Problem 1 :
Of two numbers, 1/5th of a the greater equal to 1/3rd of the smaller and their sum is 16. Find the numbers.
Solution :
Let x and y be the two numbers such that
x > y
Given : The sum of the two number is 16.
So, we have
x + y = 16 ------(1)
Given : 1/5th of a the greater equal to 1/3rd of the smaller
So, we have
1/5 ⋅ x = 1/3 ⋅ y
Simplify.
3x = 5y ------(2)
Solving (1) and (2), we get
x = 10 and y = 6
Hence, the two numbers are 10 and 6.
Problem 2 :
The wages of 8 men and 6 boys amount to $33. If 4 men earn $4.50 more than 5 boys, determine the wages of each man and boy.
Solution :
Let x and y be the wages of each man and boy.
Given : The wages of 8 men and 6 boys amount to $33.
So, we have
8x + 6y = 33 ------(1)
Given : 4 men earn $4.50 more than 5 boys.
So, we have
4x - 5y = 4.50 ------(2)
Solving (1) and (2), we have
x = 3 and y = 1.5
Hence, the wages of each man and each boy are $3 and $1.50 respectively.
Problem 3 :
A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed. Find the number.
Solution :
Because the number is between 10 and 100, it has to be a two digit number.
So, let xy be the number between 10 and 100.
Given : The number between 10 and 100 is five times the sum of its digits.
xy = 5(x + y) ------(1)
xy is a two digit number.
And x is in tens place and y is in ones place.
So, we have
xy = 10 ⋅ x + 1 ⋅ y
xy = 10x + y
Then, we have
(1)------> 10x + y = 5(x + y)
10x + y = 5x + 5y
Simplify.
5x - 4y = 0 -------(2)
Given : If 9 be added to it the digits are reversed.
xy + 9 = yx
10 ⋅ x + 1 ⋅ y + 9 = 10 ⋅ y + 1 ⋅ x
10x + y + 9 = 10y + x
Simplify.
9x - 9y = - 9
Divide both sides by 9.
x - y = - 1 ------(3)
Solving (2) and (3), we get
x = 4 and y = 5
Hence, the required number is 45.
Problem 4 :
The age of a man is three times the sum of the ages of his two sons and 5 years hence his age will be double the sum of their ages. Find the present age of the man.
Solution :
Let x be the present age of the man and "y" be the sum of the present ages of his two sons.
Given : Present age of the man is 3 times the sum of the ages of 2 sons.
So, we have
x = 3y ------(1)
5 years hence,
Age of the man = x + 5
Sum of the ages of his two sons = y + 5 + 5 = y + 10
(There are two sons in y. So 5 is added twice)
Given : 5 years hence, age of the man will be double the sum of the ages of his two sons.
So, we have
x + 5 = 2(y + 10)
Simplify.
x + 5 = 2y + 20
x = 2y + 15 ------(2)
From (1), we can plug x = 3y in (2).
(2)------> 3y = 2y + 15
Subtract 2y from both sides.
y = 15
Substitute y = 15 in (1).
(1)------> x = 3 ⋅ 15
x = 45
Hence the present age of the man is 45 years.
Problem 5 :
A trader has 100 units of a product. He sells some of the units at $6 per unit and the remaining units at $8 per units. He receives a total of $660 for all 100 units. Find the number units sold in each category.
Solution :
Let x be the no. of units sold at $6/unit and y be the no. of units sold at $8/unit.
Given : The trader sells 100 units in all.
So, we have
x + y = 100 ------(1)
Given : He receives a total of $660 for all 100 units.
So, we have
6x + 8y = 660
Divide both sides by 2.
3x + 4y = 330 ------(2)
Solving (1) and (2), we get
x = 70 and y = 30
Hence, the no. of tickets sold at $6 per unit is 70 and the no. of tickets sold at $8 per unit is 30.
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