Aptitude AreaPage 2
9. | | A circular path runs all around and outside a circular garden of diameter 42 m. If the difference between the outer circumference of the path and inner circumference of the garden is 88 m, find the width of the path ? | | (a)13 m | | (b)14 m | | (c)15 m | | (d)16 m |
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Answer is: BCircumference of inner circle = 2 x 22/7 x 42 = 264 m.
Circumference of the outer circle = 264 + 88 = 352 m.
⇒ 2 x 22/7 x R = 352, (R = outer radius of the path).
R = (352 x 7)/(2 x 22) = 56 m.
width of the path = outer radius – inner radius = 56 – 42 = 14 m.
10. | | A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road? | | (a)3 m | | (b)4 m | | (c)5 m | | (d)6 m |
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Answer is: AArea of the park = (60 x 40) m² = 2400 m².
Area of the lawn = 2109 m².
Area of the crossroads = (2400 - 2109) m² = 291 m².
Let the width of the road be X metres. Then,
60X + 40X - X² = 291
X² - 100X + 291 = 0
(X - 97)(X - 3) = 0
X = 3.
11. | | A circular garden of radius 15 m is surrounded by a circular path of width 7 m. If the path is to be covered with tiles at a rate of Rs 10 per square meter, then find the total cost of the work ? | | (a)Rs 7140 | | (b)Rs 8440 | | (c)Rs 8450 | | (d)Rs 8140 |
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Answer is: DArea of the ring (circular path) = π(R² - r²).
= π(22² - 15²) = π(22 + 15)(22 - 15) = π(37) x 7 = 814 m².
∴ Total cost at Rs 10 per sq. m = 814 x 10 = Rs 8140.
12. | | The area of a parallelogram is 72 sq.cm and its height is 8 cm. Find its base ? | | (a)7 cm | | (b)9 cm | | (c)10 cm | | (d)12 cm |
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Answer is: BThe area of parallelogram = base x height
∴ height = 72/8 = 9 cm.
13. | | In a square of side 6 cm, find the length of the diagonal ? | | (a)6√3 | | (b)6√2 | | (c)5√3 | | (d)5√2 |
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Answer is: BLength of the diagonal = √2a, where a is side of the square.
∴ diagonal = √2 x 6 = 6√2 cm.
14. | | A rectangle has twice the area of a square. The length of the rectangle is 8 cm greater than the side of the square and the breadth is equal to the side of the square. Find the perimeter of the square ? | | (a)32 cm | | (b)16 cm | | (c)28 cm | | (d)30 cm |
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Answer is: ALet the side of the square be a.
⇒ area of the square = a²
∴ area of the rectangle = 2a²
∴ (a + 8)a = 2a²
⇒ a² + 8a = 2a²
⇒ a = 8
∴ perimeter of the square = 4 x 8 = 32 cm
15. | | The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares ? | | (a)26 cm | | (b)28 cm | | (c)24 cm | | (d)30 cm |
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Answer is: CSide of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm
Area of third square = (10² - 8²) cm²
= (100 – 64) = 36 cm²
Side of the square = √36 = 6 cm
∴ Required Perimeter = 6 x 4 = 24 cm.
16. | | The diagonals of two squares are in the ratio of 2 : 5. Find the ratio of their areas ? | | (a)25 : 4 | | (b)4 : 25 | | (c)26 : 4 | | (d)4 : 26 |
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Answer is: BLet the diagonals of the squares of the to be 2X and 5X respectively.
∴ Ratio of their areas = ½ x (2X)² : ½ x (5X)² = 4X² : 25X²
∴ Required ratio = 4 : 25.
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