WBBSE Solutions For Class 6 Maths Arithmetic Chapter 3 Logical Approximation Of Number

Arithmetic Chapter 3 Logical Approximation Of Number

 

Question 1. Express the following numbers to the nearest multiple of 10 in integers:

1. 4

2. 8

3. 12

4. 69

5. 347

6. 1324

7. 5968

8. 24795

Solution :


1. The units place digit of 4 is 4 and its tens place digit is 0.

Now, the units place digit 4 < 5.

The tens place digit will remain the same and the units place digit will be 0. So the number will be 00 or 0.

To express the number 4 to the nearest multiple of 10 in integers we get 0.

The required answer = 0.

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Alternative method :

The number 4 lies between 0 and 10 and they are the multiples of 10.

Now 4-0 = 4, 10 – 4 = 6, and 4 < 6.

The number 4 is near 0.


2. If we express the number 4 as nearest to the multiple of 10 in an integer.

we get,

The units place digit of the number 8 is 8 which is greater than 5 and its tens place digit is 0. ,

The tens place digit increases by 1 and its units place digit will be 0.

The required number is 10 ( the original tens place digit is 0, and it increases by The present tens place digit is 1 and the units place digit is 0 The required number is 10).

Alternative method :

The number 8 lies between 0 and 10.

Now 8-0 = 8 and 10 – 8 = 2. 8 > 2.

The number 8 is nearer to 10 than 0.

The required number which is nearest to the multiple of 10 in an integer is 10.


3. The given number 12 lies between 10 and 20.

The units place digit is 2 and 2 < 5. The tens place digit is 1.

The tens place digit will remain the same (here 1) and the units place digit will be 0.

∴ The required integer =10.

Alternative method :

Since the given number 12 lies between 10 and 20, and 12 – 10 = 2, 20 – 12 = 8;

12 is nearer to 10. the required integer =10.


4. The units place digit of 69 is 9 and 9 > 5. 

The tens place digit is 6 which increases by 1 and so the tens place digit will be 6+1 = 7 and the units place digit will be 0.

The required integer = 70.

Alternative Method:

The given number 69 lies between 60 and 70.

Now, 69 – 60 = 9, 70 – 69 = 1, and 9 > 1.

The number 69 is nearer to 70.

The required integer = 70.


5. The units place digit of the number 347 is 7 and 7 > 5.

The tens place digit of 347 is 4 which increases by 1.

and so the tens place digit will be 4 + 1 = 5 and the units place digit will be 0.

∴ The required integer = 350.

Alternative method :

The number 347 lies between 340 and 350.

Now 347 – 340 = 7, 350 – 347 = 3, and 7 > 3.

347 is nearer to 350.

The required integer = 350.


6. The units place digit of 1324 is 4 which is less than 5.

Again the tens place digit of the given number is 2.

It will remain the same,

i.e., 2, and the units place digit will be 0.

∴ The required integer = 1320.

The number 1324 lies between 1320 and 1330.

Again, 1324 — 1320 = 4, 1330 – 1324 = 6, and 6 > 4.

So, 1324 is near 1320.

The required integer is 1320.


7. The units place digit of 5968 is 8 and 8 > 5.

The tens place digit of 5968 is 6 which increases by 1 and so the tens place digit will be 6 + 1 = 7 and the units place digit will be zero.

∴ The required integer = 5970.

Alternative method :

The number 5968 lies between 5960 and 5970.

Now, 5968 – 5960 = 8, 5970 – 5968 = 2, and 8 > 2.

So, 5968 is nearer to 5970.

The required integer = 5970.


8. The units place digit of 24795 is 5.

The tens place digit of 24795 is.

9 which increases by 1.

The tens place digit will be 9 + 1 = 10 and the units place digit will be 0.

Here the tens place digit is 10; so the hundred place digit will increase by 1 and the hundreds place digit will be 7 + 1 = 8 and the tens place digit will be 0.

The required integer = 24800.

Alternative method:

The number 24795 lies between 24790 and 24800

As 24795 – 24790 = 5, 24800 – 24795 = 5, and 5 = 5, the number 24759 is equidistant from 24790 and 24800.

In this case, as a rule, we take the greater number between 24790 and 24800, which is 24800.

The required integer = 24800.

In order to express a number in a multiple of 10 in an integer, we shall only consider the units place digit and tens place digit. Here the special case is Questions 1 and 8.

 

Question 2. Express the following numbers to the nearest multiple of 100 in integer:

1. 22

2. 78

3. 621

4. 483

5. 2178

6. 6521

7. 80345

8. 674312

Solution:


1. The given number is 22.

The tens place digit is 2 and 2 < 5.

Here the hundreds place digit is 0 (if a number does not contain the hundreds place digit, it should be taken 0) and it remains the same.

The tens place digit and units place digit both will be 0.

The required integer = 000 or 0.


2. The given number is 78.

The tens place digit is 7 which is greater than 5.

The hundreds place digit is 0 and it increases by 1.

So the hundreds place digit will be 0+1 = 1 and both the digits in the tens place and units place digit will be 0.

The required integer = 100.


3. The given number is 621.

The tens place digit is 2 < 5.

So the hundreds place digit will remain the same which is.6 here and both the digits in the tens place and units place will be 0.

∴ The required integer = 600.


4. The given number is 483.

The tens place digit is 8 and 8 > 5.

The hundreds place digit is 4 which increases by 1.

So the hundreds place digit is 4 + 1 = 5 and both the digits of both the tens place and units place should be 0.

The required integer = 500.


5. The given number is 2178.

The hundred place digit is 7 and 7 > 5.

The hundreds place digit is 1 which increases by 1.

So the hundreds place digit is 1 + 1 = 2

and both the digits of tens place and units place should be 0.

The required integer = 2200.


6. The given number is 6521.

The tens place digit is 2 and 2 < 5.

So the hundreds place digit will remain the same which is 5 here and both the digits of the tens place and units place should be 0.

The required integer = 6500.


7. The given number is 80345.

Its tens place digit is 4 and 4 < 5.

So the hundreds place digit will remain the same which is 3 here and both the digits in the tens place and units place should be 0.

∴ The required integer = 80300.


8. The given number is 674312.

Its tens place digit is 1 and 1 < 5.

So the hundreds place digit will remain the same which is 3 and both the digits in the tens place and the units place should be zero.

The required integer is 674300.

To express any number to the nearest multiple of 100 in integers we have to consider only the digits in the hundreds, the tens place and the units place.

 

Question 3. Express the following numbers to the nearest multiple of 1000 in

integers :

1. 2

2. 88

3. 346

4. 827

5. 6719

 6. 8394

7. 27985

8. 499957

Solution :


1. The given number is 2.

Its hundreds place digit is 0 and 0 < 5.

So the thousands place digit will remain the same which is 0 here and all the digits in the hundreds place, tens place, and units place should be zero.

∴ The required integer = 0000 or 0.


2. The given number is 88.

Its hundreds place digit is 0 and 0 < 5.

So the thousands place digit will remain the same which is 0 here and all the digits in the hundreds place, tens place, and units place should be zero.

∴ The required integer = 0000 or 0.


3. The given number is 346.

Its hundreds place digit is 3 and 3 < 5.

So the thousands place digit will remain the same which is 0 here and all the digits in 1

hundreds place, tens place, and units place should be zero.

∴ The required integer = 00000 or 0.


4. The given number is 827.

Its hundreds place digit is 8 and 8 > 5.

The thousands place digit increases by 1.

The thousands place digit in the given number is 0 and so the thousands place digit in the required number will be 0 + 1 = 1.

All the digits in the hundreds place, tens place, and units place are zero.

∴ The required integer = 1000.


5. The given number is 6719.

Its hundreds place digit is 7 and 7 > 5.

The thousands place digit increases by 1.

The thousands place digit in the given number is 6.

So the thousands place digit in the required number will be 6 + 1 – 7 and all the digits in the hundreds place, tens place, and units place should be zero.

∴ The required integer = 7000.


6. The given number is 8394.

Its hundreds place digit is 3 and 3 < 5.

So the thousands place digit will remain the same which is 8 here. All the digits in the

hundreds place, tens place, and units place should be zero.

∴  The required integer = 8000.


7. The given number is 27985.

Its hundreds place digit is 9 and 9 > 5.

∴ The thousands place digit increases by 1 and the thousands place digit in the given number is 7.

So the thousands place digit in the required number will be 7 + 1 = 8 and all the hundreds place digit, tens place digit, and units place digit is zero.

∴ The required integer = 28000.


8. The given number is 499957.

Its hundreds place digit is 9 and 9 > 5.

The thousands place digit increases by 1.

The thousands place digit in the given number is 9.

So the thousands place digit in the required number = 9 + 1 = 10.

Again if the thousands place digit is 10, then the ten thousand place digit will increase by 1 and it becomes 9 + 1 = 10.

Again if the ten thousand place digit is 10, then the lacs place digit will increase by 1 and so it becomes 4+1 = 5 and all the digits in the ten thousand place, thousands place, hundreds place, tens place, units place should be zero.

∴ The required integer = 500000.

 

Question 4. All the following cases after expressing the numbers to the nearest. multiple of 10 in integers:

1. 37 + 54,

2. 66 + 73,

3. 251 + 175,

4. 24 + 59 ,

Solution :


1. 37 + 54.

Here for 37, the required integer = 40 as 7 > 5

For 54, the required integer = 50 as 4 < 5.

∴  The required sum = 40 + 50 = 90.


2. 66 + 73.

Here for 66, the required integer = 70 as 6 > 5

For 73, the required integer = 70 as 3 < 5 –

∴ The required sum = 70 + 70 = 140.


3. 251 + 175.

Here for 251, the required integer = 250 as 1 < 5

For 175, the required integer = 180 as 5 = 5

The required sum is 250 + 180 = 430.


4. 24 + 59.

Here for 24, the required integer = 20 as 4 < 5

For 59, the required integer = 60 as 9 > 5

The required sum = 20 + 60 = 80.

 

Question 5. Subtract the following cases after expressing the numbers to the nearest multiple of 10 in integers :

1. 73 – 48,

2. 97 – 38,

3. 76 – 29,

4. 462 – 271

Solution :


1. 73-48

Here for 73, the required integer = 70 as 3 < 5

For 48, the required integer = 50 as 8 > 5

∴ The required result of subtraction = is 70 – 50 = 20.


2. 97 – 38

Here for 97, the required integer = 100 as 7 > 5

and so the tens place digit 9 increases by 1 and it becomes 9 + 1 = 10.

For 38, the required integer = 40 as 8 > 5

and so the tens place digit 3 increases by 1 and it becomes 3 + 1=4.

∴ The required result of subtraction = is 100 – 40 = 60.


3. 76 – 29.

Here for 76, the required integer = 80 as 6 > 5

and so the tens place digit increases by 1 and it becomes 7 + 1 = 8.

For 29, the required integer = 30 as 9 > 5

and so the tens place digit increases by 1 and it becomes 2+1=3.

∴ The result of subtraction = 80 – 30 = 50.


4. 462 – 271.

Here for 462, the required integer = 460 as 2 < 5.

For 271, the required integer = 270 as 1 < 5.

∴ The result of subtraction = 460 – 270 = 190.

 

Question 6. Add the following after expressing the numbers nearest to the multiple of 100 in integers :

1. 426 + 589,

2. 356 + 435,

3. 1248 + 4329,

4. 170 + 895,

5. 947 + 448,

6. 5612 + 2095

Solution :


1. 426 + 589 

Here for 426, the required integer = 400

as the tens place digit is 2 < 5

and the hundreds place digit will remain the same as 2.

For 589, the required integer = 600

as the tens place digit is 8 > 5

and so the hundreds place digit increases by 1.

The hundred-place digit will be 5 + 1 = 6.

∴ The required sum = 400 + 600 = 1000.


2. 356 + 435

Here for 356, the required integer = 400

as the tens place digit is 5 = 5.

So the hundreds place digit which is 3 increases by 1.

It becomes 3 + 1=4.

For 435, the required integer = 400

as the tens place digit is 3 < 5.

So the hundreds place digit will remain the same.

∴ The required sum = 400 + 400 = 800.


3. 1248 + 4329.

Here for 1248, the required integer = 1200

as the tens place digit of the given number is 4 and 4 < 5.

So the hundreds place digit will remain the same which is 2.

For 4329, the required integer = 4300

as the tens place digit of the given number is 2, and 2 < 5.

So the hundreds place digit will remain the same which is 3.

∴ The required sum = 1200 + 4300 = 5500.


4. 170 + 895.

Here for 170, the required integer = 200

as the tens place digit, is 7 which is greater than 5.

So the hundreds place digit increases by 1 and it becomes 1 + 1=2.

For 895, the required integer = 900

as the tens place digit is 9 and 9 > 5.

So the hundreds place digit increases by 1 and it becomes 8 + 1 =9.

∴ The required sum = 200 + 900 = 1100.


5. 947+ 448.

Here for 947, the required integer = 900 as the tens place digit is 4 and 4 < 5.

So the hundreds place digit remains the same which is 9.

For 448, the required integer = 400 as the tens place digit is 4 and 4 < 5.

So the hundreds place digit remains the same which is 4.

∴ The required sum = 900 + 400 = 1300


6. 5612 + 2095.

Here for 5612, the required integer = 5600,

as the tens place digit is 1 and 1 < 5.

So the hundreds place digit remains the same which is 6.

For 2095, the required integer = 2100

as the tens place digit if 9 and 9 > 5.

So the hundreds place digit increases by 1 and it becomes 0 + 1 = 1.

∴ The required sum = 5600 + 2100 = 7700.

 

Question 7. Subtract the following cases after expressing the numbers to the nearest multiple of 100 in integers:

1. 678 – 125

2. 4258 – 2436

Solution:


1. 678 – 125

Here for 678, the required integer = 700

as the tens place digit of the given number is 7, and 7 > 5.

So the hundreds place digit increases by 1 and it becomes 6+1=7.

For 125, the required integer = 100

as the tens place digit of the given number is 2, and 2 < 5.

So the hundreds place digit remains the same which is 1.

∴ The required result of subtraction = is 700 – 100 = 600.


2. 4258 – 2436

Here for 4258, the required integer is 4300

as the tens place digit is 5 = 5.

So the hundred-place digit increases by 1 and it becomes = 2 + 1 =3.

Again for the number 2436, the required integer = 2400

as the tens place digit is 3 and 3 < 5.

So the hundreds place digit remains the same and it is 4.

∴ The required result of subtraction = 4300 – 2400 = 1900.

 

Example 8.

1. Add the following cases after expressing the numbers to the nearest multiple of 1000 in integers:

1. 2836 + 7466,

2. 3076 + 5731,

3. 7767 + 3685,

4. 8005 + 7483

5. 1375 + 6307,

6. 8643 + 5285.

Solution :


1. 2836 + 7466

Here for 2836, the required integer = 3000

as the hundreds place digit of the given number is 8 and 8 > 5.

So the thousands place digit increases by 1.

It becomes 2 + 1 = 3.

Again for 7466, the required integer = 7000

as the digit in the hundred places is 4 and 4 < 5.

So the thousands place digit will remain the same as 7.

The required sum = 3000 + 7000

= 10000.

The required sum = 10000.


2. 3076 + 5731

Here for 3076, the required integer = 3000

as the hundred place digit is 0 and 0 < 5.

So the thousand place digit will remain the same as 3.

’ For the number 5731, the required integer = 6000

as the hundreds place digit is 7 and 7 > 5.

So the thousands place digit increases by 1 and it will be 5 + 1 = 6.

The required sum = 3000 + 6000

= 9000.

The required sum = 9000.

 

3. 7767 + 3685

Here for 7767, the required integer = 8000

as the hundreds place digit is 7 and 7 > 5.

So the thousands place digit increases by 1 and it becomes 7 + 1 = 8.

For the number 3685, the required integer = is 4000

as the hundreds place digit is 6 and 6 > 5.

The thousands place digit increases by 1 and it becomes 3 + 1=4.

The required sum = 8000 + 4000

= 12000.

The required sum = 12000.


4. 8005 + 7483

Here for 8005, the required integer = 8000

as the hundreds place digit is 0 and 0 < 5.

So the thousands place digit will remain the same as 8.

For the number 7483, the required integer = is 7000

as the hundred place digit is 4 and 4 < 5.

So the thousands place digit will remain the same as 7.

∴ The required sum = 8000 + 7000

= 15000.

The required sum = 15000.


5. 1375 + 6307

Here for 1375, the required integer = 1000

as the hundreds place digit is 3 and 3 < 5.

So the thousands place digit will remain the same as 1.

For the number 6307, the required integer = 6000

as the hundreds place digit = 3 and 3 < 5.

So the thousands place digit will remain the same as 6.

∴ The required sum = 1000 + 6000

= 7000.

The required sum = 7000.


6. 8643 + 5285

Here the integer 8643 will become 9000

as the hundreds place digit is 6 and 6 > 5.

So the thousands place digit will increase by 1 and it becomes 8 + 1 = 9000.

For the number 5285, the required integer = 5000

as the hundreds place digit is 2 and 2 < 5.

So the thousands place digit will remain the same as 5.

∴ The required sum = 9000 + 5000

= 14000.

The required sum = 14000.

 

2. Subtract the following cases after expressing the number to the nearest multiple of 1000 in integers :

1. 8493 – 7362 ;

2. 9852 – 4526

Solution :


1. 8493 – 7362

Here for the number 8493, the required integer = 8000

as the hundreds place digit is 4 and 4 < 5.

So the thousands place digit will remain the same as 8.

For the number 7362, the required integer = 7000

as the hundreds place digit is 3 and 3 < 5.

So the thousands place digit will remain the same as 7.

The result of subtraction = 8000 – 7000

= 1000.

The result of subtraction = 1000.


2. 9852 – 4526 

Here for the number 9852, the required integer = 10000

as the hundreds place digit is 8 and 8 > 5.

So the thousands place digit will increase by 1 and it becomes 9 + 1 = 10.

Again for the number 4526, the required integer = 5000

as the hundreds place digit is 5 and 5 = 5.

So the thousands place digit will increase by 1 and it becomes 4+1 = 5.

The. result of subtraction = 10000 – 5000

= 5000.

The. result of subtraction = 5000.

 

Question 9. In Parliament Election 2014, party A got 897436 votes and party B got 684745 votes. Which party did get more votes than the other and how many more votes did the party get? Give your answer nearest to multiples of 1000 in integers.

Solution:

Party A got = 897436 votes.

The required integer = 897000

as the hundreds place of the number 897436 is 4 and 4 < 5.

So the thousands place digit will remain the same as 7.

Party B got = 684745 votes. Here the required integer = 685000

as the hundreds place digit is 7 and 7 >5.

So the thousands place digit will increase by 1 and it becomes 4+1=5.

∴ Part A got more votes than party B.

Again, 897000 – 685000 = 212000 votes

∴ Party A got 212000 more votes than party B.

 

Question 10. This year in puja, Bulganin purchased a pant costing Rs. 1275, a shirt costing Rs. 685 and a pair of shoes costing Rs. 415. What was her total cost? Give your answer to the nearest multiple of 100 in integers.

Solution: Cost of the pant = Rs. 1275,

Cost of the shirt = Rs. 685

Cost of the shoes = Rs. 415

They are to be converted into integers to the nearest multiple of 100.

The required integers for pant = Rs. 1300 because the tens place digit is 7 and 7 > 5.

So the hundreds place digit will increase by 1 and it will become 2+1 = 3.

The required integral cost of the shirt = Rs. 700

as the tens place digit is 8 which is greater than 5 and so the hundreds place digit will be 1 more and it will be 6 + 1 – 7.

Again the required integral cost of shoes = Rs. 400 because the tens place digit is 1 and 1 < 5.

So the hundreds place digit will remain the same as 4.

∴ Total cost = Rs. (1300 + 700 + 400) = Rs. 2400.

∴ Bulganin purchased pants, shirts, and shoes for Rs. 2400.

 

Question 11. Find the approximate whole number of the number 24986 nearest to multiple by 10n when

1. n = 1,

2. n = 2,

3. n = 3.

Solution :


1. When n = 1,

then 10n = 10

= 10.

The required integer. = 24990, because the unit place digit is 6 and 6 > 5.

So the tens place digit will be 1 more than 8.

∴ The received tens place digit = 8 + 1=9.

∴ The answer is 24990.


2. When n = 2,

then 10 = 100

The required integer = 25000.

Because the tens place digit is 8 which is greater than 5.

So the hundreds place digit will increase by 1 and it becomes 9 + 1 = 10 and for this, a thousand places will also be 1 more than 4.

The required thousands place digit = 4 + 1 = 5.

The required answer is 25000.


3. When n = 3, 

then 10n= 10³ = 1000.

The required integer = 25000.

Because the hundreds place digit = 9 and 9 > 5.

So the thousands place digit will be 1 more than 4.

It becomes 4 + 1 =5.

The required answer = 25000.

 

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