Word Problems and Applications

If the smaller integer is (x), then (x^2+(x+1)^2=841), giving (x=20). Write consecutive numbers as (x) and (x+1).

Let the smaller integer be \(x\). The next consecutive odd integer is \(x+2\). Thus, \(x^2+(x+2)^2=1570\), which gives \(2x^2+4x-1566=0\) or \(x^2+2x-783=0\). Factoring, \((x-27)(x+29)=0\), so \(x=27\) or \(-29\). Since the integers are positive, the smaller one is 27. Check: \(27^2+29^2=729+841=1570\). Exam tip: consecutive odd integers always differ by 2.

Let the consecutive even numbers be (x) and (x+2). From (x^2+(x+2)^2=1924), (x=30), so the larger number is (32).

Taking breadth as (x), (x(x+14)=1152), giving (x=32). For area, multiply length and breadth.

Let the breadth of the rectangle be \(x\) cm. Its length is then \(x+18\) cm, so \(x(x+18)=1240\). Thus, \(x^2+18x-1240=0\), giving \(x=-9\pm\sqrt{1321}\). Since a dimension must be positive, the breadth is \(\sqrt{1321}-9\) cm and the length is \(x+18=9+\sqrt{1321}\) cm. Therefore, option D is correct. In an exam, select only the positive root for a physical dimension; 31 cm is not correct because \(31\times49=1519\), not 1240.

If breadth is (x), then (x^2+(x+10)^2=74^2), giving (x=48). Use Pythagoras for a rectangle with a diagonal.

Let the breadth be \(x\) cm. Then the length is \(x+13\) cm. Using the area, \(x(x+13)=888\), so \(x^2+13x-888=0\). Factoring gives \((x-24)(x+37)=0\). The positive root gives a breadth of 24 cm and a length of 37 cm. Hence, the perimeter is \(2(24+37)=122\) cm. The negative root \(-37\) cannot represent a side length. Exam tip: In rectangle word problems, assign \(x\) to one side and express the other side using the stated difference before forming the quadratic equation.

If one side is (x), the other is (77-x), and (x(77-x)=1480). The solutions are (37) and (40), so the shorter side is (37).

Since the areas are equal, \(x(x+20)=42^2=1764\). Therefore, \(x^2+20x-1764=0\). Using the quadratic formula, \(x=\frac{-20\pm\sqrt{20^2+4(1764)}}{2}=-10\pm2\sqrt{466}\). Because a breadth must be positive, the valid value is \(x=-10+2\sqrt{466}\approx33.17\) m; the negative root is not physically meaningful. In such problems, first form the area equation and then reject any root that makes a length negative.

The area of a square is (side)², so \((x+6)^2=1225=35^2\). Since a side length is positive, \(x+6=35\), giving \(x=35-6=29\). The negative root, \(x+6=-35\), is not valid for a length. Exam tip: Take the positive square root when finding a geometrical length from its area.

The area of a square is (side)², so \((x-8)^2=900\). Thus, \(x-8=\pm 30\), but a side length must be positive; hence \(x-8=30\) and \(x=38\). Option A incorrectly treats 30, the side length, as the value of \(x\). Exam tip: reject a negative root when the expression represents a length.

The direct answer is option B, 54 cm. By the Pythagorean theorem, x^2+(x+18)^2=90^2. Expanding gives 2x^2+36x+324=8100, so x^2+18x-3888=0. Factoring gives (x-54)(x+72)=0. A length cannot be negative, so x=54. The other leg is 72, and 54^2+72^2=90^2, confirming the answer. Option A, 48, would pair with 66 and fails the theorem. Option B, 54, pairs with 72 and works exactly. Option C, 60, would pair with 78 and gives a sum of squares larger than 90^2. Option D, 72, is the larger leg, not the smaller one; it also makes x+18=90 if treated as x, which is inconsistent with the required pair. Exam cue: use the hypotenuse as the largest side and discard the negative algebraic root because lengths are positive.

Let the shorter leg be \(x\) cm. Then the longer leg is \(x+19\) cm. By the Pythagorean theorem, \(x^2+(x+19)^2=109^2\), which simplifies to \(x^2+19x-5760=0\). Its positive solution is \(x=\frac{\sqrt{23401}-19}{2}\approx66.99\) cm, so option A is correct. Option B is incorrect because legs of 60 cm and 79 cm do not produce a hypotenuse of 109 cm. Exam tip: for a length problem, reject the negative root and verify the result using the Pythagorean relation.

From the right triangle, (x^2+60^2=(x+15)^2), giving (x=105). In ladder problems, the ladder is the hypotenuse.

The direct answer is B, 135 m. Let the foot distance be x metres. The vertical height is 72 m and the ladder, being the hypotenuse, is x+18 m. Pythagoras gives \\(x^2+72^2=(x+18)^2\\). Expanding and cancelling \\(x^2\\): \\(5184=36x+324\\), so \\(36x=4860\\) and x=135. Option B is correct; the ladder is 153 m long. Option A, 126, would make the ladder 144, but the equation fails. Option C, 144, would make the ladder 162 and also fails. Option D, 153, is the ladder length, not the ground distance; it would imply a ladder of 171. The hypotenuse must be longer than either perpendicular side.

Using distance = speed × time, we get \(x(x-65)=750\). Thus, \(x^2-65x-750=0\), which factors as \((x-75)(x+10)=0\). Therefore, \(x=75\) or \(x=-10\). Since speed cannot be negative, the valid answer is 75 km per hour. In the exam, first form the quadratic equation from distance = speed × time, then reject any negative or otherwise physically invalid root.

The correct answer is A, 7 hours. Use the basic relation distance = speed × time. Here, distance is 490 km, speed is x km/h, and time is x − 63 hours. Thus x(x − 63) = 490, giving x² − 63x − 490 = 0. This factors as (x − 70)(x + 7) = 0. A speed cannot be negative, so x = 70 km/h; x = −7 is rejected. The time is therefore x − 63 = 70 − 63 = 7 hours. Option A is correct, and direct checking gives 70 × 7 = 490. Option B, 8 hours, would give 70 × 8 = 560, not 490. Options C and D similarly do not satisfy the given distance with the resulting speed. Always reject a negative speed in a physical word problem.

Downstream speed is (x+3), and (5(x+3)=90), so (x=15). In boat problems, speed changes with direction.

Using distance = speed × time, we get \(x(x+12)=640\). Rearranging gives \(x^2+12x-640=0\), which factors as \((x+32)(x-20)=0\). Thus, \(x=20\) or \(x=-32\); since time cannot be negative, the valid value is 20 hours. Exam tip: For quantities such as time and distance, reject negative roots.

If the student studies for x days, he solves x + 11 questions each day. Thus, x(x + 11) = 900, so x² + 11x − 900 = 0 = (x − 25)(x + 36). The roots are 25 and −36; a number of days cannot be negative, so x = 25.

The total number of books gives the equation \(x(x+13)=1014\). Thus, \(x^2+13x-1014=0\), which factors as \((x-26)(x+39)=0\). Hence, \(x=26\) or \(x=-39\). Since the number of shelves cannot be negative, \(x=26\) is correct. The value 39 represents the number of books per shelf when \(x=26\), not the number of shelves. Exam tip: In word problems, reject any algebraic root that is not meaningful in the given context.

The total-chair condition gives x(x+17)=1568. Thus, x²+17x−1568=0, which factors as (x−32)(x+49)=0. Hence x=32 or x=−49. Since the number of rows cannot be negative, x=32 is the valid answer. Exam tip: In arrangement problems, reject negative roots and retain the positive integer root.

The total number of plants gives the equation \(x(x-9)=850\). Thus, \(x^2-9x-850=0\), which factors as \((x-34)(x+25)=0\). Therefore, \(x=34\) or \(x=-25\). Since the number of rows cannot be negative, the valid answer is 34. Check: \(34\times(34-9)=34\times25=850\). Exam tip: Substitute the answer back into the original situation and reject any negative value that is not meaningful in context.

The total number of questions gives \(x(x-7)=800\). Thus, \(x^2-7x-800=0\), which factors as \((x-32)(x+25)=0\). Hence, \(x=32\) or \(x=-25\). Since the number of students cannot be negative, the valid answer is \(32\). Exam tip: In such word problems, first express the total as ‘number of groups × items per group’, then reject any root that is not meaningful in the given context.

Let the original number be \(x\). The required equation is \((x+4)(x+10)=952\). On expanding, \(x^2+14x-912=0\), which factors as \((x-24)(x+38)=0\). Thus, \(x=24\) or \(x=-38\); since the number is specified as positive, the correct answer is 24. Verification: \((24+4)(24+10)=28\times34=952\). In an exam, substitute the chosen value back into the original product to verify it.