FPSC · Other · N/A
Solved area problems MCQs — 49 questions on rectangle, square, triangle, parallelogram, trapezium and circle areas, paving, carpeting, fencing, cross roads and circular plates, with verified answers and explanations. Free online practice for FPSC, PPSC, NTS and all one-paper exams. Covers area problems.
Paper Snapshot
Questions
49
Time
49
Minutes
Subjects
1
Topics
49
Read mode shows correct answers. Attempt mode hides answers, starts timer, and shows your result after submission.
About this paper
This is the Arithmetic: Area Problems solved MCQ bank from Advanced FPSC & Other Federal Model Papers (Edition 32): 49 published questions (book pp. 767-770, of 50 printed) on rectangle, square, triangle, parallelogram, trapezium and circle areas, including paving, carpeting, fencing, cross roads, wall papering and circular plates. Every answer was re-solved from scratch and checked against the book's printed Answer Key on p. 770; one key error was corrected (150 marble slabs, not the keyed 225), one item was dropped because its stem prints no height and cannot be solved, and the remaining print defects were retained at the key and flagged.
Paper Controls
Subject 1
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Correct Answer: Rs.16500
The floor area is 5.5 x 3.75 = 20.625 sq. m, and 20.625 x 800 = Rs.16500.
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Correct Answer: 7.2 cm
The area is 18 x 10 = 180 sq. cm, so the new breadth is 180 / 25 = 7.2 cm.
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Correct Answer: 56
The perimeter is 2 x (90 + 50) = 280 m, and 280 / 5 = 56 poles.
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Correct Answer: 2560 sq. cm
The length is 1.6 times the breadth, so 0.6 x breadth = 24 cm gives breadth 40 cm and length 64 cm, and the area is 64 x 40 = 2560 sq. cm.
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Correct Answer: 37500 sq. m
With length L, 2 x (L + 0.6L) = 800 m gives L = 250 m and breadth 150 m, so the area is 250 x 150 = 37500 sq. m.
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Correct Answer: 30 m
With breadth 25 m, length 25 + 5 = 30 m, and 30 x 25 = 750 m².
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Correct Answer: 153600
In 8 minutes at 12 km/hr (200 m/min) he covers 1600 m, so 2 x (3x + 2x) = 1600 gives x = 160, length 480 m, breadth 320 m, and area 480 x 320 = 153600 sq. m.
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Correct Answer: 20 m x 5 m
For 20 m x 5 m the area is 100 sq. m and the three fenced sides are 20 + 5 + 5 = 30 m, with the 20 m side along the wall.
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Correct Answer: 315
The average of the two parts is 700 / 2 = 350 hectares, so their difference is 350 / 5 = 70 hectares, and the smaller part is (700 - 70) / 2 = 315 hectares.
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Correct Answer: 140 cm²
If the sides are a and b, a + 2b = 34 and 2a + b = 38 give a = 14 cm and b = 10 cm, so the area is 140 cm².
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Correct Answer: 20000
The courtyard area is 25 x 16 = 400 sq. m and each brick covers 0.2 x 0.1 = 0.02 sq. m, so 400 / 0.02 = 20000 bricks.
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Correct Answer: 7.5 m
The carpet length is 810 / 4.50 = 180 m, so its area is 180 x 0.75 = 135 sq. m, and the breadth is 135 / 18 = 7.5 m.
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Correct Answer: 17 m
Length + breadth = 23 m and length x breadth = 120 sq. m give sides 15 m and 8 m, so the diagonal is sqrt(15² + 8²) = 17 m.
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Correct Answer: 60 m²
In 15 seconds A covers 52 x 0.25 = 13 m (the diagonal) and B covers 68 x 0.25 = 17 m (length + breadth), so length x breadth = (17² - 13²) / 2 = 60 m².
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Correct Answer: 40 meters
With length 3x and breadth 2x, (3x + 5) x 2x = 2600 gives x = 20, so the breadth is 2 x 20 = 40 meters.
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Correct Answer: 28, 20
With breadth 20 cm and length 28 cm, the new dimensions 35 cm by 16 cm give the same area, 560 cm².
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Correct Answer: 40 m
The two changes give length + breadth = 20 m and length - breadth = 4 m, so length 12 m and breadth 8 m, and the perimeter is 2 x (12 + 8) = 40 m.
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Correct Answer: 64
The typed area is (20 - 4) x (30 - 6) = 16 x 24 = 384 cm² out of 600 cm², which is 64%.
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Correct Answer: Data inadequate
The pathway area depends on the plot's length and breadth, which are not given, so it cannot be found from the 96 sq. m total alone and the data are inadequate.
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Correct Answer: 2 m
With walk width 2 m the inner flower area is (20 - 4) x (10 - 4) = 16 x 6 = 96 m².
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Correct Answer: Rs.258
The roads cover 55 x 4 + 35 x 4 - 4 x 4 = 344 sq. m, and 344 x 0.75 = Rs.258.
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Correct Answer: 3 m
The roads cover 2400 - 2109 = 291 sq. m, and 60w + 40w - w² = 291 gives w = 3 m.
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Correct Answer: 16
There are 10 x 5 = 50 tiles in all, 26 black border tiles, and 24 remaining, of which one-third (8) are white and the rest 16 are blue.
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Correct Answer: Rs.73800
The perimeter is 10080 / 20 = 504 m, so the side is 126 m, and the pavement area is 126² - 120² = 1476 sq. m, costing 1476 x 50 = Rs.73800.
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Correct Answer: 600 meters
With outer side S, S² - (S - 6)² = 1764 gives S = 150 m, so the outer perimeter is 4 x 150 = 600 meters.
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Correct Answer: 6 : 5
The rectangle area is 1.2 x breadth² and the square area is breadth², so the ratio is 1.2 : 1 = 6 : 5.
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Correct Answer: 100 sq. cm
On the intended reading that the length is increased by 5 cm in total, (s + 5) / s = 3 / 2 gives s = 10 cm, so the original area is 100 sq. cm.
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Correct Answer: 9 cm
With the shorter segment x cm, (x + 2)² - x² = 32 gives x = 7 cm, so the greater segment is 9 cm.
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Correct Answer: Rs.1900
Doubling each dimension multiplies every wall area by 2 x 2 = 4, so the cost is 475 x 4 = Rs.1900.
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Correct Answer: 4 m
Paper costing Rs.2 per meter at 50 cm width covers area at Rs.4 per sq. m, so 260 / 4 = 65 sq. m are papered and the wall area is 80 sq. m; with height 2x and semi-perimeter 5x, 2 x 5x x 2x = 80 gives x = 2, so the height is 4 m.
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Correct Answer: 5 cm
The other two sides sum to 17 cm and Heron's formula gives their product as 60, so they are 12 cm and 5 cm, and the smallest side is 5 cm.
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Correct Answer: 2 : 1
The triangle has the same base and height as the rectangle, so its area is half the rectangle's, and the ratio is 2 : 1.
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Correct Answer: 1890
Each triangle on the diagonal has area 1/2 x 70 x 27 = 945 cm², so the parallelogram area is 2 x 945 = 1890 cm².
Topic Section
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Correct Answer: P + T = 2R
P = R and T = (1/2) R are true, so P + T = 1.5R, not 2R, and statement (b) is the false one.
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Correct Answer: 75 m
With parallel sides 5x and 3x, 1/2 x (5x + 3x) x 24 = 1440 gives x = 15, so the longer side is 75 m.
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Correct Answer: 84 m
The area is 1/2 x (12 + 8) x depth = 840, so the depth is 84 m.
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Correct Answer: 6 days
The grazed circle has area (22/7) x 14² = 616 sq. ft, and 616 / 100 = 6.16 days, approximately 6 days.
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Correct Answer: 66 cm/sec
The wheels approach at 7πn + 14πn = 21πn cm/sec, and 21πn x 10 = 1980 gives πn = 66/7, so the smaller wheel's speed is 7πn = 66 cm/sec.
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Correct Answer: 60π
For distance D, D/(2π) - D/(3π) = 10 gives D/(6π) = 10, so D = 60π feet.
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Correct Answer: 44 cm
The square side is sqrt(784) = 28 cm, so each plate's diameter is 14 cm and its circumference is (22/7) x 14 = 44 cm.
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Correct Answer: 117 cm²
By Pythagoras AB² + BC² = AC², and semi-circle areas are proportional to the squares of their diameters, so the area on AC is 81 + 36 = 117 cm².
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Correct Answer: 50 min
One round originally takes 40 / 8 = 5 minutes, and a 10-times diameter means a 10-times circumference, so one round takes 5 x 10 = 50 minutes.
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Correct Answer: 50.5 m
The side is sqrt(2550.25) = 50.5 m.
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Correct Answer: Rs.900
The field is 1215 / 135 = 9 hectares = 90000 sq. m, so its side is 300 m, its perimeter is 1200 m, and fencing costs 1200 x 0.75 = Rs.900.
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Correct Answer: 124 cm
The sides are 6, 8, 10, 19 and 20 cm, their areas sum to 36 + 64 + 100 + 361 + 400 = 961 cm², so the new side is 31 cm and its perimeter is 124 cm.
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Correct Answer: 150
The floor area is 9 sq. m and each slab covers 0.2 x 0.3 = 0.06 sq. m, so 9 / 0.06 = 150 slabs. The book key (c) 225 is wrong and is overridden here.
Topic Section
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Correct Answer: 120 cm
Each slab covers 72 / 50 = 1.44 sq. m, so its side is sqrt(1.44) = 1.2 m = 120 cm.
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Correct Answer: 814
The largest square tile side is HCF(1517, 902) = 41 cm, so the count is (1517 / 41) x (902 / 41) = 37 x 22 = 814 tiles.
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Correct Answer: 98 m²
Each square room has side 7 m, so the rectangle is 14 m by 7 m and its area is 98 m².
Source & verification
Built by GK360 from verified bank questions — every answer rechecked before publishing.
Questions
49 questions, published of 50 printed in the book. One item was dropped, never inferred, because its stem gives no height for the room, so the four-walls area cannot be found.
Rectangle and square areas, perimeters and diagonals, areas that stay constant or change with the sides, paving and carpeting costs, fencing poles and pavement, cross roads and inner pathways, wall papering and room heights, triangle, parallelogram and trapezium areas, grazing circles, rolling wheels and circular plates cut from a square sheet.
Yes. Every answer was re-solved from scratch — side and diagonal relations, perimeter and area identities, trapezium and circle formulas, and the HCF tile method — and checked against the book's printed Answer Key on p. 770. One key error was corrected to the standard answer, one unanswerable item was dropped, and the other print defects are flagged in the working notes.
Area and mensuration problems appear in the arithmetic and general-ability sections of FPSC, PPSC, NTS, PTS and CSS MPT one-paper exams, so the same methods apply across all of them.
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