@soumitrabengeri said:12?
approach bhi bata dijiye bhai
@soumitrabengeri I like ur concept, but I was wondering if I can do it thru concept of percentage...like 1/3 increase in speed will reduce time by 1/4 etc...
@dipz10 said:@soumitrabengeri I like ur concept, but I was wondering if I can do it thru concept of percentage...like 1/3 increase in speed will reduce time by 1/4 etc...
Yeah you can..it all depends on what method you feel comfortable using
@dipz10 said:@soumitrabengeri the problem is what if I start with option D), how will I solve it then?
Always go alphabetically..:P
@pankaj1988 said:On a straight line, five points are located. You choose any two points and note down the distance between them. The ten distances so obtained are written down in an increasing order as follows: 2, 4, 5, 7, 8, n, 13, 15, 17 19. What is the value of n?
a,b,c,d,e points from left to right
a---b----c----d----e
a to e=19
=>ab+bc+cd+de=19
=>2+5+8+4=19
given
ab=2
bc=5
cd=8
de=4
be=17
ad=15
bd=13
ac=7
so missing = ce=12
a---b----c----d----e
a to e=19
=>ab+bc+cd+de=19
=>2+5+8+4=19
given
ab=2
bc=5
cd=8
de=4
be=17
ad=15
bd=13
ac=7
so missing = ce=12
@pankaj1988 said:On a straight line, five points are located. You choose any two points and note down the distance between them. The ten distances so obtained are written down in an increasing order as follows: 2, 4, 5, 7, 8, n, 13, 15, 17 19. What is the value of n?
LET the line be A1A2A3A4A5
distance between subsequent points will be
A1A2=2
A2A3=5
A3A4=8
A4A5=4
DISTANCE NOT GIVEN IS A3A5=12
@pankaj1988 said:approach bhi bata dijiye bhai
Diff in weights is 2,4,5,7,8,n,13,15,17,19
So i assumed 3 numbers
Lowest number = 10
For difference of 2..it could be anything, but i assumed it to be lowest number+2..so that the diff of 17 and 19 gets adjusted automatically if we assume the next...
Then i assumed that the max diff of 19 would be between the lowest and highest number..so i got highest number as 29..thus diff of 17 is also adjusted
Now to adjust diff of 5 and 7..if i selected a number greater than 10, then the diff of 5 and 7 gets automatically adjusted
So as of now i have 10,12,17,29 and the differences satisfied are..2,5,7,17,19 and 12..which eventually became n
Last number remains..this no should satisfy diff of 4,8,13,15
So the number cannot be 14 or 16 or 33
Next i assumed the number to be 21..but it does not satisfy all conditions..
Last assumption 25 (29-4)..which ended up satisfying all conditions
@dipz10 said:Pia and Ria simultaneously started running from point A towards point B at a distance of 10 km on a linear track. Pia on reaching point B, took a rest for 30 minutes, turned back and started running towards point A. Ria on reaching point B, immediately turned back and started running towards point A. Both of them reached point A simultaneously. If the difference between their speeds is 2 kmph, in how much time will Ria cover a distance of 24 km?A) 3 hoursB)2.4 hoursC)2 hoursD)3.5 hoursIf anyone knows how to solve this question without using equation will be welcomed...
20/p+1/2=20/r
=>20/r-20/p=1/2
=>20p-20r=rp/2
=>40p-40r-rp=0
=>(40+p)(40-r)=1600
also p=r+2
=>(42+r)(40-r)=1600
=>1680-2r-r^2=1600
=>r^2+2r-80=0
=>(r-8)(r+10)=0
r=8
so time=24/8=3 hours
A and B are two points in a plane which are 10 m apart from each other. The angle of elevation of the top of a tower as seen from the points A and B are 30째 and 60째 respectively. If AB subtends an angle of 60째 at the foot of the tower, find the height of the tower.
@bullseyes said:A and B are two points in a plane which are 10 m apart from each other. The angle of elevation of the top of a tower as seen from the points A and B are 30째 and 60째 respectively. If AB subtends an angle of 60째 at the foot of the tower, find the height of the tower.
s is a six digit number beginning with 1. if the digit is moved from the leftmost place to the rightmost, the number obtained is three times s. sum of digits of s is:
a.21
b.24
c.26
d.27
with approach please
@realslimshady said:s is a six digit number beginning with 1. if the digit is moved from the leftmost place to the rightmost, the number obtained is three times s. sum of digits of s is: a.21b.24 c.26 d.27with approach please
3*1abcde=abcde1
3(10^5+abcde)=abcde*10+1
299999=7abcde
abcde=42857
1+4+2+8+5+7=27
@realslimshady said:s is a six digit number beginning with 1. if the digit is moved from the leftmost place to the rightmost, the number obtained is three times s. sum of digits of s is: a.21b.24 c.26 d.27with approach please
1 a b c d e * 3 = a b c d e 1
=>
300000+a*30000+b*3000+c*300+d*30+e*3=a*100000+b*10000+c*1000+d*100+e*10+1
==>299999=7*(a*10000+b*1000+c*100+d*10+e)
==>42857=abcde
sum= 26
=>
300000+a*30000+b*3000+c*300+d*30+e*3=a*100000+b*10000+c*1000+d*100+e*10+1
==>299999=7*(a*10000+b*1000+c*100+d*10+e)
==>42857=abcde
sum= 26
We need more questions here.
@krum said:1 a b c d e * 3 = a b c d e 1=>300000+a*30000+b*3000+c*300+d*30+e*3=a*100000+b*10000+c*1000+d*100+e*10+1==>299999=7*(a*10000+b*1000+c*100+d*10+e)==>42857=abcdesum= 26 + 1
One desi method of doing it::
altho it can be done a bit fast...idhar likhne mien jyada time laga. :D
above method is good. but i=you could also do it like
3*1bcdef = bcdef1
a=1, f has to be 7 then onli it will give unit digit as 1 on multiplication,
3*1bcde7 = bcde71 >> 3e+2 =7...onli possible for e=5, one carry,
3*1bcd57 = bcd571 >>3d +1 =5 possible for d =8...2 carry,
3*1bc857 = bc8571 >> 3c + 2 = 8 ..possible for c =2 no carry...
3*1b2857 = b28571 >>> b= 4
above method is good. but i=you could also do it like
3*1bcdef = bcdef1
a=1, f has to be 7 then onli it will give unit digit as 1 on multiplication,
3*1bcde7 = bcde71 >> 3e+2 =7...onli possible for e=5, one carry,
3*1bcd57 = bcd571 >>3d +1 =5 possible for d =8...2 carry,
3*1bc857 = bc8571 >> 3c + 2 = 8 ..possible for c =2 no carry...
3*1b2857 = b28571 >>> b= 4
142857 = 27
How many subsets of {1, 2, 3 €Ś 11} contain at least one even integer?(1) 1900 (2) 1964(3) 1984 (4) 2048
2^11 - 2^6
2048 - 64
1984?
@Ashmukh said:How many subsets of {1, 2, 3 €Ś 11} contain at least one even integer?(1) 1900 (2) 1964(3) 1984 (4) 2048