Hey guys this is...
Mon, 2010-06-28 22:11
Hey guys this is a question from the arun sharma alligations chapter ...do let me know how to solve this using the alligations formula :-
A bag contains a total of 105 coins of Rs 1,50 p and 25 p denominations .Find the number of coins of Re 1 if there are a total of 50.5 rupees in the bag and it is known that the number of 25 paise coins are 133.33 % more than the number of 1 rupee coins .
a) 56
b)25
c)24
d)none of these
A bag contains a total of 105 coins of Rs 1,50 p and 25 p denominations .Find the number of coins of Re 1 if there are a total of 50.5 rupees in the bag and it is known that the number of 25 paise coins are 133.33 % more than the number of 1 rupee coins .
a) 56
b)25
c)24
d)none of these
Tue, 2010-06-29 12:44
#3
the answer is C
C) ANSWER
the number of coins of Re 1 are 24, that of 50p are 25 and 25p are 56.
i) 24+25+56= 105 ( first condition)
ii) (1x24) + ( 0.5 x25) + (0.25 x 56) = 50.5 ( second condition)
iii) 56/24 = 2.3333 ( third condition)
to solve this, make three simultaneous equations for all the three conditions given and solve them, you shall get the required answer.
( @ the incognito solver: notice that the question says 133.33 % more... that means that the 25p coins are 233.33 % of the number of Re1 coins, be careful with the language always)
Tue, 2010-07-13 16:50
#6
Correct answer is
It is 100% --------------->>>>>>>>>>>>> 24
The 1 Re coins are 24
50Paise cions are 25
25Paise cions are 56
The ans is none.
from the given data we will get two equations 2.33X+Y=105 and
5.33X+y=210
from equation 1 we can determine that 2.33X is odd(any no. multiplied by odd no. results odd no) so y must be even.
but in 2nd equation Y should be even then only we will get the sum as a even no.
so 1 and 2 are contraversial to each other
so ans is d