Question 1

The length of a rectangle is halved while its breadth is tripled. What is the percentage change in area?

Accepted Answer

Problem restatementFind how the area changes when L → L/2 and B → 3B. Given data Original area = L × B. New area = (L/2) × (3B). Concept/ApproachArea scales by the product of the scale factors. Step-by-Step calculation New area = (L × B) × (1/2) × 3 = (3/2)LB = 1.5 × original area ⇒ 50% increase Common pitfalls Adding percentage changes (−50% and +200%) incorrectly; use multiplicative scaling. Final Answer50% increase

Question 2

A towel, when bleached, loses 20% of its length and 10% of its breadth. What is the percentage decrease in its area?

Accepted Answer

Problem restatementCompute the net percentage change in area when dimensions shrink by given percentages. Given data Length factor = 1 − 0.20 = 0.8 Breadth factor = 1 − 0.10 = 0.9 Concept/ApproachArea scales as length × breadth, so multiply the scale factors. Step-by-Step calculation New area factor = 0.8 × 0.9 = 0.72 Percentage decrease = (1 − 0.72) × 100% = 28% Common pitfalls Adding 20% and 10% to get 30% decrease (incorrect; must multiply factors). Final Answer28%

Question 3

The length of a rectangular plot is 20 m more than its breadth. If fencing costs ₹26.50 per metre and the total fencing cost is ₹5,300, what is the length of the plot (in metres)?

Accepted Answer

Problem restatementFrom fencing cost and unit rate, find the perimeter. Use the length–breadth relation to determine the dimensions. Given data Length L = B + 20. Fencing rate = ₹26.50 per metre; total cost = ₹5,300. Concept/ApproachPerimeter P = 2(L + B). Cost = P × rate ⇒ P = Cost / rate. Step-by-Step calculation P = 5300 / 26.5 = 200 m 2(L + B) = 200 ⇒ L + B = 100 But L = B + 20 ⇒ (B + 20) + B = 100 2B = 80 ⇒ B = 40 m L = B + 20 = 60 m Verification/AlternativePerimeter = 2(60 + 40) = 200 m; cost = 200 × 26.5 = ₹5,300 (checks). Final Answer60 m

Question 4

The diagonal of a rectangular closet floor is 7½ ft and its shorter side is 4½ ft. What is the area of the closet (in sq ft)?

Accepted Answer

Problem restatementUse the Pythagorean relation to find the other side, then compute area. Given data Diagonal d = 7.5 ft. Shorter side s = 4.5 ft. Concept/ApproachFor rectangle sides a and b: d² = a² + b². Let the unknown side be x. Step-by-Step calculation x = √(d² − s²) = √(7.5² − 4.5²) = √(56.25 − 20.25) = √36 = 6 ft Area = s × x = 4.5 × 6 = 27 sq ft Verification/AlternativeCheck: 4.5² + 6² = 20.25 + 36 = 56.25 = 7.5² (consistent). Final Answer27 sq ft

Question 5

A rectangular field is to be fenced on three sides, leaving one side of 20 ft uncovered. If the area is 680 sq ft, how many feet of fencing are required?

Accepted Answer

Problem restatementOne side of the rectangle (length 20 ft) is left unfenced; the other three sides are fenced. Given the area, find the fenced length. Given data Uncovered side (take as length) = 20 ft. Area = 680 sq ft. Concept/ApproachLet breadth be b. Then area = length × breadth = 20 × b. Step-by-Step calculation 20 × b = 680 ⇒ b = 680 / 20 = 34 ft Fenced sides = other length (20 ft) + two breadths (2 × 34 ft) Total fencing = 20 + 68 = 88 ft Verification/AlternativeIf instead the uncovered side were a breadth, you would first compute the corresponding length, but the statement specifies the uncovered side is 20 ft, matching the calculation above. Common pitfalls Using the full perimeter 2(L + B); remember only three sides are fenced. Final Answer88 ft

Question 6

A square's side is measured 2% too large. What is the resulting percentage error in the computed area?

Accepted Answer

Problem restatementWhen a square's side is measured 2% in excess, the calculated area uses the inflated side. Find the percentage error in area. Given data True side = s. Measured side = 1.02s (2% excess). Concept/ApproachArea scales with the square of the side. Percentage error in area = [(measured area − true area) ÷ true area] × 100%. Step-by-Step calculationTrue area = s2Measured area = (1.02s)2 = 1.0404 s2Percentage error = (1.0404 − 1) × 100% = 4.04% Verification/AlternativeUsing binomial approximation: (1 + x)2 ≈ 1 + 2x + x2. With x = 0.02 ⇒ 1 + 0.04 + 0.0004 = 1.0404, confirming 4.04%. Common pitfallsDoubling the 2% to 4% and ignoring the small x2 term, which here contributes 0.04%. Final Answer4.04%

Question 7

A man walks diagonally across a square instead of along two edges. Approximately what percentage distance is saved?

Accepted Answer

Problem restatementCompare the diagonal path to the L-shaped edge path around one corner of a square and find the percentage saving in distance. Given data Square side = a. Edge path distance = a + a = 2a. Diagonal distance = a√2. Concept/ApproachPercent saved = [(edge path − diagonal) ÷ edge path] × 100%. Step-by-Step calculationPercent saved = [(2a − a√2) ÷ 2a] × 100%= (1 − √2 ÷ 2) × 100% ≈ (1 − 0.7071) × 100%≈ 29.3% Verification/AlternativeFor a = 1: edge = 2, diagonal ≈ 1.4142, saving ≈ 0.5858 ⇒ 0.5858 ÷ 2 ≈ 29.29%. Common pitfallsUsing 1 − (1/√2) instead of 1 − (√2/2); both are equal numerically here, but ensure the reference distance is the edge path (2a). Final Answer29.3%

Question 8

The length and breadth of a rectangular park are in the ratio 3:2. A cyclist completes one boundary lap in 8 minutes at 12 km/h. What is the area of the park (in m²)?

Accepted Answer

Problem restatementFrom speed and time, get the perimeter, then use the 3:2 ratio to find actual sides and hence the area. Given data Speed = 12 km/h. Lap time = 8 min = 8/60 h = 2/15 h. Perimeter P = distance in one lap. Length:breadth = 3:2. Concept/ApproachDistance = speed × time ⇒ P. With 2(L + B) = P and L:B = 3:2, solve for L and B, then area = L × B. Step-by-Step calculationP = 12 × (2/15) = 24/15 = 8/5 km = 1.6 km = 1600 mLet L = 3x, B = 2x ⇒ 2(L + B) = 2(5x) = 10x = 1600 ⇒ x = 160L = 480 m, B = 320 mArea = 480 × 320 = 153,600 m² Verification/AlternativePerimeter from sides: 2(480 + 320) = 2 × 800 = 1600 m, matching the lap distance. Common pitfallsForgetting to convert km to meters and minutes to hours before multiplying. Final Answer153,600 m²

Question 9

A rectangle has diagonal √41 cm and area 20 cm². What is its perimeter?

Accepted Answer

Problem restatementGiven diagonal d and area A of a rectangle with sides a and b, find the perimeter 2(a + b). Given data a2 + b2 = d2 = 41 ab = 20 Concept/ApproachUse (a + b)2 = a2 + b2 + 2ab to find a + b directly without solving for a and b separately. Step-by-Step calculation(a + b)2 = 41 + 2(20) = 81a + b = 9Perimeter = 2(a + b) = 18 cm Verification/AlternativeIf a and b are roots of t2 − 9t + 20 = 0 ⇒ (t − 4)(t − 5) = 0 ⇒ sides 4 and 5 cm. Perimeter = 2(4 + 5) = 18 cm. Common pitfallsAttempting to use Pythagoras directly to find one side without leveraging the area; the symmetric identity is faster. Final Answer18 cm

Question 10

A 60 m × 40 m rectangular park has two equal-width concrete roads crossing through its centre (one along length, one along breadth). The rest is lawn. If the lawn area is 2109 m², what is the width of each road?

Accepted Answer

Problem restatementTwo perpendicular roads of equal width w run through the middle of a 60 m by 40 m park. Lawn area (park minus roads) is 2109 m². Find w. Given data Total area = 60 × 40 = 2400 m² Road area = 60w + 40w − w2 (subtracting the overlapping square once) Lawn area = 2109 m² Concept/ApproachSet total area minus road area equal to the lawn area and solve the quadratic in w. Step-by-Step calculation2400 − (100w − w2) = 2109w2 − 100w + 2400 − 2109 = 0w2 − 100w + 291 = 0Discriminant = 10000 − 1164 = 8836 = 942w = [100 ± 94] / 2 ⇒ w = 3 or 97Feasible width < 40 ⇒ w = 3 m Verification/AlternativeRoad area = 60×3 + 40×3 − 3×3 = 180 + 120 − 9 = 291; Lawn = 2400 − 291 = 2109. Common pitfallsForgetting to subtract the intersection area w2 that gets double-counted. Final Answer3 m

Question 11

What is the least number of equal square tiles needed to pave a floor 15 m 17 cm by 9 m 2 cm?

Accepted Answer

Problem restatementFind the largest possible square tile size (in cm) that exactly divides both floor dimensions; then compute the number of such tiles needed. Given data Length = 15 m 17 cm = 1517 cm Breadth = 9 m 2 cm = 902 cm Concept/ApproachLargest square tile side = GCD of the two dimensions (in the same unit). Number of tiles = (L/gcd) × (B/gcd). Step-by-Step calculationGCD(1517, 902): 1517 − 902 = 615; 902 − 615 = 287; 615 − 2×287 = 41; 287 − 7×41 = 0 ⇒ gcd = 41 cmTiles along length = 1517 / 41 = 37Tiles along breadth = 902 / 41 = 22Total tiles = 37 × 22 = 814 Verification/Alternative41 cm divides both lengths exactly; any larger square would fail to tile one dimension evenly. Common pitfallsLeaving dimensions in mixed metres–centimetres or using LCM instead of GCD. Final Answer814

Question 12

In a rectangle, the ratio of perimeter to breadth is 5:1 and the area is 216 cm². What is its length?

Accepted Answer

Problem restatementGiven P : B = 5 : 1 and area, determine the rectangle's length. Given data Perimeter P : breadth B = 5 : 1 ⇒ Let B = x, P = 5x Perimeter P = 2(L + B) Area = L × B = 216 Concept/ApproachUse P = 2(L + B) to link L and x; substitute B = x into area to solve for x and then L. Step-by-Step calculation2(L + x) = 5x ⇒ L + x = 2.5x ⇒ L = 1.5xArea: L × B = 1.5x × x = 1.5x2 = 216x2 = 216 / 1.5 = 144 ⇒ x = 12 cm (breadth)Length L = 1.5 × 12 = 18 cm Verification/AlternativePerimeter = 2(18 + 12) = 60; P:B = 60:12 = 5:1 as required. Common pitfallsTaking the 5:1 as L:B instead of P:B. Final Answer18 cm

Question 13

The difference between the length and breadth of a rectangle is 23 m, and its perimeter is 206 m. What is its area?

Accepted Answer

Problem restatementGiven L − B = 23 and perimeter 206, find L, B and hence area. Given data L − B = 23 2(L + B) = 206 ⇒ L + B = 103 Concept/ApproachSolve the two linear equations for L and B, then multiply. Step-by-Step calculationAdding: (L + B) + (L − B) = 103 + 23 ⇒ 2L = 126 ⇒ L = 63Then B = 103 − 63 = 40Area = 63 × 40 = 2,520 m² Verification/AlternativeCheck: Perimeter = 2(63 + 40) = 206; difference = 23. Common pitfallsDividing the perimeter by 2 at the wrong step or swapping L and B in the difference equation; either way, area is unchanged. Final Answer2,520 m²

Question 14

A tank measures 25 m × 12 m × 6 m. If its walls and bottom are plastered at 75 paise per m², what is the total cost?

Accepted Answer

Problem restatementFind the interior surface area of the tank (bottom + four walls) and multiply by the rate to get cost. Given data Length = 25 m, width = 12 m, depth = 6 m. Rate = 75 paise per m² = Rs. 0.75 per m². Concept/ApproachInterior area to plaster = bottom area (L × W) + wall areas (2Lh + 2Wh). Step-by-Step calculationBottom = 25 × 12 = 300 m²Walls = 2(25 × 6) + 2(12 × 6) = 300 + 144 = 444 m²Total area = 300 + 444 = 744 m²Cost = 744 × 0.75 = Rs. 558 Verification/AlternativeUnit check: m² × Rs./m² ⇒ Rs.; arithmetic 744 × 3/4 = 186 × 3 = 558. Common pitfallsIncluding the open top or double-counting a wall; the problem specifies walls and bottom only. Final AnswerRs. 558

Question 15

By what percentage does the area of a rectangle increase if each side is increased by 20%?

Accepted Answer

Problem restatementEach side of a rectangle increases by 20%. Determine the percentage increase in area. Given data Original sides: L, B. New sides: 1.2L, 1.2B. Concept/ApproachArea scales by the product of the side multipliers. Percentage increase = (new factor − 1) × 100%. Step-by-Step calculationOriginal area = LBNew area = (1.2L)(1.2B) = 1.44 LBIncrease factor = 1.44 ⇒ Percentage increase = (1.44 − 1) × 100% = 44% Verification/AlternativeUsing percentage product rule: 20% + 20% + (20% of 20%) = 40% + 4% = 44%. Common pitfallsAdding 20% + 20% only (giving 40%) and neglecting the interaction term. Final Answer44%

Question 16

Area of a square is 1/2 hectare. The diagonal of the square is?

Accepted Answer

Area = 1/2 hectare = 10000 / 2 m2 = 5000 m2 Again Area = 1/2 x (Diagonal)2So 1/2 x (Diagonal)2 = 5000m2 ⇒ Diagonal2= 10000 ∴ Diagonal = 100

Question 17

The base of right-angled triangle is 5 meters and hypotenuse is 13 meters . Its area will be?

Accepted Answer

Altitude = √(13)2 - (5)2 = √144 = 12m ∴ Area of the triangle = (5 x 12 ) / 2 m2 = 30m2

Question 18

The length of a rectangle is double while its breadth is halved. What is the percentage change in area?

Accepted Answer

Let length = l and breadth = b Then, area = lb New length = 2l And new breadth = b/2 ∴ New area = ( 2l ) x (b/2) = lbSo, there is no change in area .

Question 19

The cost of carpeting a room 15 meters long with a carpet 75 cm wide at 30 paisa per meter is Rs. 36. The breadth of the room is?

Accepted Answer

Length of carpet = Total Cost / Rate = 3600 / 30 = 120 mArea of carpet = (120 x 75) / 100 m2 = 90 m2 ∴ Area of the room = 90 m2Breadth of the room = Area /Length = 90 / 15 m = 6m

Question 20

If the side of a square is increased by 25%, then how much percent does its area get increased?

Accepted Answer

Let area 100 m2 Then, side = 10 m New side = 125 % of 10 = (125/100) x 10 = 12.5 m New area = 12.5 x 12.5 m2 =(12.5)2 sq. m∴ Increase in area = (12.5)2 - (10)2 m2 = 22.5 x 2.5 m2 =56.25 m2 % Increase = 56.25 %

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