Problem Solving

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Year 10 Interactive Maths - Second Edition

Problem Solving

Linear equations are often used to solve practical problems that have an unknown quantity. We use a suitable pronumeral to represent the unknown quantity, translate the information given in the problem into an equation, and then solve the equation using the skills acquired earlier in this chapter.

Example 11

If a number is increased by 8, the result is 25. Find the number.

Solution:

Example 12

If a number is decreased by 4, the result is 29. Find the number.

Solution:

Example 13

If twice a number is equal to 68, find the number.

Solution:

Let x be the number. Twice x is 2x, which is given as 68. Therefore, 2x = 68. After dividing both sides by 2, we find x = 34. So, the number is 34.

Example 14

If a number is divided by 9, the result is 12. Find the number.

Solution:

Let x be the number. Dividing x by 9 gives x / 9, which is equal to 12. So, x / 9 = 12. Multiply both sides by 9 to find x = 108. So, the number is 108.

Example 15

If three times a number decreased by 5 equals 82, find the number.

Solution:

Let x be the number. Three times x is 3x, and decreasing this by 5 gives 3x - 5, which we are told is 82. Therefore, 3x - 5 = 82. Add 5 to both sides and then divide by 3 to find x = 29. So, number is 29.

Example 16

If one-half of a certain number is added to one-third of the same number, the result is 10. Find the number.

Solution:

Let x be the number. One-half of x is x / 2 and adding this to one-third of x, x / 3, gives x / 3 + x / 2, which we are told is 10. So, x / 3 + x / 2 = 10. Lowest common multiple of 3 and 2 is 6. So, multiply both sides by 6 to obtain 2x + 3x = 60.

Collecting like terms and then dividing both sides by 5 gives x = 12.

So, the number is 12.

Example 17

A rectangular paddock is twice as long as it is wide. If it has a perimeter of 570 m, find its dimensions.

Solution:

Let the width of the paddock be x m. Then w = x, l = 2x, P = 570. Substituting for l, w and P into P = 2(l + w) and solving for x gives x = 95. So, the paddock's length and width are 190 m and 95 m respectively.

Note:

$We could have assumed that the length of the paddock is x m. Then the width of the paddock would be x / 2 m. However, working with fractions is slightly harder.$

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