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15 Jul 2020, 00:47
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There are 3 consecutive odd integers. Three times of the sum of the largest and the smallest of them is 6 less than 8 times the remaining odd integer. What is the smallest integer?
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
15 Jul 2020, 01:12
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Let the integers be x, x + 2 and x + 4
3 time the sum of the largest and the smallest integers = 3(x + x + 4) = 3(2x + 4) = 6x + 12 --- (1)
8 times the middle integer = 8(x + 2) = 8x + 16 --- (2)
The difference of (2) - (1) = 6
8x + 16 - (6x + 12) = 6
8x + 16 - 6x - 12 = 6
2x = 6 + 12 - 16
2x = 2
x = 1
Option A
Arun Kumar
3 time the sum of the largest and the smallest integers = 3(x + x + 4) = 3(2x + 4) = 6x + 12 --- (1)
8 times the middle integer = 8(x + 2) = 8x + 16 --- (2)
The difference of (2) - (1) = 6
8x + 16 - (6x + 12) = 6
8x + 16 - 6x - 12 = 6
2x = 6 + 12 - 16
2x = 2
x = 1
Option A
Arun Kumar
17 Jul 2020, 00:33
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Assume \(x – 2, x\), and \(x + 2\) are the three consecutive odd integers, where \(x\) is an odd number.
We have \(3(x – 2 + x + 2) = 8x – 6, 3x – 6 + 3x + 6 = 8x - 6\) or \(6x = 8x – 6.\)
Then we have \(2x – 6 = 0\) or \(x = 3.\)
Thus, the smallest integer is \(x – 2 = 3 – 2 = 1.\)
Therefore, A is the correct answer.
Answer: A
Assume \(x – 2, x\), and \(x + 2\) are the three consecutive odd integers, where \(x\) is an odd number.
We have \(3(x – 2 + x + 2) = 8x – 6, 3x – 6 + 3x + 6 = 8x - 6\) or \(6x = 8x – 6.\)
Then we have \(2x – 6 = 0\) or \(x = 3.\)
Thus, the smallest integer is \(x – 2 = 3 – 2 = 1.\)
Therefore, A is the correct answer.
Answer: A
