@Budokai001 said: Aizen bhai,OA is d)26 .... you have the first term as well na ? i proceeded like... (a+b)^2 -(a^2+b^2)= 2ab --->divisble by 2 (a+b)^3-(a^3+b^3)=3(a^2b+aB^2) --. divisible by 3 and so extending this(a+b+c+d)^13 - (a^13+b^13+c^13+d^13) will be divisble by 13 From options 26 is the only lowest multiple of 13.

EDITED: Dhanyawad @[251735:chillfactor] sir dhyan me hi nai rha ..... Agey kabhi galti nai hogi

@dementors said: options are: 252141

@dementors said: If log25 = log5y, then what is the appropriate value of y? Please help

@RaghavMittal said: Kindly help,The number 523abc is divisible by 7,8 and 9. Then the value of a*b*c isA. 10B. 60C. 180D. Cannot be determinedKindly share the approach as well..Thank you..

@Aizen said: Iska option B hoga... a^m+b^n+c^s is divisible by (a+b+c) when (m,n,s) are odd. p.s: @chevy bhai , @culdip sir __/\__

@dementors said: If log25 = log5y, then what is the appropriate value of y? Please help

@Budokai001 said: OA is A (a+b+c+d)^13 - (a^13+b^13+c^13+d^13) is always divisble by a)156b)a+b+c+dc)12d)26

@Budokai001 said: set 'S' is defined as shown .S={3,10,17,24.....1004}'A' is a subset of S such that the product of any two elements of A is greater than 603000.What is the maximum possible number of elements in A a)33 b)32c)31d)30

@pyashraj said: @yashasvi5 Q.2 Two men and a women are entrusted with a task. The second man needs three hours more to cope with the job than the second man and the woman would need working together. The first man, working alone, would need as much time as the second man and the woman working together. The first man, working alone, would spend eight hours less than the double period of time the second man would spend working alone. How much time would the two men and the women need to complete the task if they all worked together? Is the answer 1hour...

@Stoicalme said: @sumeet1489 Yes saar,method please.

@Budokai001 said: set 'S' is defined as shown .S={3,10,17,24.....1004}'A' is a subset of S such that the product of any two elements of A is greater than 603000.What is the maximum possible number of elements in A a)33 b)32c)31d)30

@Budokai001 said: set 'S' is defined as shown .S={3,10,17,24.....1004}'A' is a subset of S such that the product of any two elements of A is greater than 603000.What is the maximum possible number of elements in A a)33 b)32c)31d)30

@sumeet1489 said: option a hai kya iska

@ImCATian said: why cant we derive the solution without simplyfying..I too got the answer after getting fractions simplified.Thanks for attempt..

@Stoicalme said: One side of a staircase is to be closed in by rectangular planks from the floor to each step.The width of each plank is 9 inches and their height are successively 6 inches ,12 inches , 18 inches.... and so on.There are 24 planks required in total.Find the area in square feet.

@Budokai001 said: set 'S' is defined as shown .S={3,10,17,24.....1004}'A' is a subset of S such that the product of any two elements of A is greater than 603000.What is the maximum possible number of elements in A a)33 b)32c)31d)30

@chevy said: haan yaar,,galti kar di maine,, 1) 4/6=2/3 6/9=2/3 5/4 LCM will be= LCM of numer/HCF of denomi = 10/1=10 2) 4/6=2/3 3/9=1/3 6/5 HCF will be=HCF on nume/LCM of denomi = 1/15 P.S.--- Aaj toh mela laga hua hai yahaan pe,,,lage raho bhai log

@Stoicalme said: LCM of fractions= (LCM of numerators)/(HCF of denominators) HCF of fractions=(HCF of numerators)/(LCM of denominators)Hence, the answers :1) 60/1. 2) 1/90

@ImCATian said: Plz solve this 1) Find LCM of 4/6,6/9,5/4.2) Find HCF of 4/6,3/9,6/5.

@ImCATian said: Answers are 1) 102) 1/15

@pyashraj said: @Stoicalme 2)Find the remainder when (10^3+9^3)^752 is divided by 12^3 1...But not sure...

@allan89 said: total distance travelled together = 2nd, where n=5; d= 100 =>1000m. A travels =2/5 *1000 =400m. => A and B will be at P during 5th meeting . Hence ans should be 0m

@ImCATian said: Plz solve this 1) Find LCM of 4/6,6/9,5/4. 2) Find HCF of 4/6,3/9,6/5.

@chevy said: 1) LCM of numerator/HCF of denominator = 60/1 ?? 2)HCF of numerator/LCM of denomi = 1/90??

@Budokai001 said: set 'S' is defined as shown .S={3,10,17,24.....1004}'A' is a subset of S such that the product of any two elements of A is greater than 603000.What is the maximum possible number of elements in A a)33 b)32c)31d)30

@yudh said: bhaio please ye saval solve karna.... Distance between P and Q is 100m and speed of A and B are 20m/s and 30 m/s.initially A and B are at P.they move between P and Q.calculate distance between P and place of fifth meeting?

@pyashraj said:@yudh The answer is 0m. They both will meet at point P for the fifth time...

@chevy said: it'll be 4hrs capacity be total 24llet A take a hrs to fill 2.4lB take b hrs to fill 2.4lin 1 hr A fills 2.4/aB fills 2.4/bnwacc to conditiona/2 * 2.4/b + b/3 * 2.4/a = 5/6 * 24sry 2.4 nhi 24 hoga a/2b + b/3a=5/6from dis a:b=2:3 aaya 1/2x + 1/3x = 1/24 so A takes,,, 4hrs

@pyashraj said:@Stoicalme 1) Find the last two digits of 101*102*103*197*198*199 06 is the answer..

It can be written as (100 + 1)(100 + 2)(100 + 3)(200 - 1)(200 - 2)(200 - 3).

The last two digits of the product of the first two terms will be 06 (100n + 6)

The last two digits of the product of the last two terms will be 94 (200n - 6)

The product of 94 and 6 is 564. Hence the last two digits will be 64

@Stoicalme said: 1) Find the last two digits of 101*102*103*197*198*199. 2)Find the remainder when (10^3+9^3)^752 is divided by 12^3.

@Stoicalme said: @Enceladus Answers are correct.Method bhi bata dijiye.

@Stoicalme said: 1) Find the last two digits of 101*102*103*197*198*199. 2)Find the remainder when (10^3+9^3)^752 is divided by 12^3.

@Stoicalme said: 1) Find the last two digits of 101*102*103*197*198*199. 2)Find the remainder when (10^3+9^3)^752 is divided by 12^3.

@yashasvi5 said: Time and Work Questions Q.1 Two pipes A and B can fill up a half tank in 1.2 hours.The tank was initially empty. Pipe B was kept open for half the time required by pipe A to fill the tank by itself.Then, pipe A was kept for as much time as was required by pipe B to fill up 1/3 of the tank by itself.It was found that the tank then was 5/6 full.The least time in which any of the pipes can fill the tank fully isa.> 4.8 hrs b.> 4 hrs c.> 3.6 hrs d.> 6 hrs Q.2 Two men and a women are entrusted with a task. The second man needs three hours more to cope with the job than the second man and the woman would need working together. The first man, working alone, would need as much time as the second man and the woman working together. The first man, working alone, would spend eight hours less than the double period of time the second man would spend working alone. How much time would the two men and the women need to complete the task if they all worked together?

@Saumz said: yes its 1st option can u please elaborate i m not able to understand

AB*r=x*y

bhaio please ye saval solve karna....

Distance between P and Q is 100m and speed of A and B are 20m/s and 30 m/s.initially A and B are at P.they move between P and Q.calculate distance between P and place of fifth meeting?

@yashasvi5 said: Time and Work QuestionsQ.1 Two pipes A and B can fill up a half tank in 1.2 hours.The tank was initially empty. Pipe B was kept open for half the time required by pipe A to fill the tank by itself.Then, pipe A was kept for as much time as was required by pipe B to fill up 1/3 of the tank by itself.It was found that the tank then was 5/6 full.The least time in which any of the pipes can fill the tank fully isa.> 4.8 hrs b.> 4 hrs c.> 3.6 hrs d.> 6 hr

@Stoicalme said:1) The remainder when 10^10+10^100....10^10000000000 is divided by 7 is? 2) The remainder when 2^2+22^2+......(222...49 2s)^2 is divided by 9 is?