Question
Find the sum of all those 3-digit numbers which when
divided by 10 leave a remainder of 5.Solution
Numbers 105, 115, 125, ……985, 995 first number (a) = 105 The last number (8) = 995 All difference (d) =10 From the formula, nth term (Tn) = a + (n-1) d, 995=105+(n-1) 10  n = 90  Sum of n terms (Sn) =n/2 [2a+(n-1) d]  90/2[2 ×105+ (90-1)10] = 45 [210+890] = 45 × 1100  = 49500.
More No. System Questions
20 * 8 + 40% of 100 + 60% of 150 = ?
7292/3 = ?
Find the result of
45 ÷ 3 of 6 of [12 ÷ 4 of (10 ÷ 2 + 1)] + (8 ÷ 2.5 + 1.8)
- What will come in place of (?), in the given expression.
144 ÷ 12 + 18 × 2 = ? (2197)1/3 + (18)2 − 121 = ? − 69 × 5
Evaluate: {2 x (0.718 + 0.982) + 0.008 of 5000}
Simplify the following expressions and choose the correct option.
18² + (27 ÷ 3) × 11 − 250 = ?
If x²- 5x + 1 = 0, what is the value of x² + 1/x2?
? ÷ 62 × 12 = 264
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