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A box had some pens in it.When the pen are distributed equally to 18 children, none remained but while distributing to 19 children,5 children got 1 less than the others.How many pens were there in the box?
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A box had some pens in it.When the pen are distributed equally to 18 c...
Problem:
A box had some pens in it. When the pens are distributed equally to 18 children, none remained. But while distributing to 19 children, 5 children got 1 less pen than the others. How many pens were there in the box?
Solution:
Let's assume that there were 'x' pens in the box.
Equal Distribution to 18 Children:
When the pens are distributed equally to 18 children, none remain. This means that the number of pens is divisible by 18.
So, we can write the equation as:
x = 18 × y (where y is a positive integer)
Unequal Distribution to 19 Children:
When the pens are distributed equally to 19 children, 5 children get 1 less pen than the others. This means that the number of pens is 1 less than a multiple of 19.
So, we can write the equation as:
x = 19 × z - 1 (where z is a positive integer)
Finding the Value of 'x':
To find the value of 'x', we need to find a common value of 'x' that satisfies both equations.
Since 'x' is both divisible by 18 and 1 less than a multiple of 19, we can find the least common multiple (LCM) of 18 and 19.
Calculating LCM of 18 and 19:
To find the LCM of 18 and 19, we can list the multiples of both numbers and find the least common multiple.
Multiples of 18: 18, 36, 54, 72, 90, ...
Multiples of 19: 19, 38, 57, 76, 95, ...
From the list, we can see that the LCM of 18 and 19 is 342.
Calculating the Value of 'x':
Substituting the value of 'x' as 342 in the equations, we can find the value of 'y' and 'z'.
For the equation x = 18 × y:
342 = 18 × y
y = 342 ÷ 18
y = 19
For the equation x = 19 × z - 1:
342 = 19 × z - 1
343 = 19 × z
z = 343 ÷ 19
z = 18
Conclusion:
Therefore, there were 342 pens in the box.
A box had some pens in it. When the pens are distributed equally to 18 children, none remained. But while distributing to 19 children, 5 children got 1 less pen than the others. How many pens were there in the box?
Solution:
Let's assume that there were 'x' pens in the box.
Equal Distribution to 18 Children:
When the pens are distributed equally to 18 children, none remain. This means that the number of pens is divisible by 18.
So, we can write the equation as:
x = 18 × y (where y is a positive integer)
Unequal Distribution to 19 Children:
When the pens are distributed equally to 19 children, 5 children get 1 less pen than the others. This means that the number of pens is 1 less than a multiple of 19.
So, we can write the equation as:
x = 19 × z - 1 (where z is a positive integer)
Finding the Value of 'x':
To find the value of 'x', we need to find a common value of 'x' that satisfies both equations.
Since 'x' is both divisible by 18 and 1 less than a multiple of 19, we can find the least common multiple (LCM) of 18 and 19.
Calculating LCM of 18 and 19:
To find the LCM of 18 and 19, we can list the multiples of both numbers and find the least common multiple.
Multiples of 18: 18, 36, 54, 72, 90, ...
Multiples of 19: 19, 38, 57, 76, 95, ...
From the list, we can see that the LCM of 18 and 19 is 342.
Calculating the Value of 'x':
Substituting the value of 'x' as 342 in the equations, we can find the value of 'y' and 'z'.
For the equation x = 18 × y:
342 = 18 × y
y = 342 ÷ 18
y = 19
For the equation x = 19 × z - 1:
342 = 19 × z - 1
343 = 19 × z
z = 343 ÷ 19
z = 18
Conclusion:
Therefore, there were 342 pens in the box.
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A box had some pens in it.When the pen are distributed equally to 18 children, none remained but while distributing to 19 children,5 children got 1 less than the others.How many pens were there in the box?
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A box had some pens in it.When the pen are distributed equally to 18 children, none remained but while distributing to 19 children,5 children got 1 less than the others.How many pens were there in the box? for Class 5 2024 is part of Class 5 preparation. The Question and answers have been prepared according to the Class 5 exam syllabus. Information about A box had some pens in it.When the pen are distributed equally to 18 children, none remained but while distributing to 19 children,5 children got 1 less than the others.How many pens were there in the box? covers all topics & solutions for Class 5 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A box had some pens in it.When the pen are distributed equally to 18 children, none remained but while distributing to 19 children,5 children got 1 less than the others.How many pens were there in the box?.
A box had some pens in it.When the pen are distributed equally to 18 children, none remained but while distributing to 19 children,5 children got 1 less than the others.How many pens were there in the box? for Class 5 2024 is part of Class 5 preparation. The Question and answers have been prepared according to the Class 5 exam syllabus. Information about A box had some pens in it.When the pen are distributed equally to 18 children, none remained but while distributing to 19 children,5 children got 1 less than the others.How many pens were there in the box? covers all topics & solutions for Class 5 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A box had some pens in it.When the pen are distributed equally to 18 children, none remained but while distributing to 19 children,5 children got 1 less than the others.How many pens were there in the box?.
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