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29 Aug 2020, 04:46
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
67% (02:36) correct
33%
(02:16)
wrong
based on 117
sessions
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Date
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Not Attempted Yet
Q1. (Algebra)
PS - Algebra.jpg [ 12.34 KiB | Viewed 2380 times ]
The positive integers from 1 up are arranged, as shown in the table above. Nine numbers in the form of a square are chosen from the arrangement. x is the upper-left most number in the square, and the sum of the 9 numbers is 171. What is the value of x?
A. 10
B. 11
C. 18
D. 19
E. 23
Attachment:
PS - Algebra.jpg [ 12.34 KiB | Viewed 2380 times ]
The positive integers from 1 up are arranged, as shown in the table above. Nine numbers in the form of a square are chosen from the arrangement. x is the upper-left most number in the square, and the sum of the 9 numbers is 171. What is the value of x?
A. 10
B. 11
C. 18
D. 19
E. 23
29 Aug 2020, 07:12
Kudos
Bookmarks
So we pick 9 numbers that are a part of the 3X3 square and add them up
Let us say we pick R1 {4,5,6}, R2 {11,12,13} and R3 {18,19,20} or R1 {3,4,5}, R2 {10,11,12} and R3 {17,18,19) (R1,R2,R3 are the rows in the square}
What we can observe is that within each row the numbers differ by 1. Let x be the left most one of the numbers in the Row R1 and our R1 will be {x, x+1, x+2}
Similarly if we observe that the difference between R1 and R2 (11-4 = 7 or 10-3=7) is 7 , so each rows differ by 7
Which implies that my R2 will be {x+7, x+8, x+9}
similarly R3 {x+14, x+15, x+16}
Adding all the constituents together we get
x+x+1+x+2+x+7+x+8+x+9+x+14+x+15+x+16 = 171
9x+72=171
9x=99
x=11
hence the OA is B
Kudos if it helps!
Let us say we pick R1 {4,5,6}, R2 {11,12,13} and R3 {18,19,20} or R1 {3,4,5}, R2 {10,11,12} and R3 {17,18,19) (R1,R2,R3 are the rows in the square}
What we can observe is that within each row the numbers differ by 1. Let x be the left most one of the numbers in the Row R1 and our R1 will be {x, x+1, x+2}
Similarly if we observe that the difference between R1 and R2 (11-4 = 7 or 10-3=7) is 7 , so each rows differ by 7
Which implies that my R2 will be {x+7, x+8, x+9}
similarly R3 {x+14, x+15, x+16}
Adding all the constituents together we get
x+x+1+x+2+x+7+x+8+x+9+x+14+x+15+x+16 = 171
9x+72=171
9x=99
x=11
hence the OA is B
Kudos if it helps!
31 Aug 2020, 03:17
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MathRevolution
Q1. (Algebra)
The positive integers from 1 up are arranged, as shown in the table above. Nine numbers in the form of a square are chosen from the arrangement. x is the upper-left most number in the square, and the sum of the 9 numbers is 171. What is the value of x?
A. 10
B. 11
C. 18
D. 19
E. 23
Attachment:
PS - Algebra.jpg
The positive integers from 1 up are arranged, as shown in the table above. Nine numbers in the form of a square are chosen from the arrangement. x is the upper-left most number in the square, and the sum of the 9 numbers is 171. What is the value of x?
A. 10
B. 11
C. 18
D. 19
E. 23
Solution:
Square form: An Equal number of rows and columns make a square. To form a square of 9 numbers: 3 by 3 square has to be selected.
When we select the first row, it will have 3 numbers as per the table. So,
- We will select the 1st, 2nd, and 3rd numbers.
The next row selection will give us the 8th, 9th, and 10th numbers.
The third row will give us the 15th, 16th, and 17th numbers.
Given: x is the upper-left most number, so the 1st number is x = 1, and the sum of the 9 numbers is 171.
=> 1st number: x
=> 2nd number: x + 1
=> 3rd number: x + 2 ….. and so on.
So, we get x + (x + 1) + (x + 2) + (x + 7) + (x + 8) + (x + 9) + (x + 14) + (x + 15) + (x + 16) = 171.
=> 9x + 72 = 171
=> 9x = 99
=> x = 11
Therefore, B is the correct answer.
Answer: B
