An alloy of gold,silver and bronze contains 90% bronze,7% gold and 3%silver. A second alloy of bronze and silver only is melted with the first and the mixture contains 85% of bronze, 5% of gold and 10% of silver. Find the percentage of bronze in the second alloy.. : ) Thanks in advance
Options are a) 75% b) 72.5 % c) 70 % d) 67.5%
And is 72.5 %
@ramirez said:An alloy of gold,silver and bronze contains 90% bronze,7% gold and 3%silver. A second alloy of bronze and silver only is melted with the first and the mixture contains 85% of bronze, 5% of gold and 10% of silver. Find the percentage of bronze in the second alloy.. : ) Thanks in advance
quantity of gold will remain same after mixing.
Let x gm of alloy1 and y gm of alloy2 are mixed, so .07x=.05(x+y)
=>x/y=5/2
Lets assume 50 gm of alloy1 and 20 gm of alloy2 is added (since ratio is 5/2)
now bronze in alloy1 + bronze in alloy2 = 85% of total
Let B be bronze in alloy2
90% of 50 + B = 85% of 70
=>45+B=59.5
B=14.5
therefore 14.5*100/20 = 72.5%
Let x gm of alloy1 and y gm of alloy2 are mixed, so .07x=.05(x+y)
=>x/y=5/2
Lets assume 50 gm of alloy1 and 20 gm of alloy2 is added (since ratio is 5/2)
now bronze in alloy1 + bronze in alloy2 = 85% of total
Let B be bronze in alloy2
90% of 50 + B = 85% of 70
=>45+B=59.5
B=14.5
therefore 14.5*100/20 = 72.5%
Alligation lod 1
A man purchased a cow and calf for Rs 1300.He sold the calf at a profit of 20% and the cow at a profit of 25%.In this way his total profit was 23 1/13%.Find the cost price of cow..
Pls help me understand this prob
@Raghu5jan said:Alligation lod 1 A man purchased a cow and calf for Rs 1300.He sold the calf at a profit of 20% and the cow at a profit of 25%.In this way his total profit was 23 1/13%.Find the cost price of cow..Pls help me understand this prob
This is a simple profit lose question
total profit is 23 1/13 =>300/13
now profit is 1300*300/13 = 300
this Rs 300 profit he got from selling calf at 20% profit and cow at 25% profit
Let cow be x
.25x + .2(1300-x) = 300
x=800
total profit is 23 1/13 =>300/13
now profit is 1300*300/13 = 300
this Rs 300 profit he got from selling calf at 20% profit and cow at 25% profit
Let cow be x
.25x + .2(1300-x) = 300
x=800
@sonic_boom said:This is a simple profit lose questiontotal profit is 23 1/13 =>300/13now profit is 1300*300/13 = 300this Rs 300 profit he got from selling calf at 20% profit and cow at 25% profitLet cow be x.25x + .2(1300-x) = 300x=800
OR
u can solve it by allegation
total profit is 300/13%, it will be between 20% and 25%
now ratio of cost of calf : cow = 25-300/13 : 300/13-20
=>5:8
=>cost of cow = 1300*8/13 = 800
u can solve it by allegation
total profit is 300/13%, it will be between 20% and 25%
now ratio of cost of calf : cow = 25-300/13 : 300/13-20
=>5:8
=>cost of cow = 1300*8/13 = 800
A sum of 36.90 is made up of 90 coins that are either 20 paise coins or 50 paise coins.. Find how many 20 paise coins are there in the total amount...??
Can u explain how u got 27??
@Raghu5jan : Let x be the no 20 paise coins and y be the no of 50 paise coins..
So x + y = 90 ---> (1)
Now 36.90 rs = 3690 paise....
So 20x + 50y = 3690 ---> (2)
Solving (1) and (2) we get x = 27...correct me if I am wrong :)
So x + y = 90 ---> (1)
Now 36.90 rs = 3690 paise....
So 20x + 50y = 3690 ---> (2)
Solving (1) and (2) we get x = 27...correct me if I am wrong :)
Q)The distance to which a boy can throw a stone is inversely proportional to its weight. He breaks the stone into 3 pieces whose weights are in the ratio 1:3:2. He then throws the stones one by one. the sum of the distances they cover is 22metres. to what distance can he throw the unbroken stone..? thanks in advance : )
@ramirez said:Q)The distance to which a boy can throw a stone is inversely proportional to its weight. He breaks the stone into 3 pieces whose weights are in the ratio 1:3:2. He then throws the stones one by one. the sum of the distances they cover is 22metres. to what distance can he throw the unbroken stone..? thanks in advance : )
12?? pata nahi aaj kyu sare answeres 12 aa rahe hai 😛
ans is 2 metre
@ramirez said:Q)The distance to which a boy can throw a stone is inversely proportional to its weight. He breaks the stone into 3 pieces whose weights are in the ratio 1:3:2. He then throws the stones one by one. the sum of the distances they cover is 22metres. to what distance can he throw the unbroken stone..? thanks in advance : )
distance is inversely proportional to weight
ratio of weight 1:3:2
=>ratio of distances 6:2:3
now total distance 6x+2x+3x=22 => x=2
distances are 12, 4, 6
now weight=constant/distance
w1=c/d1
let the weights be y,3y,2y
so
y=c/12
=>c=12y
now total weight y+3y+2y=6y
6y=c/D
6y=12y/D
=>D=2
ratio of weight 1:3:2
=>ratio of distances 6:2:3
now total distance 6x+2x+3x=22 => x=2
distances are 12, 4, 6
now weight=constant/distance
w1=c/d1
let the weights be y,3y,2y
so
y=c/12
=>c=12y
now total weight y+3y+2y=6y
6y=c/D
6y=12y/D
=>D=2
Q) The cost of manufacturing a car is made up of three items: cost of material, labour and overheads. In a year, the cost of these items were in the ratio 4:3:2. Next year, the cost of material rose by 10%, cost of labour increased by 8%, but the overheads reduced by 5%. Find the increase per cent in the price of the car. (ans is increase in price of car in percent = 44/9%)
thanks in advance : )
@ramirez said:Q) The cost of manufacturing a car is made up of three items: cost of material, labour and overheads. In a year, the cost of these items were in the ratio 4:3:2. Next year, the cost of material rose by 10%, cost of labour increased by 8%, but the overheads reduced by 5%. Find the increase per cent in the price of the car. (ans is increase in price of car in percent = 44/9%)thanks in advance : )
Let the costs be : 400, 300 and 200 -> Price of the car = 900.
Increased costs : 440, 324 and 190 -> Price of the car = 954.
Increase percent : 54/900 * 100 = 6 %.