Question
Four numbers are arranged in ascending order. When the
first two numbers are increased by the number equal to their position, and the last two numbers i.e. the 3rd and 4th number are increased by 6 and 5 respectively, the series thus formed is in A.P. If the first number is 3 less than the 2nd number, and the sum of all the numbers is 28, then find the average of the last three numbers.Solution
ATQ, Let the numbers be a, b, c, and d. According to the question, (b + 2) – (a + 1) = (c + 6) – (b + 2) Or, 2b = a + c + 3…… (1) Also, (c + 6) – (b + 2) = (d + 5) – (c + 6) Or, 2c = b + d – 5…… (2) Let a = x Then, b = x + 3 From (1), c = x + 5 And, from (2), d = x + 11 According to the question, (x + x + 3 + x + 5 + x + 11) = 28 Or, 4x + 19 = 28 Or, 4x = 9 Or, x = 2.25 Therefore, average of the last three numbers = (x + 3 + x + 5 + x + 11)/3 = (2.25 + 3 + 2.25 + 5 + 2.25 + 11)/3 = 25.5/3 = 8.5
?% of (168 ÷ 8 × 20) = 126
20% of 1500 – 75% of 200 = 125% of ?
Find the value of 16 X [(8 - 5) of 12 ÷ 4].
√196 + (0.25 × 144) + 19 = ? + 72
22 * 6 + 45% of 90 + 65% of 180 = ?
52% of 400 + √(?) = 60% of 600 - 25% of 400
(25 × 12 + 30 × 8 – 22 × 10) = ?
What will come in the place of question mark (?) in the given expression?
(240% of 175 ÷ √16) X 6 + 80% of 400 = ?3 + 179 + 42
What will come in the place of question mark (?) in the given expression?
(144 × 16 ÷ 12) × 6 = ?
808 ÷ (128)1/7 + 482 = 4 × ? + 846
Relevant for Exams:
