SATHEE: Quantitative Aptitude Ques 2326

Quantitative Aptitude Ques 2326

Question: AB is a straight line, C and D are points the same side of AB such that CA perpendicular to AB and DB is perpendicular to AB. Let AD and BC meet at E. What is $ \frac{AE}{AD}+\frac{BE}{BC} $ equal to?

Options:

A) 2

B) 1.5

C) 1

D) None of these

Show Answer

Answer:

Correct Answer: D

Solution:

Since, AB is a straight line and C and D are points such that $ AC\bot AB $ and $ BD\bot AB. $

$ \therefore $ $ AC\parallel BD $ So, ABCD forms trapezium. Now, by property of trapezium diagonals intersect each other in the ratio of lengths of parallel sides.

$ \therefore $ $ \frac{AE}{ED}=\frac{BE}{CE} $ $ \frac{AE}{AD-AE}=\frac{BE}{BC-BE} $ $ \frac{BC-BE}{BE}=\frac{AD-AE}{AE} $ $ \frac{BC}{BE}-1=\frac{AD}{AE}-1 $
$ \Rightarrow $ $ \frac{BC}{BE}=\frac{AD}{AE} $ $ \frac{AE}{AD}=\frac{BE}{BC} $ But the value of $ \frac{AE}{AD} $ or $ \frac{BE}{BC} $ cannot be determined. So, we cannot find the value of $ \frac{AE}{AD}+\frac{BE}{BC}. $

Quantitative Aptitude Ques 2325

Quantitative Aptitude Ques 2327

Quantitative Aptitude Ques 2326

Question: AB is a straight line, C and D are points the same side of AB such that CA perpendicular to AB and DB is perpendicular to AB. Let AD and BC meet at E. What is $ \frac{AE}{AD}+\frac{BE}{BC} $ equal to?

Options:

Answer:

Solution:

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