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Exercise 5.1 · Q26

Q.A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second? [Hint: If xx is the length of the shortest board, then xx, (x+3)(x + 3) and 2x2x are the lengths of the second and third piece, respectively. Thus, x+(x+3)+2x≤91x + (x + 3) + 2x \le 91 and 2x≥(x+3)+52x \ge (x + 3) + 5].

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The problem reduces to a system of two inequalities: the total length constraint and the condition that the third piece is at least 5 cm longer than the second. Solving these gives the shortest board length in the range 8≤x≤228 \le x \le 22, and since lengths are positive, the possible values are all real numbers in that interval.

The key insight here is that we are not solving a single equation, but a pair of inequalities that must hold simultaneously. The hint gives us the natural variables: let the shortest piece be xx cm. Then the second piece is x+3x+3 cm, and the third is 2x2x cm.

Why does this work? Because the problem gives two separate constraints — one about the total length available (the board is only 91 cm long), and one about the relative lengths of the third and second pieces. Each constraint translates into an inequality, and the shortest board must satisfy both at once.

Let’s work through it step by step.

  1. Write the total length constraint. The sum of the three pieces cannot exceed the board’s length:

x+(x+3)+2x≤91x + (x+3) + 2x \le 91

Simplify:

4x+3≤914x + 3 \le 91

Subtract 3:

4x≤884x \le 88

Divide by 4:

x≤22x \le 22

  1. Write the condition on the third piece relative to the second. The third piece must be at least 5 cm longer than the second:

2x≥(x+3)+52x \ge (x+3) + 5

Simplify the right side:

2x≥x+82x \ge x + 8

Subtract xx:

x≥8x \ge 8

  1. Combine the two inequalities. From step 1 we have x≤22x \le 22, and from step 2 we have x≥8x \ge 8. Together:

8≤x≤228 \le x \le 22

  1. Consider practical constraints. …

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