Find the cost of polishing a circular table-top of diameter \(1.6\) \(m\), if the rate of polishing is Rs \(15/m^2\) . (Take \(\pi\) = \(3.14\))
Solution and Explanation
Diameter = \(1.6\) \(m\)
Radius = \(\frac{1.6}{2}\)
= \(0.8\) \(m\)
Area = \(3.14 × 0.8 × 0.8\)
= \(2.0096 \;m^2\)
Cost for polishing \(1\) \(m^2\) area = Rs \(15\)
Cost for polishing \(2.0096\) \(m^2\) area = \(15 × 2.0096 \) = \(30.14\)
Therefore, it will cost Rs \(30.14\) for polishing such circular table.
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Questions Asked in CBSE Class VII exam
- What letters of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
- a vertical mirror
- a horizontal mirror
- both horizontal and vertical mirrors
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- In a class test (+3) marks are given for every correct answer and (–2) marks are given for every incorrect answer and no marks for not attempting any question. (i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly? (ii) Mohini scores –5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
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- Properties Of Division Of Integers
Match the items given in Column I with one or more items of Column II.
Column I Column II (a) A plane mirror (i) Used as a magnifying glass. (b) A convex mirror (ii) Can form image of objects spread over a large area. (c) A convex lens (iii) Used by dentists to see enlarged image of teeth. (d) A concave mirror (iv) The image is always inverted and magnified. (e) A concave lens (v) The image is erect and of the same size as the object. - (vi) The image is erect and smaller in size than the object. - CBSE Class VII
- Spherical Mirrors
- Identify the numerical coefficients of terms (other than constants) in the following expressions:
(i) 5 – 3t2
(ii) 1 + t + t2 + t3
(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) – p2q2 + 7pq
(vi) 1.2 a + 0.8 b
(vii) 3.14 r2
(viii) 2 (l + b)
(ix) 0.1 y + 0.01 y2- CBSE Class VII
- Algebraic Expressions
- Find the value of:
(i) -4÷\(\frac{2}{3}\)
(ii) -\(\frac{3}{5}\)÷2
(iii) -\(\frac{4}{5}\)÷(-3)
(iv) -\(\frac{1}{8}\)÷\(\frac{3}{4}\)
(v) -\(\frac{2}{13}\)÷\(\frac{1}{7}\)
(vi) -\(\frac{7}{12}\)÷(-\(\frac{2}{13}\))
(vii) \(\frac{3}{13}\)÷(-\(\frac{4}{65}\))- CBSE Class VII
- Operations On Rational Numbers
Concepts Used:
Circle
A circle can be geometrically defined as a combination of all the points which lie at an equal distance from a fixed point called the centre. The concepts of the circle are very important in building a strong foundation in units likes mensuration and coordinate geometry. We use circle formulas in order to calculate the area, diameter, and circumference of a circle. The length between any point on the circle and its centre is its radius.
Any line that passes through the centre of the circle and connects two points of the circle is the diameter of the circle. The radius is half the length of the diameter of the circle. The area of the circle describes the amount of space that is covered by the circle and the circumference is the length of the boundary of the circle.
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