Question
Quantity-I: The total sum of all
perfect square numbers in the range from 1 to 1000. Quantity-II: The total sum of all perfect cube numbers in the range from 1 to 2500. In the question, two Quantities I and II are given. You have to solve both the Quantity to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.Solution
ATQ, Quantity I: Largest perfect square before 1000 = 312 = 961 So, required sum = sum of squares of all natural numbers up to 31. So, required sum = {n(n + 1)(2n + 1)} ÷ 6 = (31 × 32 × 63)/6 = 10416 So, Quantity I = 10416 Quantity II Largest perfect cube before 2500 = 133 = 2197 So, required sum = [n2 (n + 1)2 ]/4 = {(13)2 × (14)2 }/4 = 8281 So, Quantity II = 8281 So, Quantity I > Quantity II
70.14% of 799.95 - 240.12 = ? + 40.17% of 299.95
(?)2 + 4.113 = 24.92 – 32.03Â
The greatest number that will divide 398,436, and 542 leaving 7, 11, and 15 as remainders, respectively, is:
58.03% of 1499.99 - ? % of 699.95 = 394.04
?% of (112.31 ÷ 13.97 × 90.011) = 359.98
What will be the approximate value of the following questions.
(√143.74 + 29.89% of 720.27) × (5/9 of 539.79) = ?
11.06 2 – 7.12 × 4.88 + 9.96 = 12.22 × ?Â
? = 49.97% of 38.09% of 1998.95
5.55% of 8120 – 66.66% of 540 = ? – 28% of 5500
(5/9 of 2699.81) + (49.88% of 144.18) - (2/7 of 489.77) = ?
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