@Amrofa C is 10% faster than B.
@Amrofa said:In a 200 m sprint, A and B start running from the Start line but A starts running, four seconds after B starts running. C starts running at the same time as B but from a point which is 20 m behind the Start line. If all three of them finish the race together, which of the following is definitely true?In a 100 m race, A beats C by 1 m.In a 100 m race, C beats A by 1 m.A and C have the same speed.C is 10% faster than B.
c is 10% faster than b?
@audiq7 @mailtoankit
please Share Your Approach
There are 10 stations in between two stations A and B. The number of different kinds of tickets to be printed so that a passenger can make a reservation between any two stations is
please Share Your Approach
There are 10 stations in between two stations A and B. The number of different kinds of tickets to be printed so that a passenger can make a reservation between any two stations is
@Amrofa said:@audiq7@mailtoankitplease Share Your Approach
lets say speed of B is 10m/s. to A and C ka speed aayega 12.5 and 11. options check karlo.
@Amrofa said:@audiq7@mailtoankitplease Share Your ApproachThere are 10 stations in between two stations A and B. The number of different kinds of tickets to be printed so that a passenger can make a reservation between any two stations is
2*(12C2) ?
there are in total 12 stations.
we have to select 2 stations out of these 12 and since the tickets can be either way, we have to multiply it by 2.
@Amrofa said:@audiq7@mailtoankitplease Share Your ApproachThere are 10 stations in between two stations A and B. The number of different kinds of tickets to be printed so that a passenger can make a reservation between any two stations is
11 ! * 2 ??
@Amrofa said:In a 200 m sprint, A and B start running from the Start line but A starts running, four seconds after B starts running. C starts running at the same time as B but from a point which is 20 m behind the Start line. If all three of them finish the race together, which of the following is definitely true?In a 100 m race, A beats C by 1 m.In a 100 m race, C beats A by 1 m.A and C have the same speed.C is 10% faster than B.
(speed of b)/(speed of c)=220/200=11/10
therefore, C is 10% faster than B.
Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c) exist such that the given 18 digit number is divisible by 4?
@Amrofa said:Farook marks up the price of an article by 50% and then offers a discount of 20% to Shahrukh. Shahrukh sells it for र20 more than the price at which he purchased it. If Shahrukh's selling price is 30% more than the original cost price of the article, then Shahrukh's profit percentage is
Let original price be x
Marked price be 1.5x
Selling price of farook be 1.2x
cost price of Shahruk be 1.2x
selling price of shahruk be 1.2x+20
1.2x+20=1.3(x)
.1x=20
x=200
Cost Price of shahruk=240
Selling price of shahruk=260
profit percentage=20/240*100=8.33%
@ravi.theja said:Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c) exist such that the given 18 digit number is divisible by 4?
for 123abc231bca312cab to be divisible by 4, ab has to be divisible by 4.
And out of 0,1,2,3,4,5,6, (which are the possible digits as it's in base 7).. the following could be the "ab": 04, 12,16,20,24,32,36,40,52,56,60,64.
(total: 12)
While "c" can be any digit from 0-6. (total: 7)
so total ordered triplets: (no of c possible)*(no of a,b possible): 7*12= 84
@Zedai said:for 123abc231bca312cab to be divisible by 4, ab has to be divisible by 4. And out of 0,1,2,3,4,5,6, (which are the possible digits as it's in base 7).. the following could be the "ab": 04, 12,16,20,24,32,36,40,52,56,60,64. (total: 12)While "c" can be any digit from 0-6. (total: 7)so total ordered triplets: (no of c possible)*(no of a,b possible): 7*12= 84
how did u reach this ab has to be divisble by 4
its a number in base 7 not base 10
@ravi.theja said:Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c) exist such that the given 18 digit number is divisible by 4?
ab div by 4 :
b can take 0,2,4,6 values.
if b is 0 ,a can be 0,2,4,6
b is 2,a can be 1,3,5
b is 4,a can be 0,2,4,6
b is 6 ,a can be 1,3,5
total ,a,b combos =4+3+4+3=14 c can take 7 values
hence ,ans is 98 (14*7)
.
@Zedai,err this number is in base 7....so why are you just taking the last digits?... for example...
232 in base 7..is not divisible by 4.
@Zedai said:for 123abc231bca312cab to be divisible by 4, ab has to be divisible by 4. And out of 0,1,2,3,4,5,6, (which are the possible digits as it's in base 7).. the following could be the "ab": 04, 12,16,20,24,32,36,40,52,56,60,64. (total: 12)While "c" can be any digit from 0-6. (total: 7)so total ordered triplets: (no of c possible)*(no of a,b possible): 7*12= 84
bhai..hw abt considering 7a+b to be divisible by 4??is dat the right approach?
@Subhashdec2 said:how did u reach this ab has to be divisble by 4its a number in base 7 not base 10
holy cow! i took it in base of 10!! 
@ravi.theja said:Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c) exist such that the given 18 digit number is divisible by 4?
123abc231bca312cab mod 4
-1+2-3+a-b+c-2+3-1+b-c+a-3+1-2+c-a+b
-2+a+b+c=4n
-1+2-3+a-b+c-2+3-1+b-c+a-3+1-2+c-a+b
-2+a+b+c=4n
n=1
a+b+c=2
4c2=6
n=1
a+b+c=6
no of sol=8!/6!2!=28
n=2
a+b+c=10
no of sol=12C2=66
subtract where any of them is more than 7
a'+7+b+c=10
a'+b+c=3
5c2*3=30
66-30=36
n=3
a+b+c=14
(6,6,2)=3
(6,5,3)=6
(6,4,4)=3
(5,5,4)=3
total 15
n=3
a+b+c=18
1 case
total=6+28+36+15+1=84
How many three-digit natural numbers have their units digit neither less than the hundreds digit nor the tens digit?