Question
Find the difference between minimum and maximum value of
'k' such that '6k8194' is always divisible by 3.Solution
A number is divisible by 3 when the sum of its digits is divisible by 3.
Sum of digits of '6k8194' = (6 + k + 8 + 1 + 9 + 4) = k + 28.
So, k + 28 should be divisible by 3.
Possible values of 'k' = 2, 5, 8
Minimum value of 'k' = 2
Maximum value of 'k' = 8
Required difference = 8 – 2 = 6
Solve the given equation for ?. Find the approximate value.
[(49.88% of 320.11) × (34.85% of 460.24)] ÷ √783.94 = ?
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
79.99% of (84.89 × 5.99) - (3.89)2 × 21.87 = ?
(29.892 × √290) + 32.98 × 6.91 = ?
44.84% of 799.94 + (625.21 ÷ 24.91) – √(224.77) = ?
What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)...
(20.98 ÷ 2.91) + (15.12 – 5.96) = ?
56.05 2 – 24.24 2 + (63.98) 3/2 – 32.28% of 1500 = ? 2 + 113.03 × 5.09
[54.96 × √99.96 – {(25.02/6.84)% of 280.24}]/(3.032 × 19.87) = ?
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