Fifty litres of a mixture of milk and water contains 30 percent of water. This mixture is added to eighty litres of another mixture of milk and water that contains 20 percent of water. Then, how many litres of water should be added to the resulting mixture to obtain a final mixture that contains 25 percent of water?
A1
B2
C3
D4
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The Setup: Mixture problems just require tracking the absolute volumes of the components like inventory. We find the exact litres of water and milk we currently hold, merge them, then solve for the extra water needed to hit the target 25%. The key discipline is that adding pure water changes the water column *and* the total, but leaves the milk column frozen.
Step 1: Audit Mixture 1. Total =50 L, water is 30%.
Water=0.30×50=15 L,Milk=50−15=35 LStep 2: Audit Mixture 2. Total =80 L, water is 20%.
Water=0.20×80=16 L,Milk=80−16=64 LStep 3: Merge the inventory. Tip both into the same barrel:
Total volume=50+80=130 L,Water=15+16=31 L,Milk=35+64=99 LStep 4: Calculate the required additive. Let w litres of pure water go in. Both the water and the total grow by w, and a 25% water mix means water is 41 of the total:
130+w31+w=41
Cross-multiply and solve:
4(31+w)=130+w⟹124+4w=130+w⟹3w=6⟹w=2Step 5: Verify from the milk side. The milk column never moved, so it is an independent check. Final total =130+2=132 L, of which milk is still 99 L:
13299=43=75% milk, hence 25% water
The two routes agree, so w=2 is right.
Final Answer: 2
In a room, there are n persons whose average height is 160cm. If m more persons, whose average height is 172cm, enter the room, then the average height of all persons in the room becomes 164cm. Then m:n is
A1:2
B1:3
C3:1
D2:1
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The Setup: This is a weighted average problem mapping two groups of distinct heights into a combined new average height. We can isolate the ratio of group sizes using the Alligation method.
Step 1: Define the parameters for Alligation.
Group 1 (n persons): Average height = 160 cm.
Group 2 (m persons): Average height = 172 cm.
Mixture (Combined group): Average height = 164 cm.
Step 2: Apply the Alligation cross-difference formula.
The ratio of the quantities (n:m) is inversely proportional to the distances their specific averages sit from the combined mixture mean.
n:m=(172−164):(164−160)n:m=8:4n:m=2:1Step 3: Evaluate the requested ratio.
We must explicitly provide the ratio m:n.
Since n:m=2:1, the ratio m:n is exactly 1:2.
Final Answer:1:2
In a bowl containing 60 ml orange juice, 40 ml of water is poured. Thereafter, 100 ml of apple juice is poured to make a fruit punch. Madhu drinks 50 ml of this fruit punch and comments that the proportion of orange juice needs to be higher for better taste. How much orange juice should be poured into the fruit punch that remained, in order to bring up the level of orange juice to 50 percentage?
A100 ml
B40 ml
C80 ml
D60 ml
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The Setup: A mixed volume fruit punch starts out with unequal ratios. After some is consumed, pure orange juice is added back to strictly raise its proportion to 50% of the newly defined mixture.
Step 1: Establish the baseline volume ratios of the mixture.
Initial ingredients: 60ml orange juice (OJ), 40ml water (W). Subtotal = 100ml.
Add 100ml apple juice (AJ). Grand total = 200ml.
The proportional composition of the punch is:
OJ=20060=30%W=20040=20%AJ=200100=50%Step 2: Calculate volumes remaining after consumption.
Madhu drinks 50ml of the punch.
Remaining total volume = 200ml−50ml=150ml.
Because mixtures are perfectly homogenous, the 30% ratio is maintained in the leftover fluid.
Volume of OJ remaining = 30% of 150=45ml.
Step 3: Calculate the required OJ addition.
Let x be the volume in ml of pure OJ poured back into the punch.
The new total volume of OJ is 45+x. The new total volume of the entire punch is 150+x.
We require the new OJ ratio to be exactly 50%:
150+x45+x=0.545+x=0.5(150+x)45+x=75+0.5x0.5x=30⟹x=60mlFinal Answer: 60 ml
An alloy P has copper and zinc in the proportion of 5:2 (by weight), while another alloy Q has the same metals in the proportion of 3:4 (by weight). If these two alloys are mixed in the proportion of a:b (by weight), a new alloy R is formed, which has equal contents of copper and zinc. Then, the proportion of copper and zinc in the alloy S, formed by mixing the two alloys P and Q in the proportion of b:a (by weight) is
A7:9
B9:7
C9:5
D5:9
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The Setup: We've got an alchemy minigame. First, we need to unlock the exact ratio (a:b) that balances the metals 1:1. Once we have those initial mix weights, we hit them with an Uno reverse card, swap the weights (b:a) for a new batch, and calculate the final drop rates for copper and zinc.
**Step 1: Find the Initial Mix Ratio (a:b)**
In alloy P, Copper is 5/7 and Zinc is 2/7.
In alloy Q, Copper is 3/7 and Zinc is 4/7.
Mix them in parts a and b to get equal Copper and Zinc:
75a+73b=72a+74b
Multiply by 7 to drop the denominators and balance the equation:
5a+3b=2a+4b3a=b⟹ba=31
So the ratio a:b is strictly 1:3.
**Step 2: Form Alloy S (Ratio b:a)**
Now we mix them using the reverse ratio: 3 parts of P and 1 part of Q.
Calculate the total Copper:
Cu=3(75)+1(73)=715+73=718
Calculate the total Zinc:
Zn=3(72)+1(74)=76+74=710Step 3: Final Proportion
The ratio of Cu to Zn in alloy S is:
718:710
Multiply by 7 and simplify by dividing by 2:
18:10=9:5Final Answer:9:5