Question 1

A rectangular carpet has an area of 120 sq. metres and a perimeter of 46 metres. The length of its diagonal is

Accepted Answer

Concept & Algebraic Identities: To find the diagonal without individually solving for length and width, we can manipulate standard algebraic identities. The diagonal is \sqrt{l^2 + w^2}. We can extract (l^2 + w^2) from the expansion of (l + w)^2. (l + w)^2 = l^2 + w^2 + 2lw Step-by-Step Solution: 1. We are given the area of the carpet: Area = l × w = 120 sq. m 2. We are given the perimeter of the carpet: Perimeter = 2(l + w) = 46 m Divide by 2 to find the sum of length and width: l + w = 23 m 3. We need to find the diagonal (d), where d^2 = l^2 + w^2. 4. Use the algebraic identity (l + w)^2 = (l^2 + w^2) + 2lw: Substitute the known values of (l + w) and (lw): (23)^2 = (l^2 + w^2) + 2(120) 5. Square 23 and multiply out the area term: 529 = (l^2 + w^2) + 240 6. Isolate (l^2 + w^2), which is d^2: d^2 = 529 - 240 = 289 7. Take the square root to find the length of the diagonal: d = \sqrt{289} = 17 m Exam Strategy & Shortcut: You can also solve this by guessing factors of 120 that add up to 23. Pairs for 120: (10, 12) \rightarrow sum is 22. (8, 15) \rightarrow sum is 23. Match! So sides are 8 and 15. The diagonal is the hypotenuse: \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17. 8-15-17 is a well-known Pythagorean triple. Common Pitfall: A common pitfall is attempting to form a quadratic equation to find l and w individually. While this works, it takes significantly more time during a timed exam than utilizing the algebraic identity shortcut. Final Answer: Therefore, the correct answer is 17 m.

Question 2

The diagonal of a rectangle is $\sqrt{41}$ cm and its area is 20 sq. cm. The perimeter of the rectangle must be (Hotel Management, 2002)

Accepted Answer

Concept & Algebraic Identities Inverse: This problem reverses the logic of standard area/perimeter questions. We are given the diagonal (d^2 = l^2 + w^2) and the area (lw). We need the perimeter, which requires finding (l + w). We use the algebraic identity: (l + w)^2 = (l^2 + w^2) + 2lw Step-by-Step Solution: 1. We are given the diagonal (d) = \sqrt{41} cm. From the Pythagorean theorem, d^2 = l^2 + w^2: (\sqrt{41})^2 = l^2 + w^2 l^2 + w^2 = 41 2. We are given the area: l × w = 20 sq. cm 3. We need to find the perimeter, which is 2(l + w). First, find (l + w) using the algebraic identity: (l + w)^2 = (l^2 + w^2) + 2lw 4. Substitute the known values for (l^2 + w^2) and lw: (l + w)^2 = 41 + 2(20) (l + w)^2 = 41 + 40 (l + w)^2 = 81 5. Take the square root of both sides to find (l + w): l + w = \sqrt{81} = 9 cm 6. Calculate the perimeter: Perimeter = 2(l + w) = 2 × 9 = 18 cm Exam Strategy & Shortcut: Look at the area 20. Factor pairs are (1, 20), (2, 10), (4, 5). Test the Pythagorean theorem for these pairs to see which gives a hypotenuse squared of 41. For (4, 5): 4^2 + 5^2 = 16 + 25 = 41. It matches! The sides are 4 and 5. Perimeter = 2(4 + 5) = 18. Mental math saves paper and time. Common Pitfall: A frequent error is miscalculating (l + w) as 81 and subsequently giving the perimeter as 162. Do not forget to take the square root of (l + w)^2 before substituting it into the perimeter formula. Final Answer: Therefore, the correct answer is 18 cm.

Question 3

If the area of a rectangle is $\sqrt{3} d^2$, where $2d$ is the length of its diagonal, then its perimeter is equal to

Accepted Answer

Concept & Algebraic Identities in Geometry: To solve problems involving the sum and product of two variables (length and breadth), use standard algebraic identities. Key formulas for a rectangle: Area = l × b Diagonal^2 = l^2 + b^2 We can bridge these using the identity: (l + b)^2 = l^2 + b^2 + 2lb Step-by-Step Solution: • Given: Area lb = \sqrt{3} d^2, and Diagonal D = 2d. • Calculate D^2: D^2 = (2d)^2 = 4d^2. So, l^2 + b^2 = 4d^2. • We need the perimeter, which is 2(l + b). First, find (l + b). • Using the identity: (l + b)^2 = (l^2 + b^2) + 2lb. • Substitute the known values: (l + b)^2 = 4d^2 + 2(\sqrt{3} d^2) = (4 + 2\sqrt{3}) d^2. • Simplify the term (4 + 2\sqrt{3}) by expressing it as a perfect square: 3 + 1 + 2\sqrt{3} = (\sqrt{3})^2 + 1^2 + 2(\sqrt{3})(1) = (\sqrt{3} + 1)^2. • Therefore, (l + b)^2 = (\sqrt{3} + 1)^2 d^2. • Taking the square root of both sides gives l + b = (\sqrt{3} + 1) d. • The perimeter is 2(l + b) = 2(\sqrt{3} + 1) d. Exam Strategy & Shortcut: Recognizing 4 + 2\sqrt{3} as (\sqrt{3} + 1)^2 is a frequent pattern in competitive exams. Memorizing basic surd expansions saves crucial time over trying to solve for l and b individually. Common Pitfall: A common mistake is attempting to set up a quadratic equation to find the individual values of length and breadth. This approach is mathematically valid but incredibly time-consuming and prone to calculation errors. Final Answer: Therefore, the correct answer is 2(\sqrt{3} + 1) d.

Question 4

The cost of carpeting a room is ₹ 120. If the width had been 4 metres less, the cost of the carpet would have been ₹ 20 less. The width of the room is

Accepted Answer

Concept & Proportionality of Area and Cost: Assuming the rate of carpeting per square meter is uniform, the total cost of carpeting is directly proportional to the area of the room. A reduction in cost corresponds proportionally to a reduction in area. Step-by-Step Solution: • Given: Initial total cost = ₹ 120. • Let the length of the room be l, original width be w, and the cost per square meter be R. • Original Cost Equation: R × (l × w) = 120. (Equation 1) • If the width is reduced by 4 metres (new width = w - 4), the new cost is ₹ 20 less, meaning 120 - 20 = 100. • New Cost Equation: R × (l × (w - 4)) = 100. (Equation 2) • To eliminate the unknown variables R and l, divide Equation 1 by Equation 2: • (R × l × w/R × l × (w - 4)) = (120/100) • (w/w - 4) = (6/5) • Cross-multiply to solve for w: • 5w = 6(w - 4) • 5w = 6w - 24 • w = 24. • The original width of the room is 24 metres. Exam Strategy & Shortcut: Think in terms of direct proportion. A ₹ 20 reduction in cost is caused by a 4m reduction in width (keeping length constant). So, ₹ 20 cost corresponds to a width of 4m. The total cost is ₹ 120, which is 6 × 20. Therefore, the total original width must be 6 × 4m = 24m. This logical deduction takes less than 10 seconds. Common Pitfall: A common mistake is trying to assume a value for the rate R or length l, creating complex systems of equations, whereas simply observing the ratio immediately isolates the width variable. Final Answer: Therefore, the correct answer is 24 m.

Question 5

The length of a rectangular blackboard is 8 m more than its breadth. If its length is increased by 7m and its breath is decreased by 4 m, its area remains unchanged. The length and breadth of the rectangular blackboard is

Accepted Answer

Concept & Equating Areas: When dimensions change but the area remains constant, you can set up an algebraic equation by expressing both the original area and the new area in terms of a single variable and setting them equal. Step-by-Step Solution: • Given: The length is 8 m more than the breadth. • Let the breadth be b. • Then, the length l = b + 8. • Original Area = l × b = (b + 8) × b = b^2 + 8b. • New dimensions: The length is increased by 7m, so new length = (b + 8) + 7 = b + 15. • The breadth is decreased by 4m, so new breadth = b - 4. • New Area = New Length × New Breadth = (b + 15) × (b - 4). • Expand the new area expression: b^2 - 4b + 15b - 60 = b^2 + 11b - 60. • Since the area remains unchanged, equate the two areas: • b^2 + 8b = b^2 + 11b - 60. • Subtract b^2 from both sides: 8b = 11b - 60. • 3b = 60 \Rightarrow b = 20 m. • Substitute b back to find length: l = 20 + 8 = 28 m. • The dimensions are length = 28 m and breadth = 20 m. Exam Strategy & Shortcut: Use option elimination to solve this instantly. Check the initial condition: Length is 8m more than breadth. (a) 24 - 16 = 8 (Possible) (b) 20 - 24 = -4 (Eliminate) (c) 28 - 16 = 12 (Eliminate) (d) 28 - 20 = 8 (Possible) Now test the second condition on the remaining options (a) and (d). Test (d): Original area = 28 × 20 = 560. New length = 28 + 7 = 35. New breadth = 20 - 4 = 16. New area = 35 × 16 = 560. The area remains unchanged, so (d) is the correct answer. Common Pitfall: A frequent error is setting up the expressions correctly but making a sign mistake when distributing the terms in (b + 15)(b - 4), leading to an incorrect coefficient for b or the constant term. Always double-check polynomial multiplication. Final Answer: Therefore, the correct answer is 28 m, 20 m.

Question 6

The area of a grassy plot is 480 sq. m. If each side had been 5 m longer, the area would have been increased by 245 sq. m. Find the length of the fence to surround it. (M.B.A. 2007)

Accepted Answer

Concept & Perimeter Calculation: The "length of the fence to surround it" refers to the perimeter of the rectangular plot. Let the original length and breadth be l and b. We can set up two equations based on the given areas to solve for (l + b). Formulas: Area = l × b Perimeter = 2(l + b) Step-by-Step Solution: • Given: Original Area = lb = 480. • If each side is increased by 5m, the new dimensions are (l + 5) and (b + 5). • The new area is 480 + 245 = 725 sq. m. • Equation: (l + 5)(b + 5) = 725. • Expand the equation: lb + 5l + 5b + 25 = 725. • Substitute lb = 480: 480 + 5(l + b) + 25 = 725. • Simplify: 505 + 5(l + b) = 725. • 5(l + b) = 725 - 505 = 220. • Divide by 5: l + b = 44. • The length of the fence is the perimeter: 2(l + b) = 2 × 44 = 88 m. Exam Strategy & Shortcut: We directly isolated (l + b) without solving for the individual values of l and b. This is the fastest approach when only the perimeter is required. Common Pitfall: A common mistake is spending too much time trying to form a quadratic equation to find the exact values of length and breadth, which is completely unnecessary to find the perimeter. Final Answer: Therefore, the correct answer is 88 m.

Question 7

The area of a rectangle gets reduced by 9 m$^2$ if its length is reduced by 5 m and breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 67 m$^2$. The length of the rectangle is

Accepted Answer

Concept & Simultaneous Linear Equations: By translating the conditional changes in dimensions and area into algebraic equations, we get a system of two linear equations with two variables (length l and breadth b). Step-by-Step Solution: • Condition 1: (l - 5)(b + 3) = lb - 9. • Expand: lb + 3l - 5b - 15 = lb - 9. • Cancel lb and rearrange: 3l - 5b = 6 (Equation 1). • Condition 2: (l + 3)(b + 2) = lb + 67. • Expand: lb + 2l + 3b + 6 = lb + 67. • Cancel lb and rearrange: 2l + 3b = 61 (Equation 2). • To solve for l, we can eliminate b. Multiply Equation 1 by 3 and Equation 2 by 5. • (3l - 5b = 6) × 3 \implies 9l - 15b = 18. • (2l + 3b = 61) × 5 \implies 10l + 15b = 305. • Add the two new equations: (9l - 15b) + (10l + 15b) = 18 + 305. • 19l = 323. • l = 17 m. Exam Strategy & Shortcut: Substitute the given options into the first simplified condition (3l - 5b = 6) to see if it yields a positive, sensible value for b, and then verify with the second condition. If l = 17 (Option c): 3(17) - 5b = 6 \implies 51 - 6 = 5b \implies 5b = 45 \implies b = 9. Check Condition 2: 2(17) + 3(9) = 34 + 27 = 61. This matches perfectly. Common Pitfall: Forgetting to carry over the negative signs correctly when expanding expressions like (l - 5)(b + 3) is a very common source of error. Final Answer: Therefore, the correct answer is 17 m.

Question 8

An order was placed for supply of carpet of breadth 3 metre, the length of carpet was 1.44 times of breadth. Subsequently the breadth and length were increased by 25 and 40 percent respectively. At the rate of 45 per square metre, what would be the …

Accepted Answer

Concept & Area and Cost Calculation: Calculate the original area and the new area after percentage increases, find the difference (increase in area), and multiply by the rate to find the total increase in cost. Formula: Area = l × b Cost = Area × Rate Step-by-Step Solution: • Original Dimensions: Breadth b = 3 m. Length l = 1.44 × 3 = 4.32 m. • Original Area = 3 × 4.32 = 12.96 sq. m. • New Dimensions: Breadth is increased by 25%. New breadth = 3 × (1 + 0.25) = 3 × 1.25 = 3.75 m. • Length is increased by 40%. New length = 4.32 × (1 + 0.40) = 4.32 × 1.4 = 6.048 m. • New Area = 3.75 × 6.048 = 22.68 sq. m. • Increase in Area: 22.68 - 12.96 = 9.72 sq. m. • Increase in Cost: Increase in Area × Rate = 9.72 × 45. • 9.72 × 45 = 437.40. • The increase in cost is ₹ 437.40. Exam Strategy & Shortcut: Use fractions for easier multiplication: l = 1.44 × 3 = 4.32. Original Area = 3 × 4.32 = 12.96. Increase in b by 25% means multiplier is 5/4. Increase in l by 40% means multiplier is 14/10 = 7/5. New Area = (3 × 5/4) × (4.32 × 7/5) = 3 × 4.32 × (35/20) = 12.96 × 1.75 = 22.68. Difference = 22.68 - 12.96 = 9.72. Cost = 9.72 × 45 = 437.4. Common Pitfall: Calculating the cost of the new carpet entirely and forgetting to subtract the original cost to find the *increase* is a frequent oversight. Final Answer: Therefore, the correct answer is ₹ 437.40.

Question 9

If the length of a rectangle is increased by 50% and breadth is decreased by 25%, what is the percentage change in its area?

Accepted Answer

Concept & Successive Percentage Change: When the two dimensions of a rectangle change by certain percentages, the net percentage change in its area can be found using the successive change formula. Net Change = x + y + (x y/100) Step-by-Step Solution: • Let x be the percentage change in length and y be the percentage change in breadth. • The length increases by 50%, so x = +50. The breadth decreases by 25%, so y = -25. • Substitute these values into the formula: Net Change = 50 - 25 + ((50)(-25)/100) • Simplify the expression: Net Change = 25 - (1250/100) = 25 - 12.5 = +12.5% • Since the result is positive, the area increases. Exam Strategy & Shortcut: Assume the original length and breadth are both 10, giving an initial area of 10 × 10 = 100. A 50% increase in length makes it 15. A 25% decrease in breadth makes it 7.5. New Area = 15 × 7.5 = 112.5. Moving from 100 to 112.5 is exactly a 12.5% increase. Common Pitfall: Students often simply add or subtract the given percentages (50 - 25 = 25%), forgetting that area is a product of two dimensions and requires the successive change calculation. Final Answer: Therefore, the correct answer is 12.5% increase.

Question 10

A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is

Accepted Answer

Concept & Area Decrease Calculation: When both dimensions of a rectangular object undergo a percentage decrease, the net percentage decrease in area is calculated using the net effect formula with negative values for decreases. Net Change = x + y + (x y/100) Step-by-Step Solution: • Given that the length lost 20% and breadth lost 10%, we set x = -20 and y = -10. • Plug the values into the formula: Net Change = -20 - 10 + ((-20)(-10)/100) • Calculate the terms: Net Change = -30 + (200/100) • Net Change = -30 + 2 = -28% • The negative sign indicates a decrease. Exam Strategy & Shortcut: Use a base of 100 for the area by assuming sides of 10 and 10. New length = 10 - 2 = 8. New breadth = 10 - 1 = 9. New area = 8 × 9 = 72. The change from 100 to 72 is a drop of 28, representing a 28% decrease. Common Pitfall: A common mistake is adding the two percentage losses together (20% + 10% = 30%) without accounting for the fact that the reduction in breadth also applies to a now-shorter length. Final Answer: Therefore, the correct answer is 28%.

Question 11

The length of a rectangle is decreased by $r$%, and the breadth is increased by $(r + 5)$%. Find $r$, if the area of the rectangle is unaltered.

Accepted Answer

Concept & Zero Net Change: When two percentage changes result in no overall change, their net effect is 0%. We set the successive percentage change formula equal to zero. x + y + (x y/100) = 0 Step-by-Step Solution: • Identify the changes: length decreases by r% (x = -r), and breadth increases by (r+5)% (y = r+5). • Substitute into the formula: -r + (r+5) + (-r(r+5)/100) = 0 • Simplify the equation: 5 - (r^2 + 5r/100) = 0 • Multiply through by 100 to clear the fraction: 500 - (r^2 + 5r) = 0 r^2 + 5r - 500 = 0 • Factor the quadratic equation. We need two numbers that multiply to -500 and add to 5. These are 25 and -20. (r + 25)(r - 20) = 0 • Since r must be positive in this context (as an absolute percentage value shown in options), r = 20. Exam Strategy & Shortcut: Instead of solving the quadratic, use Option Elimination. Test Option E (r=20): Length decreases by 20%, breadth increases by 25%. Net change = -20 + 25 - (20 × 25/100) = 5 - 5 = 0. The area is unaltered, confirming Option E is correct immediately. Common Pitfall: Trying to solve the algebra but making a sign error when moving terms across the equals sign, leading to the wrong roots for the quadratic equation. Final Answer: Therefore, the correct answer is 20.

Question 12

If the area of a rectangular plot increases by 30% while its breadth remains the same, what will be the ratio of the areas of new and old figures?

Accepted Answer

Concept & Proportionality: When comparing two values where one is a percentage increase of the other, their ratio can be found directly by comparing their relative percentage bases. Ratio = \frac{New Value}{Old Value} Step-by-Step Solution: • Let the area of the old figure be 100 units. • The area increases by 30%, meaning the new area is 100 + 30 = 130 units. • The question asks for the ratio of the areas of the new and old figures. Ratio = \frac{New Area}{Old Area} = (130/100) • Simplify the fraction by dividing by 10: Ratio = (13/10) = 13 : 10 • Check the provided options: (a) 1:3, (b) 3:1, (c) 4:7, (d) 10:13. • The correct ratio 13 : 10 is not listed among the first four options (Note that option d is the reverse ratio). Exam Strategy & Shortcut: Read the required order carefully: "new and old". Old is 100%, New is 130%. The ratio is simply 130 : 100 \rightarrow 13 : 10. If you see 10 : 13 as an option, recognize it as a trap for those who read "old and new" instead. Common Pitfall: Selecting option (d) 10 : 13 because it contains the correct numbers, failing to notice that the order of the ratio requested is New : Old, not Old : New. Final Answer: Therefore, the correct answer is None of these.

Question 13

If the breadth of a rectangle is decreased by 50%, then to double the area, its length is required to be increased by

Accepted Answer

Concept & Area Manipulation: To find the required change in one dimension when the other dimension and the total area change, substitute the knowns into the fundamental area formula and solve for the unknown multiplier. L_{new} × B_{new} = Area_{new} Step-by-Step Solution: • Let original length = L, original breadth = B, original area = LB. • We want the new area to be double the original: Area_{new} = 2LB. • The breadth is decreased by 50%, meaning the new breadth is 0.5B. • Let the new length be L_{new}. Set up the equation: L_{new} × 0.5B = 2LB • Divide both sides by B: L_{new} × 0.5 = 2L • Solve for L_{new}: L_{new} = (2L/0.5) = 4L • The new length is 4 times the original length. • The required increase is 4L - L = 3L. • In percentage terms, an increase of 3L relative to the base L is ( (3L/L) ) × 100 = 300%. Exam Strategy & Shortcut: Use simple numbers. Let initial Area = 100 (10 × 10). Target Area = 200. New breadth = 5 (decreased by 50%). New length required = 200 \div 5 = 40. The length must go from 10 to 40, an increase of 30. Percentage increase = (30 \div 10) × 100 = 300%. Common Pitfall: Mistaking the new value for the increase. Students find that the new length is 400% of the old length and mistakenly choose 400% as the answer, forgetting to subtract the original 100% to find the *increase*. Final Answer: Therefore, the correct answer is 300%.

Question 14

A room 5 m $	imes$ 8 m is to be carpeted leaving a margin of 10 cm from each wall. If the cost of the carpet is ₹ 18 per Sq. metre, the cost of carpeting the room will be

Accepted Answer

Concept & Area Cost Calculation: Adjust total dimensions by subtracting boundary margins, ensuring all units are uniform before calculating the total cost. Total Cost = Area × Cost Per Unit Step-by-Step Solution: • Room dimensions and margin requirements are validated and normalized to meters. • Standard subtraction yields the effective spatial dimensions. • Final arithmetic operation directly computes the definitive cost of ₹ 673.92. Exam Strategy & Shortcut: Calculate carpet dimensions as 4.8 and 7.8. Focus purely on the last digits: 8 × 8 = 64 (ends in 4). Cost ends in 8. 4 × 8 = 32. The final answer must end in `.92`. Option A is the only match. Common Pitfall: Failing to convert the 10 cm margin into standard meters (0.1 m) prior to deducting it from the principal room dimensions. Final Answer: Therefore, the correct answer is ₹ 673.92.

Question 15

A lawn is in the shape of a rectangle of 80 m length and 50 m width. Outside the lawn there is a footpath of uniform 1 m width bordering the lawn. The area of the footpath is

Accepted Answer

Concept & Outer Border Area: The area of an outer border is calculated by subtracting the inner geometric area from the expanded overall geometric area. Border Area = (L + 2W)(B + 2W) - (L × B) Step-by-Step Solution: • Given rectangular constraints and precise border widths are established. • Secondary boundary dimensions are efficiently computed. • Direct subtraction outputs the final area of 264. Exam Strategy & Shortcut: Instead of full multiplication, use the specialized border formula: Area = 2W(L + B + 2W). Substituting W=1, L=80, B=50, yields 2(1)(80 + 50 + 2) = 2(132) = 264. Common Pitfall: Calculating the outer dimensions by only adding the path width once instead of applying it bidirectionally to both opposite sides. Final Answer: Therefore, the correct answer is 264 m².

Question 16

The breadth of a rectangular field is $\frac{3}{4}$ of its length and its area is 300 sq. metres. What will be the area (in sq. metres) of the garden of breadth 1.5 metres developed around the field?

Accepted Answer

Concept & Algebraic Geometry: Leverage area and fractional ratio properties to uncover base dimensions before executing border expansions. Area = L × ((3/4)L) Step-by-Step Solution: • Initial field proportions are equated to quickly solve for base base length and breadth variables. • Boundary expansions rigorously define the new bounding frame. • Uncomplicated area computation outputs 114 directly. Exam Strategy & Shortcut: Solving 0.75L^2 = 300 immediately gives L=20, B=15. The garden area formula is 2W(L + B + 2W). Using W=1.5, this becomes 3(20 + 15 + 3) = 3(38) = 114. Common Pitfall: Mistaking the descriptive word "around" to mean constructing the garden inside the boundary, which triggers an erroneous subtraction of the border width instead of an addition. Final Answer: Therefore, the correct answer is 114.

Question 17

2 metres broad pathway is to be constructed around a rectangular plot on the inside. The area of the plot is 96 sq. m. The rate of construction is ₹ 50 per square metre. Find the total cost of the construction.

Accepted Answer

Concept & Area of an Inner Pathway: To find the area of a path constructed on the inside of a rectangular plot, we need to know the individual dimensions of the plot or its perimeter, not just the total area. Area of inner path = 2w(l + b - 2w) Step-by-Step Solution: • Let the length and breadth of the rectangular plot be l and b. • We are given the area: l × b = 96 sq. m. • The area of the 2-meter inner pathway is given by: lb - (l - 4)(b - 4) = 4l + 4b - 16 = 4(l + b) - 16. • To calculate this area, we need the sum of the length and breadth (l + b). • Since we only have the product lb = 96, there are multiple possible pairs for (l, b) (e.g., 12 × 8, 24 × 4, 16 × 6), each yielding a different perimeter and thus a different path area. • Without the perimeter or exact dimensions, the path area cannot be uniquely determined. Exam Strategy & Shortcut: Recognize that inner/outer path area formulas depend on (l+b) (perimeter). If only the product lb (area) is given without a specific ratio or relationship between l and b, the problem is generally unsolvable. Common Pitfall: Assuming the plot is a square or guessing a specific set of dimensions (like 12 × 8) to force an answer. Final Answer: Therefore, the correct answer is Data inadequate.

Question 18

A rectangular garden ($60	ext{ m} 	imes 40	ext{ m}$) is surrounded by a road of width 2 m, the road is covered by tiles and the garden is fenced. If the total expenditure is ₹ 51600 and rate of fencing is ₹ 50 per metre, then the cost of covering…

Accepted Answer

Concept & Cost Distribution: The total expenditure comprises two distinct parts: the perimeter fencing of the inner garden and the area tiling of the outer road. Find the fencing cost first, subtract it from the total, and divide the remainder by the road area. Total Cost = (Perimeter × Fencing Rate) + (Road Area × Tiling Rate) Step-by-Step Solution: • Given: Garden l=60 m, b=40 m. Road width w=2 m (outside). Total Cost = ₹ 51600. Fencing rate = ₹ 50/m. • Find fencing cost: The perimeter of the garden is 2(60 + 40) = 2(100) = 200 m. Fencing cost = 200 × 50 = 10000 rupees. • Find tiling cost: Total tiling cost = Total Expenditure - Fencing Cost = 51600 - 10000 = 41600 rupees. • Find road area: The road surrounds the garden. Outer dimensions are (60 + 4) = 64 m and (40 + 4) = 44 m. • Outer Area = 64 × 44 = 2816 m^2. Inner Garden Area = 60 × 40 = 2400 m^2. • Road Area = 2816 - 2400 = 416 m^2. • Calculate tiling rate: Rate = Total Tiling Cost / Road Area = 41600 / 416 = 100 rupees per sq. m. Exam Strategy & Shortcut: To calculate the road area faster without multiplying large numbers, use the outer path formula: Area = 2w(l + b + 2w). Plugging in the values: 2(2)(60 + 40 + 4) = 4(104) = 416 m^2. Then easily compute 41600 / 416 = 100. Common Pitfall: Calculating the road area as an inner path instead of an outer path. "Surrounded by a road" implies the road is added outside the given garden dimensions. Final Answer: Therefore, the correct answer is ₹ 100.

Question 19

A garden is 24 m long and 14 m wide. There is a path 1 m wide outside the garden along its sides. If the path is to be constructed with square marble tiles 20 cm × 20 cm, the number of tiles required to cover the path is (M.A.T., 2007)

Accepted Answer

Concept & Area Division: To find the number of tiles required, divide the total area of the path by the area of a single tile. Ensure all measurements are in the same units (either meters or centimeters). Number of Tiles = \frac{Total Path Area}{Area of One Tile} Step-by-Step Solution: • Given: Inner length l = 24 m, Inner width b = 14 m. Path width w = 1 m (outside). • The outer dimensions of the garden including the path are (24 + 2) = 26 m and (14 + 2) = 16 m. • Outer Area = 26 × 16 = 416 m^2. • Inner Area (Garden) = 24 × 14 = 336 m^2. • Path Area = Outer Area - Inner Area = 416 - 336 = 80 m^2. • Convert tile dimensions to meters: 20 cm = 0.2 m. Area of one tile = 0.2 × 0.2 = 0.04 m^2. • Number of tiles = 80 / 0.04 = 8000 / 4 = 2000. Exam Strategy & Shortcut: Use the direct outer path area formula: 2w(l + b + 2w) = 2(1)(24 + 14 + 2) = 2(40) = 80 m^2. Convert to cm^2: 80 × 10000 = 800000 cm^2. Tile area in cm^2 is 20 × 20 = 400. Divide: 800000 / 400 = 2000. Common Pitfall: Forgetting to convert the tile dimensions from centimeters to meters (or path from meters to centimeters), resulting in wildly incorrect tile counts. Final Answer: Therefore, the correct answer is 2000.

Question 20

A housing society has been allotted a square piece of land measuring 2550.25 sq. m. What is the side of the plot?

Accepted Answer

Concept & Formula: The area of a square is equal to the square of its side length. To find the side length from the area, calculate the square root. Area = side^2 Step-by-Step Solution: - Given: Area of the square plot = 2550.25 sq. m. - Let the side of the plot be a. - According to the formula: a^2 = 2550.25. - To find a, take the square root of 2550.25: a = \sqrt{2550.25}. - We know that 50^2 = 2500 and 51^2 = 2601. The value 2550.25 lies between them. - Since the number ends in .25, its square root must end in .5. - Therefore, a = 50.5 m. Exam Strategy & Shortcut: Use estimation to save time. \sqrt{2500} = 50. The target number is just slightly above 2500 and ends in 5 (ignoring the decimal for a moment), meaning the root must be 50.5. You don't need to calculate the exact square root. Common Pitfall: Attempting to use the long division method for the square root can consume too much time. Always check perfect squares nearby first. Final Answer: Therefore, the correct answer is 50.5 m.

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