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ATQ, Let diameter of the Pear fruit = ‘2d’ cm Diameter/Height of the seed = 2d × 1/5 = (2d/5) cm Volume of the seed = π(d/5)2(2d/5) = {2πd3/125} cm3 Volume of the Pear fruit = {(4/3)π(d)3} cm3 Required percentage = [{(4/3)π(d)3 – 2πd3/125 }/((4/3)π(d)3)] × 100 = {(4/3 – 2/125)/(4/3)} × 100 = 98.8%
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