261⟩ What is the minimum number of comparisons required to find the second smallest element in a 1000 element array?
999 comparisons are required to find the second smallest element in an array.
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999 comparisons are required to find the second smallest element in an array.
distancetravelled by the bird=2000m
time taken=187.5 sec
=375/2 sec
speed is in the ratio of 1:2
dividing the time in the ratio of 1:2
time to=375/2(2/3)=125 sec
speed=distane/time=1000/125=8 m/sec
speed in thefro direction=16m/sec
Sqrt of ((d/a)^2+(d/c)^2)
d=18 a=2 b=9 c=-3
(d/a)^2=81 (d/c)^2=36
Sqrt(81+36)=58.5
2xy/xy= average speed. its very simple
Let 'x' be the no of book that we buy with 5 rupee less from the original book price and we buy 5 more books with the rest money each costing original book price.
so the no of cases arises are
Case 1:
x(15-5)+5(15)=300
Solving this equation we get the value of x =22.5 (which is a fraction value)
so this case is not possible.
Case 2:
x(20-5)+5(20)=300
Solving this equation we get the value of x =13.33 (which is a fraction value)
so this case is not possible.
Case 3:
x(25-5)+5(25)=300
Solving this equation we get the value of x =8.75 (which is a fraction value)
so this case is not possible.
Case 4:
x(30-5)+5(30)=300
Here we get the value of x =6 (which can be possibly a no. of book, so the original price of the book is 30).
Fill the 7L bucket. Empty water from the 7L bucket into the 3L bucket, until the 3L bucket is full. Empty the 3L bucket and repeat. Now there is 1L of Water in the 7L bucket. Empty the 3L bucket and transfer the 1L of water from the 7L into it. Now fill the 7L bucket again. This time fill the 3L bucket until it is full, you will remove 2L of water from the 7L bucket because there is already 1L of water in the 3L bucket. Now you have 5L remaining in the 7L bucket.
The given question is not fully correct .
I will give u another eg ...
If a cube is painted red and cut into 64 pieces .
Approach this question as ,,
64 pieces = 4 *4* 4 so take n=4
just use this formula
1. no sided painted =(n-2)^3
2. 1 sided painted =6 * (n-2)^2
3. 2 sided painted = 12 *(n-2 )
4. 3 sided painted = always 8 (a cube has 8 corners)
All three are football players
We have 2 equations
n*10 = x and (n-10)*20 = x which gives n=20, where n is number of men to begin with.
ans= 25
bcoz, 3*3=9
3*6=18
3*7=21
5*5=25
5*4=20
so like(3*6 and 3*7 it will be 5*4 and 5*5 mens increment by 1)
It is not possible all the time, that is, A No. is divided by three is a power of three.
4/5 full tank= 12 miles
1 full tank= 12/(4/5)
1/3 full tank= 12/(4/5)*(1/3)= 5 miles
1/3+1/5+1/x=1/8
solution:
1/3+1/5-1/8=1/x
40+24-15 =1/x
120
49/120=1/x
x=120/49
or
x=2 22/49 in fraction
30 games
13
20 students
If there are 5 questions, and only 3 question are slected, that is 10 different possible combinations. There 3 section, and with each section having 10 possible combinations, give 1000 different possible combinations.
245 pieces have no color on them
98 pieces have black on one side
among 3 socks she has to choose 2 black socks from 9 pair(=18) of black socks, that can be done in 18C2 ways. and the third sock can be chosen from among the remaining socks that can be done in ((8+9-1)x2 )C 1 ways i.e. in 32 ways.
ans. 32 X 18C2
south