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QUESTION
Q
Sum of three consecutive numbers is 2262. What is 41% of the highest numbers ?
Sum of three consecutive numbers is 2262. What is 41% of the highest numbers ?
A
301.51
B
303.15
C
308.73
D
306.35
E
309.55
A
301.51
B
303.15
C
308.73
D
306.35
E
309.55
Solution
Let the three consecutive numbers be (x − 1), x ,(x + 1)
Sum of the three consecutive numbers =...
Let the three consecutive numbers be (x − 1), x ,(x + 1)
Sum of the three consecutive numbers = (x − 1) + x + (x + 1) = 3x
Given sum of three consecutive numbers = 2262
⇒ 3x = 2262
⇒ x = 754
The highest number is x + 1 = 754 + 1 = 755
41% of 755 = 309.55
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