case 1 : x + x/5 = 3000 CP = 2500 Profit =500 case 2 : (3000-x)/3000 = 1/5 CP = 3000 Profit = 600 so diff 100..hope i understood the problem correctly...
Up to 20 she is allowed, need not be exactly 20.Bhai dekh na....the question is asking how many sweets did she buy....if she is buying 20 sweets exactly then the question is a waste?regardsscrabbler
yeah you can take any number of sweets, the bottomline is the shop costs 3 rs more and her uncle's shop costs her 3rs less.. jus had a look at your solution. Aapke solution mein she is spending 3rs more at her uncle's shop
P.S While typing the previous msg i thought finally i have been able to make @scrabbler agree to my solution :(
Profit for case 1: 500 (20% on CP)Profit for case 2 : 600 (20% on SP)Hence difference = 100 ?@scrabbler Sir good morning Aap bhi bhai waali tone mein aa gaye ..mast
Main to humesha hi aisa tha! Good morning :) regards scrabbler
@Exodia said:A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dices. How many ways are possible of having 1 before 6?1) 1202) 3603) 2404) 280 No of Ways = 120 + 96 + 72 + 48 + 24 = 360..
A person calculate his % profit on cp and another on SP.what will be their diff in profit if both claims a profit of 20% on goods sold for rs 3000.
Bhai general way mai profit cp pe nikalta hai....is case mai dono pe nikalna hai... case 1:profit on cp sp-cp/cp =20/100 cp comes out to be 2500 profit p1=500 Case 2:profit on sp sp-cp/sp=20/100 cp comes out to be 2400 profit p2=600 hence p1-p2=100
A train aftr travelling 50 km frm A meets with an accident and proceeds at 4/5th of the former speed and reaches B,45 min late.had the accident happened 20 km further on ,it would hav arrived 12 min sooner.find the original speed & the distance
A train aftr travelling 50 km frm A meets with an accident and proceeds at 4/5th of the former speed and reaches B,45 min late.had the accident happened 20 km further on ,it would hav arrived 12 min sooner.find the original speed & the distance
@rnishant231@Exodia said:A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dices. How many ways are possible of having 1 before 6?1) 1202) 3603) 2404) 280No of Ways = 120 + 96 + 72 + 48 + 24 = 360..cn u plz xplain ur approach???
1 _ _ _ _ _ _ 120 ways A 1 _ _ _ _ _ A can be any thing other than 1 and 6 and then arrange the others in 4! : 4*4! A B 1 _ _ _ 4C2 * 2! *3!= 72 A B C 1 _ _ 4C3 * 3! * 2!= 48 A B C D 1 6 : 4!=24
@rnishant231@Exodia said:A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dices. How many ways are possible of having 1 before 6?1) 1202) 3603) 2404) 280No of Ways = 120 + 96 + 72 + 48 + 24 = 360..cn u plz xplain ur approach???
Taking 1 in the first place - no. of possiblity = 5! =120 For 2nd position - 4.4.3.2.1 = 96.. For 3rd Position - 4.3.3.2.1 = 72 and so on..
bro what is the mistake in my approach, see the breakdown in the first case occurs at 50kms and in the second case it occurs at 70kms So in first case the distance 50-70 is travelled at 4s/5 and in the second case it is travelled at normal speed s. The time taken in the two cases constitutes to the time difference which is 33 mins hence 20/(4s/5) - 20/s = 33/60 i am getting s=100/11 :(
Taking 1 in the first place - no. of possiblity = 5! =120 For 2nd position - 4.4.3.2.1 = 96..For 3rd Position - 4.3.3.2.1 = 72and so on..However a better approach was posted by @ExodiaNo. of ways to place 1 and 6= 6c2 No. of arrangements of the rest of the numbers= 4! Answer= 6c2. 4! =360
Even less effort required if we realise that either 1 comes before 6 or 6 before 1. By symmetry neither is more likely - and hence exactly half of the 6! i.e 720 ways should have 1 before 6. Hence 360. regards scrabbler
Even less effort required if we realise that either 1 comes before 6 or 6 before 1. By symmetry neither is more likely - and hence exactly half of the 6! i.e 720 ways should have 1 before 6. Hence 360.regardsscrabbler
yessssssss i realised after doing all the hard work... is pe confidence nahi tha...
bro what is the mistake in my approach,see the breakdown in the first case occurs at 50kms and in the second case it occurs at 70kmsSo in first case the distance 50-70 is travelled at 4s/5 and in the second case it is travelled at normal speed s.The time taken in the two cases constitutes to the time difference which is 33 minshence20/(4s/5) - 20/s = 33/60i am getting s=100/11
The time difference between the two cases has to be seen, which is 12 min. I quote:
...had the accident happened 20 km further on ,it would hav arrived 12 min sooner... regards scrabbler
Even less effort required if we realise that either 1 comes before 6 or 6 before 1. By symmetry neither is more likely - and hence exactly half of the 6! i.e 720 ways should have 1 before 6. Hence 360.regardsscrabbler
Thanks sir...dimaag ki batti haule haule jalne lagegi...
Two trains A & B start frm stations X & y towards each other.B leaves station Y half an hour aftr train A leaves station X.two hrs aftr train A has strtd,the dist between the trains A & B is 19/30th of the dist between stations X & y.how much time would it take each train (A & B) to cover the distance X to Y,if train A reaches half an hr ltr to its dest as compared to B.
@scrabbler bhai zara yeh bhi orally solve kardoTwo trains A & B start frm stations X & y towards each other.B leaves station Y half an hour aftr train A leaves station X.two hrs aftr train A has strtd,the dist between the trains A & B is 19/30th of the dist between stations X & y.how much time would it take each train (A & B) to cover the distance X to Y,if train A reaches half an hr ltr to its dest as compared to B.
So, let's look at what we know.
A starts 0.5 hours before and finishes 0.5 hours fter. So the times taken are 1 hour apart.
Abhi tak 2 hours ho gaya for A and 1.5 and total distance is just over 1/3, which means total time should be around 6 times (1.5 to 2) which is 9-12 ke range mein
Denominator 30 hai, so I want two consecutive numbers which will lend themselves to 30...is it too much of a stretch to take 9 and 10?
So if B takes 9 hours and A 10, let us check - assume 90 ka distance, so B going 10 per hour, A going 9, and in 2 hours A will have done 18 and in 1.5 hours B will have done 15 so total is 33/90 = 11/30 leaving 19/30.
Hence 9 hrs and 10 hrs is fine. (Yes, did it orally. Painful though) regards scrabbler
A train aftr travelling 50 km frm A meets with an accident and proceeds at 4/5th of the former speed and reaches B,45 min late.had the accident happened 20 km further on ,it would hav arrived 12 min sooner.find the original speed & the distance
@scrabbler bhai zara yeh bhi orally solve kardoTwo trains A & B start frm stations X & y towards each other.B leaves station Y half an hour aftr train A leaves station X.two hrs aftr train A has strtd,the dist between the trains A & B is 19/30th of the dist between stations X & y.how much time would it take each train (A & B) to cover the distance X to Y,if train A reaches half an hr ltr to its dest as compared to B.
@rnishant231@Exodia said:A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dices. How many ways are possible of having 1 before 6?1) 1202) 3603) 2404) 280No of Ways = 120 + 96 + 72 + 48 + 24 = 360..cn u plz xplain ur approach???
simplest approach
_ 1___ 6_
now these 3 marks have to be filled by 4 numbers
a+b+c=4 --solutions 6c2
but these 4 numbers can also be arranged in 4! ways so
The time difference between the two cases has to be seen, which is 12 min. I quote: