Mensuration
- The length and breadth of a rectangle are increased by 12% and 15% respectively. Its area will be increased by :
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Using Rule 10, Percentage increase in the area of rectangle
= 
12 + 15 + 12 × 15 
% 100 = 27 + 9 % 5 = 28 4 % 5 Correct Option: B
Using Rule 10, Percentage increase in the area of rectangle
= 12 + 15 + 12 × 15 % 100 = 27 + 9 % 5 = 28 4 % 5
- The radius of a circle is a side of a square. The ratio of the area of the circle and the square is
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Using Rule 10 and 14,
Radius of circle = Side of square = r units
⇒ Area of circle : Area of square = πr² : r²
= π : 1Correct Option: B
Using Rule 10 and 14,
Radius of circle = Side of square = r units
⇒ Area of circle : Area of square = πr² : r²
= π : 1
- If the length of a rectangle is increased by 25% and the width is decreased by 20%, then the area of the rectangle :
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Using Rule 10,
According to question,
Area of original rectangle = xy
Area of new rectangle = 1.25x × 0.80y = xy∴ Effective change = 25 - 20 - 25 × 20 % = 0% 100
i.e. Hence, the area of the rectangle remains unchanged.Correct Option: C
Using Rule 10,
According to question,
Area of original rectangle = xy
Area of new rectangle = 1.25x × 0.80y = xy∴ Effective change = 25 - 20 - 25 × 20 % = 0% 100
i.e. Hence, the area of the rectangle remains unchanged.
- The length of a rectangle is decreased by 10% and its breadth is increased by 10%. By what per cent is its area changed ?
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Using Rule 10,
Let the length and breadth of a rectangle are changed by x and y per cent respectively, then the net change= x + y + xy % 100
, where positive and negative signs show increase and decrease respectively.∴ Net change = - 10 + 10 - 10 × 10 = - 1% 100 Correct Option: B
Using Rule 10,
Let the length and breadth of a rectangle are changed by x and y per cent respectively, then the net change= x + y + xy % 100
, where positive and negative signs show increase and decrease respectively.∴ Net change = - 10 + 10 - 10 × 10 = - 1% 100
- If the circumference of a circle is reduced by 50%, its area will be reduced by
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Let radius = r
∴ Circumference = 2πr
Reduced circumference = πr = 2π × r/2
∴ New radius = r/2∴ Reduced area = π × r ² = πr² 2 4
It is 25% of πr² (the original area)
∴ Area is reduced by 75%.Correct Option: D
Let radius = r
∴ Circumference = 2πr
Reduced circumference = πr = 2π × r/2
∴ New radius = r/2∴ Reduced area = π × r ² = πr² 2 4
It is 25% of πr² (the original area)
∴ Area is reduced by 75%.
