2* pi* r ( h + r )
r=3cm
h=7cm
2 * 22/7 *7 ( 10 ) = 440 sq cm
4, 5, 15, 49, 201, 1011, 6073
Correct option is A)
2nd term =(1stterm×1+2)=(4×1+2)=6
3rd term =(2ndterm×2+3)=(6×2+3)=15
4th term =(3rdterm×3+4)=(15×3+4)=49
5th term =(4thterm×4+5)=(49×4+5)=201 and so on
∴5 is wrong.
Ans = 8
Use simple box method
[1][2][2][2]= 1x2x2x2 =8
Logic >
once place have 2 number (2,6)
Tens place have also 2 number (7,3)
So that number is divisible by 4
Now, 2 numbers fixed at once and tens place left 2 numbers which is choice at hundredth place
Now 3 number fix left 1 number which is placed at thousands place
Total number form is 4×4 = 16
But divisible by 4 is 1x2x2x2 =8
ans : 101 bcoz after every two matches one team is eliminated…so to eliminate 50 team there will be require 100 matches…+ 1 for deciding winner…
900
Let x and y be the two numbers. And x > y.
Then,
x + y = 80—–(1)
y=80-y
Also,
x = 4y – 5
Solving the two equations,
80 – y = 4y – 5
85 = 5y
Therefore, y = 17
put the value of y in equation(1)
then
x+y=80
x+17=80
x=63
the numbers are 17&63
I would expect progression in my career through acquiring new skills
only if the results are greater than my input
30%
suppose the area is 100 and it was increased by 69%, then
the area is 169 which it indicates the side of the square
is 13.. which means 30% increase in its side.
Blue
If n = no. of revolutions for back wheel
9n = 7n + 10*7
n=315
Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1 cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2 i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61,P(B)=61,P(C)=65,P(A)=61,P(B)=65,P(C)=61 and P(A)=65,P(B)=61,P(C)=61
Hence the required probability = 61×61×65×6×3=21690=125
Collective performance
Postive attitude towards working
Shaping team player
Turning individuals into team players