TCS
Company
Numerical Ability
Geometry
two identical circle so that their centre,and the point at which they intersect , form a square of side 1cm.Find the area in sq. cm of the portion that is common to the two circles?
Read Solution (Total 6)
-
- required area= 2*(area of sector) - area of square
= (pi/2)-1 cm^2
- 10 years agoHelpfull: Yes(13) No(0)
- area of one sector is (90/360)*pi*1 cm^2
area of sq is 1cm^2
common area = 1-(2*pi/4)=1-pi/2 - 10 years agoHelpfull: Yes(5) No(7)
- (area of 2 quad - area of square)/2
(pi/4+pi/4-1)/2
answer will be (pi-2) cm^2 - 10 years agoHelpfull: Yes(2) No(6)
- area of square=1 cm^2
area of one sector=90/360*3.14*1 cm^2 =0.785 cm^2
area of remaining portion=area of sq.-area of one sector=.215cm^2
common area of both the circle= 1-(2*0.215)=0.572cm^2 - 9 years agoHelpfull: Yes(2) No(0)
- sorry! gopikant is 100% correct. coz pi/2 is always greater than 1
- 10 years agoHelpfull: Yes(1) No(1)
- as the angle formed by each portion of the two circle is 90 .so the aarea of that part will be (pi*r^2)/4
now the area common between two circle is double of this .. i.e. 2*(pi*r^2)/4=(pi*r^2)/2.
r=1. so the area will be pi/2 - 10 years agoHelpfull: Yes(0) No(2)
TCS Other Question
Distance between 2 places A and B is 180 km. A 3rd place C is between A and B, about 120 km from A. Ram started from A and Shyam started from B simultaneously at 11AM. However Shyam waits for 15mins at C to eat mangoes. Speed of Ram and Shyam are 70kmph and 50kmph resp. Find the time at which they meet..
a. 12:36pm, b. 12:30pm, c. 12:45pm, d. 12:25pm, e. 12:40pm
A, B, C, D, E, F and G are seven positive integers. And given
a. AB is odd.
b. C+ DE is odd.
c. AD is odd.
d. FACG is odd.
How many of the above integers are odd?