Word Problems and Applications

Correct answer: B. (11)

Consecutive odd numbers are (x) and (x+2), so (x(x+2)=143) gives (x=11). Consecutive odd numbers differ by (2).

Correct answer: C. (14)

The direct answer is option C: 14. Consecutive even positive numbers differ by 2, so let them be x and x+2. Their product is x(x+2)=168, giving x^2+2x−168=0. Factoring gives (x−12)(x+14)=0. Thus x=12 or x=−14. Since the numbers must be positive, x=12 is the valid smaller number. The pair is 12 and 14, so the larger number is 14. Option A, 10, is wrong because the consecutive even pair 10 and 12 has product 120. Option B, 12, is the smaller number, not the larger one. Option C is correct because 12×14=168 and 14 is larger. Option D, 16, is wrong because 14×16=224. The key distinction is that consecutive even numbers do not differ by 1; they differ by 2. After solving, verify both the product and the requested larger value.

Correct answer: B. (15)

(x(x+2)=195) gives (x=13), so the larger number is (15). For odd numbers, (x) and (x+2) is the correct form.

Correct answer: B. 14

Let the smaller number be \(x\). The next consecutive even number is \(x+2\). Thus, \(x(x+2)=224\), giving \(x^2+2x-224=0\). Factoring gives \((x-14)(x+16)=0\), so \(x=14\) or \(x=-16\). Since the numbers are specified as positive, \(x=14\) is valid, and the numbers are 14 and 16. Exam tip: consecutive even numbers always differ by 2.

Correct answer: B. 12 m

Apply the rectangle area formula A = length × breadth. Let the breadth be x metres. The length is then x + 8 metres, so x(x + 8) = 240. This gives x² + 8x − 240 = 0. Factoring yields (x + 20)(x − 12) = 0, with solutions x = −20 and x = 12. Since a classroom dimension cannot be negative, the breadth is 12 m. Its length is 20 m, and the check 12 × 20 = 240 confirms both the equation and the units. Therefore option B is correct. The value 20 is the length, not the breadth. Values 10 and 15 would give lengths 18 and 23 respectively, producing areas 180 m² and 345 m², so they do not satisfy the stated area condition.

Correct answer: B. 15 cm

Let the breadth be \(x\) cm. Then the length is \(x+10\) cm, so the area condition gives \(x(x+10)=375\), or \(x^2+10x-375=0\). Factoring gives \((x-15)(x+25)=0\). Thus \(x=15\) or \(x=-25\); a dimension cannot be negative, so the breadth is 15 cm. The value 25 cm is the length, not the breadth. Exam tip: verify the result using \(15\times25=375\).

Correct answer: A. (16) m

Taking breadth as (x), (x(x+9)=400), giving (x=16). Do not accept the wrong negative value as a dimension.

Correct answer: C. 7

Let the number be \(x\). Then \(x(x+2)=63\), so \(x^2+2x-63=0\). Factoring gives \((x-7)(x+9)=0\), yielding \(x=7\) or \(x=-9\). Since the question asks for a positive number, the correct answer is 7. For comparison, choosing 9 gives \(9\times11=99\), not 63. Exam tip: In word problems, first represent the unknown number by \(x\) and express the related number in terms of \(x\).

Correct answer: B. 11

Let the number be \(x\). The other number is \(x+4\), so \(x(x+4)=165\). Thus, \(x^2+4x-165=0\), which factors as \((x-11)(x+15)=0\). The roots are \(x=11\) and \(x=-15\), but the number is specified as positive; therefore, the correct answer is 11. Exam tip: “4 more than a number” means \(x+4\), not \(4x\).

Correct answer: B. 14

Let the larger number be x. The phrase “6 less than itself” refers to x − 6, so the condition becomes x(x − 6) = 112. Expanding gives x² − 6x − 112 = 0. Factoring, (x − 14)(x + 8) = 0, so x = 14 or x = −8. The pair associated with x = 14 is 14 and 8, and its product is 112; therefore the larger number is 14. Option B is correct. The negative solution represents the ordered pair −8 and −14, whose larger member is −8, but the intended positive-number interpretation and the listed choices identify 14. A check shows 14 × (14 − 6) = 14 × 8 = 112. The other positive options do not produce 112 when multiplied by six less than themselves.

Correct answer: C. (12)

Let the smaller number be (x) and the larger be (x+5), then (x(x+5)=84) gives (7) and (12). The larger number is (12).

Correct answer: A. (9)

Putting the smaller number as (x) and larger as (x+3), (x(x+3)=108), giving (x=9). Write the difference in the correct direction.

Correct answer: B. 5 m

This is a quadratic word problem based on the area of a rectangle. Let the breadth be x metres. The length is one metre greater, so it is x + 1 metres. Using area = length × breadth gives x(x + 1) = 30, or x² + x − 30 = 0. Factorising, (x + 6)(x − 5) = 0, so x = −6 or x = 5. A physical breadth cannot be negative, so x = 5 m is the valid solution. Verification gives length = 6 m and area = 6 × 5 = 30 m². Hence option B is correct. The other choices do not satisfy both the one-metre difference and the required area, even if their numerical values appear plausible.

Correct answer: B. 13 m

Let the breadth be \(x\) metres. Then the length is \(x+7\) metres. Using area, \(x(x+7)=260\), so \(x^2+7x-260=0\). Factoring gives \((x-13)(x+20)=0\), hence \(x=13\) or \(x=-20\). Since a length cannot be negative, the breadth is 13 metres. Exam tip: for a rectangle, always use area = length × breadth.

Correct answer: C. 8 cm

Let the side of the square be x centimetres. The governing formula is area of a square = side × side = x². The given area is 64 square centimetres, so x² = 64. Taking square roots gives x = ±8, but a geometrical length must be positive; therefore x = 8 cm. Substitution verifies the result because 8 × 8 = 64 square centimetres. Hence option C is correct. The values 6, 7 and 9 are not suitable because their squares are 36, 49 and 81 respectively, not 64. The negative square-root value is an algebraic possibility but cannot represent the side of a tile.

Correct answer: D. (48) cm

(x^2=144) gives side (12) cm and perimeter (4x=48) cm. If perimeter is asked, do not stop after finding the side.

Correct answer: B. (10)

The direct answer is B: the smaller number is 10. Let the smaller number be x. Since the sum is 21, the other number is 21−x. Their product is 110, so x(21−x)=110. Expanding gives 21x−x^2=110, or x^2−21x+110=0. Factoring gives (x−10)(x−11)=0, so x=10 or x=11. The two numbers are therefore 10 and 11, and the smaller one is 10. Option A, 9, is wrong because 9+12=21 but 9×12=108, not 110. Option B is correct. Option C, 11, is one of the numbers but not the smaller one. Option D, 12, would require the other number to be 9, again giving product 108. A useful check is that 10+11=21 and 10×11=110. Memory cue: for sum and product questions, write x and S−x, form the product equation, then choose the smaller root.

Correct answer: B. 13

Let the numbers be x and 25 − x. Their product gives x(25 − x) = 156, so x² − 25x + 156 = 0, or (x − 12)(x − 13) = 0. The numbers are 12 and 13, so the larger number is 13.

Correct answer: B. 12

Let the smaller number be \(x\). The other number is \(x+3\), so \(x(x+3)=180\), or \(x^2+3x-180=0\). Factoring gives \((x-12)(x+15)=0\), so \(x=12\) or \(x=-15\). Since the number is positive, \(x=12\) is the valid answer. Check: \(12\times15=180\). Exam tip: In such word problems, represent the smaller number by \(x\) and express the other using the given difference, here \(x+3\).

Correct answer: C. 17

Let the positive number be x. The number that is 5 greater is x + 5, and their product is 204. Thus x(x + 5) = 204, which gives x² + 5x − 204 = 0. Factorising yields (x + 17)(x − 12) = 0, so x = −17 or x = 12. Because the question specifies a positive number, x = 12 is the valid smaller number. The larger number is therefore x + 5 = 17. Substitution confirms the condition: 12 × 17 = 204. Option A is the smaller number rather than the requested larger one; 15 and 20 do not form a pair differing by 5 with product 204. Hence option C is correct.

Correct answer: C. (11)

Total marks are (x \times x=x^2), so (x^2=121) and (x=11). In context, the number of questions is positive.

Correct answer: B. (10)

The equation is (x(x+2)=120), giving (x=10). For total quantity, multiply number of groups by number in each group.

Correct answer: B. (8)

The direct answer is B: there are 8 rows. If there are x rows and x+4 students in each row, total students equal rows multiplied by students per row. Therefore x(x+4)=96. Expanding gives x^2+4x−96=0. Factorization gives (x+12)(x−8)=0, so x=−12 or x=8. A number of rows cannot be negative, so x=8. Then each row has 12 students and 8×12=96, confirming the answer. Option A, 6, would give 6×10=60, not 96. Option B is correct because 8×12=96. Option C, 10, would give 10×14=140. Option D, 12, would give 12×16=192. The negative algebraic root is rejected because it has no physical meaning in this arrangement. Memory cue: total = number of rows × students per row; after solving, keep only a positive whole-number answer.

Correct answer: B. 9

The total number of students gives \(x(x-1)=72\), or \(x^2-x-72=0\). Factoring, \((x-9)(x+8)=0\), so \(x=9\) or \(x=-8\). Since the number of benches cannot be negative, \(x=9\) is the valid answer. In word problems, remember to reject roots that have no practical meaning.

Correct answer: B. 11

The total number of plants gives \(x(x+5)=176\), or \(x^2+5x-176=0\). Factoring, \((x-11)(x+16)=0\), so \(x=11\) or \(x=-16\). Since the number of rows cannot be negative, the answer is 11. Exam tip: In such application problems, retain only the positive root. Option 16 is incorrect because it uses the magnitude of the negative root \(-16\).