Problem 1 :
Twice a number is 500 more than six times the number. What is the number?
Solution :
Let x be the required number.
From the ngiven information,
2x = 6x + 500
Subtract 6x from both sides.
-4x = 500
Divide both sides by -4.
x = -125
Therefore, the number is -125.
Problem 2 :
Three-sevenths of a number is 24. Find the number.
Solution :
Let x be the required number.
From the ngiven information,
Multiply both sides by 7.
3x = 168
Divide both sides by 3.
x = 56
Therefore, the number is 56.
Problem 3 :
What number increased by ¼ of itself is equal to 30?
Solution :
Let x be the required number.
From the ngiven information,
Multiply both sides by 4 and simplify.
4x + x = 120
5x = 120
Divide both sides by 5.
x = 24
Therefore, the number is 24.
Problem 4 :
If 6 times a number is decreased by 6, the result is the same as when 3 times the number is increased by 12. Find the number.
Solution :
Let x be the required number.
From the ngiven information,
6x - 6 = 3x + 12
Subtract 3x from both sides.
3x - 6 = 12
Add 6 to both sides.
3x = 18
Divide both sides by 3.
x = 6
Therefore, the number is 6.
Problem 5 :
If 3 times a number is increased by 22, the result is 14 less than 7 times the number. What is the number?
Solution :
Let x be the required number.
From the ngiven information,
3x + 22 = 7x - 14
Subtract 3x from both sides.
22 = 4x - 14
Add 14 to both sides.
36 = 4x
Divide both sides by 4.
9 = x
Therefore, the number is 9.
Problem 6 :
Separate 84 into two parts such that one part will be 12 less than twice the other.
Solution :
Let x be one part of 84.
If y be the other part, then
x = 2y - 12 ----(1)
Since x and y are the two parts, we have
x + y = 84
Substitute x = 2y - 12.
2y - 12 + y = 84
3y - 12 = 84
Add 12 to both sides.
3y = 96
Divide both sides by 3.
y = 32
Substitute y = 32 into (1).
x = 2(32) - 12
x = 64 - 12
x = 52
Therefore, the two parts of 84 are 52 and 32.
Problem 7 :
The difference between two numbers is 24. Find the numbers, if their sum is 88.
Solution :
Let x and y be the two numbers.
From the given information,
x - y = 24 ----(1)
x + y = 88 ----(2)
(1) + (2) :
2x = 112
Divide both sides by 2.
x = 56
Substitute x = 56 into (2).
56 + y = 88
Subtract 56 from both sides.
y = 32
Therefore, the two numbers are 56 and 32.
Problem 8 :
One number is 3 times another number. If 17 be added to each, the first resulting number is twice the second resulting number. Find the two numbers.
Solution :
Let x be one of the numbers.
Then, the other number is 3x.
From the information,
3x + 17 = 2(x + 17)
3x + 17 = 2x + 34
Subtract 2x from both sides.
x + 17 = 34
Subtract 17 from both sides.
x = 17
3x = 51
Therefore, the numbers are 17 and 51.
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