1.Shaurya and Ajit take a straight route to the same terminal point and travel with constant speeds. At the initial moment, the positions of the two and the terminal point form an equilateral triangle. When Ajit covers a distance of 80 km, the triangle becomes right-angled. When Arjit was at a distance of 120 km from the terminal point, Shaurya arrived at the point. Find the distance between them at the initial moment assuming that there are integral distances throughout the movements described. (a) 300 km (b) 240 km (c) 200 km (d) 225 km (e) 275 km
2.Mrinalini and Neha travel to Connaught place along two straight roads with constant speeds. At the initial moment, the positions of Mrinalini, Neha and Connaught place form a right triangle. After Mrinalini travelled 30 km,the triangle between the points became equilateral. Find the distance between Mrinalini and Neha at the initial moment if at the time Mrinalini arrived at Connaught place, Neha had to cover 6.66 km to reach Connaught place a 10*3^1/2 b 12*3^1/2 c.30*5^1/2 d none
1.Shaurya and Ajit take a straight route to the same terminal point and travel with constant speeds. At the initial moment, the positions of the two and the terminal point form an equilateral triangle. When Ajit covers a distance of 80 km, the triangle becomes right-angled. When Arjit was at a distance of 120 km from the terminal point, Shaurya arrived at the point. Find the distance between them at the initial moment assuming that there are integral distances throughout the movements described.(a) 300 km(b) 240 km(c) 200 km(d) 225 km(e) 275 km2.Mrinalini and Neha travel to Connaught place along two straight roads with constant speeds. At the initial moment, the positions of Mrinalini, Neha and Connaught place form a right triangle. After Mrinalini travelled 30 km,the triangle between the points became equilateral. Find the distance between Mrinalini and Neha at the initial moment if at the time Mrinalini arrived at Connaught place, Neha had to cover 6.66 km to reach Connaught placea 10*3^1/2 b 12*3^1/2 c.30*5^1/2 d none
1. b) 240 ?
s1/80 = s2/(x) s1/(d - 120 - 80) = s2/(d - x)
80/(d - 200) = x/(d - x) 80d - 80x = dx - 200x 80d = dx - 120x x = 80d/(d -120) Put d = 240 x = 160
2. None? Distance between Neha and Connaught place = d 30 = 2d - d + a = d + a s1/a = s2/30 s1/(d-a-6.66) = s2/(d - a) a/(d-a-6.66) = 30/(d-a) a(d-a) = 30(d-a) - 200 (30-a)(d-a) = 200 (d)(d-a) = 200
Required distance will be d*sqrt(3) From options: d = 10 => a = -10 d = 12 => a = 12 - 200/12 = 12 - 16.66 = -4.66 So none of the given options match.
(1) All the divisors of 360, including 1 and the number itself, are summed up. The sum is 1170. What is the sum of the reciprocals of all the divisors of 360?
take a small case of 12= 2^2 * 3
no f factors = 3 * 2 = 6 (1,2,3,4,6,12)
sum f reciprocals = 1+1/2+1/3+1/4+1/6+1/12 = 1+2+3+4+6+12/12= 28/ (12)
(1) All the divisors of 360, including 1 and the number itself, are summed up. The sum is 1170. What is the sum of the reciprocals of all the divisors of 360?
divisor of 360 are 1,2,3,4......360 now, sum of reciprocal of divisor is =>1/1 +1/2 +1/3 + 1/4 +................................ + 1/360 =>1/360 * (360/1 +360/2 +360/3 + .......................360/360) =>1/360* (360 +180 +120 + + 3 + 2 + 1) =>1170/360=3.25
@poornaveturi either data insufficient or question is wrong or area is infinite..........solve it any way you get all equations in 'x' i.e. all lines parallel to y axis and no line parallel or inclined to axis this makes an open figure and area of open figure is infinite.................
I hope this won't be counted as spamming. All CAT 2013 aspirants please join this thread. We need to run this thread for overall prep of CAT. And please quant masters post some questions there also. 😃 http://www.pagalguy.com/forums/prep-resources/cat-2013-prep-t-94066/p-3622166?page=1
A person sells 20 kg s at 400 .A customer asks for 20 % discount and he agrees to it,but instead of 1 kg he gives 4 % less.what is the effective discount that the customer gets?
