@pavimai said:what can be the maximum number of acute angles in an octagon in which all the interior angles are less than 180 degrees??1 234pls explain
If angle is acute, corresponding exterior angle is obtuse i.e. > 90. But total of all exterior angles is 360. So not more than 3 can exceed 90. Hence 3.
regards
scrabbler
regards
scrabbler
A salesman is appointed on the basic salary of rs 1200 per month and the condition that for every sales of rs 10000 above rs 10000 he will get 50% of basic salary and 10% of the sales as reward. this incentive scheme dos not operate for the first rs 10000 of sales . then which of the following income cannot be achieved in a month?
a. rs 6000
b. rs 9000
c.both a and b
d any income can be achieved .
b) in the above question despite a 5 percentage point increment on the commission from 20% the total commission remain unaltered . find the change in the volume of the transaction.
a -10% b -16% c -25% d. -20%
please suggest.
@IIM-A2013 : my one question ... every sale is in multiple of 10000??
anyways for first ... every 10000 will give him 1600= 50% of 1200=600 & 10% of 10000=1000
so 6000-1200=4800 multiple of 1600 but 9000 -12= 7800 which is not possible
2nd = didnt get it
@soham2208 said:Consider the total number of numbers to be 5NNow, each number can be represented as 5k, 5k+ 1, 5k + 2, 5k +3, 5k + 4So, every form (i.e. 5k + 1 and so on) has N numbers in it So, probability that the product of three numbers is not divisible by 5 = (4N/5N)^3 = 0.512So, Probability required = 1 - 0.512 = 0.488 ?
can u explain a lil more ???
@wovfactorAPS said:5k 5k+1 5k+2 5k+3 5k+4 be the numbersprobability=1-(4c1+4c2*2+4c3)/(5c1+5c2*2+5c3)=1-(20/35)=3/7?
can u explain wat logic u hav used? i used d same logic to segregate nos
@soham2208 said:Consider the total number of numbers to be 5NNow, each number can be represented as 5k, 5k+ 1, 5k + 2, 5k +3, 5k + 4So, every form (i.e. 5k + 1 and so on) has N numbers in it So, probability that the product of three numbers is not divisible by 5 = (4N/5N)^3 = 0.512So, Probability required = 1 - 0.512 = 0.488 ?
i have a slight confusion pls help me
suppose we have 5 numbers 1,2,3,4,5 ,product of 3 numbers can be formed in 125 ways
for divisibility we can have one two or all 3 of them factors of 5
for one =1×4×4=16 cases,for two =1×1×4=4 cases and for all 1 case so the total becomes 21 favourable ones ,prob=21/125 ......what wrong i have done
pls help,
@gautam22 said:507??
........................................ Ans nahi pata ....................................................
......................................??APPROACH 
>??..................................................
>??..................................................@Brooklyn For all integers n ‰Ä˝ 1, a sequence of positive integers ai is defined by .what is the minimum value of a1 + a2 ?
1) 213 2) 407 3)507 4)613
@antodaya said:........................................ Ans nahi pata .................................................... ......................................????..................................................
wats d q??
@Brooklyn said:can u explain a lil more ???
@jaiswalicsi said:i have a slight confusion pls help mesuppose we have 5 numbers 1,2,3,4,5 ,product of 3 numbers can be formed in 125 waysfor divisibility we can have one two or all 3 of them factors of 5for one =1×4×4=16 cases,for two =1×1×4=4 cases and for all 1 case so the total becomes 21 favourable ones ,prob=21/125 ......what wrong i have donepls help,
@Brooklyn @jaiswalicsi See, there are 4N integers which when used in the forming three numbers' product will not form a number which is divisible by 5. Say 1,2,3,4,5 -> Now, probability that a number selected is not divisible by 5 = 4/5 = 0.8
Now, If I keep on doing this three times, with replacement -> (0.8)^3 = 0.512
So, product will be divisible by 3 if I don't do this -> 1 - 0.512 = 0.488
Also @jaiswalicsi , if you are directly trying for favorable cases, you need to take into account the order also
Case 1: one -> 1*4*4 * ( 3!/2!) = 48
Case 2: two -> 1*1*4 *(3!/2!) = 12
Case 3: three -> 1*1*1*(3!/3!) = 1
total = 61
=> Prob = 61/125 = 0.488
@soham2208 said:@Brooklyn@jaiswalicsiSee, there are 4N integers which when used in the forming three numbers' product will not form a number which is divisible by 5. Say 1,2,3,4,5 -> Now, probability that a number selected is not divisible by 5 = 4/5 = 0.8Now, If I keep on doing this three times, with replacement -> (0.8)^3 = 0.512 So, product will be divisible by 3 if I don't do this -> 1 - 0.512 = 0.488Also @jaiswalicsi, if you are directly trying for favorable cases, you need to take into account the order alsoCase 1: one -> 1*4*4 * ( 3!/2!) = 48Case 2: two -> 1*1*4 *(3!/2!) = 12Case 3: three -> 1*1*1*(3!/3!) = 1 total = 61 => Prob = 61/125 = 0.488
thx a lot bro
@gautam22 said:507??
soln plz...(not just the ans)
i am not able to see any constraints on a1 and a2.....
edit:
there are conditions on a1 and a2-->they are integers.....
i am not able to see any constraints on a1 and a2.....
edit:
there are conditions on a1 and a2-->they are integers.....
starting wid an easy one (was missing pg badly)
widout stoppage a train travels at an avg speed of 75km/hr & wid stoppage it covers d same distance at an avg speed of 60km/hr. hw many minute per hr does d train stops?
10, 12, 14, 18 minutes
better if u share ur approach.
@hiteshkhurana82 said:starting wid an easy one (was missing pg badly)widout stoppage a train travels at an avg speed of 75km/hr & wid stoppage it covers d same distance at an avg speed of 60km/hr. hw many minute per hr does d train stops?10, 12, 14, 18 minutesbetter if u share ur approach.
since this doesn't have to do anything with time, let us assume that all of it is just 1hr....so it has the capacity to cover 75km in 1hr. but it covered only 60km....which is done in 60/75=4/5hr = 48mins....so it was idle for 1hr-48mins = 12 mins
Minimum and max value of (3/2)sinx + 2cosx + 5
OA: 2.5 and 7.5.....c-rt(a^2 + b^2) and c+rt(a^2 + b^2)
was just remembering some formulae
OA: 2.5 and 7.5.....c-rt(a^2 + b^2) and c+rt(a^2 + b^2)
was just remembering some formulae
@gautam22 said:sir sinx:cosx=3:4......put this in sin^2x+cos^2x=1.....u will get dis anssir vaise ye ques aapne poocha tha ke aap kuch bata rahe the.....vo c-rt(a^2+b^2) vala part kuch samajh nahi aaya
-sqrt(a^2+b^2)
@gautam22 said:sir sinx:cosx=3:4......put this in sin^2x+cos^2x=1.....u will get dis anssir vaise ye ques aapne poocha tha ke aap kuch bata rahe the.....vo c-rt(a^2+b^2) vala part kuch samajh nahi aaya
I was going thru formulae...and i thought ki this formula was worth sharing...no more
€Ž4 dies are thrown. find the probability that the sum of them will be 18?
@hiteshkhurana82 said:starting wid an easy one (was missing pg badly)widout stoppage a train travels at an avg speed of 75km/hr & wid stoppage it covers d same distance at an avg speed of 60km/hr. hw many minute per hr does d train stops?10, 12, 14, 18 minutesbetter if u share ur approach.
12?
lets say 300kms is the total distance so it will take 4hrs
in case of 60kmph , it will take 5 hrs
so 1/5 is for stoppage
so 12 😁
welcome back 😃