Math, asked by pushpa10sisodia, 1 month ago

Mr. A and Mr. B are supposed to fill a rectangular array of 16 columns and 10 rows, with the numbers 1, 2, 3, .., 160. Mr. A chose to do it row wise, so that first row is numbered 1, 2, 3, ..., 16 (in order), then second row is numbered 17, 18, ..., 32 and so on. Mr. B chose to do it column wise so that first column is numbered 1, 2, 3, ..., 10 (in order), then second column is numbered 11, 12, 13,..., 20 and so on. Let N be the sum of numbers which occupy same position in both arrangements, find the value of n/7​

Answers

Answered by amitnrw
0

Given: Mr. A and Mr. B are supposed to fill a rectangular array of 16 columns and 10 rows, with the numbers 1, 2, 3, .., 160. Mr. A chose to do it row wise, so that first row is numbered 1, 2, 3, ..., 16 (in order), then second row is numbered 17, 18, ..., 32 and so on. Mr. B chose to do it column wise so that first column is numbered 1, 2, 3, ..., 10 (in order), then second column is numbered 11, 12, 13,..., 20 and so on.

Let N be the sum of numbers which occupy same position in both arrangements

To Find : , find the value of n/7​

Solution:

Number which occupy the same positions are

1 , 54 , 107 , 160

Sum of numbers  = 1 + 54 + 107 + 160  = 322

322/7  = 46

value of n/7​ = 46

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