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Updated on: 11 Oct 2015, 23:27
Originally posted by AryamaDuttaSaikia on 11 Oct 2015, 23:24.
Last edited by Bunuel on 11 Oct 2015, 23:27, edited 1 time in total.
Last edited by Bunuel on 11 Oct 2015, 23:27, edited 1 time in total.
Edited the question.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
48% (02:00) correct
52%
(02:14)
wrong
based on 248
sessions
History
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Time
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Not Attempted Yet
As shown in the figure above, 20 equal squares are arranged to form a large rectangle. How many total rectangles does the large rectangle contain?
(a) 20
(b) 40
(c) 90
(d) 150
(e) 190
Attachment:
1.jpg [ 11.69 KiB | Viewed 6860 times ]
Updated on: 03 Dec 2015, 21:34
Originally posted by AryamaDuttaSaikia on 03 Dec 2015, 02:55.
Last edited by AryamaDuttaSaikia on 03 Dec 2015, 21:34, edited 1 time in total.
Last edited by AryamaDuttaSaikia on 03 Dec 2015, 21:34, edited 1 time in total.
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Jamboree way
If we choose any two horizontal lines and any two vertical lines the enclosed quadrilateral formed will be a rectangle.
The given figure has 5 horizontal lines and 6 vertical lines.
The number of ways we can select 2 horizontal lines out of 5 horizontal lines = 5C2 = 10
The number of ways we can select 2 vertical lines out of 6 vertical lines = 6C2 = 15
Total number of rectangles formed = (The number of ways we can select 2 horizontal lines) x (The number of ways we can select 2 vertical lines)
Total number of rectangles formed = 10 x 15 = 150
If we choose any two horizontal lines and any two vertical lines the enclosed quadrilateral formed will be a rectangle.
The given figure has 5 horizontal lines and 6 vertical lines.
The number of ways we can select 2 horizontal lines out of 5 horizontal lines = 5C2 = 10
The number of ways we can select 2 vertical lines out of 6 vertical lines = 6C2 = 15
Total number of rectangles formed = (The number of ways we can select 2 horizontal lines) x (The number of ways we can select 2 vertical lines)
Total number of rectangles formed = 10 x 15 = 150
12 Oct 2015, 11:59
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I don't think that such kind of problems appear in GMAT.
We can use this formula to solve this one: \(\frac{m(m+1)*n(n+1)}{4}\)==> \(\frac{4*5*5*6}{4}\)=150
Alternative solution:
In a rectangle m × n there are (m+1) vertical grid lines
and (n+1) horizontal grid lines (5 and in the example here).
To define any rectangle within the grid,
we must choose 2 of each and there are
( (m+1) choose 2 ) × ( (n+1) choose 2 ) ways to do that.
For 5 × 6 that gives us
2C5 × 2C6 = 10 * 21 = 150
What would the experts say about this question, isn't it too specific/hard for GMAT ?
We can use this formula to solve this one: \(\frac{m(m+1)*n(n+1)}{4}\)==> \(\frac{4*5*5*6}{4}\)=150
Alternative solution:
In a rectangle m × n there are (m+1) vertical grid lines
and (n+1) horizontal grid lines (5 and in the example here).
To define any rectangle within the grid,
we must choose 2 of each and there are
( (m+1) choose 2 ) × ( (n+1) choose 2 ) ways to do that.
For 5 × 6 that gives us
2C5 × 2C6 = 10 * 21 = 150
What would the experts say about this question, isn't it too specific/hard for GMAT ?
