integer

Question 1.

Evaluate the following, using the number line:

(i) 4 – (-2)

(ii) -4 – (-2)

(iii) 3 – 6

(iv) -3 – (-5)

Solution:

(i) Start from 4 on the number line.

Move 2 units to the digits we reach at 6

∴ 4 – (-2) = 4 + 2 = 6

 

(ii) Start from -4 on the number line.

Move 2 units to the right, we reach at -2

∴ -4 – (-2) = —4 + 2 = -2

(iii) Start from 3 on the number line.

Move 6 units to the left, we reach at -3

3 – 6 = -3

(iv) Start from -3 on the number line.

Move 5 units to the right, we reach at 2

-3 – (-5) = -3 + 5 = 2

Question 2.

Subtract :

(i) -6 from 9

(ii) 6 from -9

(iii) -6 from -9

(iv) -725 from -63

(v) -376 from 10

(vi) 92 from -620

Solution:

(i) 9 – (-6) = 9 + 6 = 15

(ii) -9 – 6 = -15

(iii) -9 – (-6) = -9 + 6 = -3

(iv) -63 – (-725) = -63 + 725 = +662

(v) 10 – (-376) = 10 + 376 = 386

(vi) -620 – 92 = -712

Question 3.

Evaluate the following:

(i) -237 – (+ 1884)

(ii) -346 – (- 1275)

(iii) -190 – (-3512)

(iv) -2718 – (+ 6827)

Solution:

(i) -237 – (+ 1884)

= -237 – 1884

= -(237 + 1884) = -2121

 

(ii) -346 -(- 1275)

= -346 + 1275

= 1275 – 346 = 929

(iii) -190 – (-3512)

= -190 + 3512

= 3512- 190 = 3322

(iv) 2718 – (+ 6827)

= -2718 – 6827

= -(2718 + 6827) = -9545

Question 4.

The sum of two integers is 17. If one of them is -35, find the other.

Solution:

One number = -35

Sum of two integers =17

Second number = Sum of integers – (The given number)

= 17 – (-35)

= 17 + 35 = 52

Question 5.

What must be added to -23 to get -9?

Solution:

Let the number to be added = x

∴ -23 + x = -9

∴ The required number = -9 – (-23)

= -9 + 23 = 14

Question 6.

Find the predecessor of 0.

Solution:

Predecessor of 0 = 0 – 1 = -1

Question 7.

Find the successor and the predecessor of the following integers:

(i) -31

(ii) -735

(iii) -240

Solution:

(i) Successor of -31 = -31 + 1 = -30

Predecessor of -31 = -31 – 1 = -32

(ii) Successor of -735 = -735 + 1 = -734

Predecessor of-735 = -735 – 1 = -736

(iii) Successor of -240 = -240 + 1 = -239

Predecessor of -240 = -240 – 1 = -241

Question 1.

Find the value of:

(i) 6 – 9 + 4

(ii) -5 – (-3) + 2

(iii) 7 + (-5) + (-6)

(iv) 6 – 3 – (-5)

Solution:

(i) 6 – 9 + 4

= (6 + 4) – 9 = 10 – 9= 1

 

(ii) -5 – (-3) + 2

= -5 + 3 + 2 = -5 + 5 = 0

(iii) 7 + (-5) + (-6)

= 7 – 5 – 6 = 2 – 6 = -4

(iv) 6 – 3 – (-5)

= 6 – 3 + 5 = 8

Question 2.

Evaluate the following:

(i) -77 + (-84) + 318

(ii) 54 + (-218) – (-76)

(iii) -121 – (-78) + (-193) + 576

(iv) -65 + (-76) – (-28) + 32

Solution:

(i) -77 + (- 84) + 318

= -77 – 84 + 318

= -(77 + 84)+ 318

= -(161) + 318

= -161 +318

= 318 – 161 = 157

(ii) 54 + (-218) – (-76)

= 54 – 218 + 76

= (54 + 76) – 218

= 130 – 218 = – 88

(iii) -121 – (-78) + (-193) + 576

= -121 + 78 – 193 + 576

= -121 – 193 + 78 + 576

= -(121 + 193) + 78 + 576

= -(314) + 654

= 654 – 314 = 340

(iv) -65 + (-76) – (-28) + 32

= -65 – 76 + 28 + 32

= -(65 + 76) + 60

= -141 + 60 = -81

Question 3.

Find the value of:

(i) 8 – 6 + (-2) – (-3) + 1

(ii) 31 + (-23) – 35 + 18 – 4 – (-3)

Solution:

(i) 8 – 6 + (-2) – (-3) + 1

= 8 – 6 – 2 + 3 + 1

=-6 – 2 + 8 + 3 + 1

= -6 – 2 + 12

=-8 + 12 = 4

(ii) 31 + (-23) – 35 + 18 – 4 – (-3)

= 31 – 23 – 35 + 18 – 4 + 3

= -23 – 35 – 4 + 31 + 18 + 3

= -23 – 35 – 4 + 52

= -62 + 52 = -10

Question 4.

Rashmi deposited ₹ 4370 in her account on Monday and then withdrew ₹ 2875 on Tuesday. Next day she deposited ₹ 1550. What was her balance on Thursday?

Solution:

Rashmi deposited in her account on Monday = ₹ 4370

Less withdrawal on Tuesday = ₹ 2875

So the Balance on Tuesday

= ₹ 4370 – ₹ 2875

= ₹ 1495

Again she deposited on Wednesday = ₹ 1550

Balance on Thursday

= ₹ 1495 + ₹ 1550 = ₹ 3045

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Question 1.

Use the appropriate symbol < or > to fill in the following blanks:

(i) (-3 + ……… (-6) (-3) – (-6)

(ii) (-21) – (-10) ……. (-31)+ (-11)

(iii) 45 – (-11) ……….. (57) + (-4)

(iv) (-25) – (-42) …………. (-42) – (-25)

Solution:

(i) (-3 + (-6) < (-3) – (-6)

(ii) (-21) – (-10) > (-31) + (-11)

(iii) 45 – (-11) > (57) +(-4)

(iv) (-25 – (-42) > (-42) – (-25)

Question 2.

Find the value of:

(i) 12 + ( -3) + 5 – (-2)

(ii) 39 – 35 + 7-(-4) + 21

(iii) -15- (-2) – 71 – 8 + 6

Solution:

(i) 12 + (-3) + 5 – (-2)

= 12 – 3 + 5 + 2

= 9 + 7= 16

(ii) 39 – 35 + 7 – (-4) + 21

= 39 – 35 + 7 + 4 + 21

= 4 + 11 + 21

= 15 + 21 =36

(iii) -15 – (-2) – 71 – 8 + 6

= -15 + 2 – 71 – 8 + 6

= -13 – 79 + 6

= 92 + 6 = -86

Question 3.

Evaluate:

(i) |-13| – |-15|

(ii) |35 – 41| – |7-(-2)|

Solution:

(i) |-13| – |-15|

= +13 – 15 =-2

(ii) |35 – 41| – |7 – (-2)|

= 6 – 9 = -3

Question 4.

Arrange the following integers in ascending order:

-39, 35, -102, 0, -51, -5, -6, 7

Solution:

-102, -51, -39, -6, -5, 0, 7, 35

Question 5.

Find the successor and the predecessor of -199.

Solution:

Successor = -199 – 1 = -198

Predecessor = -199 – 1 = -200

Question 6.

Subtract the sum of -235 and 137 from -152.

Solution:

Sum of (-235 and 137)

= -235 + 137

= 137 – 235 = -98

Now, subtract the sum of (-235 and 137) from -152

= 152 – (-98)

= -152 + 98 = -54

Question 7.

What must be added to -176 to get -95?

Solution:

Let the number to be added = x

∴ -176 + x = -95

x = -95 + 176 = 81

Question 8.

What is the difference in height between a point 270 m above sea level and 80 m below sea level?

Solution:

Height above sea level = +270 m

Height below sea level = -80 m

Difference = +270 – (-80)

= 270 + 80 = 350 m

whole number

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Question 1.

Fill in the blanks to make each of the following a true statement:

(i) 378 + 1024 = 1024 + …….

(ii) 337 + (528 + 1164) = (337 + ……..) + 1164

(iii) (21 + 18) + ……….. = (21 + 13) + 18

(iv) 3056 + 0 = ……….. = 0 + 3056

Solution:

(i) 378 + 1024= 1024 + 378 (Commutative property of addition)

(ii) 337 + (528 + 1164) = (337 + 528) + 1164 (Associative law of addition)

(iii) (21 + 18) + 13 = (21 + 13) + 18 (Associative law of addition)

(iv) 3056 + 0 = 3056 = 0 + 3056

 

Question 2.

Add the following numbers and check by reversing the order of addends :

(i) 3189 + 53885

(ii) 33789 + 50311.

Solution:

(i) 3189 + 53885 = 57074

Check 53885 + 3189 = 57074

∴57074

(ii) 33789 + 50311 = 84100

Check 50311 + 33789 = 84100

∴ 84100

Question 3.

By suitable arrangements, find the sum of:

(i) 311,528,289

(ii) 723, 834, 66, 277

(iii) 78, 203, 435, 7197, 422.

Solution:

(i) 311, 528, 289

Sum (311 +289)+ 528

= 600+ 528= 1128

(ii) 723 + 834 + 66 + 277

= (723 + 277) + (834 + 66)

= 1000 + 900 = 1900

(iii) 78, 203, 435, 7197, 422

Sum = (78 + 422) + (203 + 7197) + 435

= 500 + 7400 + 435

= 7900 + 435 = 8335

 

Question 4.

Fill in the blanks to make each of the following a true statement:

(i) 375 × 57 = 57 × ……….

(ii) (33 × 16) × 25 = 33 × (…….. × 25)

(iii) 37 × 24 = 37 × 18 + 37 × …………

(iv) 7205 × 1 = …………. = 1 × 7205

(v) 366 × 0 =

(vi) …………… × 579 = 0

(vii) 473 × 108 = 473 × 100 + 473 × ………….

(viii) 684 × 97 = 684 × 100 – …………… × 3

(ix) 0 ÷= 5 =

(x) (14 – 14) ÷ 7 = ………….

Solution:

(i) 375 × 57 = 57 × 375 (Commutative property of multiplication)

(ii) 33 × 16) × 25 = 33 × (16 × 25) (Associative law of multiplication)

(iii) 37 × 24 = 37 × 18 + 37 × 6 (Distributive law of multiplication)

(iv) 7205 × 1 = 7205 = 1 × 7205

(v) 366 × 0 = 0

(vi) 0 × 579 = 0

(vii) 473 × 108 = 473 × 100 + 473 × 8

(viii) 684 × 97 = 684 × 100 – 684 × 3

(ix) 0 ÷ 5 = 0

(x) (14 – 14) ÷ 7 = 0

Question 5.

Determine the following products by suitable arrangement:

(i) 4 × 528 × 25

(ii) 625 × 239 × 16

(iii) 125 × 40 × 8 × 25

Solution:

(i) 4 × 528 × 25 = 4 × 25 × 528

= 100 × 528 = 52800

(ii) 625 × 239 × 16 = 625 × 16 × 239

= 10000 × 239 = 2390000

(iii) 125 × 40 × 8 × 25 = 125 × 8 × 40 × 25

= 1000 × 1000 = 1000000

Question 6.

Find the value of the following:

(i) 54279 × 92 + 54279 × 8

(ii) 60678 × 262 – 60678 × 162

Solution:

(i) 54279 × 92 + 54279 × 8

= 54279 (92 + 8)

= 54279 × 100 = 5427900

 

(ii) 60678 × 262 – 60678 × 162

= 60678 (262 – 162)

= 60678 × 100 = 6067800

Question 7.

Find the following products by using suitable properties:

(i) 739 × 102

(ii) 1938 × 99

(iii) 1005 × 188

Solution:

(i) 739 × 102

= 739 × (100 + 2)

= 739 × 100 + 739 × 2

= 73900 + 1478 = 75378

(ii) 1938 × 99

= 1938 × (100- 1)

= 1938 × 100 – 1938 × 1

= 193800 – 1938 = 191862

(iii) 1005 × 188

= (1000 + 5) × (100 + 88)

= 1000 × 100 + 1000 × 88 + 5 × 100 + 88 × 5

= 100000 + 88000 + 500 + 440 = 188940

Question 8.

Divide 7750 by 17 and check the result by division algorithm.

Solution:

7750 ÷ 17

On dividing 7750 by 17, we get

Quotient = 455 and Remainder = 15

Check by division algorithm:

Divident = Divisior × Quotient + Remainder

= 17 × 455 + 15 = 7750

Question 9.

Find the number which when divided by 38 gives the quotient 23 and remainder 17.

Solution:

Divisor = 38,Quotient = 23

Remainder = 17

Dividend = divisor × quotient + remainder

= 38 × 23 + 17 = 874 + 17 = 891

 

Question 10.

Which least number should be subtracted from 1000 so that the difference is exactly divisible by 35.

Solution:

On dividing 1000 by 35

we get quotient = 28 and remainder 20

So, 20 should be subtracted from 1000.

Question 11.

Which least number should be added to 1000 so that 53 divides the sum exactly.

Solution:

On dividing 1000 by 53, we get quotient = 18 and remainder = 46. To get the remainder 0, we should add 53 – 46 = 7 to 1000.

∴ 7

Question 12.

Find the largest three-digit number which is exactly divisible by 47.

Solution:

Largest three digit no. = 999

On dividing 999 by 47, we get

Quotient = 21 and Remainder =12

So on subtracting 12 from 999, we get

999 – 12 = 987

Question 13.

Find the smallest five-digit number which is exactly divisible by 254.

Solution:

Smallest 5 digit number = 10000

On dividing 10000 by 254, we get

Remainder = 94

So 254 – 94 = 160 should be added to 10000 to get the smallest 5 digit number divisible by 254.

∴ 10000 + 160 = 10160

Question 14.

A vendor supplies 72 litres of milk to a student’s hostel in the morning and 28 litres of milk in the evening every day. If the milk costs?39 per litre, how much money is due to the vendor per day?

Solution:

Supply of milk in morning = 72 litres

Supply of milk in evening = 28 litres

Cost of per litre milk = ₹ 39

Money of per day = ₹ 39 (72 l + 28 l)

= ₹ 39 × 100 = ₹ 3900

Question 15.

State whether the following statements are true (T) or false (F):

(i) If the product of two whole numbers is zero, then atleast one of them will be zero.

(ii) If the product of two whole numbers is 1, then each of them must be equal to 1.

(iii) If a and b are whole numbers such that a ≠ 0 and b ≠ 0, then ab may be zero.

Solution:

(i) True

(ii) True

(iii)False

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Question 1.

Using shorter method, find

(i) 3246 + 9999

(ii) 7501 + 99999

(iii) 5377 – 999

(iv) 25718 – 9999

(v) 123 × 999

(vi) 203 × 9999

Solution:

(i) 3246 + 9999

= (3246 – 1) + (9999 + 1) (Adding and subtracting 1)

= 3245 + 10000 = 13245

 

(ii) 7501 + 99999

= (7501 – 1) + (99999 + 1) (Adding and subtracting 1)

= 7500+ 100000 = 107500

(iii) 5377 – 999

= 5377 – (1000- 1)

= 5377 – 1000 + 1 = 4377 + 1 = 4378

(iv) 25718 – 9999

= 25718 – (10,000 – 1)

= 15718 + 1 = 15719

(v) 123 × 999

= 123 × (1000 – 1) (By subtracting 1)

= 123 × 1000 – 1 × 123 = 123000 – 123 = 122877

(vi) 203 × 9999

= 203 × (10,000 – 1) (By subtracting 1)

= 203 × 10,000 – 203 × 1 = 2030000 – 203 = 2029797

 

Question 2.

Without using a diagram, find

(i) 9th square number

(ii) 7th triangular number

Solution:

(i) 9th square number = ?

The first square number is 1 × 1 = 1

The second square number is 2 × 2 = 4

The third square number is 3 × 3 = 9

Similarly 9th square number is 9 × 9 = 81

(ii) 7th triangular number = ?

First triangular number = 1

Second triangular number = 1 + 2 = 3

Third triangular number = 1 + 2 + 3 = 6

Fourth triangular number = 1 + 2 + 3 + 4 = 10

Similarly 7th triangular number = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28

Question 3.

(i) Can a rectangular number be a square number?

(ii) Can a triangular number be a square number?

Solution:

(i) Yes, 9 is a square as well as rectangular number.

(ii) Yes, 8th triangular number = 36, which is a square number.

Question 4.

Observe the following pattern and fill in the blanks:

1 × 9 + 1 = 10

12 × 9 + 2= 110

123 × 9 + 3 = 1110

1234 × 9 + 4 = ……….

12345 × 9 + 5 = …………..

Solution:

1 × 9 + 1 = 10

12 × 9 + 2= 110

123 × 9 + 3 = 1110

1234 × 9 + 4 = 11110

12345 × 9 + 5 = 111110

Question 5.

Observe the following pattern and fill in the blanks:

9 × 9 + 7 = 88

98 × 9 + 6 = 888

987 × 9 + 5 = 8888

9876 x 9 + 4 = …………

98765 × 9 + 3 = ……….

Solution:

9 × 9 + 7 = 88

98 × 9 + 6 = 888

987 × 9 + 5 = 8888

9876 × 9 + 4 = 88888

98765 × 9 + 3 = 888888

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Question 1.

Write next three consecutive whole numbers of the number 9998.

Solution:

The next three consecutive whole number of 9998 are:

9998 + 1 = 9999

9999 + 1 = 10000

10000 + 1 = 10001

∴ Numbers are = 9999, 10000, 10001

 

Question 2.

Write three consecutive whole numbers occurring just before 567890.

Solution:

The three consecutive whole numbers just before 567890 are:

567890 – 1 = 567889 – 1 = 567888 – 1

= 567887

∴ These are : 567889, 567888, 567887

Question 3.

Find the product of the successor and the predecessor of the smallest number of 3-digits.

Solution:

Smallest number of 3-digits = 100

Successor = 100 + 1

Predecessor =100 – 1

∴ Product = 100 + 1 × 100 – 1

= 101 × 99 = 9999

Question 4.

Find the number of whole numbers between the smallest and the greatest numbers of 2-digits.

Solution:

Smallest number of 2-digits = 10

Greatest number of 2-digits = 99

Numbers between 10 and 99

11, 12, …………, 98

= 98 – 10 = 88

Question 5.

Find the following sum by suitable arrangements:

(i) 678 + 1319 + 322 + 5681

(ii) 777 + 546 + 1463 + 223 + 537

Solution:

(i) 678 + 1319 + 322 + 5681

= (678 + 322) + (5681 + 1319)

= 1000 + 7000 = 8000

(ii) 777 + 546 + 1463 + 223 + 537

= (777 + 223) + (1463 + 537) + 546

= 1000 + 2000 + 546 = 3546

 

Question 6.

Determine the following products by suitable arrangements:

(i) 625 × 437 × 16

(ii) 309 × 25 × 7 × 8

Solution:

(i) 625 × 437 × 16

= 437 × (625 × 16)

= 437 × 10000 = 4370000

(ii) 309 × 25 × 7 × 8

= (309 × 7) × (25 × 8)

= 2163 × 200 = 432600

Question 7.

Find the value of the following by using suitable properties:

(i) 236 × 414 + 236 × 563 + 236 × 23

(ii) 370 × 1587 – 37 × 10 × 587

Solution:

(i) 236 × 414 + 236 × 563 + 236 × 23

= 236 × (414 + 563 + 23)

= 236 × (1000) = 236000

(ii) 370 × 1587 – 37 × 10 × 587

= 37 × 10(1587 – 587)

= 370 × 1000 = 370000

Question 8.

Divide 6528 by 29 and check the result by division algorithm.

Solution:

6528 ÷ 29

∴ Quotient = 225

and Remainder = 3

Question 9.

Find the greatest 4-digit number which is exactly divisible by 357.

Solution:

Largest 4 digit number is 9999

Dividing 9999 by 357, we get

Remainder = 3

Subtracting 3 from 9999, 9999 – 3 = 9996,

we get the required number divisible by 357.

So 9996

Question 10.

Find the smallest 5-digit number which is exactly divisible by 279.

Solution:

Smallest 5-digit number is 10000

Dividing it by 279, we get remainder = 235

To make the smallest 5-digit number exactly divisible by 279, we have to add 279 – 235 = 44 in 10000

∴ 10000 + 44 = 10044.

Icse class 6 chapter 1 math

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Question 1.

Round off each of the following numbers to their nearest tens:

(i) 77

(ii) 903

(iii) 70 1205

(iv) 999

Solution:

(i) 77

The digit at unit place is 7, which is greater than 5.

Hence, the rounded off number to nearest tens = 80.

 

(ii) 903

The digit at unit place is 3, which is less than 5.

Hence, the rounded off number to nearest tens = 900.

(iii) 1205

The digit at unit place is 5, which is equal to 5.

Hence, the rounded off number to nearest tens = 1210

(iv) 999

The digit at unit place is 9, which is greater than 5.

Hence, the rounded off number to nearest tens = 1000.

Question 2.

Estimate each of the following numbers to their nearest hundreds:

(i) 1246

(ii) 32057

(iii) 53961

(iv) 555555

Solution:

(i) 1246

The digit at tens place is 4, which is less than 5.

Hence, the rounded off number to nearest hundreds = 1200.

(ii) 32057

The digit at tens place is 5, which is equal to 5.

Hence, the rounded off number to nearest hundreds = 32100.

 

(iii) 53961

The digit at tens place is 6, which is greater than 5.

Hence, the rounded off number to nearest hundreds = 54000.

(iv) 555555

The digit at tens place is 5, which is equal to 5.

Hence, the rounded off number to nearest hundreds = 555600.

Question 3.

Estimate each of the following numbers to their nearest thousands:

(i) 5706

(ii) 378

(iii) 47,599

(iv) 1,09,736

Solution:

(i) 5706

The digit at hundred place is 7, which is greater than 5.

Hence, the rounded off number to nearest thounsands = 6000.

(ii) 378

The digit at hundred place is 3, which is less than 5.

Hence, the rounded off number to nearest thousands = 0.

(iii) 47,599

The digit at hundred place is 5, which is equal to 5.

Hence, the rounded off number to nearest thousands = 48000.

(iv) 1,09,736

The digit at hundred place is 7, which is greater than 5.

Hence, the rounded off number to nearest thousands = 1,10,000.

Question 4.

Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):

(i) 439 + 334 + 4317

(ii) 8325 – 491

(iii) 1,08,734-47,599

(iv) 4,89,348 – 48,365

Solution:

(i) Rounding off to nearest hundreds 439 + 334 + 4317

= 400 + 300 + 4300 = 5000 Rounding off to nearest tens 439 + 334 + 4317

= 440 + 330 + 4320 = 5090

 

(ii) Rounding off to nearest hundreds 8325 – 491

= 8300 – 500 = 7800 Rounding off to nearest tens 8325 – 491

= 8330 – 490 = 7840

(iii) 1,08,734 – 47,599

Rounding off to nearest hundreds 1,08,734 – 47,599

= 1,08,700 – 47,600 = 61,100 Rounding off to nearest tens 1,08,734 – 47,599

= 1,08,730 – 47,600 = 61,130

(iv) 4,89,348 – 48,365

Rounding off to nearest hundreds 4,89,348 – 48,365

= 4,89,300 – 48,400 = 4,40,900 Rounding off to nearest tens 4,89,348 – 48,365

= 4,89,350 – 48,370 = 4,40,980

Question 5.

Estimate each of the following by rounding off each number nearest to its greatest place:

(i) 730 + 998

(ii) 5,290 + 17,986

(iii) 796-314

(iv) 28,292 – 21,496

Solution:

(i) 730 + 998

Rounding off 730 to its greatest place i.e. hundred place = 700

Rounding off 998 to its greatest place i.e. hundreds place = 1000

Hence, estimated sum = 700 + 1000 = 1700

(ii) 5,290 + 17,986

Rounding off 5,290 to its greatest place i.e. thousands place = 5000

Rounding off 17,986 to its greatest place i. e. thousands place = 18,000 Hence, estimated sum = 5,000 + 18,000 = 23,000

(iii) 796 – 314

Rounding off 796 to its greatest place i.e. hundreds place = 800 Rounding off 314 to its greatest place i.e. hundreds place = 300 Hence, estimated difference = 800 – 300 = 500

 

(iv) 28,292 – 21,496

Rounding off28,292 to its greatest place i.e. thousands place = 28,000 Rounding off 21,496 to its greatest place i. e. thousands place = 21,000 Hence, estimated difference = 28,000 – 21,000 = 7,000

Question 6.

Estimate the following products by rounding off each of its factors nearest to its greatest place:

(i) 578 × 161

(ii) 9650 × 27

Solution:

(i) 578 × 161

Rounding off 578 to its greatest place i.e. hundreds place = 600

Rounding off 161 to its greatest place i.e. hundreds place = 200

Hence, estimated product = 600 x 200 = 1,20,000

(ii) 9650 × 27

Rounding off 9650 to its greatest place i.e. thousands place = 10000

Rounding off 27 to its greatest place i.e. tens place = 30

Hence, estimated product = 10000 × 30 = 3,00,000

Question 7.

Estimate the following products by rounding off each of its factors nearest to its hundreds place:

(i) 5281 × 3491

(ii) 1387 × 888

Solution:

(i) 5281 x 3491

Rounding off 5281 to its hundreds place = 5300

Rounding off 3491 to its hundreds = 3500

Hence, estimated product = 5300 × 3500 = 1,85,50,000

(ii) 1387 × 888

Rounding off 1387 to its hundreds place = 1400

Rounding off 888 to its hundreds place = 900

Hence, estimated product = 1400 × 900 = 12,60,000

ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 1 Knowing Our Numbers Objective Type Questions

Mental Maths

Question 1.

Fill in the blanks:

(i) The digit …………… has the highest place value in the number 2309.

(ii) The digit …………… has the highest face value in the number 2039.

(iii) The digit …………… has the lowest place value in the number 2039.

(iv) Both Indian and International systems of numeration have …………… period in common.

(v) In the International system of numeration, commas are placed from …………… after every …………… digits.

(vi) The bigger number from the numbers 57,631 and 57,361 is ……………

(vii) 1 crore = …………… million

(viii)The smallest 4-digit number with 3 different digits is ……………

(ix) The greatest 4-digit number with 3 different digits is ……………

(x) 15 km 300 m = …………… m

(xi) 7850 cm = …………… m …………… cm

(xii) The number 5079 when estimated to the nearest hundreds is ……………

Solution:

(i) The digit 2 has the highest place value in the number 2309.

(ii) The digit 9 has the highest face value in the number 2039.

(iii) The digit 0 has the lowest place value in the number 2039.

(iv) Both Indian and International systems of numeration have ones period in common.

(v) In the International system of numeration, commas are placed from right after every 3 digits.

(vi) The bigger number from the numbers 57,631 and 57,361 is 57,631.

(vii) 1 crore = 10 million (viii)The smallest 4-digit number with 3 different digits is 1002.

(ix) The greatest 4-digit number with 3 different digits is 9987.

(x) 15 km 300 m = 15300 m

(xi) 7850 cm = 78 m 50 cm

(xii) The number 5079 when estimated to the nearest hundreds is 5100.

 

Question 2.

State whether the following statements are true (T) or false (F):

(i) The difference between the place value and the face of the digit 7 in the number 2701 is 693.

(ii) The smallest 4-digit number -1 = the greatest 3-digit number.

(iii) The place of a digit is independent of whether the number is written in the Indian system or International system of numeration.

(iv) In the International system, a number having less number of digits is always smaller than the number having more number of digits.

(v) The estimated value of 9999 to the nearest tens is 10000.

Solution:

(i) The difference between the place value and the face of the digit 7 in the number 2701 is 693. True

(ii) The smallest 4-digit number-1 = the greatest 3-digit number. True

(iii) The place of a digit is independent of whether the number is written in the Indian system or International system of numeration.

True

(iv) In the International system, a number having less number of digits is always smaller than the number having more number of digits.

True

(v) The estimated value of 9999 to the nearest

Question 2.

State whether the following statements are true (T) or false (F):

(i) The difference between the place value and the face of the digit 7 in the number 2701 is 693.

(ii) The smallest 4-digit number -1 = the greatest 3-digit number.

(iii) The place of a digit is independent of whether the number is written in the Indian system or International system of numeration.

(iv) In the International system, a number having less number of digits is always smaller than the number having more number of digits.

(v) The estimated value of 9999 to the nearest tens is 10000.

Solution:

(i) The difference between the place value and the face of the digit 7 in the number 2701 is 693. True

(ii) The smallest 4-digit number-1 = the greatest 3-digit number. True

(iii) The place of a digit is independent of whether the number is written in the Indian system or International system of numeration.

True

(iv) In the International system, a number having less number of digits is always smaller than the number having more number of digits.

True

(v) The estimated value of 9999 to the nearest

Multiple Choice Questions

Choose the correct answer from the given four options (3 to 17):

Question 3.

The face value of the digit 5 in the number 36503 is

(a) 5

(b) 503

(c) 500

(d) none of these

Solution:

The place value of 5 at hundred’s place

= 5 × 100 = 500 (c)

Question 4.

The difference between the place values of 6 and 3 in 76834 is

(a) 3

(b) 5700

(c) 5930

(d) 5970

Solution:

The place value of 6 at thousand’s place

= 6 × 1000 = 6,000

The place value of 3 at ten’s place

= 3 × 10 = 30

The difference between the place value of 6 and 3 = 6000 – 30 = 5970 (d)

Question 5.

The sum of the place values of all the digits in 5003 is

(a) 8

(b) 53

(c) 5003

(d) 8000

Solution:

The place value of 3 at one’s place

=3 × 1 = 3

The place value of 0 at ten’s place = 0 × 10 = 0

The place value of 0 at hundred’s place = 0 × 100 = 0

The place value of 5 at thousand’s place = 5 × 1000 = 5000

The sum of the place value of all the digits

in 5003 = 3 + 0 + 0 + 5000 = 5003 (c)

 

Question 6.

The total number of 4-digit numbers is

(a) 9000

(b) 9999

(c) 10000

(d) none of these

Solution:

The greatest 3-digit number = 999 The greatest 4-digit number = 9999 .’. The total number of 4-digit numbers

= 9999 – 999 = 9000 (a)

Question 7.

The product of the place values of two-threes in 73532 is

(a) 9000

(b) 90000

(c) 99000

(d) 1000

Solution:

The place value of 3 at ten’s place = 3 × 10 = 30

The place value of 3 at thousand’s place = 3 × 1000 = 3000

The product of place value of two threes = 30 × 3000 = 90000 (b)

Question 8.

The smallest 4-digit number having distinct digits is

(a) 1234

(b) 1023

(c) 1002

(d) 3210

Solution:

The smallest 4-digit number having distinct digits is 1002. (c)

Question 9.

The largest 4-digit number having distinct digits is

(a) 9999

(b) 9867

(c) 9786

(d) 9876

Solution:

The largest 4-digit number having distinct digits is 9867. (b)

 

Question 10.

The largest 4-digit number is

(a) 9999

(b) 9876

(c) 9990

(d) none of these

Solution:

The largest 4-digit number is 9999. (a)

Question 11.

The difference between the largest number of 3-digit and the largest number of 3-digit with distinct digits is

(a) 0

(b) 10

(c) 12

(d) 14

Solution:

The largest number of 3-digit = 999

The largest number of 3-digit with distinct digits = 987

∴ Their difference = 999 – 987 = 12 (c)

Question 12.

If we write natural numbers from 1 to 100, the number of times the digit 5 has been written is

(a) 11

(b) 15

(c) 19

(d) 20

Solution:

If we write natural numbers from 1 to 100, the number of times the digit 5 has been written is 20. (d)

Question 13.

The number 28,549 when rounded off to the nearest hundreds is

(a) 28,000

(b) 28,500

(c) 28,600

(d) 29,000

Solution:

28,549

The digit at tens place is 4, which is less than 5.

Hence, the rounded off number to nearest hundreds = 28,500. (b)

Question 14.

The smallest natural number which when rounded off to the nearest hundreds as 500 is

(a) 499

(b) 501

(c) 450

(d) 549

Solution:

The smallest natural number which when rounded off to the nearest hundreds as 500 is 450. (c)

This is so because the digit at tens place is 5, which is equal to 5.

Question 15.

The greatest natural number which when rounded off to the nearest hundreds as 500 is

(a) 549

(b) 599

(c) 450

(d) none of these

Solution:

The greatest natural number which when rounded off to the nearest hundreds as 500 is 549. (a)

This is so because the digit at tens place is 4, which is less than 5.

Question 16.

The greatest 5-digit number formed by the digits 3, 0, 7 is

(a) 33077

(b) 77730

(c) 77330

(d) none of these

Solution:

The greatest 5-digit number formed by the digits 3, 0, 7 is 77730. (b)

Question 17.

In the International place value system, we write 1 billion for

(a) 10 lakh

(b) 1 crore

(c) 10 crore

(d) 100 crore

Solution:

In the International place value system, we write 1 billion for 100 crore. (d)

Value Based Questions

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Question 1.

The distance between Anu’s home and her school is 4 km 850 m. Everyday she cycles both ways. Find the distance covered by her in a week. (Sunday being a holiday).

What are the advantages of cycling?

Solution:

Distance between Anu’s home and her school = 4 km 850 m = 4 x 1000+ 850 = 4850 m Distance travelled by Anu per day = 4850 m x 2 = 9700 m Since, in a week there are 7 days but Sunday is off.

Hence, distance travelled by Anu for 6 days (a week) = 9700 × 6 = 58200 m

= 58 km 200 m

Cycling is good for health and it saves fuel and helps in reducing pollution.

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Higher Order Thinking Skills (HOTS)

Question 1.

Is there any digit whose place value is always equal to its face value irrespective of its position in any number?

Solution:

Yes, the digit is 0.

Question 2.

Write all 4-digit numbers that can be formed with the digits2 and 5, using both digits equal number of time. Also find their sum.

Solution:

Possible numbers are : 2255, 2552, 2525, 5225, 5252, 5522

and their sum = 2255 + 2552 + 2525 + 5225 + 5252 + 5522 = 23331

Thousand Hundred Tens Ones

2 2 5 5

2 5 5 2

2 5 2 5

5 2 2 5

5 2 5 2

5 5 2 2

Question 3.

What is the difference between the smallest 6-digit number with five different digits and the greatest 5-digit number with four different digits?

Solution:

The smallest 6-digit number with five different digits = 100234.

The greatest 5-digit number with four different digits = 99876.

Their difference = 100234 – 99876 = 358

Question 4.

How many times does the digit 3 occur at tert’s place in natural numbers from 100 to 1000?

Solution:

90 times i.e. 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 and upto 983, 993

 

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ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 1 Knowing Our Numbers 

Question 1.

Write the numeral for each of the following numbers and insert commas correctly:

(i) Six crore nine lakh forty seven.

(ii) One hundred four million seven hundred twenty two thousand three hundred ninety four.

Solution:

(i) 6,09,00,047

(ii) 104,722,394

 

Question 2.

Insert commas suitably and write the numebr 30189301 in words in Indian and International system of numeration.

Solution:

International system : 30,189,301

Three crore one lakh eighty nine thousand three hundred one Thirty million one hundred eighty nine thousand three hundred one

Question 3.

Find the difference between the place value and the face value of the digit 6 in the number 72601.

Solution:

Place value of 6

Question 4.

Write all possible two-digit number using the digits 4 and 0. repetition of digits is allowed.

Solution:

Possible digit numbers = 40, 44

Question 5.

Write all possible natural numbers using the digits 7, 0, 6. Repetition of digits is not allowed.

Solution:

The given digits are 7,0, 6 and repetition of digits is not allowed.

The one- digit numbers that can be formed are 7 and 6.

We are required to write 2-digit numbers.

Out of the given digits, the possible ways of choosing the two digits are

7, 0; 6, 0; 6, 7

Using the digits 7 and 0, the numbers are 70.

Similarily, Using the digits 6 and 0, the numbers are 60

Using the digits 6 and 7, the numbers are 67 and 76.

Hence, all possible 2-digit numbers are

60, 70, 67, 76

Now, We are required to write 3-digit numbers using the digits 7, 0, 6 and the repetition of the digits is not allowed. Keeping 0 at unit’s place, 3-digit number obtained are 670 and 760.

 

Keeping 6 at unit’s place, 3-digit number obtained are 706.

Keeping 7 at unit’s place, 3-digit number obtained are 607.

Hence, all possible 3-digit numbers are : 670, 760, 706 and 607.

All possible numbers using the digits 7, 0 and 6 are

6, 7, 76, 67, 70, 60, 706, 607, 760, 670.

Question 6.

Arrange the following numbers in ascending order:

3706, 58019, 3760, 59801, 560023

Solution:

3706, 3760, 58019, 59801, 560023

Question 7.

Write the greatest six-digit number using four different digits.

Solution:

Greatest six-digits number using four different digit is 999876.

Question 8.

Write the smallest eight-digit number using four different digits.

Solution:

Smallest eight-digit number = 10000023

Question 9.

Find the difference between the greatest and the smallest 4-digit numbers formed by the digits 0, 3, 6, 9.

Solution:

The greatest 4-digit number using 0, 3, 6, 9 = 9630

The smallest 4-digit number using 0, 3, 6, 9 = 3069

∴ Their Difference = 9630 – 3069 = 6561

Question 10.

Find the sum of the four-digit greatest number and the five-digit smallest number, each number having three different digits.

Solution:

Four digit greatest number with three different digits = 9987

Five digit smallest number with three different digit = 10002

∴ Their Sum = 9987 + 10002 = 19989

 

Question 11.

Write the greatest and the smallest four-digit numbers using four different digits with the conditions as given:

(i) Digit 3 always at hundred’s place.

(ii) Digit 0 always at ten’s place.

Solution:

(i) 9387; 1302

(ii)9807; 1203

Question 12.

A mobile number consists of ten digits. First four digits are 9, 9, 7 and 9. Make the smallest mobile number by using only one digit twice from the digits 8, 3, 5, 0, 6.

Solution:

The mobile number is 9979003568.

Question 13.

Two stitch a uniform, 1 m 75 cm cloth is needed. Out of 153 m cloth, how many uniforms can be stitched and how much cloth will remain?

Solution:

Total cloth 153 m = 15300 cm

To stich a uniform, cloth needed

= 1 m 75 cm = 175 cm

Total uniforms can be stiched = 

Hence, 87 uniforms can be stiched 75 cm cloth will remain extra.