10
16
12
3, 7, 15, 39, 63, 127, 255, 511
C. 39
3*2+1 = 7
15*2+1 = *31*
31*2+1 = 63
63*2+1 = 127
127*2+1 = 255
255*2+1 = 511
17,19,23,29
I solve it as the 280m train being stationary… and the 120m train moving at (42+30) = 72kph…. or 20mps
The front of the train will have to go 280m to pass the stationary train plus another 120m for the backs to clear… that is 400m in all.
The time to travel 400m at 20mps is 20secs
let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y-3’
and 4th digit is ‘Y+4’
then the no is ‘(X)(Y)(Y-3)(Y+4)(3X)’
from the above we can say 3X<=9
so X<=3 and any of the digit in the number is <=9
and also given that 3 pairs sum is 11...
so make trial and error..
if X=1...any of the no is 10 which is wrong trial....
if X=2...then let Y+4=9 ==> Y=5
then no is 25296
first pair 2+9=11
second pair 2+9=11
third pair 5+6=11
now the answer is 25296
10000
one-legged =5% of 10000=500
remaining=10000-500=9500
barefooted=9500/2=4750
remaining people= 9500-4750=4750
hence required number of shoes= 4750*2+500*1=100005% of 10000 = 500 one legged
9500 / 2 = 4750 bare foot
minium no of shoes = 4750*2 + 500*1 = 10000
The numbers that lie between 100 and 1000 which are divisible by 14 are 112, 126,140 …,994
a = 112; l = 994, d = 14
n= (l−a)/d+1
= (994-112)/14+1
= 64
Sn=n/2(l+a)
= 64/2(994+112)
= 32*1106
= 35392
Let the numbers be x and x + 2.
Then, (x + 2)2 – x2 = 84
⇒ 4x + 4 = 84
⇒ 4x = 80
⇒ x = 20.
∴ The required sum
= x + (x + 2)
= 2x + 2
= 42
Ratio is 3:2:1
Total=6
Reverse of ratio is 1/3:1/2:1/1
Portion of first part is=1/3*6=>2
Portion of second part is=1/2*6=>3
Portion of third part is 1*6=>6
∴So the new ratio is 2:3:6
40
u c 5 numbers…..
we add the combination of digits…..
a.17->1+7=8(even)
b.23->2+3=5(odd)!!!!!
c.37->3+7=10(even)
d.13->1+3=4(even)
e.73->7+3=10(even)
all combinations are even xept 23
ans:23
I am working institute
16.25
5:6::7:10::6:5
=((5/6)/(7/10)/(6/5))
=((5/6)*(6/5)/(7/10))
=(1/(7/10))
=10/7
d
2, 6, 12, 72, 824
2
6*2=12
12*6=72
72*12=864
answer = 16.5Rs
A=3*9=27Rs
B=7*15=105Rs
132 R for 10 kg
therefore,
1kg amount = 132/10=13.2 so,
5kg mixture price = 13.2*5=66Rs
mixture will sell 25%profit then the price =66*(25/10)=33/2=16.5Rs
The formula to find number of diagonals (D) given total number of vertices or sides (N) is
N * (N – 3)
D = ———–
2
Using the formula, we get
1325 * 2 = N * (N – 3)
N2 – 3N – 2650 = 0
Solving the quadratic equation, we get N = 53 or -50
It is obvious that answer is 53 as number of vertices can not be negative.
Alternatively, you can derive the formula as triange has 0 diagonals, quadrangel has 2, pentagon has 5, hexagon has 9 and so on……
Hence the series is 0, 0, 0, 2, 5, 9, 14, …….. (as diagram with 1,2 or 3 vertices will have 0 diagonals).
Using the series one can arrive to the formula given above.