Question 1

Find circumference from area of a circle The area of a circle is 24.64 m². Compute its circumference (take π ≈ 3.14).

Accepted Answer

Introduction / Context:Converting between a circle’s area and circumference requires finding the radius from area and then applying the circumference formula. Given Data / Assumptions: Area A = 24.64 m^2 π ≈ 3.14 Concept / Approach:Area A = πr^2 ⇒ r = sqrt(A/π). Circumference C = 2πr. Step-by-Step Solution: r = sqrt(24.64 / 3.14) ≈ sqrt(7.8481) ≈ 2.801 m C = 2 * 3.14 * 2.801 ≈ 17.57 m ≈ 17.60 m Verification / Alternative check:Using π = 22/7 gives a close value; rounding will still land near 17.6 m, consistent with the option set. Why Other Options Are Wrong:14.64 m, 16.36 m, and 18.40 m do not match the computed circumference from the given area. Common Pitfalls:Skipping the square root stage or mixing up the formulas for area and circumference can cause major errors. Final Answer:17.60 m

Question 2

In an old Palace there are 50 stairs to go on top floor. Each is 20 cm high, 30 cm wide and 60 cm long. If all the stairs are re-plastered at the rate of Rs. 4 per 100 sq.cm. What is the expenditure ?

Accepted Answer

Total are of all stairs = 50 (20x60 + 30x60) = 50 (1200 + 1800) = 50 x 3000 = 150000 sq cm. Given rate of 100 sq cm = Rs. 4 Then for 150000 sq. cm = ? => 4 x 150000/100 = Rs. 6000

Question 3

Rectangle with given perimeter and offset length The length of a rectangle is 10 cm more than its breadth. If the perimeter is 44 cm, find the length of the diagonal (in cm).

Accepted Answer

Introduction / Context:This problem blends linear relations (perimeter and difference between sides) with the Pythagorean theorem to determine the rectangle’s diagonal. Given Data / Assumptions: Perimeter P = 44 cm Let breadth = b; length = b + 10 Diagonal d = sqrt(l^2 + b^2) Concept / Approach:Perimeter relation: 2(l + b) = 44 ⇒ l + b = 22. Combined with l = b + 10 gives a solvable linear pair for l and b. Then apply Pythagoras for the diagonal. Step-by-Step Solution: l + b = 22 and l = b + 10 ⇒ 2b + 10 = 22 ⇒ b = 6 cm, l = 16 cm d = sqrt(16^2 + 6^2) = sqrt(256 + 36) = sqrt(292) ≈ 17.088 cm Rounded to two decimals ⇒ 17.08 cm Verification / Alternative check:Perimeter check: 2(16 + 6) = 44 (correct). The diagonal magnitude is reasonable for these sides. Why Other Options Are Wrong:12.50 cm, 14.21 cm, and 15.98 cm are not the true space diagonal for sides 16 cm and 6 cm. Common Pitfalls:Mixing up perimeter with area or mistakenly using (l - b) instead of (l + b) to form equations will lead to incorrect side values. Final Answer:17.08 cm

Question 4

Isosceles triangle with altitude and perimeter In an isosceles triangle, the altitude drawn to the base is 8 cm. The perimeter of the triangle is 32 cm. Find the area of the triangle (in cm²).

Accepted Answer

Introduction / Context:For an isosceles triangle, the altitude to the base bisects the base and forms two congruent right triangles. With perimeter given, we can determine the base and hence the area quickly. Given Data / Assumptions: Altitude to base = 8 cm Perimeter = 32 cm Let equal sides be a and base be b Concept / Approach:The altitude splits the base into b/2 and forms a right triangle with legs 8 and b/2 and hypotenuse a. So a = sqrt(8^2 + (b/2)^2). Also 2a + b = 32. Area = (1/2)*b*8 = 4b. Step-by-Step Solution: 2*sqrt(64 + (b/2)^2) + b = 32 Solving gives b = 12 cm; hence Area = 4 * 12 = 48 cm^2 Verification / Alternative check:a = sqrt(64 + 36) = sqrt(100) = 10 cm. Perimeter = 10 + 10 + 12 = 32 cm (consistent). Why Other Options Are Wrong:24, 60, and 72 do not match the derived base and the formula (1/2)*b*h with h = 8 cm. Common Pitfalls:Forgetting that the altitude bisects the base in an isosceles triangle, or confusing perimeter with semiperimeter while substituting values leads to errors. Final Answer:48

Question 5

Percentage error in area from side measurement errors In measuring a rectangle, one side is taken 5% in excess and the other side 4% in deficit. Find the percentage error in the computed area.

Accepted Answer

Introduction / Context:Area of a rectangle is the product of its length and breadth. When each dimension has a percentage error, the approximate percentage error in the product equals the product of the error factors minus one. Given Data / Assumptions: First side taken as 105% of true ⇒ factor 1.05 Second side taken as 96% of true ⇒ factor 0.96 Concept / Approach:Computed area factor = 1.05 * 0.96. Percentage error = (Computed factor − 1) * 100%. Step-by-Step Solution: Computed factor = 1.05 * 0.96 = 1.008 Percentage error = (1.008 − 1) * 100% = 0.8% Verification / Alternative check:Using small-error approximation: +5% and −4% roughly add to about +1%, but the exact multiplicative method yields +0.8%, which is precise. Why Other Options Are Wrong:0.7%, 0.9%, and 0.3% are near misses that arise from additive or rounding approximations rather than correct multiplication of error factors. Common Pitfalls:Simply adding +5% and −4% gives +1% which is not exact; products of measurements require multiplicative treatment. Final Answer:0.8%

Question 6

Four equal circles touching pairwise; area of enclosed central region Four circular cardboard pieces of radius 7 cm are placed so that each piece touches two others, forming a square of centres. Find the area of the region enclosed by the four piece…

Accepted Answer

Introduction / Context:When four equal circles of radius r touch pairwise, their centres form a square of side 2r. The central enclosed region’s area equals the square’s area minus four quarter-circles (i.e., one full circle). Given Data / Assumptions: Radius r = 7 cm Centres form a square of side 2r = 14 cm π ≈ 22/7 or 3.14 (either yields same integer result here) Concept / Approach:Area(enclosed) = Area(square of centres) − Area(4 quarter-circles) = (2r)^2 − πr^2. Step-by-Step Solution: Square area = 14^2 = 196 cm^2 Circle area = π * 7^2 ≈ 22/7 * 49 = 154 cm^2 Enclosed area = 196 − 154 = 42 cm^2 Verification / Alternative check:Using π = 3.14: πr^2 = 3.14 * 49 = 153.86 → 196 − 153.86 ≈ 42.14 ≈ 42 (rounding consistent with option). Why Other Options Are Wrong:12, 32, and 52 cm² do not match the precise geometric decomposition of the region. Common Pitfalls:Confusing the arrangement (thinking of a ring or lens) or subtracting two circles instead of one full circle (formed by four quarters). Final Answer:42 cm²

Question 7

Dividing 700 hectares into two parts with a constraint on difference A large field of 700 hectares is split into two parts. The difference between the areas equals one-fifth of their average. Find the area of the smaller part (in hectares).

Accepted Answer

Introduction / Context:Problems involving sum and difference can be solved using the identities for two numbers: sum S = x + y and difference D = |x − y|, then numbers are (S ± D)/2. Given Data / Assumptions: Total S = 700 Difference D = (1/5) * average = (1/5) * (S/2) = (1/5) * 350 = 70 Concept / Approach:Let larger = (S + D)/2 and smaller = (S − D)/2. Compute both and choose the smaller. Step-by-Step Solution: Larger = (700 + 70)/2 = 770/2 = 385 Smaller = (700 − 70)/2 = 630/2 = 315 Verification / Alternative check:Average = 350; one-fifth = 70; difference between 385 and 315 is indeed 70, satisfying the condition. Why Other Options Are Wrong:Other values do not maintain both the fixed sum of 700 and the required difference of 70 simultaneously. Common Pitfalls:Confusing “one-fifth of the average” with “one-fifth of the sum” or misplacing which number is larger vs smaller is common; always compute both explicitly. Final Answer:315

Question 8

Rhombus with side 8 cm and ∠PQR = 120° If rhombus PQRS has each side 8 cm and ∠PQR = 120°, find the length of diagonal QS (in cm).

Accepted Answer

Introduction / Context:In a rhombus (all sides equal), the diagonals depend on the included angle between sides. There are compact formulas for the diagonals in terms of side s and angle θ between adjacent sides. Given Data / Assumptions: Side s = 8 cm Angle at Q (∠PQR) = 120° Diagonals are PR and QS; one is shorter, one longer. Concept / Approach:For a rhombus with side s and included angle θ, diagonals are: d1 = s * sqrt(2 + 2*cosθ) and d2 = s * sqrt(2 − 2*cosθ) The diagonal opposite the obtuse angle is shorter. With θ = 120°, cosθ = −1/2. Step-by-Step Solution: d_short = s * sqrt(2 + 2*(-1/2)) = s * sqrt(1) = 8 d_long = s * sqrt(2 − 2*(-1/2)) = s * sqrt(3) = 8√3 QS corresponds to the shorter diagonal for the given vertex angle ⇒ 8 cm Verification / Alternative check:In a rhombus, the diagonal through the obtuse angle is shorter; 8√3 (~13.86) and 8 are consistent magnitudes for the pair. Why Other Options Are Wrong:4√5 (~8.94), 6, and 12 do not match either diagonal as per the exact formulas for θ = 120° and s = 8 cm. Common Pitfalls:Mixing up which diagonal is shorter/longer at an obtuse angle or using law of cosines on the wrong triangle can cause confusion. Final Answer:8

Question 9

Rectangle with diagonal and one side in mixed numbers The diagonal of a rectangular closet floor is 7½ feet. The shorter side is 4½ feet. Find the area of the closet (in square feet).

Accepted Answer

Introduction / Context:Using the Pythagorean theorem on a rectangle (right triangle formed by sides and diagonal), we can find the unknown side and then compute the area as product of the two sides. Given Data / Assumptions: Diagonal d = 7.5 ft Shorter side s1 = 4.5 ft Other side s2 unknown Concept / Approach:For rectangle sides s1 and s2, d^2 = s1^2 + s2^2 ⇒ s2 = sqrt(d^2 − s1^2). Area = s1 * s2. Step-by-Step Solution: s2 = sqrt(7.5^2 − 4.5^2) = sqrt(56.25 − 20.25) = sqrt(36) = 6 ft Area = 4.5 * 6 = 27 sq ft Verification / Alternative check:Pythagoras check: 4.5^2 + 6^2 = 20.25 + 36 = 56.25 = 7.5^2 (consistent). Why Other Options Are Wrong:9, 18, and 36 sq ft do not match the exact side computed from the diagonal relation. Common Pitfalls:Misreading mixed numbers (7 1/2 as 7.12) or arithmetic mistakes when squaring/ subtracting undermine accuracy. Final Answer:27

Question 10

Square and rhombus on the same base (and between the same parallels) If a square and a rhombus stand on the same base and between the same parallels, what is the ratio of their areas?

Accepted Answer

Introduction / Context:The classical result “on the same base and between the same parallels” implies equal heights to that base. Figures with the same base and equal height have equal areas (Area = base * height for all parallelogram-type figures). Given Data / Assumptions: Square and rhombus share the same base segment. Both lie between the same pair of parallels ⇒ same altitude to that base. Concept / Approach:Area(square) = base * common height. Area(rhombus) = base * the same common height. Hence, areas are equal. Step-by-Step Solution: Area ratio = (b * h) : (b * h) = 1 : 1 Verification / Alternative check:This mirrors the familiar result for triangles and parallelograms on the same base and between the same parallels; equality of heights forces equal areas. Why Other Options Are Wrong:Any ratio other than 1:1 contradicts the shared base and equal height condition. Common Pitfalls:Overlooking the phrase “between the same parallels” (often assumed) may lead to thinking the areas must differ; that phrase is essential for equality. Final Answer:1:1

Question 11

Number of fence poles around a rectangle A rectangular plot measures 90 m by 50 m. It is to be enclosed by a wire fence with poles placed every 5 m (all around). How many poles are needed?

Accepted Answer

Introduction / Context:When equally spaced poles are placed around a closed loop with total length a multiple of the spacing, the count equals total length divided by the spacing. Given Data / Assumptions: Length = 90 m, Breadth = 50 m ⇒ Perimeter P = 2*(90 + 50) = 280 m Spacing = 5 m Perimeter is an exact multiple of spacing. Concept / Approach:If P is divisible by the spacing, number of poles N = P / spacing (because start and end coincide on a looped fence). Step-by-Step Solution: N = 280 / 5 = 56 poles Verification / Alternative check:Positions every 5 m around 280 m yield exactly 56 intervals/poles, closing the loop without overlap. Why Other Options Are Wrong:45, 55, and 65 do not equal the exact division result for a multiple-of-5 perimeter. Common Pitfalls:Double-counting the initial pole or assuming an extra pole at closure; when P is a multiple of spacing, simple division suffices. Final Answer:56

Question 12

Length of rope from grazing area of a cow A cow grazes a circular region of area 2826 m². Find the required rope length (radius) in metres (use π ≈ 3.14).

Accepted Answer

Introduction / Context:Converting a given circular area to radius is a standard application of A = πr^2, widely used in practical geometry and land measurement problems. Given Data / Assumptions: Area A = 2826 m^2 π ≈ 3.14 Concept / Approach:Radius r = sqrt(A/π). The rope length equals this radius. Step-by-Step Solution: r^2 = 2826 / 3.14 = 900 r = sqrt(900) = 30 m Verification / Alternative check:Check area with r = 30: A = 3.14 * 900 = 2826 m^2, matching perfectly. Why Other Options Are Wrong:15 m, 24 m, and 36 m would yield areas not equal to 2826 m^2 with π = 3.14. Common Pitfalls:Using circumference formula instead of area or mishandling the square root step are frequent mistakes. Final Answer:30 m

Question 13

Scaled carpet order and cost comparison An initial carpet has breadth 6 m and length 1.44 times the breadth. A second carpet has its length increased by 40% and breadth increased by 25% over the first. If cost is ₹45 per m², what is the cost of the …

Accepted Answer

Introduction / Context:This involves proportional scaling of dimensions followed by area and cost computation. Percentage increases act as multiplicative factors on the original dimensions. Given Data / Assumptions: Original breadth B = 6 m; original length L = 1.44 * 6 = 8.64 m New length = 1.4 * 8.64 = 12.096 m New breadth = 1.25 * 6 = 7.5 m Rate = ₹45 per m^2 Concept / Approach:Compute new area and multiply by rate to obtain the total cost. Step-by-Step Solution: Area(new) = 12.096 * 7.5 = 90.72 m^2 Cost = 90.72 * 45 = ₹4,082.40 Verification / Alternative check:Original area = 8.64 * 6 = 51.84 m^2. Scale factors 1.4 and 1.25 multiply area by 1.4 * 1.25 = 1.75, and 51.84 * 1.75 = 90.72 (consistent). Why Other Options Are Wrong:The other amounts do not align with the precise area scaling and rate multiplication. Common Pitfalls:Adding percentages instead of multiplying scale factors, or applying the rate to a single dimension instead of the area. Final Answer:₹4,082.40

Question 14

Perimeter change after scaling rectangle sides In a rectangle, length is three times the breadth. If the length and breadth are increased by 30% and 100% respectively, by what percentage does the perimeter increase?

Accepted Answer

Introduction / Context:Scaling linear dimensions affects perimeter linearly. Given a fixed relation between length and breadth, we can express the perimeter in terms of one variable and compare before/after values. Given Data / Assumptions: Let breadth = x, length = 3x Original perimeter P = 2(3x + x) = 8x New length = 1.3 * 3x = 3.9x New breadth = 2x Concept / Approach:Perimeter scales with the sum of side lengths; compute the new perimeter and compare to the original. Step-by-Step Solution: P(new) = 2(3.9x + 2x) = 2 * 5.9x = 11.8x Increase% = [(11.8x − 8x) / 8x] * 100% = (3.8 / 8) * 100% = 47.5% Verification / Alternative check:Pick x = 10: P(old) = 80; P(new) = 118; increase = 38 on 80 = 47.5%, confirming the ratio result. Why Other Options Are Wrong:20%, 25%, and 27% underestimate the effect because both sides scale appreciably; the exact linear computation gives 47.5%. Common Pitfalls:Confusing perimeter (linear) with area (quadratic) scaling, or adding 30% and 100% naively to get 130% which is unrelated to perimeter percent change. Final Answer:47.5%

Question 15

Gravel path inside a rectangular lawn (unit corrections applied) A rectangular grassy plot measures 110 m by 65 m. A 0.5 m wide gravel path runs all around the inside. Find the cost of gravelling the path at ₹0.80 per m².

Accepted Answer

Introduction / Context:When a uniform path runs inside a rectangle, the path area equals outer area minus inner rectangle area. Here, obvious unit typos are interpreted reasonably (metres throughout, 0.5 m path). Given Data / Assumptions: Outer dimensions: 110 m by 65 m Path width w = 0.5 m inside on all sides Rate = ₹0.80 per m^2 (80 paise per m^2) Concept / Approach:Inner rectangle dimensions reduce by 2w in each direction. Path area = Outer − Inner. Step-by-Step Solution: Outer area = 110 * 65 = 7,150 m^2 Inner dimensions = (110 − 1) * (65 − 1) = 109 * 64 Inner area = 6,976 m^2 Path area = 7,150 − 6,976 = 174 m^2 Cost = 174 * 0.80 = ₹139.20 Verification / Alternative check:Approximate path area also equals perimeter * width − 4 * (w^2) for thin paths: 2(110+65)*0.5 − 4*(0.25) = 350*0.5 − 1 = 175 − 1 = 174 m^2, same result. Why Other Options Are Wrong:₹340, ₹320, and ₹480 arise from misread unit rates or dimensions; only ₹139.20 matches the corrected interpretation and exact calculation. Common Pitfalls:Using centimetres inadvertently or treating the cost as ₹80 per m^2 instead of ₹0.80 per m^2 drastically overstates the amount. Final Answer:₹139.20

Question 16

Relating areas, bases, and altitudes of two triangles The ratio of the bases of two triangles is x:y and the ratio of their areas is a:b. Find the ratio of their corresponding altitudes.

Accepted Answer

Introduction / Context:Area of a triangle equals (1/2) * base * altitude. Comparing two triangles lets us relate the ratios of areas, bases, and altitudes algebraically. Given Data / Assumptions: Base ratio b1:b2 = x:y Area ratio A1:A2 = a:b Altitudes h1 and h2 correspond to b1 and b2 respectively. Concept / Approach:Since A ∝ base * altitude, we have A1/A2 = (b1 * h1)/(b2 * h2). Rearranging gives the altitude ratio in terms of given ratios. Step-by-Step Solution: a/b = (x * h1) / (y * h2) ⇒ h1/h2 = (a/b) * (y/x) = ay / (bx) Verification / Alternative check:Plug numbers: if x=2, y=3, a=4, b=5, then h1/h2 = (4*3)/(5*2) = 12/10 = 6/5, consistent with formula. Why Other Options Are Wrong:a:b ignores base effects; x:y ignores area differences; bx:ay is the inverse of the correct relation. Common Pitfalls:Forgetting that the 1/2 factor cancels out or inverting the ratios during cross-multiplication leads to wrong answers. Final Answer:ay:bx

Question 17

If each side of a square is increased by 25%, find the percentage change in its area.

Accepted Answer

Let each side of the square be a. Then, area = . a 2 New side = 125 a 100 = 5 a 4 . New area = 5 a 4 2 = 25 a 2 16 Increase in area = 25 a 2 16 - a 2 = 9 a 2 16 Increase% = 9 a 2 16 * 1 a 2 * 100 % = 56.25%.

Question 18

If the perimeters of a square and a rectangle are the same, then the area a & b enclosed by them would satisfy the condition:

Accepted Answer

Take a square of side 4cm and a rectangle having l=6cm and b=2cm then perimeter of square = perimeter of rectangle area of square = 16 sq.cm area of rectangle = 12 sq.cm Hence a >b

Question 19

The area of a square field is 69696sq.cm..its diagonal will be equal to

Accepted Answer

d = 2 * s = > d = 2 * 69696

Question 20

A rectangular parking space is marked out by painting three of its sides.If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, then what is the area of the parking space in square feet?

Accepted Answer

Let l = 9 ft. Then l + 2b = 37 => 2b = 37 – l = 37 – 9 = 28 b = 28/2 = 14 ft. Area = lb = 9 × 14 = 126 sq. ft.

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