@KaranGarcia said:b??
Nope..
@albiesriram said:sorry bhai . short term memory loss ho gayaa...
let the cost be Rs.100/m ... in summer the scale expands, means he ends up giving more.. 1m --> 1.1m... so S.P is 3*100/1.1m similarly for winter S.P is 7*100/0.9m
total S.P is 3000/11 + 7000/9 --> (27000 + 77000)/99 = 104000/99 and C.P is 1000
so profit = 5000/99 so profit is 5.05% ??
@albiesriram said:sorry bhai . short term memory loss ho gayaa...
CP per meter be x
summer ,
sold = 3n meter
length of scale = 1.1m
SP of 1 m = (x/1.1)
SP of 3n meter = 30nx /11
winter,
SP of 7n m = 70nx/9
total CP = 10nx
total SP = (104/99)*10nx
Profit % = (5x10nx/99x10nx)*100 = 5.05%
A shopkeeper cheats by 10% while purchasing and cheats by 12% while selling the goods. if he professes to sell the goods at cost price then find his profit percentage.?
@Asfakul said:A shopkeeper cheats by 10% while purchasing and cheats by 12% while selling the goods. if he professes to sell the goods at cost price then find his profit percentage.?
(1 + 10%)(1 + 12%) = (1 + p%)
p = 23.2 %
using a different approach,
he cheats by 10% means he gives Rs 100 but takes away 110.
Then he cheats by 12%.
112% of 110 = 123.2
he gives 110 when he should have given 123.2
total profit = 123.2 - 100 = 23.2%
Asfakul
25%
suppose he gives money for 1kg -> he got 1.1 kg
then he sells 1.1 kg as x kg where x =1.1/0.78 0.88 =1.25 kg hence he got
25% profit 
Asfakul
cheating by 10% means he is getting 110 ( instead of 100)
so next cheating he is doing on 110 so 12% of 110 is 13.2
so total 110+13.2=123.2 ( instead of 100)
profit percentage = 23.2
@Asfakul said:A shopkeeper cheats by 10% while purchasing and cheats by 12% while selling the goods. if he professes to sell the goods at cost price then find his profit percentage.?
23.2% ?
by cheating 10% , he gets 110 units
now by cheating of 12% , he is doing 110*112/100 = 123.2
so profit = 23.2 %
How Many right triangles with integral sides have one leg of length 35 cm ?
techgeek2050
juice by one apple= A
juice by one orange = O
Given (n-1)O = nA
O= nA/(n-1)
equal quantites (say x of each are mixed)
so orange juice = xO = nxA/(n-1) -> putting earlier value of O
apple juice = xA
Percentage of apple juice = xA/(xO + xA)
putting the value of O you get = (n-1)/(2n - 1)
for n= 2 ans- 33.33
n=3 ans = 40.00
@techgeek2050 said:If x[x] = 2013, then find the value of x/[x] where [x] is greatest integer les than or equal to x.
2013/2025 I guess....?
x must be something negative...
regards
scrabbler
no..@psk.becks said:techgeek2050juice by one apple= Ajuice by one orange = OGiven (n-1)O = nAO= nA/(n-1)equal quantites (say x of each are mixed)so orange juice = xO = nxA/(n-1) -> putting earlier value of Oapple juice = xAPercentage of apple juice = xA/(xO + xA)putting the value of O you get = (n-1)/(2n - 1)for n= 2 ans- 33.33n=3 ans = 40.00
it's given he required one orange less to extract the same quantity of juice as extracted from apples. But it is not given that the quantity of both used were equal.