CBSE Class 6

These NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers InText Questions

NCERT In-text Question Page No. 28

Question 1.
Write the predecessor and successor of 19; 1997; 12000; 49; 100000.
Answer:
Predecessor means the number before the given number and Successor is just opposite of it, the number after given number.

Question 2.
Is there any natural number that has no predecessor?
Answer:
Yes, the smallest natural number 1 has no predecessor.

Question 3.
Is there any natural number which has no successor? Is there a last natural number?
Answer:
(i) No, there is no natural number which has no successor.
(ii) No, there is no last natural number.

NCERT In-text Question Page No. 29

Question 1.
Are all natural numbers also whole numbers?
Answer:
Yes, all natural numbers are whole numbers.

Question 2.
Are all whole numbers also natural numbers?
Answer:
No, all whole numbers are not natural numbers. Because 0 is a whole number but it is not a natural number.

Question 3.
Which is the greatest whole number?
Answer:
Since, every whole number has a successor. .’. There is no greatest whole number.

NCERT In-text Question Page No. 30

Question 1.
Find 4 + 5; 2 + 6; 3 + 5 and 1 + 6 using the number line.
Answer:
(i) 4 + 5

Let us start from 4. Since, we have to add 5 to this number, we make 5 jumps to the right. Each jump being equal to 1 unit. After five jumps we reach at 9 (as shown above).
∴ 4 + 5 = 9

(ii) 2 + 6

Let us start from 2. Since, we have to add 6 to this number, we make 6 equal jumps, each jump being equal to 1 unit, to the right and reach to 8.
∴ 2 + 6 = 8

(iii) 3 + 5

We have to add 5 to 3.
∴ start from 3. We make 5 equal jumps. Each jump being equal to 1 unit (as shown in the figure) to the right and reach to 8.
∴ 3 + 5 = 8.

(iv) 3 + 5

As we have to add 6 to 1, therefore, we start from 1 and make 6 equal jumps to the right. Each jump being equal to 1 unit.
We reach to 7.
∴ 1 + 6 = 7

Question 2.
Find 8 -3; 6 -2; 9-6 using the number line.
Answer:
(i) 8 – 3

To subtract 3 from 8, start from 8 and make 3 equal jumps towards left. Each jump being equal
i to 1 unit.
So, we reach at 5,

(ii) 6 – 2

To subtract 2 from 6, we start from 6. Make 2 equal jumps towards left. Each jump being equal
to 1 unit.
So, we reach at 4.

(iii) 9 – 6

To subtract 6 from 9, we start from 9. Make 6 equal jumps towards left. Each jump being equal to 1 unit.
So, we reach at 3.

NCERT In-text Question Page No. 31

Question 1.
Find 2 x 6; 3 x 3; 4 x 2 using the number line.
Answer:
(i) 2 x 6

Starting from 0, move 2 units at a time to the right.
Make 6 such moves. So, we reach at 12
∴ 2 x 6 = 12

(ii) 3 x 3

Starting from 0, move 3 units at a time to the right. Make 3 such moves. So, we reach at 9,
∴ 3 x 3 = 9

(iii) 4 x 2

Starting from 0, move 4 units at a time to the right. Make 2 such moves.
So, we reach at 8,
∴ 4 x 2 = 8

NCERT In-text Question Page No. 37

Question 1.
Find: 7 + 18 + 13; 16 + 12 + 4
Answer:
(i) 7 + 18 + 13 = (7 + 13) + 18
= 20 + 18 = 38
(ii) 16 +12 + 4 = (16 + 4) + 12
= 20 + 12 = 32

Question 2.
Find:
25 x 8358 x 4; 625 x 3759 x 8
Answer:
(i) 25 x 8358 x 4 = (25 x 4) x 8358
(Using associativity of whole numbers)
= (100) x 8358 = 835800

(ii) 625 x 3759 x 8 = (625 x 8) x 3759 (Using associativity of whole numbers)
= 500 x 3759
= 5 x 1000 x 3759
= (3759 x 5) x 1000
= 18795 x 1000 = 18795000
∴ 625 x 3759 x 8 = 18795000

NCERT In-text Question Page No. 39

Question 1.
Find 15 x 68; 17 x 23; 69 x 78 + 22 x 69 using distributive property.
Answer:
(i) 15 x 68 = (10 + 5) x 68
= (10 x 68) + (5 x 68)
(By distributivity of multiplication over addition)
= 680 + 340 = 1020

(ii) 17 x 23 = 17 x (20 + 3)
= (17 x 20)+ (17 x 3)
= (17 x 20)+ (17 x 3)
(By distributivity of multiplication over addition)
= 340 + 51 = 391

(iii) 69 x 78 + 22 x 69 = 69 [78 + 22]
= 69 [100]
= 6900
Thus, 69 x 78 + 22 x 69 = 6900

NCERT In-text Question Page No. 42

Question 1.
Which numbers can be shown only as a line?
Answer:
The numbers 2, 5, 7, 11, 13, 14, 17, 19, … can be shown only as a line.

Question 2.
Which can be shown as squares?
Answer:
The numbers 4, 9, 16, 25 … can be shown as squares.

Question 3.
Which can be shown as rectangles?
Answer:
The numbers like 4,6, 8,9,10,12,… can be shown as rectangles.

Question 4.
Write down the first seven numbers that can be arranged as triangles, e.g. 3, 6,.
Answer:
We have

Thus, the first seven triangular numbers are: 3, 6, 10, 15, 21, 28 and 36.

Question 5.
Some numbers can be shown by two rectangles, for example.
There can be many such examples. Some of them are as follows:

Give at least five other such examples.
Answer:
There can be many such examples. Some of them are as follows:

These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exercise 14.6

Question 1.
Draw ∠POQ of measure 75° and find its line of symmetry.
Answer:

Steps of construction:
(a) Draw a line l and mark a point O on it.
(b) Place the pointer of the compasses at O and draw an arc of any radius which intersects the line l at A.
(c) Taking same radius, with centre A, cut the previous arc at B.
(d) Join OB, then ∠BOA = 60 .
(e) Taking same radius, with centre B, cut the previous arc at C.
(f) Draw bisector of ∠BOC. The angle is of 90 . Mark it at D. Thus, ∠DOA = 90°
(g) Draw \(\overline{\mathrm{OP}}\) as bisector of ∠DOB. Thus, ∠POA = 75°

Question 2.
Draw an angle of measure 147° and construct its bisector.
Answer:
Steps of construction:
(a) Draw a ray \(\overline{\mathrm{OA}}\) .
(b) With the help of protractor, construct ∠AOB = 147°
(c) Taking centre O and any convenient radius, draw an arc which intersects the arms \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) at P and Q respectively.
(d) Taking P as centre and radius more than half of PQ, draw an arc.
(e) Taking Q as centre and with the same radius, draw another arc which intersects the previous at R.
(f) Join OR and produce it.
(g) Thus, \(\overline{\mathrm{OR}}\) is the required bisector of ∠AOB.

Question 3.
Draw a right angle and construct its bisector.
Answer:
Steps of construction:
(a) Draw a line PQ and take a point O on it.
(b) Taking O as centre and convenient radius, draw an arc which intersects PQ at A and B.
(c) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C.
(d) Join OC. Thus, ∠COQ is the required right angle.
(e) Taking B and E as centre and radius more than half of BE, draw two arcs which intersect each other at the point D.
(f) Join OD. Thus, \(\overline{\mathrm{OD}}\) is the required bisector of ∠COQ.

Question 4.
Draw an angle of measure 153° and divide it into four equal parts.
Answer:
Steps of construction:
(a) Draw a ray \(\overline{\mathrm{OA}}\)
(b) At O, with the help of a protractor, construct ∠AOB = 153°.
(c) Draw \(\overline{\mathrm{OC}}\) as the bisector of ∠AOB.
(d) Again, draw \(\overline{\mathrm{OD}}\) as bisector of ∠AOC.
(e) Again, draw \(\overline{\mathrm{OE}}\) as bisector of ∠BOC.
(f) Thus, \(\overline{\mathrm{OC}}\), \(\overline{\mathrm{OD}}\) and \(\overline{\mathrm{OE}}\) divide ∠AOB in four equal arts.

Question 5.
Construct with ruler and compasses, angles of following measures:
(a) 60°
(b) 30°
(c) 90°
(d) 120°
(e) 45°
(f) 135°
Answer:
Steps of construction:
(a) 60°

(i) Draw a ray \(\overline{\mathrm{OA}}\).
(ii) Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{\mathrm{OA}}\) at P.
(iii) Taking P as centre and same radius, cut previous arc at Q.
(iv) Join OQ.
Thus, ∠BOA is required angle of 60°.

(b) 30°

(i) Draw a ray \(\overline{\mathrm{OA}}\) .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{\mathrm{OA}}\) at P.
(iii) Taking P as centre and same radius, cut previous arc at Q.
(iv) Join OQ. Thus, ∠BOA is required angle of 60°.
(v) Put the pointer on P and mark an arc.
(vi) Put the pointer on Q and with same radius, cut the previous arc at C. Thus, ∠COA is required angle of 30°.

(c) 90°

• Draw a ray \(\overline{\mathrm{OA}}\).
• Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{\mathrm{OA}}\) at X.
• Taking X as centre and same radius, cut previous arc at Y.
• Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.
• Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.
• Join OS and produce it to form a ray OB.
  Thus, ∠BOA is required angle of 90.

(d) 120°

• Draw a ray \(\overline{\mathrm{OA}}\).
• Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{\mathrm{OA}}\) at P.
• Taking P as centre and same radius, cut previous arc at Q.
• Taking Q as centre and same radius cut the arc at S.
• Join OS.
  Thus, ∠AOD is required angle of 120°.

(e) 45°

• Draw a ray \(\overline{\mathrm{OA}}\) .
• Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{\mathrm{OA}}\) at X.
• Taking X as centre and same radius, cut previous arc at Y.
• Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.
• Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.
• Join OS and produce it to form a ray OB. Thus, ∠BOA is required angle of 90°.
• Draw the bisector of ∠BOA.
  Thus, ∠MOA is required angle of 45°.

(f) 135°

• Draw a line PQ and take a point O on it.
• Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.
• Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R.
• Join OR. Thus, ∠QOR = ∠POQ = 90° .
• Draw OD the bisector of ∠POR.
  Thus, ∠QOD is required angle of 135°.

Question 6.
Draw an angle of measure 45° and bisect it.
Answer:
Steps of construction:
(a) Draw a line PQ and take a point O on it.
(b) Taking O as centre and a convenient radius, draw an arc which intersects PQ at two points A and B.
(c) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C.
(d) Join OC. Then ∠COQ is an angle of 90.
(e) Draw \(\overline{\mathrm{OE}}\) as the bisector of ∠COE. Thus, ∠QOE = 45°
(f) Again draw \(\overline{\mathrm{OG}}\) as the bisector of ∠QOE.
Thus, ∠QOG = ∠EOG = 22\(\frac { 1 }{ 2 }\)°

Question 7.
Draw an angle of measure 135° and bisect it.
Answer:
Steps of construction:
(a) Draw a line PQ and take a point O on it.
(b) Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.
(c) Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R.
(d) Join OR. Thus, ∠QOR = ∠POQ = 90°.
(e) Draw \(\overline{\mathrm{OD}}\) the bisector of ∠POR.
Thus, ∠QOD is required angle of 135°.
(f) Now, draw \(\overline{\mathrm{OE}}\) as the bisector of ∠QOD.
Thus, ∠QOE = ∠DOE =67\(\frac { 1 }{ 2 }\)°

Question 8.
Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.
Answer:
Steps of construction:
Step I: Draw a line 1 and mark a point O on it.
Step II: Using a protractor construct ∠AOB = 70°.
Step III: With centre O and a suitable radius, draw an arc which intersects \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) at E and F respectively.

Step IV: Draw a ray \(\overline{\mathrm{PQ}}\).

Step V: Keeping the same radius and centre P, draw an arc intersecting \(\overline{\mathrm{PQ}}\) at R.
Step VI: With centre R and radius equal to EF, draw an arc intersecting the previous arc at S.
Step VII: Join PS and produce it.

Question 9.
Draw an angle of 40° Copy its supplementary angle.
Answer:
Steps of construction:
Step I: (a) By using protractor draw ∠AOB = 40°
∠COF is the supplementary angle.
Step II: With centre O and a convenient radius, draw an arc which intersects \(\overline{\mathrm{OC}}\) and \(\overline{\mathrm{OB}}\) and E and F respectively.

Step III: Draw a ray \(\overline{\mathrm{OC}}\).

Step IV: With centre Q and same radius, draw an arc intersecting \(\overline{\mathrm{OC}}\) at L.
Step V: With centre L and radius equal to EF draw an arc which intersects the previous arc at S.
Step VI: Join QS and produce it.
Thus, ZPQS is the copy of the supplementary angle ZCOB.

These NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Exercise 2.3

Question 1.
Which of the following will not represent zero:
(a) 1 + 0
(b) 0 × 0
(c) \(\frac { 0 }{ 2 }\)
(d) \(\frac { (10-10) }{ 2 }\)
Answer:
(a) 1 + 01 Hence, it does not represent zero.

Question 2.
If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.
Answer:
Yes, if we multiply any number with zero the resultant product will be zero.
Example: 2 × 0 = 0, 5 × 0 = 0, 9 × 0 = 0 If both numbers are zero, then the result also be zero.
0 × 0 = 0
Yes, if the product of two whole numbers is zero, then both of them will be zero.

Question 3.
If the product of two whole numbers is 1, can we say that one of both of them will be 1? Justify through examples.
Answer:
If only one number be 1 then the product cannot be 1.
Examples: 5 × 1 = 5, 4 × 1 = 4, 8 × 1 = 8
If both number are 1, then the product is 1.
1 × 1 = 1

Question 4.
Find using distributive property:
(a) 728 × 101
(b) 5437 × 1001
(c) 824 × 25
(d) 4275 × 125
(e) 504 × 35
Answer:
(a) 728 × 101 = 728 × (100 + 1)
= 728 × 100 + 728 × 1 = 72800 + 728 = 73528
(b) 5437 × 1001 = 5437 × (1000 + 1)
= 5437 × 1000 + 5437 × 1 = 5437000 + 5437 = 5442437
(c) 824 × 25 = 824 × (20 + 5)
= 824 × 20 + 824 × 5 = 16480 + 4120 = 20600
(d) 4275 × 125 = 4275 × (100 + 20 + 5)
= 4275 × 100 + 4275 × 20 + 4275 × 5 = 427500 + 85500 + 21375 = 534375
(e) 504 × 35 = (500 + 4) × 35
= 500 × 35 + 4 × 35 = 17500 + 140 = 17640

Question 5.
Study the pattern:
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
12345 × 8 + 5 = 98765
Write the next two steps. Can you say how the pattern works?
(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1)
Answer:
123456 × 8 + 6 = 987654
1234567 × 8 + 7 = 9876543
Pattern works like this:
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
12345 × 8 + 5 = 98765
123456 × 8 + 6 = 987654
1234567 × 8 + 7 = 9875643

These NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
Write the next three natural numbers after 10999.
Answer:
The next three numbers after 10,999 are
10,999+ 1 = 11,000
11,000+ 1 = 11,001
11,001 + 1 = 11,002

Question 2.
Write the three whole numbers occurring just before 10001.
Answer:
10,001 – 1 =10,000
10000 10000 – 1 = 9999, 9999 – 1 = 9998

Question 3.
Which is the smallest whole number?
Answer:
‘0’ (zero) is the smallest whole number.

Question 4.
How many whole numbers are there between 32 and 53?
Answer:
53 – 32 – 1 = 20
There are 20 whole numbers between 32 and 53.

Question 5.
Write the successor of:
(a) 2440701
(b) 100199
(c) 1099999
(d) 2345670
Answer:
(a) Successor of 2440701 is 2440701 + 1 = 2440702
(b) Successor of 100199 is 100199 + 1 = 100200
(c) Successor of 1099999 is 1099999+ 1 = 1100000
(d) Successor of 2345670 is 2345670 + 1 = 2345671

Question 6.
Write the predecessor of:
(a) 94
(b) 10000
(c) 208090
(d) 7654321
Answer:
(a) The predecessor of 94 is 94 – 1 = 93
(b) The predecessor of 10000 is 10000 – 1 = 9999
(c) The predecessor of 208090 is 208090 – 1 = 208089
(d) The predecessor of 7654321 is 7654321 – 1 = 7654320

Question 7.
In each of the following pairs of numbers, state which 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
Answer:
(a) 530 > 503
So, 503 appear on left side of 530 on number line.
(b) 370 > 307; So 307 appear on left side of 370 on number line.
(c) 98765 > 56789; So 56789 appear on left side of 98765 on number line.
(d) 9830415 < 10023001; So 9830415 appear on left side of 10023001 on number line.

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 i digit number.
Answer:
(a) False
(b) False
(c) True
(d) True
(e) True
(f) False
(g) False
(h) False
(i) True
(j) False
(k) False
(l) True
(m) False

These NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Exercise 2.2

Question 1.
Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647
Answer:
(a) 837 + 208 + 363
= (837 +363)+208
= 1200 + 208 = 1408

(b) 1962 + 453 + 1538 + 647
= (1962+ 1538)+ (453+ 647)
= 3500 + 1100 = 4600

Question 2.
Find the product by suitable rearrangement:
(a) 2 × 1768 × 50
(b) 4 × 166 × 25
(c) 8 × 291 × 125
(d) 625 × 279 × 16
(e) 285 × 5 × 60
(f) 125 × 40 × 8 × 25
Answer:
(a) 2 × 1768 × 50 = (2 × 50) × 1768
= 100 × 1768 = 176800
(b) 4 × 166 × 25 = (4 × 25) × 166
= 100 × 166 = 16600
(c) 8 × 291 × 125 = (8 × 125) × 291
= 1000 × 291 = 291000
(d) 625 × 279 × 16 = (625 × 16) × 279
= 10000 × 279 = 2790000
(e) 285 × 5 × 60 = 285 × (5 × 60)
= 285 × 300 = 85500
(f) 125 × 40 × 8 × 25
= (125 × 8) × (40 × 25)
= 1000 × 1000 = 1000000

Question 3.
Find the value of the following:
(a) 297 × 17 + 297 × 3
(b) 54279 × 92 + 8 × 54279
(c) 81265 × 169 – 81265 × 69
(d) 3845 × 5 × 782 + 769 × 25 × 218
Answer:
(a) 297 × 17 + 297 × 3
= 297 × (17 + 3)
= 297 × 20 = 5940

(b) 54279 × 92 + 8 × 54279
= 54279 × (92 + 8)
= 54279 × 100 = 5427900

(c) 81265 × 169 – 81265 × 69
= 81265 × (169 – 69)
= 81265 × 100 = 8126500

(d) 3845 × 5 × 782 + 769 × 25 × 218
= 3845 × 5 × 782 + 769 × 5 × 5 × 218
= 3845 × 5 × 782 + 3845 × 5 × 218
= 3845 × 5 × (782 + 218)
= 3845 × 5 × 1000 = 19225000

Question 4.
Find the product using suitable properties.
(a) 738 × 103
(b) 854 × 102
(c) 258 × 1008
(d) 1005 × 168
Answer:
(a) 738 × 103 = 738 × (100 + 3)
= 738 × 100 + 738 × 3
(Using distributive property)
= 73800 + 2214 = 76014

(b) 854 × 102 = 854 × (100 + 2)
= 854 × 100 + 854 × 2
(Using distributive property)
= 85400 + 1708 = 87108

(c) 258 × 1008 = 258 × (1000 + 8)
= 258 × 1000 + 258 × 8
= 258000 + 2064 = 260064

(d) 1005 × 168 = (1000 +5) × 168
= 1000 × 168 + 5 × 168
= 168000 + 840 = 168840

Question 5.
A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he tilled the tank with 50 litres of petrol. If the petrol costs ₹ 44 per litre, how much did he spend in all on petrol?
Answer:
Petrol filled on Monday = 40 litres
Petrol filled on next day (i.e. Tue.) = 50 litres
Total petrol filled = 90 litres
Now, Cost of 1 litre petrol = ₹ 44
Cost of 90 litres petrol = 44 × 90 = 44 × (100 – 10)
= 44 × 100 – 44 × 10 = 4400 – 440
= ₹ 3960
Therefore, he spent ? 3960 on petrol.

Question 6.
A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs ? 41 per litre, how much money is due to the vendor per day?
Answer:
Supply of milk in morning the = 32 litres
Supply of milk in the evening = 68 litres
The cost of milk = ₹ 45 per litre
= ₹ 45 × [32+68]= ₹ 45 × 100 = ₹ 4500
Therefore, ₹ 4500 is due to the vendor per day.

Question 7.
Match the following:

Answer:
(i) → (c), (ii) → (a), (iii) → (b)

These NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers InText Questions

NCERT In-text Question Page No. 2

Question 1.
Can you instantly find the greatest and the smallest numbers in each row?

1. 382,4972,18, 59785, 750.
Answer:
59785 is the greatest and 18 is the smallest.

2. 1473, 89423, 100, 5000, 310.
Ans. ………………………..

3. 1834, 75284, 111, 2333,450 .
Ans. ………………………..

4. 2853, 7691, 9999,12002,124.
Ans. ………………………..
Was that easy? Why was it easy?
Answer:
2. 89423 is the greatest and 100 is the
smallest.
3. 75284 is the greatest and 111 is the smallest.
4. 12002 is the greatest and 124 is the smallest.
This is not very difficult. The greatest number has the most thousands and the smallest is only in hundreds or in tens.

Question 2.
Find the greatest and the smallest numbers.
(a) 4536,4892,4370,4452.
(b) 15623,15073,15189,15800.
(c) 25286,25245,25270,25210.
(d) 6895,23787, 24569, 24659.
Ans:
(a) Each of the given numbers is having four digits and their digits at the thousands place are the same.

∴ The greatest number is 4892.
The smallest number is 4370.

(b) Each of the given numbers is having five digits and their two leftmost places have the same digits.

∴ The greatest number is 15800.
The smallest number is 15073.

(c) Each of the given numbers is having five digits and their three leftmost places have the same digits.
On comparing the fourth leftmost digits, we have:

The greatest number is 25286.
The smallest number is 25210.

(d) The given numbers are:

The number 6895 is having four digits, so, it must be the smallest number.
2378 7,2456 9,2465 9, we observe that 4 > 3.
Again, the two leftmost places of 24569 and 24659 are having the same digits (i.e. 2 and 4).
∴ On comparing their third leftmost digits, we have:

∴ The greatest number is 24659.
The smallest number is 6895.

NCERT In text Question Pages Nos 3 & 4

Question 1.
Use the given digits without repetition and make the greatest and smallest 4-digit numbers.
(a) 2, 8, 7, 4
(b) 9,7,4, 1
(c) 4, 7, 5,0
(d) 1,7, 6, 2
(e) 5, 4, 0,3
(Hint: 0754 is a 3-digit number.)
Answer:
(a) greatest-8742 — smallest-2478
(b) greatest-9741 — smallest-1479
(c) c.greatest-7540 — smallest-4057
(d) greatest-7621 — smallest-1267
(e) greatest-5430 — smallest-3045

Question 2.
Now make the greatest and the smallest 4-digit numbers by using any one digit twice.
(a) 3,8,7
(b) 9,0,5
(c) 0,4,9
(d) 8,5,1
(Hint: Think in each case which digit will you use twice.)
Answer:
(a) 8873 is the greatest while 3387 is the smallest
(b) 9950 is the greatest while 5009 is the smallest
(c) 9940 is the greatest while 4009 is the smallest
(d) 8851 is the greatest while 1158 is the smallest

Question 3.
Make the greatest and the smallest 4-digit numbers using any four different digits with conditions as given.
(a) Digit 7 is always at ones place

(Note, the number cannot begin with the digit 0. Why?)

(b) Digit 4 is always at tens place

(c) Digit 9 is always at hundreds place

(d) Digit 1 is always at thousands place

Answer:
Keeping the digit ‘7’ at ones place, we have:

∴ The greatest 4-digit number = 9867.
The smallest 4-digit number = 1027.

(b) Keeping the digit 4 at tens place, we have:

∴ The greatest 4-digit number = 9847.
The smallest 4-digit number = 1042.

(c) Keeping the digit 9 at the hundreds place, we have:

∴ The greatest 4-digit number = 8976.
The smallest 4-digit number = 1902.

(d) Keeping the digit 1 at the thousands place, we have:


∴The greatest 4-digit number = 1987.
The smallest 4-digit number = 1023.

Question 4.
Take two digits, say 2 and 3. Make 4-digit numbers using both the digits equal number of times.
Which is the greatest number?
Which is the smallest number?
How many different numbers can you make in all?
Answer:
The given digit are 2 and 3. For making 4-digit numbers, we will repeat the given digits equal number of times.
∴ The possible 4-digit numbers are 3322, 2233, 2323, 3232, 3223 and 2332.
Among these, we have
(i) The greatest number = 3322.
(ii) The smallest number = 2233.
(iii) We can make six different 4-digit numbers.

NCERT In text Question Page No 4

Question 5.
Think of five more situations where you compare three or more quantities.
Answer:
Do it yourself

NCERT In-text Question Page No. 5

Question 1.
Arrange the following numbers in ascending order:
(a) 847, 9754, 8320, 571
(b) 9801,25751,36501,38802
Answer:
(a) 571 < 847 < 8320 < 8320
(b) 9801 <25751 < 25751 <38802

Question 2.
Arrange the following numbers in descending order:
(a) 5000,7500, 85400, 7861
(b) 1971,45321,88715,92547
Make ten such examples of ascending/ descending order and solve them
Answer:
(a) 85400 > 7861 > 7500 > 5000
(b) 92547 > 88715 >45321 > 1971

NCERT In-text Question Pages Nos. 6 & 7

Question 1.
Read and expand the numbers wherever there are blanks.
Answer:
I’m 28
Write five more 5 – digits numbers, read them and expand them.

NCERT In-text Question Pages No. 7

Question 1.
Read and expand the numbers wherever there are blanks.
Answer:

NCERT In-text Question Page No. 8

Question 1.
1. What is 10 – 1 = ?
2. What is 100 – 1 = ?
3. What is 10,000 – 1 = ?
4. What is 1,00,000 – 1 = ?
5. What is 1,00,00,000 – 1 = ?
(Hint: Use the said pattern.)
Answer:
Using the pattern, we have:
10 – 1 = 9 100 – 1 = 99
1.0 – 1 = 999
10.0 – 1 = 9,999
1.0. 000 – 1 = 99,999
10.0. 000 – 1 = 9,99,999
1.0. 00.000 – 1 = 99,99,999

Using this pattern, we can say:
1. 10-1
2. 100 – 1 = 99
3. 10,000 – 1 = 9,99
4. 1,00,000 – 1 = 99,999
5. 1,00,00,000 – 1 = 99,99,999.

Question 2.
Give five examples where the number of things counted would be more than 6-digit number.
Answer:

1. The number of stars seen in a clear dark night.
2. Number of cars in Delhi.
3. Number of children in a big city.
4. Number of grains in a sack full of grains.
5. Number of pages of note books of all students in a town.

Question 3.
Starting from the greatest 6-digit number, write the previous five numbers in descending order.
Answer:
The greatest 6-digit number is 999999
∴ 1st previous number = 999999 – 1 = 999998
2nd previous number 999999 – 2 = 9999997
3rd previous number = 999999 – 3 = 999996
4th previous number = 999999 – 4 = 999995
5th previous number = 999999 – 5 = 9999994
Writing these numbers in descending order, we have: 999998, 999997, 999996, 999995, 999994.

Question 4.
Starting from the smallest 8-digit number, write the next five numbers in ascending order and read them.
Answer:
The smallest 8-digit number = 1000000
1st next number is: 10000000 + 1 = 10000001
2nd next number is: 10000000 + 2 = 10000002
3rd next number is: 10000000 + 3 – 10000003
4th next number is: 10000000 + 4 = 10000004
5th next number is: 1000000 + 5 = 10000005
Reading these numbers, we have:
10000001: one crore one
10000002: one crore two
10000003: one crore three
10000004: one crore four
10000005: one crore five

NCERT In-text Question Page No. 11

Question 5.
Read these numbers. Write them using placement boxes and then write their expanded forms.
(i) 475320
(ii) 9847215
(iii) 97645310
(iv) 30458094
(a) Which is the smallest number?
(b) Which is the greatest number?
(c) Arrange these numbers in ascending and descending orders.
Answer:
(i) 475320

We read it as: four lakh seventy five thousand three hundred twenty.
Expanded form: 475320 = 4 × 100000 + 7 × 10000 + 5 × 1000 + 3 × 1oo + 2 × 10 + 0

(ii) 9847215

We read it as: ninety eight lakh forty seven thousand two hundred fifteen.
Expanded form: 9847215 = 9 × 1000000 + 8 × 100000 + 4 × 10000 + 7 × 1000 + 2 × 100 + 1 × 10 + 5 × 1

(iii) 97645310

We read it as: nine crore seventy six lakh forty five thousand three hundred ten.
Exxpanded form: 97645310 = 9 × 10000000 + 7 × 1000000 + 6 × 100000 + 4 × 10000 + 5 × 1000 + 3 × 100 + 1 × 10 + 0

(iv) 30458094

We read it as: three crore four lakh fifth eight thousand ninety four.
Expanded form: 30458094 = 3 × 10000000 + 4 × 100000 + 5 × 10000 + 8 × 1000 + 9 × 10 + 4 × 1
(a) The smallest number is 475320.
(b) The greatest number is 97645310.
(c) Ascending order: 475320, 9847215, 30458094, 97645310
Descending order: 97645310, 30458094, 9847215, 475320.

Question 6.
Read these numbers.
(i) 527864 (ii) 95432 (iii) 18950049 (iv) 70002509
(a) Write these numbers using placement boxes and then using commas in Indian as well as International System of Numeration.
(b) Arrange these in ascending and descending order.
Answer:
Reading these numbers, we have:
(i) 5, 27, 864: Five lakh twenty seven thousand eight hundred sixty four
(ii) 95, 432: Ninety five thousand four hundred thirty two
(iii) 1,89,50,049: One crore eighty nine lakh fifty thousand forty nine
(iv) 7,00,02,509: Seven crore two thousand five hundred nine
(a) Using the placement boxes, we have:

International System of Numeration
Using commas, we can re-write these numbers as:
(i) 527864 = 5,27,864 ← (Indian System) = 527,864 ← (International System)
(ii) 95432 = 95,432 ← (Indian System) = 95,432 ← (International System)
(iii) 18950049 = 1,89,50,049 ← (Indian System) = 18,950, 049 ← (International System)
(iv) 70002509 = 7,00,02,509 ← (Indian System) = 70,002,509 ← (International System)
(b) Ascending order: 95, 432; 5,27,864; 1,89,50,049; 7,00,02,509
Descending order: 7,00,02,509; 1,89,50,049; 5,27,864; 95,432

Question 7.
Take three more groups of large numbers and do the exercise given above.
Answer:
Do it yourself.

Question 8.
You have the following digits 4, 5, 6, 0, 7 and 8. Using them, make five numbers each with 6 digits.
(a) Put commas for easy reading.
(b) Arrange them in ascending and descending order.
Answer:
Five numbers each with 6 digits using the given digits are:
(i) 876540
(ii) 876450
(iii) 867540
(iv) 867405
(v) 876045

(a) Rewriting the above numbers using commas,
(i) 8,76,540
(ii) 8,76,450
(iii) 8,67,540
(iv) 8,67,405
(v) 8,76,045

(b) Ascending order: 8, 67, 405; 8, 67, 540; 8,76, 045; 8,76,450; 8,76,540
Descending order: 8,76,5402; 8,76, 450; 8,76,045; 8,67,540; 8,67,405

Question 9.
Take the digits 4, 5, 6, 7, 8 and 9. Make any three numbers each with 8 digits. Put commas for easy reading.
Answer:
Three numbers each with 8 digits using given digits are:
(i) 9,88,77,456
(ii) 9,88,77,465
(iii) 9,88,77,654

Question 10.
From the digits 3, 0 and 4, make five numbers each with 6 digits. Use commas.
Answer:
Five number each with 6 digits using the digits 3, 0 and 4 are:
(i) 4,44,330
(ii) 3,43,340
(iii) 4,40,340
(iv) 3,00,343
(iv) 3,03,403

NCERT In-text Question Page No. 12

Question 1.
How many centimetres make a kilometre?
Answer:
∵ 1 km = 1000 m and 1 m = 100 cm
∴ 1 km = 1000 × 100 cm =1,00,000 cm
= one lakh cm
There are 1 lakh centimetres in a kilometre.

Question 2.
Name five large cities in India. Find their population. Also, find the distance in kilometres between each pair of these cities.
Answer:
Do it yourself

NCERT In-text Question Page No. 13

Question 1.
How many milligrams make one kilogram?
Answer:
∵ 1 kg = 1000 g and 1 g = 1000 mg
∴ 1 kg = 1000 × 1000 mg = 10,0,000 mg
= 10 lakh mg

Question 2.
A box contains 2,00,000 medicine tablets each weighing 20 mg. What is the total weight of all the tablets in the box in grams and in kilograms?
Answer:
Number of tablets = 2,00,000
Weight of one tablet = 20 mg
∴ Total weight of all tablets
= 20 x 200000 mg
= 4000000 mg
Since 1 kg = 1000 g and 1 g = 1000 mg
∴ 4000000 mg = \(\frac{4000000}{1000}\)g = 4000 g
Thus, the total weight of all the tablets in grams = 4000 g.
Also, 4000000 mg = \(\frac{4000000}{1000 \times 1000}\) = 4 kg
Thus, the total weight of all the tablets in kilograms = 4 kg

NCERT In-text Question Pages Nos. 13 & 14

Question 1.
A bus started its journey and reached different places with a speed of 60 km/ hour. The journey is shown below.

(i) Find the total distance covered by the bus from A to D.
(ii) Find the total distance covered by the bus from D to G.
(iii) Find the total distance covered by the bus, if it starts from A and returns back to A.
(iv) Can you find the difference of distances from C to D and D to E?
(v) Find out the time taken by the bus to reach
(i) A to B
(ii) A to B
(iii) A to B
(iv) Total journey
Answer:
(i) Distance covered by the bust for going from A to D
= 4170 km + 3410 km + 2160 km
= [4170 + 3410+ 2160] km
= 9740 km

(ii) Total distance covered by the bus for going from D to G
= 8140 km + 4830 km + 2550 km
= [8140 + 4830 + 2550] km
= 15520 km

(iii) Total distance covered by the bus for going round from A to A
= [Distance between A and D] + [Distance between D and G] + [Distance between G and A]
= [9740 km] + [15520 km] + [1290 km]
= [9740 + 15520 + 1290] km
= 26550 km

(iv) [Distance between D and E]
– [Distance between C and D]
= [8140 km] – [2160 km]
= [8140 – 2160] km = 5980 km

(v) Since time = \(\frac{\text { dis tan ce }}{\text { speed }}\) and the speed of the bus is 60 km/hour.
(i) Time taken by the bus to each from A to B
= \(\frac{4170 \mathrm{~km}}{60 \mathrm{~km} / \text { hour }}=\frac{4170}{60} \text { hour }=69 \frac{1}{2}\)hours
60km /hour 60 . 2
(ii) Time taken by the bus to reach from C to D = \(\frac{2160}{60}\) hours = 36 hours
(iii) Time taken by the bus to reach from E to G

(iv) Total time taken by the bus for the total journey

Question 2.
Raman’s Shop

The sales during the last year

(a) Can you find the total weight of apples and oranges Raman sold last year?
Weight of apples = ________ kg
Weight of oranges = _______kg
Therefore, total weight = kg + _____ kg= kg
Answer – The total weight of oranges and apples = _______ kg.
(a) Can you find the total money Raman got by selling apples?
(b) Can you find total money Raman got by selling apples and oranges together?
(c) Can you find total money Raman got by selling apples and oranges together?
(d) Make a table showing how much money Raman received from selling each item. Arrange the entries of amount of money received in descending order. Find the item which brought him the highest amout. How much is this amount?
Answer:
(a) Quantity of oranges and apples sold by Raman last year:
Weight of apples = 2457 kg
Weight of oranges = 3004 kg
Therefore, total weight = 2457 kg + 3004 kg
= 5461 kg
The total weight of oranges and apples = 5461 kg.

(b) Selling price of 1 kg of apples = ₹ 40
Weight of apples sold = 2457 kg
∴ Total money Raman got by selling apples
= ₹ 2457 × 40 = ₹ 98,280

(c) Selling price of 1 kg of oranges = ₹ 30
Weight of oranges sold = 3004 kg
Total money Raman got by selling oranges
= ₹ 3004 × 30 = ₹ 90,120
∴ Total money Raman got by selling apples and oranges together = ₹ 98,280 + ₹ 90, 120 = ₹ 1,88,400


Arranging the entries of amount of money received in descending order, we have:
₹ 253670; ₹ 240012; ₹ 160040;
₹ 98120; ₹ 90120; ₹ 68280; 38530
Obviously, the highest amount of money is received against the item “tooth brushes”. This highest amount of money is ₹ 2,53,670.

NCERT In-text Question Page No. 19

Question 1.
Round these numbers to the nearest tens.

Answer:
Number 5 is rounded off to 10.

NCERT In-text Question Page No. 20

Question 2.
Round off the given numbers to the nearest tens, hundreds and thousands.
Answer:

NCERT In-text Question Page No. 22

Question 1.
Estimate the following products :
(a) 87 × 313
(b) 9 × 795
(c) 898 × 785
(d) 958 × 387
Make five more such problems and solve them
Answer:
(a) 87 × 313
87 → 90 [Rounding to tens]
313 → 300 [Rounding to hundreds]
.’. Estimated product = 90 × 300 = 27000

(b) 9 × 795
9 → 10 [Rounding to tens]
795 → 800 [Rounding to hundreds]
.’. Estimated product = 10 × 800 = 8000

(c) 898 × 900
898 → 900 [Rounding to tens]
785 → 800 [Rounding to hundreds]
.’. Estimated product = 900 × 800 = 720000

(d) 958 × 387
958 → 1000 [Rounding to tens]
387 → 400 [Rounding to hundreds]
∴ Estimated product = 1000 × 400 = 400000

NCERT In-text Question Page No. 23

Question 2.
Write the expressions for each of the following using brackets.
(a) Four multiplied by the sum of nine and two.
(b) Divide the difference of eighteen and six by four.
(c) Forty five divided by three times the sum of three and two.
Answer:
(a) four multiplied by the sum of nine and two 4 × (9+2)
(b) Divide the difference of eighteen and six by four (18-6) ÷ 4
(c) 45 ÷ [3(3+2)]

Question 3.
Write three different situations for (5 + 8) × 6. (One such situation is : Sohani and Reeta work for 6 days; Sohani works 5 hours a day and Reeta 8 hours a day. How many hours do both of them work in a week?)
Answer:
Rahul pays ? 5 for a morning tea and ? 8 for the evening coffee daily. He takes tea and coffee for six days in a week. What is his weekly expanses for tea and coffee? Prema and Shanti work for six days. Prema earns ? 5 per day and Shanti earns ? 8 per day. What do they earn together in six days?

Prateek read 5 pages of a novel in the morning and 8 pages in the evening. How many pages did Prateek read in 6 days?

Question 4.
Write five situations for the following where brackets would be necessary.
(a) 7(8-3) (b) (7+ 2) (10-3).
Answer:
(a) 7 (8 – 3) = 7 × (8) + 7 × (-3)
= 56 – 21
= 35

(b) (7 + 2) (10-3) = 7 (10-3)+ 2 (10-3)
Using distributive property
= 7 × (7) + 2 × (7)
= 49+14
= 63

NCERT In-text Question Page No. 25

Question 1.
Write in Roman numerals.
1. 73
2. 92
Answer:
1. 73 = LXXIII
2. 92 = XCII