I know the answer. Please let me know the logic u followed.
@jain4444 said:SP of shirt = x SP of coat = (4000 - x) let CP of coat = 300y CP of shirt = 175y x - 175y = 4000 - x - 300y => 2x + 125y = 4000 from option only 1 value will satisfy this i.e. 2100
if price of shirt is 2100
then price of coat is ~ 3318....
then the guy cannot have profit on 4000 coz his CP in total is 5418 and SP is 4000
then price of coat is ~ 3318....
then the guy cannot have profit on 4000 coz his CP in total is 5418 and SP is 4000
@Shrutim90 said:@DexianI know the answer. Please let me know the logic u followed.
eliminate from the options... and the reply above will help :)
@VJ12 said:@DexianHow did you get 133?
took each number from 2 to 19 and counted....
not sue may be wrong...
dont know the OA...
not sue may be wrong...
dont know the OA...
What was your approach for the co prime question?
Did you count or did you follow any other approach?
@hesse said:@The_LoserD1 + D2 +D3 +D4 = 17D represents diceFind number of ways. ?d1 d2 d3 d46 6 4 1 - 12 ways6 6 3 2 - 12 ways6 5 4 2 - 24 ways6 5 5 1 - 12 ways6 5 3 3 - 12 ways5 5 5 2 - 4 ways5 5 4 3 - 12 ways5 4 4 4 - 4 ways
Also 6 4 4 3 - 12 ways
So I get 104.
OA kya hai?
regards
scrabbler
@The_Loser said:could you rectify this. ?a+b+c+d = 17a'+b'+c'+d' = 13now 16c3 - 4 * 10c3 = 80
How about this logic: When you do 10C3 you are removing cases when 1 die goes above 6 right? But what about the cases where 2 dice go above 6? These will be subtracted twice in the above situation...so need to be added back.
Each such case occurs 4c3 times. and there are 4C2 i.e. 6 such cases where 2 of the 4 dice cross 6. So we need to add back 4C3 * 6 = 24.
3 of the dice cannot cross 6 simultaneously ad the total is 17 only. So we can say it should be 16C3 - 4 * 10C3 + 6 * 4C3 = 104.
regards
scrabbler
@rkshtsurana
Each such case occurs 4c3 times. and there are 4C2 i.e. 6 such cases where 2 of the 4 dice cross 6. So we need to add back 4C3 * 6 = 24.
3 of the dice cannot cross 6 simultaneously ad the total is 17 only. So we can say it should be 16C3 - 4 * 10C3 + 6 * 4C3 = 104.
regards
scrabbler
@rkshtsurana
@Shrutim90 said:@DexianI know the answer. Please let me know the logic u followed.
the approach can be
let the cost price of coat =x
cost price of shirt =58.33% of x =(50%+8.33%) of x=(1/2+1/12)x =7/12x
total cost price =x +(7/12)x =19/12x
now if the shirt's price is 2000 then
7/12 part is 2000 so obviously 19/12 parts wil be greater than 4000 and hence by sellinh the shirt and coat at 4000 there wil be a loss
oly one option is below 2000. hence D
@The_Loser said:Triangle having perimeter = 100.all side integeral.How many triangle possible. ?
Options hai? Mera 1173 ke aaspaas aa raha hai...
regards
scrabbler
regards
scrabbler
@The_Loser said:Triangle having perimeter = 100.all side integeral.How many triangle possible. ?
Options hai? Mera 1173 ke aaspaas aa raha hai...
regards
scrabbler
regards
scrabbler
How many pairs of integers (not necessarily positive) are there such that both a^2 + 6b^2 and b^2 + 6a^2 are both squares ?
@Shrutim90 said:Selling price of a shirt and a coat is Rs. 4000. The cost price of a shirt is 58.33% of the cost price of a coat and so amount of profit on both the shirt and coat is same, then the price of the shirt could be:a) 2100b) 2525c) 2499d) 1120
58.33% = 7/12. So Price of shirt : coat is 7 : 12 and so shirt is 7/19 of total CP. Since there is a profit, total CP
regards
scrabbler
@scrabbler said:Options hai? Mera 1173 ke aaspaas aa raha hai...regardsscrabbler
1176?sorry ab pata chala y u getting 1173..
@jain4444 said:How many pairs of integers (not necessarily positive) are there such that both a^2 + 6b^2 and b^2 + 6a^2 are both squares ?
4?
@iLoveTorres said:1176?sorry ab pata chala y u getting 1173..
Paper nahin hai so solved on my hand
...started with natural solutions of a + b + c = 100 and removed cases when any of the 3 became 50 or more...but I think I did whole number instead of natural solutions so kuch galat ho gaya...didn't have patience/paper to redo hence "approx" answer bola.Edit: 1176 hi correct hai....99C2 - 3 * 50C2 = 99*49 - 3*25*49 = 24 * 49 = 1176
Dubara edit: Ab paper leke aaya :D
regards
scrabbler