Word Problems and Applications

If the numbers are (x) and (29-x), then (x(29-x)=204) gives (12) and (17). The smaller number is (12).

Let one number be \(x\); then the other number is \(31-x\). Thus, \(x(31-x)=238\), which gives \(x^2-31x+238=0\). Factoring, \((x-14)(x-17)=0\), so the two numbers are 14 and 17, and the larger number is 17. Exam tip: When the sum and product are given, look for a pair of numbers satisfying both conditions; the larger member of the pair is the answer.

If breadth is (x), then (x(x+12)=448), giving (x=16). Do not forget to write the unit of dimension in the answer.

(x(x+13)=300) gives breadth (12) and length (25). Choose the answer according to the part asked.

By Pythagoras, (x^2+(x+5)^2=25^2), giving (x=15). Always take the hypotenuse as the largest side.

By the Pythagorean theorem, \(x^2+(x+4)^2=20^2\). Simplifying gives \(2x^2+8x-384=0\), or \(x^2+4x-192=0\). Factoring yields \((x-12)(x+16)=0\), so \(x=12\) or \(x=-16\). Since a side length cannot be negative, \(x=12\) m is the correct answer. Exam tip: In geometry word problems, reject any negative algebraic root when it represents a length.

((x+5)^2=225) and (x+5=15), so (x=10). If the positive root is given, do not take the negative root.

((x-7)^2=100) and (x-7=10), so (x=17). In square-based questions, read the condition carefully.

The equation is (x^2-9x=90), or (x^2-9x-90=0), which gives (x=15). Keep the subtraction order correct in difference statements.

(x^2=11x+60) gives (x^2-11x-60=0), whose positive solution is (15). More than means addition in the equation.

(x^2+x=156) gives (x^2+x-156=0), so (x=12). Write the sum of the number and its square directly.

(x^2+x=210) gives (x^2+x-210=0), whose positive solution is (14). While factoring, check factors of (210).

Consecutive odd numbers are (x) and (x+2), so (x(x+2)=255) gives (x=15). Consecutive odd numbers differ by (2).

Consecutive even numbers are (x) and (x+2), then (x(x+2)=360) gives (x=18). So the larger number is (20).

Let the smaller odd number be \(x\). The next consecutive odd number is therefore \(x+2\). Thus, \(x(x+2)=399\), giving \(x^2+2x-399=0\). Factoring, \((x-19)(x+21)=0\), so the positive value is \(x=19\), and the larger number is \(19+2=21\). Option 19 is the smaller number, not the larger one. Exam tip: consecutive odd numbers always differ by 2.

Let the smaller number be x. The next consecutive even number is x+2, so x(x+2)=528. This gives x²+2x−528=0, which factors as (x−22)(x+24)=0. Since the numbers are positive, x=22; hence the numbers are 22 and 24. Exam tip: consecutive even integers always differ by 2.

Let the breadth be x metres. Then the length is (x + 14) metres. Using area = length × breadth, we get x(x + 14) = 480, or x² + 14x − 480 = 0. Factoring gives (x − 16)(x + 30) = 0, so x = 16 or x = −30. Since a length cannot be negative, the breadth is 16 m. Exam tip: In rectangle word problems, first use area = length × breadth and reject any negative dimension.

Let the breadth of the board be \(x\) cm. Its length will then be \(x+15\) cm. Using the area, \(x(x+15)=250\), so \(x^2+15x-250=0\). Factoring gives \((x-10)(x+25)=0\), so \(x=10\) or \(x=-25\). Since a dimension cannot be negative, the breadth is 10 cm. Check: \(10\times25=250\) square cm. Exam tip: In rectangle word problems, assign \(x\) to one side first and form the area equation before solving.

Let the number be \(x\). Then \(x(x+12)=405\), which gives \(x^2+12x-405=0\). Factoring, \((x-15)(x+27)=0\), so \(x=15\) or \(x=-27\). Since the question asks for a positive number, the correct answer is 15. Exam tip: translate “12 greater” as \(x+12\) and check both roots before applying the stated condition.

Let the larger number be \(x\). Then the other number is \(x-7\), so \(x(x-7)=198\). Thus, \(x^2-7x-198=0\), which factors as \((x-18)(x+11)=0\). The roots are \(x=18\) and \(x=-11\). Since the numbers are positive, the valid larger number is 18, and the other number is 11. Exam tip: In such problems, represent the larger number by \(x\) and the number 7 less than it by \(x-7\).