Crux of the problem: many of us are tempted to make all the variables equal. This technique would help only when expression is cyclical in nature. for example (ab,bc,ca) is cyclical, i.e if we replace a by b, b by c and c by a ...we get back the original expression. But a closer look reveals that all the terms inside the bracket are not fully cyclical. Hence variables are not interchangeable out of all the five values i.e a+b+c, b+c+d, c+d+e, d+e+f, e+f+g we observe that terms containing a,d,g are disjoint. i.e if any expression contains a then it does not contain d and g. so we set a=d=g=1/3 and rest all the other variables = 0.ATDH.
__/\__ Thanks a lot sirjee...wasn't getting a solution for this problem.. Nothing wonders me more about you except your timing :P
Nothing else wonders me because now i am used to being in awe!!
A railway line passes through (in order) the 11 stations A, B, ... , K. The distance from A to K is 56. The distances AC, BD, CE, ... , IK are each = 17. The distance between B and G is
A railway line passes through (in order) the 11 stations A, B, ... , K. The distance from A to K is 56. The distances AC, BD, CE, ... , IK are each = 17. The distance between B and G is (1) 21 (2) 23 (3) 27 (4) 29 (5) 31
A railway line passes through (in order) the 11 stations A, B, ... , K. The distance from A to K is 56. The distances AC, BD, CE, ... , IK are each = 17. The distance between B and G is (1) 21 (2) 23 (3) 27 (4) 29 (5) 31
A railway line passes through (in order) the 11 stations A, B, ... , K. The distance from A to K is 56. The distances AC, BD, CE, ... , IK are each = 17. The distance between B and G is (1) 21 (2) 23 (3) 27 (4) 29 (5) 31
A B C D E F G H I J K let A to D = 17 A to B = 5 B to C = 6 C to D = 6 B to E = 17 B to C = 6 C to D = 6 D to E = 5 C to F = 17 C to D = 6 D to E = 5 E to F = 6 D to G = 17 D to E = 5 E to F = 6 F to G = 6 distance from B to G = 29
Crux of the problem: many of us are tempted to make all the variables equal. This technique would help only when expression is cyclical in nature. for example (ab,bc,ca) is cyclical, i.e if we replace a by b, b by c and c by a ...we get back the original expression. However, in the problem under consideration a closer look reveals that all the terms inside the bracket are not fully cyclical. Hence variables are not interchangeable out of all the five values i.e a+b+c, b+c+d, c+d+e, d+e+f, e+f+g we observe that terms containing a,d,g are disjoint. i.e if any expression contains a then it does not contain d and g. so we set a=d=g=1/3 and rest all the other variables = 0.ATDH.
__/\__.pranam sirji..sir aap kaise soch lete ho is tarah..???..amazing hai sirji...thoda hmein bhi bataein..
Thanks for complement mate.On a lighter note: Q: aap kaise soch lete ho is tarah..???Ans: I dont know"thoda hmein bhi bataein.."Ans: Sir, last 8 years se PG pe share kar raha hun jitna mujhe aata hai.ATDH.
are sir aap "sir" na bolo mujhe..abhi deserve nai karta hu...honestly...but seriously..hats off to ur thinking.......app kamaal hain..
"thoda hmein bhi bataein.."Ans: Sir, last 8 years se PG pe share kar raha hun jitna mujhe aata hai.ATDH.
aur sir jii hum apse sikhte aa rahe hai
Sohan and Rohan cut a regular hexagon and regular octagon out of circular papers of radius 20 each. find difference between perimeters of hexagon and octagon
...wats ur approach??
