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Three types of tea the a,b,c costs Rs. 95/kg,100/kg and70/kg respectively.How many kgs of each should be blended to produce 100 kg of mixture worth Rs.90/kg, given that the quantities of band c are equal
a)70,15,15 b)50,25,25 c)60,20,20 d)40,30,30
Read Solution (Total 7)
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- let qty of b and c are x
then qty of a is 100 - 2x
cost of each,
a -> 95(100-2x)
b -> 100x
c -> 70*x = 70x
and total cost of the mixture is 90*100
therefore
95(100 -2x) + 100x + 70x = 90*100
=> 9500 - 190x + 170X = 9000
=> 9500 - 9000 = 20x
=> x = 500/20
=> x = 25
therefore qty of B and C is 25 kg each
and qty of A is 50 kg (100 - 25 -25) - 11 years agoHelpfull: Yes(27) No(1)
- ans b)50,25,25 satisfies the condition.
given quantities of b and c are equal .
therefore instead of considering them as a different quantities take average of both and consider it as a single entity .
so the cost of the mixture "D" (equal quantities of "B" and "C") is 170/2=85rs/kg
now the tea contains only "A" of 95 rs/per and "D" of 85 rs/kg.
it is clearly seen that "A" is 5 more and "D" is 5 less than the required cost per kg .
therefore 50/50 will go well .
A=50kg and D=50kg. (D= B&C in equal ratio). therefore B&C=25kg each .
(A:B:C)=(50,25,25) - 13 years agoHelpfull: Yes(20) No(3)
- option b is correct because 50kg of a 50*95=4750
25kg of b 25*100=2500
25kg of c 25*70=1750
total =9000 which is require price for 100 kg
- 13 years agoHelpfull: Yes(14) No(3)
- here (50*95)+(25*100)+(25*70)=90*100
so ANS is b)50,25,25 - 13 years agoHelpfull: Yes(6) No(3)
- b)50,25,25 satisfies the condition
- 13 years agoHelpfull: Yes(3) No(4)
- total amount will be=9000RS.
b)50,25,25 satisfied the value.... - 13 years agoHelpfull: Yes(2) No(7)
- let the weight of a ,b,c tea be m,n,n kg now we have 2 equations
m+n+n=100
95m=100n=70n=9000
solving these we get m=50
n=25
so answer is( b)
- 8 years agoHelpfull: Yes(1) No(0)
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