tcs

1. If n is the sum of two consecutive odd integers and less than 100, what is the greatest possibility of n? Solution: Taking greatest possibility we take 51 + 49 but 51 + 49 = 100,  we need x + y < 100 so we take  47 + 49 = 96 2.  If the roots of the equation (x + 1) (x + 9) + 8 = 0 are a and b, then the roots of the equation (x + a) (x + b) - 8 = 0 are Solution: (x + 1) (x + 9) + 8 = 0  => (x^2) + 10x + 17 = 0  Hence a+b = -10 and ab=17 (x + a) (x + b) - 8 = 0  => (x^2) + (a+b)x + ab - 8 = 0  (x^2) + 10x + 17 - 8 = 0  => x = 1,9 3. If F(X) = AF(X)4 + B(X)2 + X + 5 , F(– 3) = 2, then F(3) = ? Solution: f(-3)=af(-3)4+bf(-3)+(-3)+5 2=a*2*4+b*2+2 8a+4b=0 f(3)=af(3)4+bf(3)2+3+5 f(3)[1-4a-2b]=_8 multiply by 2 both sides f(3)[2-(8a+4b)]=16 f(3)=8 (or) 81a+9b-3+5=2 =>81a+9b+5=5 now f(3) 81a+9b+3+5 =5+3=8 4.  4^2 + 2* 5^2 + 3* 6^2 +4* 7^2+...........+27*30^2 ? Solution: ∑ r^2 = n(n+1)(2n+1)/6 ∑ r^3 = [n(n+1)/2] ∴ 4^2+(5^3+6^3+

1. T he two sides of a triangle are 32 and 68. the area is 960 sq.cm. find the third side of triangle Solution: we see 68^2-32^2 =(68+32)*(68-32) =100*36=60^2 now (1/2)*60*32 =960(match with given options) (i.e area of a right angled triangle whose sides are 32,60,68) third side= 60 2.  Smita was making a cube with dimensions 5*5*5 using 1*1*1 cubes. What is the number of cubes needed to make a hollow cube looking of the same shape? Solution:  first are so (5*5*5)=125 then to make hollow cube v muz take out the outr most cubes  so 5-2 and 5-2 and 5-2 so 125-[(5-2)*(5-2)*(5-2)]  = 98 3. Arun makes a popular brand of icecream in a rectangular shaped bar 6cm long, 5cm wide and 4cm thick. To cut costs,the comapny had decided to reduce the volume of bar by19%. The thickness will remain the same, but the length and width will be decreased by the same percentage. the new width will be? Solution: volume=l*b*h =6*5*2 =60cm^3 volume is reduced by 19% new volume =(100-