Quantitative Ability Questions
Quantitative Ability Questions:
1. A chord of length 32 cm is drawn inside a circle of radius 20 cm. And a point, whose distance from the centre of circle is 13 cm, is marked on the chord. Calculate the lengths of the segment of the chord made by this point.
a) 21 cm and 11 cm
b) 19 cm and 13 cm
c) 16 cm each
d) 18 cm and 14 cm
e) 17 cm and 15 cm
2. Tiwari sells samosas in boxes of different sizes. The samosas are priced at Rs. 2 / samosa up to 200 samosas. For every additional 20 samosas the price of the whole lot goes down by 10 paise / samosa. What should be the maximum size of the box that would maximize the revenue?
a) 240
b) 300
c) 400
d) 350
e) None of these
Related posts:


December 7th, 2009 at 12:26 am
1.c..16cm
2.e..
December 7th, 2009 at 1:00 am
Social comments and analytics for this post…
This post was mentioned on Twitter by sudiptaroy: Published a new post: Quantitative Ability Questions http://tinyurl.com/ye8ymkw...
December 7th, 2009 at 2:44 am
1. A
2. B
December 7th, 2009 at 2:56 am
Explanation:-
1. Distance of the Permendicular from the center to cord which devides cord equal distance (32/2=16) = sqroot of (20^2-16^2)=12.
Now the point on the cord having distance = 13 from the center is situated from the point of intersection between cord and perpendicular drawn from center = squroot of (13^2-12^2)=5.
so required answer=16+5 and 16-5
or 21 and 11.
2. Say x is the extra somosa after 200.
Revenew=(200+x)*(2-x/200)=400+x-x^2/200
For Max(Revenew), d/dx(Revinew)=0
so we get x=100
so required answer=200+100=300.
December 7th, 2009 at 4:59 am
uttam i didn’t read the first question correctly so i agree with your answer,,although the texts are not very clear..
And in second question your method seems to be oK but have you ysed 10 paise / samosa any where?
December 7th, 2009 at 5:25 am
Hey Govind,
Let say u are using extra x somosa.
For every additional 20 samosas the price of the whole lot goes down by 10 paise / samosa.
so for x somosa it will (x/20) * 10 paisa
=(x/20)*(10/100) rupies
=x/200
so the cost per somosa will be 2-(x/200)
Now u have (200+x) somosa
So Revenie =F(x)=(200+x)*{2-(x/200)}
now rest is same as above.
Is it clear now????
December 8th, 2009 at 9:35 pm
Agreed dude..Thanks for the lucid explanation..
December 10th, 2009 at 5:44 am
thats a nice explanation for 2nd qs…thanks 4 dat..