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Nine men went to a restaurant. Eight of them spent ₹3 each on their meals and the ninth spent ₹2 more than the average expenditure of all the nine. Determine the total money spent by them.
₹29.25
₹30.25
₹32
₹35
₹29.25
Let the average be A. Total sum is 9A. The 9th man spends A+2. Therefore, 8×3+(A+2)=9A. Solving 26+A=9A gives 8A=26, so A=3.25. The total spent is 9×3.25=29.25.
There are 9 men in total. Each of the first 8 men spends ₹3. The 9th man spends ₹2 more than the average expenditure of all 9 men.
A=n∑x
Let the average be A. Total sum is 9A. The 9th man spends A+2. Therefore, 8×3+(A+2)=9A. Solving 26+A=9A gives 8A=26, so A=3.25. The total spent is 9×3.25=29.25.
Many students mistake the 9th man's expenditure as 2 instead of A+2, or forget to include the 9th person when calculating the average denominator (9 instead of 8).
Define the variables
Let A be the average expenditure of all nine men. The total expenditure of all nine men is 9A.
Total=9A
Formulate the equation
Eight men spent ₹3 each, so their sum is 8×3=24. The ninth man spent A+2. The total sum is the sum of these parts.
24+(A+2)=9A
Solve for A
Simplify the equation: 26+A=9A. Subtracting A from both sides gives 26=8A, which leads to A=3.25.
A=826=3.25
Calculate the total money
The total money spent is 9×A=9×3.25=29.25.
Total=9×3.25=29.25
A is correct because the calculated total expenditure of the nine men is ₹29.25.
This type of 'average plus/minus' problem is frequently used in mixture and alligation problems as well; always focus on equating the sum of parts to the total represented by the average.
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