WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Miscellaneous Examples

Chapter 1 Simplification Miscellaneous Examples

Miscellaneous Examples

Example 1. Find the greatest number by which 10019 and 10621 will be exactly divisible.

Solution: 

Given: 10019 and 10621

The required greatest number will be the H. C. F. of 10019 and 10621.

Wbbse Class6 Math Solution

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 1

 

∴ The H.C.F. of 10019 and 10621 is 43.

So the required greatest number = is 43.

Read And Learn More: WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Problems

Wbbse Class6 Math Solution

Example 2. Find the greatest number that will divide 1347, 2046, and 2568 leaving the remainder 8, 5, and 7 respectively.

Solution:

Given:

1347, 2046, and 2568 leaving the remainder 8, 5, and 7 respectively

We have to find the greatest number that will divide 1347, 2046, and 2568 leaving the remainder 8, 5, and 7 respectively.

∴ 1347 – 8 = 1339; 2046 – 5 = 2041; 2568 – 7= 2561.

Therefore, the numbers 1339, 2041, and 2561 must be divisible by the required greatest number and so the H. C. F. of these numbers will be the required greatest number.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 2 Q 1

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 2 Q 2

 

∴ H.C.F. 1339, 2041, and 2561 = 13.

Wbbse Class6 Math Solution

Example 3. Find the least number which when divided by 12, 18, 24, 36, and 45 will leave a remainder of 8 in each case.

Solution:

Given:

12, 18, 24, 36, and 45 will leave a remainder of 8 in each case

The L. C. M. of 12, 18, 24 36, and 45 will be the least number, which must be divisible by these numbers.

So the required least number exceeds the L. C. M. by 8 as it leaves a remainder of 9 in each case.

Now first we shall find the L. C. M. of the given numbers.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 3

 

.. L. C. M. 2 x 2 x 3 x 3 x 2 x 5 = 360.

So the required least number = 360+ 8 = 368.

 

Example 4. Find the greatest number which will divide 98, 137, and 202 so as to leave the same remainder in each case.

Solution:

Given:

98, 137, and 202 so as to leave the same remainder in each case

Since the required number when divides 98, 137, and 202 leaves the same remainder in each case, therefore (137 – 98) or 39 and (202 – 137) or 65 must be divisible by the required greatest number. 

So the required greatest number will be the H. C. F. of 39 and 65.

Now the H. C. F. of 39 and 65 = 13.

The required greatest number = is 13.

 

Example 5. Find the least number which when added to 5 will be exactly divisible by 36, 54, 66, and 72.

Solution:

Given:

36, 54, 66, and 72.

Here we shall first find the L. C. M. of 36, 54, 66, and 72.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 5

 

L.C. M. = 2 x 2 x 3 x 3 x 3 x 11 x 2 = 2376.

So 2376 is the least number that must be divisible by 36, 54, 66, and 72.

But the required number will be 5 less than 2376.

∴ The required least number is 2376 – 5 = 2371.

 

Example 6. From what least number should 7 be subtracted to make it divisible by 65, 91, 104, and 195?

Solution:

Given:

65, 91, 104, and 195

Here we shall first find the L. C. M. of 65, 91, 104, and 195.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 6

 

∴ So L. C. M. = 5 x 13 x 7 x 8 x 3 = 10920.

So 10920 is the least number that must be divisible by 65, 91, 104, and 195.

The required least number will be 7 more than the L. C. M.

Hence the required least number = 10920 + 7

= 10927.

 

Example 7. Find the greatest number of 5 digits which is exactly divisible by 24, 36, 54, and 72.

Solution:

Given:

24, 36, 54, and 72

First, we shall find the L. C. M. of 24, 36, 54, 72.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 7 Q 1

 

 

∴ L. C. M. = 2 x 2 x 2 x 3 x 3 x 3 = 216.

Since L. C. M. of the given numbers is the least number that is divisible by the numbers individually, the required number will also be divisible by the L. C. M.

The greatest number of 5 digits = 99999.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 7 Q 2

 

∴ Greatest number of 5 digits divisible by 216 = 99999 – 207

= 99792

Hence the required number = 99792.

 

Example 8. Find the least number of 6 digits which is exactly divisible by 33, 55, 66, and 88.

Solution:

Given:

33, 55, 66, and 88

First, we shall find the L. C. M. of 33, 55, 66, 88.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 8 Q 1

 

∴ L. C. M. = 11 x 2 x 3 x 5 x 4 = 1320.

So L. C. M. of the given numbers is the least number divisible by the numbers individually and the required number will also be divisible by 1320.

The least number of 6 digits = 100000.

wbbse class 6 math solution

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 8 Q 2

 

1320 – 1000 = 320

∴ 100000 + 320 = 100320.

So the least 6-digit number divisible by the given numbers = 100320.

 

Example 9. Find the least number between 800 and 900 which when divided by 46 and 69 will leave a remainder of 5 in each case.

Solution:

Given:

The least number between 800 and 900 which when divided by 46 and 69 will leave a remainder of 5 in each case

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 9 Q 1

 

L. C. M. of 46 and 69 = 23 x 2 x 3 = 138.

wbbse class 6 math solution

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 9 Q 2

 

138 – 110 = 28

∴ 800 + 28 = 828.

 828 is a number lying between 800 and 900 which is divisible by the L. C. M. of 46 and 69.

So 828 is divisible by 46 and 69 individually also.

But there is a remainder 5.

Hence the required number = 828 + 5

= 833.

The least number is = 833.

 

Example 10. Find the least number which when divided by 56, 70, 84, and 140 will leave the remainder of 49, 63, 77, and 133 respectively.

Solution:

Given:

56, 70, 84, and 140 will leave the remainder of 49, 63, 77, and 133 respectively

We have, 56 497; 7063 = 7; 84 77 = 7; 140 133 7.

The required number when added to 7 becomes exactly divisible by 56, 70, 84, and 140.

wbbse class 6 math solution

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 10

 

∴ L. C. M. of 56, 70, 84, and 140 = 7 x 2 x 2 x 5 x 2 x 3 = 840.

So 840 is the least number exactly divisible by 56, 70, 84, and 140.

The required number = is 840 – 7

= 833.

The least number  is= 833.

class 6 maths solutions wbbse

Example 11. What least number of 6 digits has 233 as a factor?

Solution:

Given:

6 digits has 233 as a factor

Here the required least number of 6 digits should be divisible by 233. 

The least number of 6 digits = 100000.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 11

233 – 43 = 190.

∴ 100000 + 190 = 100190

So the required least number = 100190.

 

Example 12. Find the least number which when divided by 6, 8, and 10 will leave a remainder of 3 in each case but will leave no remainder when divided by 11. 

Solution:

Given:

6, 8, and 10 will leave a remainder of 3 in each case but will leave no remainder when divided by 11

L.C. M. of 6, 8, 10 = 120.

∴ The required number will be 3 more than the least multiple of 120 and is divisible by 11.

120 x 1+3 = 123, it is not divisible by 11.

120 x 2 + 3 = 243, it is not divisible by 11.

120 x 3 + 3 = 363, which is divisible by 11.

Hence the required least number = is 363.

 

Example 13. Four bells toll together and then begin to toll at intervals of 12, 15, 18, and 20 seconds respectively. When will they toll together again?

Solution:

Given:

Four bells toll together and then begin to toll at intervals of 12, 15, 18, and 20 seconds respectively.

The time between the two consecutive simultaneous tolls of the four bells will be exactly divisible by 12, 15, 18, and 20 seconds i.e., will be exactly divisible by the L. C. M. of 12, 15, 18, and 20.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 13.

wbbse math solution class 6

∴ L. C. M. = 2 x 2 x 3 x 5 x 3 = 180.

So the required time = 180 seconds 3 min.

∴ The bells will toll together again after 3 min.

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