Algebra
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Q.

Ramesh has two examinations on Wednesday - Engineering Mathematics in the morning Engineering drawing in the afternoon. He has a fixed amount of time to read the textbooks of both these subjects on Tuesday. During this time he can read 80 pages of Engineering Mathematics and 100 pages of Engineering Drawing. Alternatively, he can also read 50 pages of Engineering Mathematics and 250 pages of Engineering Drawing. Assume that the amount of time it takes to read one page of the textbook of either subject is constant. Ramesh is confident about Engineering Drawing and wants to devote full time in reading Engineering Mathematics. The number of Engineering Mathematics text book pages he can read on Tuesday is:

 A.

500

 B.

300

 C.

100

 D.

60

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Solution:
Option(C) is correct

In the given time Ramesh can read 80 pages of Engineering maths and 100 pages of Engineering drawing.
Or
He can read 50 pages of Engineering Maths and 250 pages of Engineering drawing.

⇒ 30 pages of Engineering Maths ≈ 150 pages of Engineering Drawing. 
⇒ 10 pages of Engineering Maths ≈ 50 pages of Engineering Drawing.

So in the given time Ramesh can read:

\(=80+\left(\dfrac{100}{50}\times 100\right)\)

\(=80+20\)

\(=100\) pages of Engineering Maths.


(4) Comment(s)


ABHIJEET
 ()

time to read 1 em page = em , similarly ed

we have to find number of em pages read in time T = T/em

80em + 100ed = T..........(1)

50em + 250ed = T..........(2)

subtracting both we get em = 5 ed => em/5 = ed

substituting back in any eq (1) or (2) we get 100em =T

so T/em = 100

thats all



Surabhi
 ()

Whats this solution?Its equivalent to having no soln.



Ritu
 ()

One more solution..

N = Total time allocated for study ; T=total number of pages read in EM; x=time taken to read 1 page of EM;y=time taken to read 1 page of ED

80 x + 100 y = N

50 x + 250 y = N

=> x = 5 y

Now y=0 since he has decided to study only EM

then T * x = N

T * 5 y = 500 y

T = 100

Answer is 100 pages of EM



Anu
 ()

Please explain this solution not clear