Problem 1 :
Find the ratio between 150 grams and 2 kilograms.
Solution :
The given two quantities are in different units (grams and kilograms).
To do ratio between two quantities, both of them must be in same units.
2 kilograms = 2 ⋅ 1000 grams
2 kilograms = 2000 grams
Ratio between 150 grams and 2 kilograms :
= 150 grams : 2000 grams
= 150 : 2000
Divide both terms by 50.
= 3 : 40
Problem 2 :
Find the ratio between 25 minutes and 45 seconds.
Solution :
The given two quantities are in different units (minutes and seconds).
To do ratio between two quantities, both of them must be in same units.
25 minutes = 25 ⋅ 60 seconds
25 minutes = 1500 seconds
Ratio between 25 minutes and 45 seconds :
= 1500 seconds : 45 seconds
= 1500 : 45
Divide both terms by 15.
= 100 : 3
Problem 3 :
Find which ratio is greater :
2⅓ : 3⅓, 3.6 : 4.8
Solution :
Convert the given ratios to like fractions and compare the numerators to find which ratio is greater.
2⅓ : 3⅓ = 7/3 : 10/3 2⅓ : 3⅓ = 7 : 10 2⅓ : 3⅓ = 7/10 |
3.6 : 4.8 = 36 : 48 3.6 : 4.8 = 3 : 4 3.6 : 4.8 = 3/4 |
Convert 7/10 and 3/4 to like fractions.
Least common multiple of the denominators (10, 4) = 20.
7/10 = 14/20
3/4 = 15/20
2⅓ : 3⅓ = 14/20 |
3.6 : 4.8 = 15/20 |
In 14/20 and 15/20, the denominator is same.
Compare numerators.
14 < 15
So,
14/20 < 15/20
Therefore, 3.6 : 4.8 is greater 2⅓ : 3⅓.
Problem 4 :
Find the value of x, if the ratios 5 : 2 and 15 : x are in proportion.
Solution :
Because the ratios 5 : 2 and 15 : x are in proportion, they are equal.
5 : 2 = 15 : x
Use cross product rule.
5x = 2 ⋅ 15
5x = 30
Divide each side by 5.
x = 6
Problem 5 :
The ratio of the no. of boys to the no. of girls in a school of 720 students is 3 : 5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2 : 3.
Solution :
Sum of the terms in the given ratio is
= 3 + 5
= 8
So, no. of boys in the school is
= 720 ⋅ (3/8)
= 270
No. of girls in the school is
= 720 ⋅ (5/8)
= 450
Given : Number of new girls admitted in the school is 18.
Let x be the no. of new boys admitted in the school.
After the above new admissions,
No. of boys in the school = 270 + x
No. of girls in the school = 450 + 18 = 468
Given : The ratio after the new admission is 2 : 3.
Then, we have
(270 + x) : 468 = 2 : 3
Use cross product rule.
3(270 + x) = 468 ⋅ 2
810 + 3x = 936
3x = 126
x = 42
So, the number of new boys admitted in the school is 42.
Problem 6 :
The monthly incomes of two persons are in the ratio 4 : 5 and their monthly expenditures are in the ratio 7 : 9. If each saves $50 per month, find the monthly income of the second person.
Solution :
From the given ratio of incomes (4 : 5),
income of the 1^{st} person = 4x
income of the 2nd person = 5x
(Expenditure = Income - Savings)
Then, expenditure of the 1st person = 4x - 50,
expenditure of the 2^{nd} person = 5x - 50
expenditure ratio = 7 : 9 (given)
So, we have
(4x - 50) : (5x - 50) = 7 : 9
Use cross product rule.
9(4x - 50) = 7(5x - 50)
36x - 450 = 35x - 350
x = 100
Then, the income of the second person is
= 5x
= 5(100)
= 500.
So, income of the second person is $500.
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