Problem 1 :
One number is greater than thrice the other number by 2. If four times the smaller number exceeds the greater by 5, find the numbers.
Solution :
Let x be the smaller number and y be the greater number.
Given : One number is greater than thrice the other number by 2.
Then, we have
y = 3x + 2 -----(1)
Given : Four times the smaller number exceeds the greater by 5.
Then, we have
4x = y + 5
Substitute (3x + 2) for y.
4x = 3x + 2 + 5
4x = 3x + 7
Subtract 3x from each side.
x = 7
Substitute 7 for x in (1).
(1)-----> y = 3(7) + 2
y = 21 + 2
y = 23
So, the numbers are 7 and 23.
Problem 2 :
The ratio of income of two persons is 1 : 2 and the ratio of their expenditure is 3 : 7. If the first person saves $2000 and the second person saves $3000, find the expenditure of each person.
Solution :
From the ratio of income 1 : 2, the incomes of the two persons can be assumed as x and 2x.
From the ratio of expenditure 3 : 7, the expenditures of the two persons can be assumed as 3y and 7y.
Savings = Income - Expenditure
The first person saves $3000.
Then, we have
x - 3y = 3000
Add 2y to each side.
x = 3000 + 3y -----(1)
The second person saves $7000.
Then, we have
2x - 7y = 7000
Substitute (3000 + 3y) for x.
(1)-----> 2(2000 + 3y) - 7y = 3000
4000 + 6y - 7y = 3000
4000 - y = 3000
Subtract 1000 from each side.
-y = -1000
Multiply each side by -1.
y = 1000
3y = 3(1000) = 3000
7y = 7(1000) = 7000
So, the expenditures of the two persons are $3000 and $7000.
Problem 3 :
Three chairs and two tables cost $700 and five chairs and six tables cost $1100. Find the cost of each chair.
Solution :
Let x and y be the costs of each chair and table respectively.
Then, we have
3x + 2y = 700 -----(1)
5x + 6y = 1700 -----(2)
Multiply (1) by 3.
(1) ⋅ 3 -----> 9x + 6y = 2100
Subtract 9x from each side.
6y = 2100 - 9x
Substitute (2100 - 9x) for 6y in (2).
(2)-----> 5x + 2100 - 9x = 1700
2100 - 4x = 1700
Subtract 2100 from each side.
-4x = -400
Divide each side ny -4.
x = 100
So, the cost of each chair is $100.
Problem 4 :
A father's age is three times the sum of the age of his two son. After 5 years, his age will be twice the sum of the ages of his two sons. Find the present age of the father.
Solution :
Let x be the present age of the father and y be the sum of the present ages of the two sons.
At present :
x = 3y -----(1)
After 5 years :
x + 5 = 2(y + 5 + 5)
x + 5 = 2(y + 10)
x + 5 = 2y + 20
Subtract 5 from each side.
x = 2y + 15
Substitute 3y for x.
3y = 2y + 15
Subtract 2y from each side.
y = 15
Substitute 15 for y in (1).
(1)-----> x = 3(15)
x = 45
So, the present age of the father is 45 years.
Problem 5 :
If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1. Again if the numerator is decreased by 4 and the denominator by 2, it becomes 1/2. Find the fraction.
Solution :
Let x/y be the required fraction.
Given : If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1.
(x + 2) / (y + 1) = 1
Multiply each side by (y + 1).
x + 2 = 1(y + 1)
x + 2 = y + 1
Subtract 2 from each side.
x = y - 1 -----(1)
Given : If the numerator is decreased by 4 and the denominator by 2, it becomes 1/2.
(x - 4) / (y - 2) = 1 / 2
Multiply each side by 2(y - 2).
2(x - 4) = 1(y - 2)
2x - 8 = y - 2
Add 8 to each side.
2x = y + 6
Substitute (y - 1) for x.
2(y - 1) = y + 6
2y - 2 = y + 6
Add 2 to each side.
2y = y + 8
Subtract y from each side.
y = 8
Substitute 8 for y in (1).
(1)-----> x = 8 - 1
x = 7
Therefore, we have
x / y = 7 / 8
So, the required fraction is 7/8.
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