Question
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.Solution
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
95.001% of 8219.99 - 4/9 % of 5399.98 + 109.99 = ?Â
25.05% of 220.05 – 10.15% of 119.99 × 2.02 = ?
16.98 × 88.05 + 1999.996% of 299.08 + 5.005 % of 4999.997 = ? × 20.98 × 40.009
(32.18% of 2399.89 - √624 × 26.25) % of 149.79 = ?
(752.09 - 43.04 x 7.94) ÷  16.9 = ?
(44/25) ÷ (154/199.5) × 419.91 = ? – (11.11)3
39.99% of 159.98 = ?2 – 319.99% of 12.5 – 49.98% of 129.99
15.22 × 9.99 + 150.15 = ?
Raju invests Rs. (p + 200) at a simple interest rate of 25% per annum for a period of 6 years and earns a total interest of Rs. (...
2660.03 ÷ 69.98 x 49.9 = ? + 10.32
Relevant for Exams:
