Alternative approach The boy is 6km away because he was late for 1/2 hour in which he would have covered 10/2= 5 km so he would have been 1 km away at...
let distance be 'd' and speed of train be x and boy be y d/(x+y)=20 d/(x-y)=30 x/d+y/d=1/20 x/d-y/d=1/30 adding both we get 2x/d=1/12 d/x=24 which i...
16/3 min
while doing reciprocal in inequality sign changes 7/3<x/(x+5)<17/3 3/7> (x+5)/x > 3/17 3/7> 1+5/x > 3/17 subtracting 1 from each si...
its a G.P sum of n terms of G.P =(arn-1)/(r-1) here a=1 ,r=2 22n – 49= 2n -1 2n (3)=48 2n =16 n=4
none ?
1 kg weight is must then for 2 kg weight we can have =1+1 or 2 0r 3-1 but 3-1 is best of them because then we can get 3 and 3+1=4 also from them so ...
n= 3 is the answer?
Answer is 4?
2nd let AB=a, BC=b, CA=c now a+b+c=2 and we have to find the max value of ab+bc which can also be written as b(a+c) from given equation a+c=2-b b(a+...
150 ?
Options ?
Let x be time taken by the cyclist while traveling from his home to crossing Earlier boy travels 10x distance, in same time train travels vx (assume ...
x^4 + x^2 +1 -y^2 =0 Since both roots are +ve c/a >0 1-y^2>0 1>y^2 So no possible value of y to be integer Hence no such pairs