There are three numbers 'a', 'b' and 'c' (a < b < c) such that 'b' is equal to the average of 'a' and 'c'. 70% of 'c' is equal to 'a' and the difference between 'b' and 'c' is 21. Find the sum of all three numbers.
ATQ, 'b' = (1/2) × (a + c) Or, 2b = a + c And, 0.7c = 'a' Or, c = 10:7 Let 'c' = '10d' and 'a' = '7d' So, 2b = 7d + 10d Or, 'b' = (17d/2) Now, c - b = 21 Or, 10d - (17d/2) = 21 Or, (3d/2) = 21 Or, 'd' = 14 Required sum = 7d + {(17/2) × 14} + (10 × 14) = 98 + 119 + 140 = 357
36.76 + 2894.713 + 34965.11 =?
3% of 842 ÷ 2% of 421 = ?
Find the HCF of 15x2 + 8x – 12, 3x² + x – 2, 3x² - 2x, 9x² - 12x + 4
60 = (? x 10 + 250)/5
779 + 136 – 334 = 270 + 121 + ?
40% of 220 × 15 ÷ 20 = ?
1024 ÷ 32 = 2(1/2)×?
{(522 – 482 ) ÷ (27 + 73)} × 35 = ?% of 175
(12% of 1250 + 85% of 400) x 10 = ?2
