Anchit 777 wrote:A parallelogram ABCD has diagonal AC intersected by segment BM at P, such that M is the midpoint of CD. If AP = 65 and PM= 30, then find the largest possible integral value of AB:
A> 124 B> 120 C> 119 D>118
Drawn the figure as directed
:
Find, ∆APB ~ ∆CPM. How?
We can now have, AP/CP = PB/PM, or CP × PB = 65 × 30, and we further have
AP/AB = CP/CM, or
65/AB = CP/ (½ AB), or
CP = 65/2, and hence PB = 60.
Now, in ∆APB, AP = 65, and PB = 60, the third side AB CANNOT get to the sum of AP and PB. Why? In other words, AB < 125, Hence, [spoiler]
124[/spoiler] is the greatest integer AB.
[spoiler]
A[/spoiler]
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
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