These are 33, 34, 35, 36, 37, 38, 39, 40, 41, 42,43, 44, 45, 46, 47,48, 49, 50, 51 and 52.

Question 5.
Write successor of:
(a) 2440701
(b) 100199
(c) 1099999
(d) 2345670
Solution :
(a) The successor of 2440701 is 24,40,702.
(b) The successor of 100199 is 1,00,200.
(c) The successor of 1099999 is 11,00,000.
(d) The successor of 2345670 is 23,45,671.

Question 6.
Write the predecessor of:
(a) 94
(b) 10000
(c) 208090
(d) 7654321.
Solution :
(a) The predecessor of 94 is 93.
(b) The predecessor of 10000 is 9,999.
(c) The predecessor of 208090 is 2,08,089.
(d) The predecessor of 7654321 is 76,54,320.

Question 7.
In each of the following pairs of numbers, the state which the whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.
(a) 530, 503
(b) 370, 307
(c) 98765, 56789
(d) 9830415, 10023001.
Solution :
(a) The whole number 503 is on the left of the whole number 530 on the number line. So, 503 < 530.
(b) The whole number 307 is on the left of the whole number 370 on the number line. So, 307 < 370.
(c) The whole number 56789 is on the left of the whole number 98765 on the number line. So, 56789 < 98765. ;
(d) The whole number 9830415 is on the left of the whole number 10023001 on the number line. So, 98,30,415 < 100,23,001.

Question 8.
Which of the following statements are true (T) and which are false (F)?
(a) Zero is the smallest natural number.
(b) 400 is the predecessor of 399.
(c) Zero is the smallest whole number.
(d) 600 is the successor of 599.
(e) All natural numbers are whole numbers.
(f) All whole numbers are natural numbers,
(g) The predecessor of a two digit number is never a single digit number.
(h) 1 is the smallest whole number.
(i) The natural number 1 has no predecessor,
(j) The whole number 1 has no predecessor.
(k) The whole number 13 lies between 11 and 12.
(l) The whole number 0 has no predecessor.
(m) The successor of a two digit number is always a two digit number.
Solution :
(a) This statement is false (F).
(b) This statement is false (F).
(c) This statement is true (T).
(d) This statement is true (T).
(e) This statement is true (T).
(f) This statement is false (F).
(g) This statement is false (F).
(h) This statement is false (F).
(i) This statement is true (T).
(j) This statement is false (F).
(k) This statement is false (F).
(l) This statement is true (T).
(m) This statement is false (F).

 

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NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-1/

BoardCBSE
TextbookNCERT
ClassClass 6
SubjectMaths
ChapterChapter 1
Chapter NameKnowing Our Numbers
Exercise Ex 1.1
Number of Questions Solved4
CategoryNCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1

Question 1.
Fill in the blanks:
(a) 1 lakh = …………………. ten thousand.
(b) 1 million = …………………. hundred thousand.
(c) 1 crore = …………………. ten lakh.
(d) 1 crore = …………………. million.
(e) 1 million = …………………. lakh.
Solution :
(a) 1 lakh = ten ten thousand.
(b) 1 million = ten hundred thousand.
(c) 1 crore = ten ten lakh
(d) 1 crore = ten million
(e) 1 million = ten lakh

Question 2.
Place commas correctly and write the numerals :
(a) Seventy-three lakh seventy’ five thousand three hundred seven.
(b) Nine crore five lakh forty-one.
(c) Seven crores fifty-two lakh twenty-one thousand three hundred two.
(d) Fifty-eight million four hundred twenty-three thousand two hundred two.
(e) Twenty-three lakh thirty thousand ten.
Solution :
(a) 73,75,307
(b) 9.05.00.041
(c) 7,52.21,302
(d) 58.423.202
(e) 23.30,010.

Question 3.
Insert commas suitably and write the names according to Indian System of Numeration :
(a) 87595762
(b) 8546283
(c)99900046
(d) 98432701.
Solution :
(a) 8, 75, 95, 762. Eight crores seventy-five lakh ninety-five thousand seven hundred and sixty-two.
(b) 85, 46, 283. Eighty-five lakh forty-six thousand two hundred and eighty-three.
(c) i 9, 99, 00, 046. Nine crore ninety-nine lakh and forty-six.
(d) 9, 84, 32, 701. Nine crore eighty-four lakh thirty-two thousand seven hundred and one.

Question 4.
Insert commas suitably and write the names according to International System of Numeration :
(a) 78921092
(b) 7452283
(c) 99985102
(d) 48049831.
Solution :
(a) 78,921,092. Seventy-eight million nine hundred twenty-one thousand and ninety-two.
(b) 7.452,283. Seven million four hundred fifty-two thousand two hundred and eighty-three.
(c) 99. 985. 102. Ninety-nine million nine hundred eighty-five thousand one hundred and two.
(d) 48. 049. 831. Forty-eight million forty-nine thousand eight hundred and thirty-one.

 

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NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.1

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.1. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-3/

BoardCBSE
TextbookNCERT
ClassClass 6
SubjectMaths
ChapterChapter 3
Chapter NamePlaying With Numbers
Exercise Ex 3.1
Number of Questions Solved4
CategoryNCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.1

Question 1.
Write all the factors of the following numbers:
(i) 24
(ii) 15
(iii) 21
(iv) 27
(v) 12
(vi) 20
(vii) 18
(viii) 23
(ix) 36.
Solution :
(i) 24
24 = 1 × 24
24 = 2 × 12
24 = 3 × 8
24 = 4 × 6
24 = 6 × 4
Stop here, because 4 and 6 have occurred earlier. Thus, all the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

(ii) 15
15 = 1 × 15
15 = 3 × 5
15 = 5 × 3
Stop here, because 3 and 5 have occurred earlier. Thus, all the factors of 15 are 1, 3, 5 and 15.

(iii) 21
21 = 1 × 21
21 = 3 × 7
21 = 7 × 3
Stop here, because 3 and 7 have occurred earlier. Thus, all the factors of 21 are 1, 3, 7 and 21.

(iv) 27
27 = 1 × 27
27 = 3 × 9
27 = 9 × 3
Stop here, because 3 and 9 have occurred earlier. Thus, all the factors of 27 are 1, 3, 9 and 27.

(v) 12
12 = 1 × 12
12 = 2 × 6
12 = 3 × 4
12 = 4 × 3
Stop here, because 3 and 4 have occurred earlier. Thus, all the factors of 12 are 1,2,3,4,6 and 12.

(vi) 20
20 = 1 × 20
20 = 2 × 10
20 = 4 × 5
20 = 5 × 4
Stop here, because 4 and 5 have occurred earlier. Thus, all the factors of 20 are 1, 2, 4, 5, 10 and 20.

(Vii) 18
18 = 1 × 18
18 = 2 × 9
18 = 3 × 6
18 = 6 × 3
Stop here, because 3 and 6 have occurred earlier. Thus, all the factors of 18 are 1,2,3,6,9 and 18.

(viii) 23
23 = 1 × 23
Thus, all the factors of 23 are 1 and 23.

(ix) 36
36 = 1 × 36
36 = 2 × 18
36 = 3 × 12
36 = 4 × 9
36 = 6 × 6
36 = 6 × 6
Stop here, because both the factors 6 and 6 have occurred earlier. Thus, all the factors of 36 are 1,2,3,4,6,9,12, 18 and 36.

Question 2.
Write the first five multiples of:
(i) 5
(ii) 8
(iii) 9.
Solution :
(i) 5
First, five multiples of 5 are 5 × 1, 5 × 2, 5 × 3, 5 × 4 and 5 × 5
i. e.,5, 10, 15, 20 and 25.

(ii) 8
First, five multiples of 8 are 8×1,8×2,8×3,8 × 4 and 8×5
i.e., 8, 16, 24, 32 and 40.

(iii) 9
First, five multiples of 9 are 9 × 1,9 × 2,9 × 3,9 × 4 and 9×5
i.e., 9, 18,27, 36 and 45.

Question 3.
Match the items in column 1 with the items in column 2 :
Solution :

Question 4.
Find all the multiples of 9 up to 100.
Solution :
The multiples of 9 are
9 × 1, 9 × 2, 9 × 3. 9 × 4, 9 × 5, 9 × 6, 9 × 7, 9 × 8, 9 × 9, 9 × 10, 9 × 11, 9 × 12……….
i.e.. 9, 18, 27, 36, 45, 54. 63, 72, 81, 90, 99, 108,
Thus, all the multiples of 9 up to 100 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.

 

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NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.1

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.1. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-4/

BoardCBSE
TextbookNCERT
ClassClass 6
SubjectMaths
ChapterChapter 4
Chapter NameBasic Geometrical Ideas
Exercise Ex 4.1
Number of Questions Solved6
CategoryNCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.1

Question 1.
Use the figure to name:
(a) Five points
(b) A line
(c) Four rays
(d) Five line segments.
Solution :
(a) O, B, C. D, E
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 1
Question 2.
Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.
Solution :
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 2
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 3

Question 3.
Use the figure to name:
(a) The line containing point E.
(b) The line passing through A.
(c) The line on which O lies
(d) Two pairs of intersecting lines.
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 4
Solution :
(a) \(\overleftrightarrow { AE }\), etc.
(b) \(\overleftrightarrow { AE }\), etc.
(c) \(\overleftrightarrow { CO } or\overleftrightarrow { OC }\)
(d) \(\overleftrightarrow { CO, } \overleftrightarrow { AE } ;\overleftrightarrow { AE } ,\overleftrightarrow { EF }.\)

Question 4.
How many lines can pass through
(a) one given point?
(b) two given points?
Solution :
(a) Countless lines can pass through one given point.
(b) One and only one line can pass through two given points.

Question 5.
Draw a rough figure and label suitably in each of the following cases :
(a) Point P lies on \(\bar { AB } \)
(b) \(\overleftrightarrow { XY }\) and \(\overleftrightarrow { PQ }\) intersect at M.
(c) Line contains E and F but not D.
(d) \(\bar { Op } \) and \(\bar { OQ } \) meet at O.
Solution :
(a)
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 5
(b)
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 6
(c)
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 7
(d)
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 8
Question 6.
Consider the following figure of line \(\bar { MN } \) Say whether following statements are true or false in context of the given figure.
(a) Q, M, O, N, P are points on the line \(\bar { MN } \)
(b) M, O, N are points on a line segment \(\bar { MN } \).
(c) M and N are end points of line segment \(\bar { MN } \).
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas 9
(d) O and N are end points of line segment \(\bar { OP } \).
(e) M is one of the end points of line segment \(\bar { QO } \).
(f) M is point on ray \(\overrightarrow { OP }\).
(g) Ray \(\overrightarrow { OP }\) is different from ray \(\overrightarrow { QP }\).
(h) Ray \(\overrightarrow { OP }\) is same as ray \(\overrightarrow { OM }\).
(i) Ray \(\overrightarrow { OM }\) is not opposite to ray \(\overrightarrow { OP }\).
(j) O is not an initial point of \(\overrightarrow { OP }\).
(k) N is the initial point of \(\overrightarrow { NP }\) and \(\overrightarrow { NM }\).
Solution :
(a) True
(b) True
(c) True
(d) False
(e) False
(f) False
(g) True
(h) False
(i) False
(j) False
(k) True.

 

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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-3/

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 3
Chapter NamePair of Linear Equations in Two Variables
ExerciseEx 3.1
Number of Questions Solved3
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Ex 3.1 Class 10 Question 1.
Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting)? Represent this situation algebraically and graphically.
Solution:
Let present age of Aftab = x years and present age of Aftab’s daughter = y years.
Ex 3.1 Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
1st Condition :
Seven years ago
x – 7 = 7(y – 7)
⇒ x – 7 = 7y – 49
⇒ x – 7y = – 42
Table :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

2nd Condition :
Three years later,
x + 3 = 3(y + 3)
x + 3 = 3y + 9
x – 3y = 6
Table :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3
Thus, the algebraic equations are
x – 7y + 42 = 0 and x – 3y – 6 = 0

Students can also visit Linear Equations In Two Variables Class 10 Practice Set 1.1 here you can see more solved problems.

Math Class 10 Ex 3.1 Question 2.
The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.
Solution:
Let cost of one bat = ₹ x
and the cost of one ball = ₹y
A.T.Q.
1st Condition :
3x + 6y = 3900
Table :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
2nd Condition :
x + 3y = 1300
Table :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5
Thus, the algebraic equations are 3x + 6y = 3900 and x + 3y – 1300
Math Class 10 Ex 3.1 NCERT Solutions Chapter 3

Class 10 Maths Chapter 3 Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.
Solution:
Let cost of one kg of apples = ₹ x and the cost of one kg of grapes = ₹y
Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1
A.T.Q.
1st Condition :
2x + y = 160
Table :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.8
2nd Condition :
4x + 2y = 300
Table :
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.9
Thus, algebraic situations are 2x + y = 160 and 4x + 2y = 300

We hope the NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1, drop a comment below and we will get back to you at the earliest.