Question 1

In two triangles, the ratio of their areas is 4:3, and the ratio of their corresponding heights is 3:4. What is the ratio of their corresponding bases?

Accepted Answer

Introduction / Context:This question tests proportional reasoning using the triangle area formula. The area of a triangle is (1/2)*base*height. When comparing two triangles, the factor 1/2 cancels in the ratio, so the area ratio depends on the product of base and height. If the area ratio and height ratio are known, the base ratio can be isolated by dividing the area ratio by the height ratio (in multiplicative terms). This is a classic “ratio of products” problem and does not require any absolute measurements. Given Data / Assumptions: Area1:Area2 = 4:3 Height1:Height2 = 3:4 Area proportionality: Area is proportional to base*height Concept / Approach:Use: Area1/Area2 = (b1*h1)/(b2*h2). Rearrange to get: b1/b2 = (Area1/Area2) * (h2/h1). Substitute the ratios and simplify. Step-by-Step Solution: Area1/Area2 = (b1*h1)/(b2*h2) So (4/3) = (b1/b2) * (h1/h2) Given h1/h2 = 3/4 (4/3) = (b1/b2) * (3/4) b1/b2 = (4/3) * (4/3) = 16/9 Base ratio = 16:9 Verification / Alternative check:Pick b1:b2 = 16:9 and h1:h2 = 3:4. Then (b1*h1):(b2*h2) = (16*3):(9*4) = 48:36 = 4:3, matching the area ratio, so the result is consistent. Why Other Options Are Wrong: 13:9, 14:9, 15:9 are close-looking but do not produce exact 4:3 when combined with 3:4 heights. 9:16 is the inverse of the correct ratio. Common Pitfalls:Forgetting that area is proportional to base*height, reversing height ratio, or incorrectly adding ratios instead of multiplying/dividing them. Final Answer:The ratio of their bases is 16:9.

Question 2

The area of a circular field is 13.86 hectares. Find the cost of fencing the field at the rate of Rs 4.40 per metre (assume pi = 22/7).

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Introduction / Context:This question tests circle mensuration with unit conversion and a practical cost calculation. The fencing required equals the circumference of the field. The area is given in hectares, so we must convert to square metres before using the area formula A = pi*r^2 to find the radius. Once radius is known, circumference C = 2*pi*r gives the fence length in metres. Finally, multiply by the rate (Rs per metre) to get the total cost. Using pi = 22/7 helps produce neat numbers if the data is designed for it. Given Data / Assumptions: Area = 13.86 hectares 1 hectare = 10000 sq m pi = 22/7 Fencing rate = Rs 4.40 per metre Concept / Approach:Convert area to sq m, solve for r from A = pi*r^2, compute circumference C = 2*pi*r, then cost = C * 4.40. Step-by-Step Solution: Area in sq m = 13.86 * 10000 = 138600 A = (22/7)*r^2 => r^2 = 138600 * 7 / 22 138600/22 = 6300, so r^2 = 6300*7 = 44100 r = sqrt(44100) = 210 m Circumference C = 2*(22/7)*210 = 1320 m Cost = 1320 * 4.40 = 5808 Verification / Alternative check:Check area: (22/7)*210^2 = (22/7)*44100 = 22*6300 = 138600 sq m, which matches the converted area, so r and C are consistent. Why Other Options Are Wrong: 2808, 3808, 4808 occur if circumference or rate multiplication is done incorrectly. 6808 is too high and suggests using a larger circumference or a wrong unit conversion. Common Pitfalls:Forgetting hectare conversion, using diameter as radius, using C = pi*r instead of 2*pi*r, or multiplying by Rs 4.40 incorrectly. Final Answer:The cost of fencing is Rs 5808.

Question 3

A semicircular shaped window has a diameter of 63 cm. Find its perimeter (in cm), including the curved part and the diameter (assume pi = 22/7).

Accepted Answer

Introduction / Context:This question tests perimeter of a semicircle. A semicircle's boundary consists of two parts: the curved arc (which is half the circumference of the full circle) and the straight diameter line. Many mistakes happen when learners compute only the curved arc and forget to add the diameter, or they use radius incorrectly. Since diameter is given, first find radius, compute the semicircular arc length as pi*r, then add the diameter to get total perimeter. Units remain in centimetres throughout, and pi = 22/7 is used for neat arithmetic. Given Data / Assumptions: Diameter d = 63 cm Radius r = d/2 = 31.5 cm Perimeter of semicircle = (arc length) + diameter Arc length of semicircle = (1/2)*2*pi*r = pi*r Concept / Approach:Compute arc = pi*r and then add the diameter. Use pi = 22/7 to evaluate pi*r accurately. Step-by-Step Solution: r = 63/2 = 31.5 cm Arc length = pi*r = (22/7)*31.5 31.5/7 = 4.5, so arc = 22*4.5 = 99 cm Perimeter = arc + diameter = 99 + 63 = 162 cm Verification / Alternative check:Half circumference approach: full circumference = pi*d = (22/7)*63 = 198 cm, half is 99 cm, then add diameter 63 cm gives 162 cm. Same result confirms correctness. Why Other Options Are Wrong: 99 (not listed) would be the arc only; 132 and 142 come from wrong pi usage or wrong r. 172 is too large and suggests adding extra length beyond arc + diameter. Common Pitfalls:Forgetting to include diameter, using d instead of r in pi*r, or calculating half circumference incorrectly. Final Answer:The perimeter of the semicircular window is 162 cm.

Question 4

A triangle and a parallelogram are constructed on the same base, and their areas are equal. If the altitude (height) of the parallelogram is 100 m, what is the altitude (height) of the triangle?

Accepted Answer

Introduction / Context:This question tests area formulas for a triangle and a parallelogram when both share the same base. The area of a parallelogram is base * height, while the area of a triangle is (1/2) * base * height. If they are built on the same base and have equal areas, the only way to compensate for the factor 1/2 in the triangle formula is for the triangle's altitude to be twice the parallelogram's altitude. This is a direct relationship question and does not require any actual base value. Given Data / Assumptions: Triangle and parallelogram have the same base Areas are equal Parallelogram altitude = 100 m Area(parallelogram) = base * height Area(triangle) = (1/2) * base * height Concept / Approach:Set areas equal: base*Hp = (1/2)*base*Ht. Cancel base and solve for Ht in terms of Hp. Step-by-Step Solution: base*Hp = (1/2)*base*Ht Cancel base (same and non-zero): Hp = (1/2)*Ht Ht = 2*Hp Ht = 2*100 = 200 m Verification / Alternative check:If base is any value b, parallelogram area = b*100. Triangle with height 200 has area = (1/2)*b*200 = b*100, exactly equal. This confirms the relationship is correct for any base. Why Other Options Are Wrong: 100 m would make the triangle area half of the parallelogram area. 300 m and 400 m make triangle area larger than the parallelogram. 150 m does not offset the 1/2 factor correctly. Common Pitfalls:Forgetting the 1/2 in triangle area, assuming triangle and parallelogram heights must be equal, or mixing up base and height roles. Final Answer:The altitude of the triangle is 200 m.

Question 5

Find the area of an obtuse-angled triangle where two sides are 8 units and 12 units, and the included angle between them is 150 degrees. What is the area (in square units)?

Accepted Answer

Introduction / Context:This question tests the triangle area formula using two sides and the included angle. When you know two sides a and b and the angle C between them, area can be found by: Area = (1/2)*a*b*sin(C). This formula works for acute, right, and obtuse triangles. Even though the angle is 150 degrees (obtuse), the sine value is still positive, so the area remains positive. A helpful fact is sin(150 degrees) = sin(30 degrees) = 1/2. Using this identity makes the calculation quick and exact. Given Data / Assumptions: Side a = 8 units Side b = 12 units Included angle C = 150 degrees Area formula: A = (1/2)*a*b*sin(C) Concept / Approach:Use A = (1/2)*a*b*sin(C). Replace sin(150) with 1/2, multiply the numbers, and express the final area in square units. Step-by-Step Solution: A = (1/2) * 8 * 12 * sin(150) sin(150) = sin(30) = 1/2 A = (1/2) * 96 * (1/2) A = 48 * (1/2) = 24 sq units Verification / Alternative check:Since 150 degrees is obtuse, the height relative to side 12 is 8*sin(150)=8*(1/2)=4. Area = (1/2)*base*height = (1/2)*12*4 = 24 sq units, matching the result. Why Other Options Are Wrong: 48 occurs if sin(150) is incorrectly taken as 1. 12 and 6 come from missing one of the (1/2) factors. 36 comes from using a wrong sine value or wrong base-height pairing. Common Pitfalls:Using cos instead of sin, treating sin(150) as negative, or forgetting the initial (1/2) factor in the formula. Final Answer:The area of the triangle is 24 sq units.

Question 6

A sector of 120 degrees cut out from a circle has an area of 66/7 sq cm. Find the radius of the circle (in cm), assuming pi = 22/7.

Accepted Answer

Introduction / Context:This question tests the area of a sector of a circle. A sector is a fraction of a full circle based on the central angle. If the central angle is theta degrees, then sector area = (theta/360)*pi*r^2. Here theta is 120 degrees, so the sector is exactly one-third of the full circle. With the sector area known, we can set up an equation in r^2 and solve for the radius. Using pi = 22/7 is intended to keep the arithmetic exact and clean. Given Data / Assumptions: Sector angle theta = 120 degrees Sector area = 66/7 sq cm pi = 22/7 Sector area formula: (theta/360)*pi*r^2 Concept / Approach:Use: (120/360)*pi*r^2 = 66/7. Simplify 120/360 to 1/3, substitute pi = 22/7, solve for r^2, then take square root to get r in cm. Step-by-Step Solution: (120/360) = 1/3 (1/3) * (22/7) * r^2 = 66/7 Multiply both sides by 3: (22/7)*r^2 = 198/7 Cancel /7: 22*r^2 = 198 r^2 = 198/22 = 9 r = sqrt(9) = 3 cm Verification / Alternative check:Full circle area would be 3 times sector area (since 120 degrees is 1/3 of 360): full area = 3*(66/7) = 198/7. Using r=3 gives pi*r^2 = (22/7)*9 = 198/7. Perfect match confirms the radius. Why Other Options Are Wrong: 1 cm and 2 cm give much smaller areas because area grows with r^2. 4 cm and 6 cm produce areas larger than 66/7 for the 120-degree sector. Common Pitfalls:Using theta as 120 radians (wrong), forgetting the (theta/360) factor, or not squaring r properly when rearranging. Final Answer:The radius of the circle is 3 cm.

Question 7

A room is 5.5 m long and 3.75 m wide. Find the cost of paving its floor with slabs at the rate of Rs 800 per sq m.

Accepted Answer

Introduction / Context:This problem tests rectangle area and a simple cost calculation. The floor of a rectangular room is a rectangle, so its area is length * width. Once area is obtained in square metres, the paving cost is found by multiplying the area by the given rate in Rs per square metre. The main care points are correct multiplication of decimals and ensuring that the units match (metres and square metres). Since the rate is per sq m, the area must be in sq m before multiplying. Given Data / Assumptions: Length = 5.5 m Width = 3.75 m Rate = Rs 800 per sq m Floor is rectangular: Area = L*W Concept / Approach:Compute area in sq m using L*W, then cost = area * 800. Keep arithmetic exact by multiplying first, then scaling by 800. Step-by-Step Solution: Area = 5.5 * 3.75 = 20.625 sq m Cost = 20.625 * 800 20.625 * 800 = 20.625 * (8*100) = (20.625*8)*100 20.625*8 = 165 Cost = 165 * 100 = Rs 16500 Verification / Alternative check:Since 3.75 is close to 4, approximate area is 5.5*4=22. Cost would be around 22*800=17600. The exact width is slightly smaller than 4, so the exact cost should be slightly less than 17600. Rs 16500 fits this logic. Why Other Options Are Wrong: Rs 15500 and Rs 16000 come from underestimating the area or multiplying 800 incorrectly. Rs 17500 and Rs 18500 come from rounding area up too much or using wrong dimensions. Common Pitfalls:Adding dimensions instead of multiplying, using sq m rate on an area mistakenly computed in sq cm, or mistakes in decimal multiplication. Final Answer:The cost of paving the floor is Rs 16500.

Question 8

What is the length of the longest pole that can be kept inside a room that is 12 m long, 4 m broad, and 3 m high? (Assume the pole can be placed diagonally in 3D space.)

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Introduction / Context:This question tests the concept of the space diagonal of a cuboid (rectangular box). The longest straight object that can fit inside a rectangular room is the space diagonal connecting two opposite corners in 3D. To find this length, we use the 3D Pythagoras theorem: space diagonal = sqrt(l^2 + b^2 + h^2). It is essentially applying Pythagoras twice: first to get the floor diagonal, then combining with height. Because the room dimensions are given in metres, the resulting pole length is also in metres. Given Data / Assumptions: Length l = 12 m Breadth b = 4 m Height h = 3 m Longest pole length = space diagonal = sqrt(l^2 + b^2 + h^2) Concept / Approach:Compute l^2 + b^2 + h^2, then take the square root. If the sum is a perfect square, the diagonal is an integer. Step-by-Step Solution: Space diagonal d = sqrt(12^2 + 4^2 + 3^2) 12^2 = 144, 4^2 = 16, 3^2 = 9 Sum = 144 + 16 + 9 = 169 d = sqrt(169) = 13 m Verification / Alternative check:Floor diagonal = sqrt(12^2 + 4^2) = sqrt(160). Then space diagonal = sqrt(160 + 3^2) = sqrt(169) = 13. Same result confirms correctness. Why Other Options Are Wrong: 14, 15, 16 are larger than the computed space diagonal and will not fit corner-to-corner. 12.5 can come from using only the floor diagonal and ignoring height. Common Pitfalls:Using only 2D diagonal of the floor, adding dimensions instead of squaring and rooting, or mixing up breadth and height values. Final Answer:The longest pole length is 13 m.

Question 9

The diagonals of two squares are in the ratio 2:5. Find the ratio of their areas.

Accepted Answer

Introduction / Context:This question tests similarity scaling for squares. In a square, the diagonal is directly proportional to the side length (diagonal = side*sqrt(2)). That means if diagonals are in a given ratio, the side lengths are in the same ratio. Area of a square is proportional to the square of its side length (Area = side^2). Therefore, the ratio of areas equals the square of the ratio of diagonals. This is a classic “linear ratio to area ratio” conversion problem. Given Data / Assumptions: Diagonal1:Diagonal2 = 2:5 For squares, side is proportional to diagonal Area is proportional to (side)^2 Concept / Approach:If d1:d2 = 2:5, then s1:s2 = 2:5. Hence Area1:Area2 = s1^2:s2^2 = 2^2:5^2 = 4:25. Step-by-Step Solution: d1:d2 = 2:5 Since d = s*sqrt(2), side ratio s1:s2 = d1:d2 = 2:5 Area ratio = s1^2:s2^2 = 2^2:5^2 Area ratio = 4:25 Verification / Alternative check:Assume diagonals are 2 and 5. Then sides are 2/sqrt(2) and 5/sqrt(2). Areas become (2^2/2)=2 and (5^2/2)=12.5. Ratio 2:12.5 simplifies to 4:25, matching the result. Why Other Options Are Wrong: 2:5 is the diagonal ratio, not the area ratio. 3:25, 3:15, and 5:25 come from wrong squaring or partial simplification errors. Common Pitfalls:Forgetting to square the ratio for areas, mixing up diagonal ratio with side ratio, or simplifying incorrectly after squaring. Final Answer:The ratio of their areas is 4:25.

Question 10

An equilateral triangle is constructed on the diagonal of a square (the diagonal is the side of the equilateral triangle). What is the ratio of the area of the triangle to the area of the square?

Accepted Answer

Introduction / Context:This question tests linking two shapes using a shared measurement and then comparing their areas. The diagonal of the square becomes the side of the equilateral triangle. For a square with side s, diagonal is s*sqrt(2). For an equilateral triangle with side a, area is (sqrt(3)/4)*a^2. Square area is s^2. The ratio of areas is found by expressing the triangle side a in terms of s (using the diagonal relationship), substituting into the triangle area formula, and then dividing by s^2. The result is a clean symbolic ratio involving sqrt(3). Given Data / Assumptions: Square side = s Square diagonal = s*sqrt(2) Equilateral triangle side a = square diagonal Area(square) = s^2 Area(equilateral triangle) = (sqrt(3)/4)*a^2 Concept / Approach:Let a = s*sqrt(2). Then triangle area = (sqrt(3)/4)*(s^2*2) = (sqrt(3)/2)*s^2. Ratio (triangle:square) = (sqrt(3)/2)*s^2 : s^2 = sqrt(3):2. Step-by-Step Solution: Let square side = s Diagonal = s*sqrt(2) Triangle side a = s*sqrt(2) Area(triangle) = (sqrt(3)/4)*a^2 a^2 = (s*sqrt(2))^2 = 2*s^2 Area(triangle) = (sqrt(3)/4)*2*s^2 = (sqrt(3)/2)*s^2 Area(square) = s^2 Ratio = (sqrt(3)/2)*s^2 : s^2 = sqrt(3):2 Verification / Alternative check:If s=1, square area is 1. Diagonal is sqrt(2), triangle area becomes (sqrt(3)/4)*2 = sqrt(3)/2. Ratio is sqrt(3)/2 : 1 which corresponds to sqrt(3):2. This confirms the symbolic ratio is consistent. Why Other Options Are Wrong: 1:2, 1:3, 2:3 are plain fractions and ignore the sqrt(3) factor from equilateral triangle area. 3:2 incorrectly replaces sqrt(3) with 3. Common Pitfalls:Using triangle area as (1/2)*base*height without correct height, forgetting diagonal = s*sqrt(2), or squaring sqrt(2) incorrectly. Final Answer:The ratio of the area of the triangle to the area of the square is sqrt3:2.

Question 11

A takes 15 seconds to cross a rectangular field diagonally while walking at 52 m/min. B takes the same 15 seconds to cross the same field along its sides (from one corner to the opposite corner via two adjacent sides) while walking at 68 m/min. What…

Accepted Answer

Introduction / Context:This problem combines speed-time-distance with rectangle geometry. Person A walks along the diagonal, so the distance covered equals the diagonal length. Person B walks along the sides from one corner to the opposite corner, so the distance covered equals length + breadth. Since both take the same time, we can calculate these distances from their speeds. With diagonal and (length + breadth) known, we use the identity: (length + breadth)^2 = length^2 + breadth^2 + 2*length*breadth. Here length^2 + breadth^2 equals diagonal^2. This allows us to solve for length*breadth, which is the area. Given Data / Assumptions: Time for both = 15 seconds A speed = 52 m/min B speed = 68 m/min Diagonal distance = A speed * time Length + breadth distance = B speed * time Concept / Approach:Convert speeds to m/s or convert time to minutes. Compute diagonal d and sum S = L + B. Use S^2 = d^2 + 2LB to find LB (area). Step-by-Step Solution: Time = 15 sec = 15/60 = 0.25 min Diagonal d = 52 * 0.25 = 13 m Sum of sides S = 68 * 0.25 = 17 m S^2 = (L + B)^2 = 17^2 = 289 d^2 = L^2 + B^2 = 13^2 = 169 (L + B)^2 = L^2 + B^2 + 2LB => 289 = 169 + 2LB 2LB = 120 => LB = 60 Area = L*B = 60 sq m Verification / Alternative check:If area is 60, one valid pair is L=12 and B=5 (since 12*5=60). Then L+B=17 and diagonal sqrt(12^2+5^2)=sqrt(169)=13. Both distances match the computed S and d, confirming the area. Why Other Options Are Wrong: 50, 70, 80, 90 do not satisfy the derived equation LB = (S^2 - d^2)/2. They would require different S or d than what the speeds and time produce. Common Pitfalls:Not converting seconds to minutes, assuming B's path is perimeter instead of L+B, or trying to solve L and B separately when only area is needed. Final Answer:The area of the field is 60 sq m.

Question 12

Find the circumference of a circle whose area is 24.64 sq m (assume pi = 22/7). What is the circumference (in m)?

Accepted Answer

Introduction / Context:This question tests the connection between a circle's area and its circumference. The area formula is A = pi*r^2 and the circumference formula is C = 2*pi*r. Since area is given, we first solve for r by rearranging the area formula. Then we substitute r into the circumference formula. Because pi is specified as 22/7, the values are designed to produce a neat radius. Units also matter: area is in square metres, so radius comes out in metres, and circumference will be in metres. Given Data / Assumptions: Area A = 24.64 sq m pi = 22/7 A = pi*r^2 C = 2*pi*r Concept / Approach:Compute r^2 = A/pi, then r = sqrt(r^2). Finally compute C = 2*pi*r. Keep calculations exact using pi = 22/7. Step-by-Step Solution: r^2 = A / pi = 24.64 / (22/7) = 24.64 * 7 / 22 24.64/22 = 1.12, so r^2 = 1.12 * 7 = 7.84 r = sqrt(7.84) = 2.8 m C = 2*pi*r = 2*(22/7)*2.8 C = (44/7)*2.8 = 44*0.4 = 17.6 m So circumference = 17.60 m Verification / Alternative check:Check area with r=2.8: A = (22/7)*2.8^2 = (22/7)*7.84 = 22*1.12 = 24.64 sq m, exactly matching the given area. So circumference is consistent. Why Other Options Are Wrong: 14.64 and 16.36 come from incorrect radius extraction or wrong pi placement. 18.40 comes from using r=2.93 approx or arithmetic slips in 2*pi*r. 15.40 is not compatible with the radius implied by the area. Common Pitfalls:Using C = pi*d without finding d, mixing up pi*r^2 and 2*pi*r, or rounding r too early before computing circumference. Final Answer:The circumference of the circle is 17.60 m.

Question 13

In an old palace there are 50 stairs to reach the top floor. Each stair 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 (plastering the tread and riser surfaces), what is the total expe…

Accepted Answer

Introduction / Context:This is a surface area and cost problem involving repeated identical objects (stairs). For re-plastering steps, the commonly plastered exposed faces are the top horizontal surface (tread) and the front vertical surface (riser). Using the given dimensions, we compute the area of these faces for one stair in square centimetres, multiply by the number of stairs to get total plaster area, then apply the rate given per 100 sq cm. The main skill is careful area calculation with consistent units, and applying the rate correctly by grouping the total area into blocks of 100 sq cm. Given Data / Assumptions: Number of stairs = 50 Height (riser) = 20 cm Width = 30 cm Length = 60 cm Plaster surfaces: tread (width*length) and riser (height*length) Rate = Rs 4 per 100 sq cm Concept / Approach:Area per stair = (tread area) + (riser area). Total area = (area per stair) * 50. Cost = (total area / 100) * 4. Step-by-Step Solution: Tread area per stair = 30*60 = 1800 sq cm Riser area per stair = 20*60 = 1200 sq cm Total plaster area per stair = 1800 + 1200 = 3000 sq cm Total plaster area for 50 stairs = 3000*50 = 150000 sq cm Number of 100 sq cm blocks = 150000/100 = 1500 Cost = 1500 * 4 = Rs 6000 Verification / Alternative check:Rate per sq cm is Rs 4/100 = Rs 0.04. Multiplying 150000 * 0.04 also gives Rs 6000, confirming the same result by an alternate method. Why Other Options Are Wrong: Rs 4680 and Rs 5480 come from missing one face (tread or riser) or using wrong stair count. Rs 7600 and Rs 7000 result from adding extra faces or applying rate per sq cm incorrectly. Common Pitfalls:Mixing up width and length, forgetting to multiply by 50 stairs, or treating Rs 4 per 100 sq cm as Rs 4 per sq cm. Final Answer:The total expenditure is Rs 6000.

Question 14

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

Accepted Answer

Introduction / Context:This question combines rectangle perimeter with the Pythagoras theorem for the diagonal. First, perimeter gives a relationship between length and breadth: P = 2*(L + B). With the extra information that L is 10 cm more than B, we can solve for both sides exactly. Then the diagonal is the hypotenuse of a right triangle whose legs are L and B, so diagonal d = sqrt(L^2 + B^2). The only approximation happens at the final square root, which is why the answer is given to two decimal places. Given Data / Assumptions: Length L = B + 10 cm Perimeter P = 44 cm Perimeter formula: 2*(L + B) = 44 Diagonal formula: d = sqrt(L^2 + B^2) Concept / Approach:Use perimeter to solve for L and B, then use Pythagoras to compute the diagonal. Keep units in cm throughout. Step-by-Step Solution: 2*(L + B) = 44 => L + B = 22 L = B + 10 Substitute: (B + 10) + B = 22 => 2B = 12 => B = 6 cm L = 6 + 10 = 16 cm Diagonal d = sqrt(16^2 + 6^2) = sqrt(256 + 36) = sqrt(292) sqrt(292) ≈ 17.08 cm Verification / Alternative check:Check perimeter: 2*(16+6)=2*22=44 cm, correct. Since diagonal must be longer than the longer side 16 cm and shorter than 16+6=22 cm, 17.08 cm is a reasonable value in the correct range. Why Other Options Are Wrong: 12.50 and 14.21 are too small (even smaller than the length 16 cm in some cases). 15.98 comes from using wrong sides or rounding errors. 16.00 would imply a square-like diagonal and does not fit L=16, B=6. Common Pitfalls:Forgetting to divide perimeter by 2, mixing up which side is 10 cm longer, or adding squares incorrectly before taking square root. Final Answer:The diagonal of the rectangle is approximately 17.08 cm.

Question 15

The diagonal of the floor of a rectangular closet is 7 1/2 feet, and the shorter side is 4 1/2 feet. What is the area of the closet floor (in square feet)?

Accepted Answer

Introduction / Context:This question tests rectangle geometry using the Pythagoras theorem. The diagonal of a rectangle forms a right triangle with the two sides as perpendicular legs. If the diagonal and one side are known, the other side can be found using: other_side = sqrt(diagonal^2 - known_side^2). Once both sides are known, area is simply side1 * side2. The values are mixed fractions (7 1/2 and 4 1/2), so converting them to decimals or improper fractions first makes calculations easier and avoids mistakes. Given Data / Assumptions: Diagonal d = 7 1/2 ft = 7.5 ft Shorter side s = 4 1/2 ft = 4.5 ft Other side = sqrt(d^2 - s^2) Area = side1 * side2 Concept / Approach:Use the right triangle relation: d^2 = a^2 + b^2. With d and one side known, solve for the other side, then multiply both sides for area. Step-by-Step Solution: d = 7.5, s = 4.5 Other side = sqrt(7.5^2 - 4.5^2) 7.5^2 = 56.25 4.5^2 = 20.25 Difference = 56.25 - 20.25 = 36 Other side = sqrt(36) = 6 ft Area = 4.5 * 6 = 27 square feet Verification / Alternative check:Check the diagonal using found sides: sqrt(4.5^2 + 6^2) = sqrt(20.25 + 36) = sqrt(56.25) = 7.5, which matches the given diagonal, so the sides are correct. Why Other Options Are Wrong: 9 and 18 are too small and would imply much smaller side lengths. 36 is too large and would require a larger second side than 6 ft. 30 comes from incorrect multiplication or wrong square subtraction. Common Pitfalls:Squaring mixed numbers incorrectly, adding squares instead of subtracting to find the missing side, or using diagonal as if it were a side. Final Answer:The area of the closet floor is 27 square feet.

Question 16

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

Accepted Answer

Introduction / Context:This question tests properties of an isosceles triangle combined with right-triangle geometry. In an isosceles triangle, the altitude from the vertex to the base bisects the base into two equal halves. That creates two identical right triangles. With altitude given, if we can find the base length, the area becomes straightforward because Area = (1/2)*base*height. The perimeter gives a relation between the equal sides and the base. Using these facts together allows us to solve for the side length and base length without guessing. Given Data / Assumptions: Altitude to base (height) = 8 cm Perimeter = 32 cm Let equal sides = a cm each Let base = b cm Altitude bisects base => half-base = b/2 Concept / Approach:Use perimeter: 2a + b = 32. Use Pythagoras in the right triangle: a^2 = 8^2 + (b/2)^2. Solve these equations to find b, then area = (1/2)*b*8. Step-by-Step Solution: Perimeter: 2a + b = 32 => b = 32 - 2a Right triangle: a^2 = 8^2 + (b/2)^2 Substitute b: a^2 = 64 + ((32 - 2a)/2)^2 = 64 + (16 - a)^2 Expand: a^2 = 64 + (a^2 - 32a + 256) = a^2 - 32a + 320 Cancel a^2: 0 = -32a + 320 => a = 10 b = 32 - 2*10 = 12 Area = (1/2)*b*height = (1/2)*12*8 = 48 Verification / Alternative check:Half-base is 6 cm. Check side: sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10 cm, matching a. Perimeter becomes 10+10+12=32, consistent. Hence area is correct. Why Other Options Are Wrong: 24 results from halving incorrectly or using base/2 as the base. 60 and 72 imply a larger base than what the perimeter allows. 36 comes from using wrong height or wrong base after solving. Common Pitfalls:Forgetting the altitude bisects the base in an isosceles triangle, using b instead of b/2 in Pythagoras, or making algebra errors when expanding (16 - a)^2. Final Answer:The area of the triangle is 48 sq cm.

Question 17

While measuring the sides of a rectangle, one side is measured 5% more than its actual length and the other side is measured 4% less than its actual length. Using these incorrect measurements, what is the percentage error in the calculated area of t…

Accepted Answer

Introduction:This problem is about percentage error in the area of a rectangle when both dimensions are not measured exactly. In aptitude tests, it is common to see questions where one side is measured in excess and the other in deficit. The goal is to understand how these small percentage errors combine and affect the final computed area. Instead of recomputing exact numeric values every time, there is a neat percentage error shortcut that helps you quickly reach the answer. Given Data / Assumptions: • One side of the rectangle is measured 5% more than the true length (that is, +5% error). • The other side is measured 4% less than the true length (that is, -4% error). • Area of a rectangle = length * breadth. • Errors are small, so percentage error combination rules can be applied accurately. Concept / Approach:When two quantities with small percentage errors a% and b% are multiplied, the approximate percentage error in the product is given by a + b + (a*b)/100. Here, the product is the area (length * breadth). The positive error increases the area, while the negative error reduces it. The interaction term (a*b)/100 accounts for the combined effect of both errors happening simultaneously, and it is very important for accurate results. Step-by-Step Solution: Step 1: Let the true length be L and the true breadth be B. Step 2: Measured length = L * (1 + 5/100) = 1.05L. Step 3: Measured breadth = B * (1 - 4/100) = 0.96B. Step 4: Measured area = 1.05L * 0.96B = 1.008L * B. Step 5: This means the measured area is 1.008 times the true area, so the percentage error in area is 0.8%. Step 6: Using the shortcut: a = +5, b = -4, error = a + b + (a*b)/100 = 5 - 4 + (5 * -4)/100 = 1 - 0.2 = 0.8%. Verification / Alternative check:Assume a simple rectangle where the true length and breadth are both 100 units. Then the true area = 100 * 100 = 10000 square units. With errors, measured length = 105 and measured breadth = 96. The measured area = 105 * 96 = 10080 square units. The increase in area = 10080 - 10000 = 80. Percentage error in area = (80 / 10000) * 100 = 0.8%, which exactly matches the error obtained using the shortcut formula. Why Other Options Are Wrong: Option 0.7%: This slightly underestimates the interaction term and does not match exact calculation. Option 0.9%: This overestimates the combined effect of the percentage errors. Option 0.3%: This assumes almost complete cancellation of the errors and ignores the multiplicative effect. Option 1.2%: This would arise if both sides had positive errors, not when one is in excess and the other is in deficit. Common Pitfalls:Candidates often simply add 5% and -4% and conclude that the error is 1%, without considering the product term (a*b)/100. Another common mistake is to ignore the sign of the deficit error or to apply the formula to sums or differences instead of products. Remember that the shortcut formula applies only when we are multiplying quantities, such as when calculating area, volume, or combined value. Final Answer:The percentage error in the calculated area is 0.8%.

Question 18

Four identical circular cardboard pieces, each having a radius of 7 cm, are placed so that each circular piece touches exactly two of the others. What is the area of the region enclosed between the four circular pieces?

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Introduction:This geometry question involves four identical circles arranged in a simple symmetric pattern. Each circle touches two others, suggesting that their centres form a square. The problem asks for the area of the empty region enclosed between the four circles. This type of question is common in aptitude and competitive exams to test a student's understanding of circle geometry, area subtraction, and visualisation of composite figures. Given Data / Assumptions: • Each circular cardboard piece is a circle of radius 7 cm. • Four such circles are arranged so that each circle touches exactly two others. • The circles are arranged in a square-like configuration, with centres at the corners of a square. • We need the area of the region enclosed in the middle between the four circles. Concept / Approach:When four equal circles are arranged so that each touches two others, their centres lie at the vertices of a square. The side of this square equals the diameter of a circle, because centres of two touching circles are separated by a distance of 2 * radius. The enclosed region is obtained by taking the area of this square and subtracting the areas of the four quarter-circles inside the square. Four quarter-circles of the same radius combine to form one complete circle. Thus, the required area is: area of square minus area of one full circle. Step-by-Step Solution: Step 1: Compute the side length of the square formed by the centres. Side of square = diameter of circle = 2 * 7 = 14 cm. Step 2: Compute area of this square. Area of square = side * side = 14 * 14 = 196 sq cm. Step 3: Compute area of one circle of radius 7 cm. Area of circle = pi * r^2 = pi * 7^2 = 49 * pi. Step 4: Use pi = 22/7 for calculation. Area of circle = 49 * (22/7) = 49 * (22 / 7) = 7 * 22 = 154 sq cm. Step 5: Compute the enclosed area. Enclosed area = area of square - area of one circle = 196 - 154 = 42 sq cm. Verification / Alternative check:We can visualise the configuration: each circle occupies one corner of the square, and inside the square four quarter-circles carve away space from the corners. These four quarter-circles exactly make one full circle of radius 7 cm. The remaining region must be smaller than the square but clearly more than zero. Numerically, 42 sq cm is significantly less than 196 sq cm but not negligible, which matches our geometric intuition. No other meaningful subtraction pattern emerges from this arrangement, so the method is reliable. Why Other Options Are Wrong: Option 12 sq cm: This is too small and would imply most of the square is removed, which is not the case. Option 32 sq cm: Still smaller than the correct area and does not match 196 - 154. Option 52 sq cm: This is larger than the correct leftover area and does not correspond to a simple combination of standard areas. Option 72 sq cm: This would require the circle area to be 196 - 72 = 124, which does not match any simple pi * r^2 value for radius 7. Common Pitfalls:Students sometimes mistakenly subtract the areas of four full circles from the square, or they subtract only one quarter-circle instead of four. Another common mistake is to take the side of the square as 7 cm instead of 14 cm by forgetting that the distance between centres of two touching circles is the diameter, not the radius. Also, using an incorrect value of pi or rounding too early can lead to approximate but wrong answers. Final Answer:The area of the region enclosed by the four circular cardboard pieces is 42 sq cm.

Question 19

A rectangular plot measures 90 metres in length and 50 metres in breadth. It is to be completely enclosed by a wire fence, and fence poles are to be placed at intervals of 5 metres along the boundary. How many fence poles are required in total?

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Introduction:This is a perimeter and spacing problem involving a rectangular plot surrounded by fence poles placed at equal intervals. Such questions are very common in aptitude exams to test understanding of perimeter, regular spacing, and careful counting of points on a closed boundary. The key is to calculate the total boundary length and then decide how many poles are needed if they are placed every fixed distance along that boundary. Given Data / Assumptions: • Length of the rectangular plot = 90 metres. • Breadth of the rectangular plot = 50 metres. • Distance between consecutive fence poles = 5 metres. • Poles are placed along the entire perimeter of the rectangle. Concept / Approach:The total length of the boundary of the rectangle is its perimeter. For a rectangle, perimeter = 2 * (length + breadth). Once we know the perimeter, we divide this length by the spacing between the poles to find how many segments of that length fit around the boundary. For equally spaced poles on a circular or closed path, each spacing corresponds to the distance between two consecutive poles, and the number of such spacings is equal to the number of poles if we complete the loop exactly. Step-by-Step Solution: Step 1: Compute perimeter of the rectangle. Perimeter = 2 * (length + breadth) = 2 * (90 + 50). Step 2: Add the sides inside the bracket. 90 + 50 = 140, so perimeter = 2 * 140 = 280 metres. Step 3: Compute the number of intervals of 5 metres along the boundary. Number of intervals = perimeter / spacing = 280 / 5. Step 4: Perform the division. 280 / 5 = 56. Step 5: Interpret the result. Each 5 metre interval corresponds to the distance between two consecutive poles. On a complete loop, the number of intervals equals the number of poles, so 56 poles are required. Verification / Alternative check:If we imagine walking along the boundary, starting at one corner and placing a pole there, then placing another pole every 5 metres, after 56 such steps we would come back to the starting point exactly because 56 * 5 = 280 metres, which matches the perimeter. This confirms that there is no need for an extra pole and that 56 is the correct total count of poles around the rectangle. Why Other Options Are Wrong: Option 65 poles: This is larger than required and would imply a total length greater than the perimeter. Option 55 poles: This would give a total covered length of 55 * 5 = 275 metres, leaving 5 metres of boundary without a pole. Option 45 poles: This is far too small and would not cover the whole perimeter. Option 60 poles: This would mean 60 * 5 = 300 metres of fencing, which exceeds the perimeter and is not possible with a 5 metre spacing on a 280 metre boundary. Common Pitfalls:A typical mistake is adding one extra pole by thinking that the starting point must be counted separately, leading to perimeter/spacing + 1. This is not correct when the total perimeter is exactly a multiple of the spacing, because the last pole placed at the final 5 metre step coincides with the starting corner. Another mistake is to miscalculate the perimeter due to incorrect addition of the sides. Final Answer:The number of fence poles required is 56 poles.

Question 20

A large field has a total area of 700 hectares and is divided into two parts. The difference between the areas of these two parts is equal to one-fifth of the average of their areas. What is the area of the smaller part, in hectares?

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Introduction:This is an algebra-based word problem that involves total, average, and difference of two quantities. The total area of the field is given, and we know a special relation between the difference of the two parts and their average. Such questions check a student's ability to translate words into equations, use averages correctly, and solve simultaneous equations to find the required quantity, in this case the smaller area. Given Data / Assumptions: • Total area of the field = 700 hectares. • The field is divided into two parts, with areas A and B hectares. • Without loss of generality, assume A is the larger part and B is the smaller part. • A + B = 700. • The difference of the two areas, A - B, is equal to one-fifth of the average of A and B. Concept / Approach:The average of the two areas is (A + B) / 2. We are given that A - B = (1/5) * ((A + B) / 2). Since A + B is known to be 700 hectares, the average becomes 350 hectares. The difference A - B is thus (1/5) * 350. After computing this difference, we can solve the system of equations: A + B = 700 and A - B = difference, using the standard method of adding and subtracting equations. Step-by-Step Solution: Step 1: Use the total area. A + B = 700. Step 2: Compute the average of the two areas. Average = (A + B) / 2 = 700 / 2 = 350 hectares. Step 3: Use the given relation for the difference. A - B = (1/5) * 350 = 70 hectares. Step 4: Solve the system of equations: A + B = 700. A - B = 70. Step 5: Add the two equations. (A + B) + (A - B) = 700 + 70 => 2A = 770. Step 6: Compute A. A = 770 / 2 = 385 hectares. Step 7: Compute B. B = 700 - 385 = 315 hectares. Thus, the smaller part has an area of 315 hectares. Verification / Alternative check:Compute the average again: (385 + 315) / 2 = 700 / 2 = 350 hectares. The difference between the two areas is 385 - 315 = 70 hectares. One-fifth of the average is (1/5) * 350 = 70 hectares. This exactly matches the given condition that the difference equals one-fifth of the average, so our values for A and B are consistent and correct. Why Other Options Are Wrong: Option 385 hectares: This is the larger part, not the smaller one. Option 415 hectares: Then the other part would be 700 - 415 = 285 hectares, giving an average of 350 and a difference of 130, which is not one-fifth of 350. Option 485 hectares: This would make the other part 215 hectares, giving a difference far larger than the required one-fifth of the average. Option 350 hectares: That would imply both parts are equal, giving difference 0, which contradicts the given condition. Common Pitfalls:Some students confuse the average with half of one of the areas instead of half of the sum. Others mistakenly set the difference equal to one-fifth of the total rather than one-fifth of the average. There can also be careless errors when solving the two equations, such as subtracting incorrectly or mixing up which area is larger. Writing down the equations carefully and solving stepwise avoids these mistakes. Final Answer:The area of the smaller part of the field is 315 hectares.

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