RD Sharma Class 8 Math Second Chapter Powers Exercise 2.3 Solution

Chapter:- Powers Exercise 2.3

1.Express the following numbers in standard form:

  1. To express a number in the standard form, move the decimal point such that there is only one digit to the left of the decimal point.(i) 6020000000000000

    Solution:

    6020000000000000 = 6.02 x 1015 (The decimal point is moved 15 places to the left.)

     (ii) 0.00000000000943

    Solution:

    0.0000000000943 = 9.43 x 10−12 (The decimal point is moved 12 places to the right.)

    (iii) 0.00000000085

    Solution:

    0.00000000085 = 8.5 x 10−10 (The decimal point is moved 10 places to the right.)

    (iv) 846 × 107

    Solution:

    846 x 107 = 8.46 x 102 x 107 = 8.46 x 109 (The decimal point is moved two places to the left.)

    (V)  3759 × 10−4

    Solution:

    3759 x 10−4 = 3.759 x 103 x 10−4 = 3.759 x 10−1 (The decimal point is moved three places to the left.)

    (vi) 0.00072984

    Solution:

    0.00072984 = 7.984 x 10−4 (The decimal point is moved four places to the right.)

    (vii) 0.000437 × 104

    Solution:

    0.000437 x 104 = 4.37 x 10−4 x 104 = 4.37 x 100 = 4.37 (The decimal point is moved four places to the right.)

    (viii)  4 ÷ 100000

    Solution:

    4/100000 = 4 x 100000−1 = 4 x 10−5 (Just count the number of zeros in 1,00,000 to determine the exponent of 10.)

2. Write the following numbers in the usual form:

(i) 4.83 × 107

Solution:

4.83×107=4.83×1,00,00,000=4,83,00,000

(ii) 3.02×106

Solution:

3.02×10−6=3.02106=3.0210,00,000=0.00000302

(iii) 4.5×104

Solution:

4.5×104=4.5×10,000=45,000

(iv) 3×108

Solution:

(v) 1.0001×109

Solution:

1.0001×109=1.0001×1,00,00,00,000=1,00,01,00,000

(vi) 5.8×102

Solution:

5.8×102=5.8×100=580

(vii)  3.61492×106

Solution:

3.61492 x 106=3.61492 x 10,00,000=3614920

(viii) 3.25×107 

Solution:

 


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