Let's analyze the problem step-by-step.
Problem Restatement
- A man has x kg of sugar costing 0p per kg (which presumably means free or zero cost).
- He mixes it with y kg of sugar costing 34p per kg.
- He sells the mixture at 44p per kg.
- He gains 1 (i.e., 100%) on the total outlay (cost price).
We need to find how many kg of sugar at 0p per kg (x) he must mix with y kg of sugar at 34p per kg.
Step 1: Define variables
- Let the quantity of sugar at 0p/kg be x kg.
- The quantity of sugar at 34p/kg be y kg.
Step 2: Calculate total cost price (CP)
- Cost of sugar at 0p/kg = 0 × x = 0 pence.
- Cost of sugar at 34p/kg = 34 × y pence.
- Total cost price = 0 + 34y = 34y pence.
Step 3: Calculate total quantity and selling price (SP)
- Total quantity = x + y kg.
- Selling price per kg = 44p.
- Total selling price = 44 × (x + y) pence.
Step 4: Use gain formula
Gain = 1 on outlay means gain = 100% of cost price. So,
Gain=SP−CP=CP\text{Gain}=\text{SP}-\text{CP}=\text{CP}Gain=SP−CP=CP
⇒SP=2×CP\Rightarrow \text{SP}=2\times \text{CP}⇒SP=2×CP
Substitute:
44(x+y)=2×34y44(x+y)=2\times 34y44(x+y)=2×34y
44x+44y=68y44x+44y=68y44x+44y=68y
44x=68y−44y=24y44x=68y-44y=24y44x=68y−44y=24y
x=24y44=6y11x=\frac{24y}{44}=\frac{6y}{11}x=4424y=116y
Final answer:
The man must mix (6/11) times the quantity y of sugar at 0p/kg with y kg of sugar at 34p/kg.
Example:
If he has 11 kg of sugar at 34p/kg, he must mix:
x=611×11=6 kgx=\frac{6}{11}\times 11=6\text{ kg}x=116×11=6 kg
of sugar at 0p/kg. If you want, I can help you with further examples or explanations!
