@Tina_angel it happens
@sonic_boom Question is from Number system P-23, Q-13.
How you arrived at the solution?
Note - Answer isn't provided
How you arrived at the solution?
Note - Answer isn't provided
@anilapex said:For the number 7200 find, the sum & no.of factors not divisible by 75?
@ashishgupta144 said:@anilapex number of factors is 18 ?
@sonic_boom said:no. of factors 12sum of factor 252whats the ans?
Solution I found is -
Sum of factors = 6489
No.of factors = 42
Strategy I used:
Sum of factors not divisible by 75 = Sum of all factors - Sum of factors divisible by 75
6489 = 25389 - 18900
Similarly,
No.of factors not divisible by 75 = Total factors - No.of factors divisible by 75
42 = 54 - 12
@anilapex said:Solution I found is -Sum of factors = 6489 No.of factors = 42Strategy I used:Sum of factors not divisible by 75 = Sum of all factors - Sum of factors divisible by 75 6489 = 25389 - 18900Similarly,No.of factors not divisible by 75 = Total factors - No.of factors divisible by 75 42 = 54 - 12
yes u r right.
Find the remainder when 51^203 is divided by 7?
a) 4 b) 2 c) 1 d) 6
Pls. Note:- Providing steps of the solution will be helpful
remainder when 51^203 is divided by 7 => (51^203) mod 7
=((7*7+2)^203 ) mod 7
=(2^203)mod7
=(2^2 * 2^201)mod 7
=(4 * 2^3*67 )mod 7
=(4 * 8^67 ) mod 7
=(4 * (7+1)^67) mod 7
=(4)mod 7
= 4
@anilapex : 51^203/7 => 2^203/7 --> Euler number of 7 = 6 => [(2^6)^33]*[2^5]/7 => 2^5/7 => Rem = 4.
@anilapex said:Find the remainder when 51^203 is divided by 7?a) 4 b) 2 c) 1 d) 6Pls. Note:- Providing steps of the solution will be helpful
51^203/7 => 2^203/7 => (2^201 * 2^2)/7 =>(1^67 * 2^2)/7 => 4
Q) There are three taps A,B,C in a tank. They can fill the tank in 10 hrs,20 hrs and 25 hours respectively. all of them are opened simultaneously. Then after 2 hours, tap C is closed and A and B kept running. Then after the 4th hour, tap B is closed. The remaining work is done by A alone. Find the percentage of work done by tap A itself. (Thanks in advance ) and is 72%
@ramirez said:Q) There are three taps A,B,C in a tank. They can fill the tank in 10 hrs,20 hrs and 25 hours respectively. all of them are opened simultaneously. Then after 2 hours, tap C is closed and A and B kept running. Then after the 4th hour, tap B is closed. The remaining work is done by A alone. Find the percentage of work done by tap A itself. (Thanks in advance ) and is 72%
pahele 2hrs me work (2/20+2/20+2/25) + agle 2 hrs me work ( 2/10+2/20)
total work in 4hrs =17/25
work left =1- 17/25 =8/25
total work done by A= 8/25 alone will be done by A + 4/10 (work done in 4hrs)
=18/25*100=72%
@ramirez said:Q) There are three taps A,B,C in a tank. They can fill the tank in 10 hrs,20 hrs and 25 hours respectively. all of them are opened simultaneously. Then after 2 hours, tap C is closed and A and B kept running. Then after the 4th hour, tap B is closed. The remaining work is done by A alone. Find the percentage of work done by tap A itself. (Thanks in advance ) and is 72%
A's 1hr work = 10%
B's 1hr work = 5%
C's 1hr work = 4%
work done by A in 4hrs = 40%
by B in 4 hrs = 20%
by C in 2 hrs = 4%
total work done = 40+20+8 = 68%
work remaining = 32 %. It is to be done by A
So total work done by A = 40+32 = 72%
B's 1hr work = 5%
C's 1hr work = 4%
work done by A in 4hrs = 40%
by B in 4 hrs = 20%
by C in 2 hrs = 4%
total work done = 40+20+8 = 68%
work remaining = 32 %. It is to be done by A
So total work done by A = 40+32 = 72%
Q)It takes 6 days for 3 women and 2 men working together to complete a work. 3 Men would do the same work 5 days sooner than 9 women. How many times does the output of am an exceed with that of a woman..? Thanks in advance:-)
Q) Two women A and B are working on an embroidery design. If A worked alone she would need 8 hr more to complete the design than if they both worked together. Now if B worked alone she would need 4.5 hr more to complete the design than they both working together. What time would it take B alone to complete the design? Thanks in advance:-) and is 10.5 hr
@ramirez :
Let, a women does 1/w in 1 day and a man does 1/m in 1 day.
3/w + 2/m = 1/6 ---------[1]
let, 9/w = 1/d => 3/m = 1/(d-5) ------- [2]
substitute, w and m in [1].
1/3d + 2/3(d-5) = 1/6 => d^2-11d+10 = 0 => d=1, 10 ==> d=10.
9/w = 1/10 -> 1/w = 1/90.
3/m = 1/5 -> 1/m = 1/15.
Men are 6 times efficient.
@ramirez : let A and B do 1/A and 1/B work in a day respectively. let A+B do 1/w in a day.
1/(w+8) = 1/A ----[1]
1/(w+4.5) = 1/B ----[2]
1/(w+8) + 1(w+4.5) = 1/w ... solve : w = 6 .. in one day B does 1/10.5 --> ans. 10.5 days.
For heap problem the answer is a