Word Problems and Applications

Interpret “11 less than a number” as x−11, where x is the original number. The equation is therefore (x−11)²=1156. Since 1156=34², both square-root cases must be considered: x−11=34 or x−11=−34. These give x=45 and x=−23. The question asks for the positive case, so x=45 and option C is correct. Substitution confirms the answer because (45−11)²=34²=1156. Option A is only the positive square-root value, not the requested original number. Options B and D produce different squared values and do not satisfy the equation. Considering both signs before applying the positivity condition is essential.

Let the side of the square be s centimetres. The area of a square is s², so s²=2025. Since a side length is positive, s=√2025=45 cm. The wire forms the complete boundary of the square, so its length equals the perimeter, not the area and not merely one side. The perimeter is 4s=4×45=180 cm. Therefore option B is correct. Option A is not obtained from either relevant formula, while option C results from an incorrect numerical operation. Option D may arise from confusing the given area or its square-root calculation with the perimeter. The essential sequence is to find the positive side from the area and then add the four equal sides. Substitution confirms that a square of side 45 cm has area 2025 square cm.

This is a word problem that is modeled by a quadratic equation. Let the units digit be u. Then the tens digit is u - 2, because it is 2 less. Their product gives u(u - 2) = 24, so u² - 2u - 24 = 0. Factoring, (u - 6)(u + 4) = 0, giving u = 6 or u = -4. A digit cannot be negative, so u = 6 and the tens digit is 6 - 2 = 4. The resulting two-digit number is 46, and its digits have product 4 × 6 = 24. Option B reverses the digit relationship, while 38 and 83 do not have digit product 24 and also fail the stated difference condition. Therefore option A is uniquely correct.

Let the unknown number be x. The wording means that the square of the number after subtracting 22 is 2500, so (x − 22)² = 2500. Taking square roots gives x − 22 = ±50. Therefore the two algebraic possibilities are x = 72 and x = −28. The question specifies the positive case, so x = 72 must be selected. A direct check confirms it: 72 − 22 = 50 and 50² = 2500. Thus option C is correct. Option A is only the positive square root before adding 22, option B does not satisfy the equation, and option D results from an incorrect adjustment after taking the square root. The sign condition is essential because squaring otherwise produces two possible roots.

Let the side of the square be s centimetres. The area formula for a square is s², so s² = 3025. Since a length is positive, s = √3025 = 55 cm, not −55 cm. The wire forms the complete boundary of the square, so its length equals the perimeter, 4s. Therefore the wire length is 4 × 55 = 220 cm, making option C correct. Option A and option B do not equal four times the side, while option D would correspond to an incorrect side length of 60 cm. The important modelling step is to distinguish area, which is measured in square centimetres, from the wire length, which is the perimeter measured in centimetres.

Let the side of the inner square field be x metres. A path 2 m wide runs around all four sides, so the outer square gains 2 m on each side at both ends; its side is therefore x + 4, not x + 2. The total outer area is 196 m², giving (x + 4)² = 196. Since a side length is positive, x + 4 = 14, and hence x = 10 m. The negative square-root possibility is not geometrically meaningful. Thus option B is correct. Option A would make the outer side 12 m and area 144 m², while C and D produce outer areas larger than 196 m². The essential modelling concept is converting the uniform surrounding width into the correct outer dimension.

The governing concept is forming a quadratic equation from the area of a rectangle. Let the photo’s breadth be x cm; its length is then x + 7 cm. Since the frame is 1 cm wide on every side, each outer dimension increases by 2 cm. Thus the outer breadth is x + 2 and the outer length is x + 9. Therefore, (x + 2)(x + 9) = 180, giving x² + 11x + 18 = 180, or x² + 11x − 162 = 0. Factoring gives (x − 9)(x + 18) = 0. The negative root x = −18 is impossible for a length, so x = 9 cm. Hence option B is correct; the other options do not satisfy the outer-area equation.

The governing idea is to translate the length condition and area formula into a quadratic equation. Let the breadth be x m. The length is three times the breadth minus 2, so it is 3x − 2 m. Using area = length × breadth gives x(3x − 2) = 161. Hence 3x² − 2x − 161 = 0. Testing the listed positive values, x = 7 gives 3(49) − 14 − 161 = 147 − 14 − 161 = 0. Algebraically, the equation factors as (x − 7)(3x + 23) = 0, producing x = 7 or x = −23/3. A breadth cannot be negative, so x = 7 m. Thus option B is correct, while the other options fail to produce the stated area with length 3x − 2.

The governing concept is the translation of a geometric word problem into a quadratic equation. Let the breadth be x cm; then the length is x + 5 cm. Since area equals length × breadth, x(x + 5) = 150, giving x² + 5x − 150 = 0. Factorising, (x − 10)(x + 15) = 0, so x = 10 or −15. A breadth cannot be negative, hence x = 10 cm and the length is 15 cm. Option A is therefore correct. Checking confirms that 10 × 15 = 150 and 15 − 10 = 5. The other options either have the wrong product or do not differ by 5 cm.

Let the smaller integer be \(x\). Then the other integer is \(x+1\). Thus, \(x(x+1)=240\), or \(x^2+x-240=0\). Factoring gives \((x-15)(x+16)=0\). Since the integers are positive, \(x=15\), so the pair is 15 and 16; indeed, \(15\times16=240\). Option A is close but \(14\times15=210\), not 240. Exam tip: For consecutive integers, represent them as \(x\) and \(x+1\) before forming the quadratic equation.

The mother's present age is \(x+20\) years. Five years ago, their ages were \(x-5\) and \(x+15\) years, respectively. Thus, \((x-5)(x+15)=189\), which gives \(x^2+10x-264=0\). Factoring, \((x-12)(x+22)=0\). Since age cannot be negative, \(x=12\) years, so option C is correct. In age problems, remember to subtract the stated number of years from both present ages.

Let the car’s original speed be \(x\) km/h. The original time is \(\frac{240}{x}\) hours, and after increasing the speed to \(x+20\) km/h, the new time is \(\frac{240}{x+20}\) hours. Therefore, \(\frac{240}{x}-\frac{240}{x+20}=1\). On simplifying, \(x^2+20x-4800=0\), which factors as \((x-60)(x+80)=0\). Since speed must be positive, \(x=60\) km/h is the valid answer; \(-80\) km/h is rejected. Exam tip: In speed–time–distance problems, write the times for both situations first and then form their difference equation.

Let the width of the path be \(x\) m. The dimensions of the remaining grassy rectangle are \((30-2x)\) m and \((20-2x)\) m. Thus, \((30-2x)(20-2x)=416\), which gives \(x^2-25x+46=0\), or \((x-2)(x-23)=0\). The root \(x=23\) m is not possible because the path width must be less than half the breadth, so the valid value is \(x=2\) m. Exam tip: for a path inside all four sides, subtract \(2x\) from both original dimensions.

Let the first odd number be \(x\). The next consecutive odd number is \(x+2\), since consecutive odd numbers differ by 2. Thus, \(x(x+2)=255\), giving \(x^2+2x-255=0\). Factoring, \((x-15)(x+17)=0\), so the positive value is \(x=15\). Therefore, the numbers are 15 and 17, and \(15\times17=255\). Exam tip: for consecutive odd numbers, use a difference of 2, not 1.

Let the other side be \(x\) cm. Then the hypotenuse is \((x+5)\) cm. By the Pythagorean theorem, \(x^2+10^2=(x+5)^2\), so \(x^2+100=x^2+10x+25\). Hence, \(10x=75\) and \(x=7.5\) cm. Therefore, option D is correct. Exam tip: in a right triangle, the hypotenuse is the longest side, so it must be represented by \(x+5\), not by \(x\).

Let the tens digit be \(x\). Then the units digit is \(x+1\). Therefore, \(x^2+(x+1)^2=61\), which gives \(2x^2+2x-60=0\) or \(x^2+x-30=0\). Thus, \((x-5)(x+6)=0\), so \(x=5\) or \(x=-6\). Since a digit cannot be negative, the tens digit is 5 and the units digit is 6; hence the number is 56. Exam tip: In digit problems, reject roots that are negative or greater than 9. The nearby option 67 has consecutive digits, but their squared sum is 36+49=85, not 61.

The combined work rate is \(\frac{1}{6}\) of the task per day, so \(\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}\). Clearing the denominators gives \(6(2x+5)=x(x+5)\), or \(x^2-7x-30=0\). Factoring, \((x-10)(x+3)=0\), so \(x=10\) or \(x=-3\). Since a number of days cannot be negative, \(x=10\), making option B correct. Exam tip: In work-rate problems, add the reciprocals of the individual completion times.

Let the breadth of the rectangle be \(x\) metres. Then its length is \(x+4\) metres. Using the area, \(x(x+4)=96\), so \(x^2+4x-96=0\). Factoring gives \((x-8)(x+12)=0\), hence \(x=8\) or \(x=-12\). Since a length cannot be negative, the breadth is 8 metres. Exam tip: In rectangle word problems, assign a variable to one dimension and express the other using the given relationship before applying the area formula.

The direct answer is C: the number can be 2 or \\(1/2\\). Let the positive number be x. Its reciprocal is \\(1/x\\), so \\(x+1/x=5/2\\). Since x is positive, multiply by 2x: \\(2x^2+2=5x\\), hence \\(2x^2-5x+2=0\\). Factoring gives \\( (2x-1)(x-2)=0\\), so x=1/2 or x=2. Both are positive and checking gives \\(2+1/2=5/2\\). Option C is correct. Option A gives 1+1=2, not 5/2. Option B gives 3+1/3=10/3, not 5/2. Option D gives 4+1/4=17/4, also incorrect. The two answers naturally come as reciprocal pairs, because replacing x by 1/x does not change the sum. Remember to reject x=0 before multiplying, although it is already excluded because the reciprocal is undefined.

Let the number be \(x\). Then \(x^2=3x+28\), so \(x^2-3x-28=0\). Factoring gives \((x-7)(x+4)=0\), hence \(x=7\) or \(x=-4\). Since the number is positive, the correct answer is 7. Exam tip: when a word problem specifies a positive number, discard any negative root.

Let the height of the triangle be \(x\) cm. Then its base is \((x+4)\) cm. Using the area formula, \(\frac{1}{2}x(x+4)=60\), which gives \(x^2+4x-120=0\). Factoring, \((x-10)(x+12)=0\), so \(x=10\) or \(x=-12\). Since a length cannot be negative, the height is \(10\) cm. Exam tip: Always reject a negative root when the variable represents a physical length.

The wall, ground and ladder form a right-angled triangle. By the Pythagorean theorem, \(x^2+(x+7)^2=13^2\). Simplifying gives \(2x^2+14x-120=0\), or \((x-5)(x+12)=0\). Since a distance cannot be negative, \(x=5\) metres is valid; \(x=-12\) is rejected. In such word problems, always choose the root that is physically meaningful.

Let the length and breadth be x and y. From the perimeter, 2(x+y)=50, so x+y=25; from the area, xy=150. Substituting y=25-x gives \(x(25-x)=150\), or \(x^2-25x+150=0\). Factoring, \((x-10)(x-15)=0\), so the dimensions are 10 m and 15 m. Option B has the correct sum of 25 but its product is 144, not 150, so it does not satisfy the area condition. Exam tip: In rectangle problems, use the perimeter to find the sum of the dimensions and the area to find their product.

Let the original price per item be \(x\) rupees. The original number of items is \(\frac{1200}{x}\), while after the price reduction it is \(\frac{1200}{x-10}\). Therefore, \(\frac{1200}{x-10}-\frac{1200}{x}=4\). Simplifying gives \(x(x-10)=3000\), or \(x^2-10x-3000=0\). Thus, \(x=60\) or \(x=-50\); a price cannot be negative, so the original price was 60 rupees. Exam tip: In price-quantity problems, express the number of items as total money divided by price and reject any impossible root.

Let the speed of the boat in still water be \(x\) km/h. Its upstream and downstream speeds are then \(x-4\) and \(x+4\) km/h, respectively. Therefore, \(\frac{48}{x-4}-\frac{48}{x+4}=3\). On simplifying, we get \(x^2=144\), so \(x=12\) or \(x=-12\). Since speed is positive and must exceed the stream speed, the valid answer is 12 km/h. Exam tip: In boat-and-stream problems, use \(x-v\) for upstream speed and \(x+v\) for downstream speed.