Problem 1 :
A class of 45 students is divided into two groups. If one group has 9 students less than the other, how many stidents are in the smaller group?
Solution :
Let x be the number of students in one of the two groups.
Number of students in the other group is (x - 9).
Given : Total number of students in the class is 45.
x + (x - 9) = 45
x + x - 9 = 45
2x - 9 = 45
Add 9 to both sides.
2x = 54
Divide both sides by 2.
x = 27
x - 9 = 27 - 9 = 18
There are 18 students in the smaller group.
Problem 2 :
Subtracting 17 from four times of a number results 35. Find the number.
Solution :
Let x be the number.
Given : When 17 is subtracted 17 from four times of the number is equal to 35.
Then, we have
4x - 17 = 35
Add 17 to both sides.
4x = 52
Divide both sides by 4.
x = 13
The number is 13.
Problem 3 :
9 less than five times of a number is equal 3 more than two times of the same number. Find the number.
Solution :
Let x be the number.
Five times of the number = 5x
9 less than five times of the number = 5x - 9 ----(1)
Two times of the number = 2x
3 more than two times of the number = 2x + 3 ----(2)
From the given information,
(1) = (2)
5x - 9 = 2x + 3
3x = 12
x = 4
The number is 4.
Problem 4 :
The length of one of the sides of a triangle is 7 cm and the lengths of the remaining two sides differ by 4 cm. If the perimeter of the triangle is 33 cm, find the lengths of the missing sides of the triangle.
Solution :
Since the lengths of the two sides of the triangle differ by 4 cm, the lengths can be assumed as x and (x + 4).
The sides of the triangle are
7 cm, x cm, (x + 4) cm
Perimeter of the triangle = 33 cm
7 + x + (x + 4) = 33
7 + x + x + 4 = 33
2x + 11 = 33
Subtract 11 from both sides.
2x = 22
Divide both sides by 2.
x = 11
x + 4 = 11 + 4 = 15
The missing sides of the triangle are 11 cm and 15 cm.
Problem 5 :
One-sixth of a number is greater than one-seventh of the same number by 1. Find the number.
Solution :
Let x be the number.
One-sixth of the number =(⅙)x = ˣ⁄₆
One-seventh of the number = (⅐)x = ˣ⁄₇
From the given information,
ˣ⁄₆ = ˣ⁄₇ + 1
Least common multiple of the denominators (6, 7) is 42.
Multiply both sides of the equation by 42 to get rid of the denominators 6 and 7.
42(ˣ⁄₆) = 42(ˣ⁄₇ + 1)
7x = 42(ˣ⁄₇) + 42
7x = 6x + 42
x = 42
The number is 42.
Problem 6 :
Three times of my age 5 years ago is equal to 5 less than two times of my age 5 years hence. How old am i now?
Solution :
Let x be the your present age.
Your age 5 years ago = x - 5
Three times of your age 5 years ago :
= 3(x - 5)
= 3x - 15----(1)
Your age 5 years hence = x + 5
Two times of my age 5 years hence = 2(x + 5)
= 2x + 10
5 less than two times of my age 5 years hence :
= 2x + 10 - 5
= 2x + 5 ----(2)
From the given information,
(1) = (2)
3x - 15 = 2x + 5
x = 20
You are 20 years old now.
Problem 7 :
One number is 4 times the another number, and they sum to -20. What is the greater of the two numbers?
Solution :
Let x be one of the numbers.
Then the other number is 4x.
Given : Sum of the two numbers is -20.
x + 4x = -20
5x = -20
Divide both sides by 5.
x = -4
4x = 4(-4) = -16
The two numbers are -4 and -16.
The greater one is -4.
Problem 8 :
Paulina owns a certain number of dishes. 24 added to 3 times the number of dishes is equal to 9 times the number of dishes. How many dishes does Palina own?
Solution :
Let x be the numbers of dishes Paulina owns.
Given : 24 added to 3 times the number of dishes is equal to 9 times the number of dishes.
3x + 24 = 9x
Subtract 3x from both sides.
24 = 6x
Divide both sides by 6.
4 = x
Paulina owns 6 dishes.
Problem 9 :
There is a monthly subscription fee of $15 for a movie rental sevice and $3 for each movie rented. Last month, Adam paid $33 for the movie rental sevice. How many movies did Adam rent last month.
Solution :
Let x be the numbers of movies Adam rented last month.
Amount paid by Adam last month = 3x + 15
It is given that Adam paid $33 for the movie rental sevice.
3x + 15 = 33
Subtract 15 from both sides.
3x = 18
Divide both sides by 3.
x = 6
Adam rented 6 movies last month.
Problem 10 :
In a fraction, the numerator is less than the denominator by 7. If 4 is added to the numerator and the denominator 2 is added to the denominator, the fraction becomes ¾. Find the fraction.
Solution :
Let x be the denominator.
Given : The numerator is less than the denominator by 7.
From the above information,
fraction = ⁽ˣ ⁻ ⁷⁾⁄ₓ ----(1)
Given : If 4 is added to the numerator and the denominator 2 is added to the denominator, the fraction becomes ¾.
From the above information, we have
⁽ˣ ⁻ ⁷ ⁺ ⁴⁾⁄₍ₓ ₊ ₂₎ = ¾
Simplify.
⁽ˣ ⁻ ³⁾⁄₍ₓ ₊ ₂₎ = 3/4
4(x - 3) = 3(x + 2)
4x - 12 = 3x + 6
x = 18
Substitute x =18 in (1).
fraction = ⁽¹⁸ ⁻ ⁷⁾⁄₁₈
= ¹¹⁄₁₈
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