Latest CTS Aptitude Question SOLUTION: The sum of nine consecutive odd number of set A is 657. What is the sum of seven consecutive odd numbers whose lowest number is 18 more than the lowest number of set A? Options 1)

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The sum of nine consecutive odd number of set A is 657. What is the sum of seven consecutive odd numbers whose lowest number is 18 more than the lowest number of set A?

Options
1) 620
2) 625
3) 635
4) 615
5) 623

Asked In CTS ABHINAV PUSHPAM (8 years ago)

Unsolved Read Solution (3)

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Read Solution (3) :

Suneena Deen Jacob  (7 years ago)

General representation of an odd number is 2x+1
so 9 consecutive odd numbers (ie set A) can be represented as
2x+1 , 2x+3 , 2x+5, 2x+7, 2x+9 , 2x+11 ,2x+13 , 2x+15 ,2x+17
It is told that sum of 9 consecutive odd numbers is 657
so upon adding and equating
(2x+1) + (2x+3) + (2x+5) +( 2x+7) + (2x+9) +( 2x+11) +(2x+13)+ ( 2x+15) +(2x+17)=657
ie 18x+81 = 657
18x =657-81
x =576/18
=32
Now that we have found that x=32
2x+1 is the 1st odd number. So upon substituting x we get
first odd number in set A as 65
We need to find the sum of 7 consecutive odd numbers whose lowest is 18 more than the lowest of set A.
So the new starting odd number of the series will be 18+65=83
so 83+85+87+89+91+93+95=623
Therefore option 5

Is this solution Helpfull? Yes (11) | No   

Avinash Nishad  (8 years ago)

n Set A, 9x + 2(1 + 2 + ..... + 8) = 657
9x + 8*9*2/2 = = 657
9x + 36 × 2 = 657
x = 65

lowest number of another set = 65 + 18 = 83

Sum of seven consecutive number of another set
= 7x + (1 + 2 + 3 + 4 + 5 + 6) × 2
= 7x + 6*7*2/2
= 7x + 21 = 83 × 7 + 42 = 623

Is this solution Helpfull? Yes (6) | No (5)  

Tuhin Barai  (7 years ago)

s=n/2[2a+(n-1)d] = 657 for set a
s=657 , n=9 , d=2 (since odd number)
find a , a=72.11
Now, add 18 according to the problem.. to the loest number...vwhich becomes 90.11
Again , S=7/2[2a+(n-1)d] = 3.5[2x90.11+(7-1)2] = 672.77

Is this solution Helpfull? Yes (2) | No (5)  

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