Word Problems and Applications

The original side of the square is \(\sqrt{225}=15\text{ cm}\), and the new side is \(\sqrt{400}=20\text{ cm}\). Thus, \(15+x=20\), giving \(x=5\text{ cm}\). Option B is incorrect because a side of \(15+4=19\text{ cm}\) would produce an area of \(361\text{ cm}^2\), not \(400\text{ cm}^2\). Exam tip: take the square root of each area to find the corresponding side, then subtract the original side from the new side.

With perimeter (20), (l+b=10) and (lb=96). The equation (x(10-x)=96) has discriminant (100-384<0), so the rectangle is impossible.

Let the actual speed be (x), then (\frac{240}{x-20}-\frac{240}{x}=1). This gives (x^2-20x-4800=0), so (x=80).

Each pair of points determines one line segment, so the total is n(n − 1)/2. Setting n(n − 1)/2 = 45 gives n² − n − 90 = 0, or (n − 10)(n + 9) = 0. The roots are 10 and −9; since a number of points cannot be negative, n = 10.

The direct answer is option B: the perimeter is 40 m. Let the breadth be x m. The length is 4 m more, so it is x+4. The area equation is x(x+4)=96, giving x²+4x−96=0. Factoring gives (x−8)(x+12)=0, so x=8 or x=−12. Breadth cannot be negative, hence breadth=8 m and length=12 m. Perimeter of a rectangle is 2(length+breadth), so P=2(12+8)=40 m. Option A, 36 m, would correspond to a different sum of side lengths and is not obtained from the calculated dimensions. Option B, 40 m, is correct and also checks because the four sides are 12, 8, 12, and 8, whose sum is 40. Option C, 44 m, and option D, 48 m, are too large for these dimensions. Do not confuse area, 96 m², with perimeter. Memory cue: solve the sides first, then use 2(l+b).

Let the original number be \(x\). Then \(x(x+6)=135\), giving \(x^2+6x-135=0\). Factoring, \((x-9)(x+15)=0\), so \(x=9\) or \(x=-15\). Since the question asks for the positive number, the correct answer is 9. Exam tip: Find both roots of a quadratic equation and then apply the condition stated in the question, such as positivity.

By the Pythagorean theorem, \(x^2+(x+7)^2=(x+8)^2\). On simplifying, \(x^2-2x-15=0\), or \((x-5)(x+3)=0\). Thus, \(x=5\) or \(x=-3\); since a length cannot be negative, the shorter side is 5 cm. In such problems, always reject roots that give a non-positive length.

Let breadth be (x), then increase is ((x+5)(x+15)-x(x+10)=350). This gives (10x+75=350), so (x=27.5), hence no listed option is exact.

Let students per room be (x), then (\frac{60}{x}-\frac{60}{x+2}=5). This gives (5x^2+10x-120=0), so (x=4).

Let the number be \(x\). Then \(x^2+x=156\), so \(x^2+x-156=0\). Factoring gives \((x+13)(x-12)=0\), hence \(x=-13\) or \(x=12\). Since the number is positive, the correct answer is 12. The nearby option 13 is incorrect because \(13+13^2=182\), not 156. Exam tip: Form the quadratic equation first, then select the root that satisfies the given condition.

Let the stream speed be (x), then (\frac{48}{10+x}+\frac{32}{10-x}=10). This gives (x^2+4x-12=0), so (x=2).

Let original breadth be (x), then the box base is ((x-4)(x+4)) and height is (2). From (2(x-4)(x+4)=240), (x^2=136), so none of the options is exact.

Let the actual speed be (x\text{ km/h}), then (\frac{300}{x}-\frac{300}{x+10}=1). This gives (x^2+10x-3000=0), so the positive root is (x=50).

Let the breadth be (x\text{ m}), then (x(x+8)=273). This gives (x=13) and length (21), so the perimeter is (2(13+21)=68\text{ m}).

Let the first worker's time be (x\text{ days}), then (\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}). This gives (x^2-7x-30=0), so (x=10).

Let the smaller integer be (x); then the next integer is (x+1). Thus, x(x+1)=552, giving x^2+x-552=0. Factoring gives (x-23)(x+24)=0, so x=23 or x=-24. Since the integers are positive, x=23 is valid, and the numbers are 23 and 24. Exam tip: represent consecutive integers as x and x+1 before forming the quadratic equation.

The direct answer is B: 13. Let the two numbers be x and y. Their sum gives x+y=31, so y=31-x. Their product gives x(31-x)=234. Expanding, 31x-x^2=234, or x^2-31x+234=0. Factoring gives (x-13)(x-18)=0, so the numbers are 13 and 18. The smaller number is therefore 13. Option A, 12, is wrong because the other number would be 19, and 12×19=228, not 234. Option B, 13, is correct because the other number is 18, 13+18=31, and 13×18=234. Option C, 14, is wrong because the other number would be 17 and the product would be 238. Option D, 15, is wrong because the other number would be 16 and the product would be 240. A useful check is to test both the sum and product, not just one condition. Memory cue: for two numbers, form the quadratic using sum and product, then choose the smaller root.

Let the breadth of the rectangle be \(x\) cm. Its length is then \(x+3\) cm. Using the area, \(x(x+3)=180\), so \(x^2+3x-180=0\). Factoring gives \((x+15)(x-12)=0\), hence \(x=12\) or \(x=-15\). Since a length cannot be negative, \(-15\) is rejected and the breadth is 12 cm. Exam tip: In geometrical word problems, discard any negative root because dimensions must be positive.

Let the breadth be \(x\) m. Then the length is \(x+5\) m, so \(x(x+5)=336\), giving \(x^2+5x-336=0\). Factoring, \((x+21)(x-16)=0\), so \(x=16\) or \(x=-21\). Since a breadth cannot be negative, the correct answer is 16 m. Exam tip: reject any negative root when the variable represents a physical length.

Let the original side be \(x\) m. The increase in area gives \((x+4)^2-x^2=96\). Expanding, \(x^2+8x+16-x^2=96\), so \(8x=80\) and \(x=10\) m. Therefore, option B is correct. Exam tip: if the side of a square increases by \(a\), the area increase is \(2ax+a^2\); here, \(2\times4\times x+4^2=96\) gives the answer directly.

Let the smaller integer be \(x\). Since the integers are consecutive and even, the other integer is \(x+2\). Thus, \(x(x+2)=360\), giving \(x^2+2x-360=0\). Factoring, \((x-18)(x+20)=0\), so \(x=18\) or \(x=-20\). Because the integers are positive, the smaller integer is 18. Exam tip: consecutive even integers always differ by 2.

Let the smaller integer be \(x\). Then the larger integer is \(x+2\). Thus, \(x(x+2)=483\), or \(x^2+2x-483=0\). Factoring gives \((x-21)(x+23)=0\), so \(x=21\) or \(x=-23\). Since the integers are positive, \(x=21\), and the larger integer is \(21+2=23\). Exam tip: consecutive odd integers always differ by 2, not 1.

Let the positive number be \(x\). Then \(x(x+7)=330\), so \(x^2+7x-330=0\). Factoring gives \((x-15)(x+22)=0\), yielding \(x=15\) or \(x=-22\). Since the number must be positive, \(x=15\) is correct. Exam tip: verify the answer using \(15\times(15+7)=15\times22=330\).

Let the number be \(x\). Then \(x^2+9x=400\), or \(x^2+9x-400=0\). Factoring gives \((x-16)(x+25)=0\), so \(x=16\) or \(x=-25\). Since the question asks for the positive number, the answer is 16. Exam tip: Find both roots of the quadratic and then apply the condition specified in the question.

The equation is (x^2-14x=51). From (x^2-14x-51=0), (x=17) or (x=-3), so the positive number is (17).