Problem Solving

Tom is on a certain diet that requires him to limit the number of calories he takes in each day. He is allowed to take in 2400 calories each day from three square meals, and 200 calories each day from snacks and dessert combined. On some days, he splurges by taking in three times the recommended number of calories from snacks and dessert. The rest of the days, he follows the calorie guidelines precisely. If his average calorie intake for a 10 day period was 2720, on how many days did he splurge?

A)3
B)4
C)5
D)6
E)7

OAA

Tom is on a certain diet that requires him to limit the number of calories he takes in each day. He is allowed to take in 2400 calories each day from three square meals, and 200 calories each day from snacks and dessert combined. On some days, he splurges by taking in three times the recommended number of calories from snacks and dessert. The rest of the days, he follows the calorie guidelines precisely. If his average calorie intake for a 10 day period was 2720, on how many days did he splurge?

A. 3
B. 4
C. 5
D. 6
E. 7

This is a weighted average/mixture question.

Ingredient 1: Average caloric intake on deprived days = meal calories + snack calories = 2400 + 200 = 2600.
Ingredient 2: Average caloric intake on splurge days = meal calories + snack calories = 2400 + 3*200 = 3000.
Mixture: Average caloric intake for the MIXTURE of deprived days and splurge days = 2720.

Let D = the number of deprived days and S = the number of splurge days.
To determine the ratio of D to S in the mixture, use ALLIGATION:

Step 1: Plot the 3 averages on a number line, with the averages for the two ingredients on the ends and the average for the mixture in the middle.
D 2600--------------2720-------------3000 S

Step 2: Calculate the distances between the averages.
D 2600------120----2720-----280-----3000 S

Step 3: Determine the ratio in the mixture.
The required ratio of D to S is the RECIPROCAL of the distances in red.
D:S = 280:120 = 7:3.

Thus, of the 10 days, D=7 and S=3.

The correct answer is A.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html

j_shreyans wrote:Tom is on a certain diet that requires him to limit the number of calories he takes in each day. He is allowed to take in 2400 calories each day from three square meals, and 200 calories each day from snacks and dessert combined. On some days, he splurges by taking in three times the recommended number of calories from snacks and dessert. The rest of the days, he follows the calorie guidelines precisely. If his average calorie intake for a 10 day period was 2720, on how many days did he splurge?

A)3
B)4
C)5
D)6
E)7

Alternate approach:

Tom is allowed to take in 2400 calories each day from three square meals, and 200 calories each day.
Thus, calories consumed on regular days = 2400 + 200 = 2600.
On splurge days, Tom takes in three times the recommended number of calories from snacks and dessert.
Thus, calories consumed on splurge days = 2400 + 3*200 = 3000.

We can PLUG IN THE ANSWERS, which represent the number of splurge days.
When the correct answer choice is plugged in, the average caloric intake over 10 days = 2720.
Since the average caloric intake (2720) is closer to the number of calories consumed on regular days (2600) than to the number of calories consumed on splurge days (3000), there must be MORE REGULAR DAYS THAN SPLURGE DAYS.
Thus, the number of regular days must be GREATER THAN 5, while the number of splurge days must be LESS THAN 5.
Eliminate C, D and E.

Answer choice B: 4 splurge days
Total calories consumed over 4 splurge days = 4*3000 = 12000.
Total calories consumed over 6 regular days = 6*2600 = 15600.
Average caloric intake over all 10 days = (12000 + 15600)/10 = 2760.
The average calories intake is too high.
Eliminate B.

The correct answer is A.

Answer choice A: 3 splurge days
Total calories consumed over 3 splurge days = 3*3000 = 9000.
Total calories consumed over 7 regular days = 7*2600 = 18200.
Average caloric intake over all 10 days = (9000 + 18200)/10 = 2720.
Success!

The correct answer is A.

With 3 experts responded above I think OP should already have a good idea how to solve the problem. But I would still like to chip in with my $0.02 on how to solve problem quickly.

Normal day: 2600
Splurge day: 3000
Total calories intake in 10 days: 27200
get rid of those trailing zeros, you will get: normal day: 26, S day 30, total 272

Since the # of days will always be an integer, and the total calories from the splurge days will also ending in 0, we really only need to concern what integer times 6 will get you a number ending with 2. So...we get 2*6=12 and 7*6=42

test 2 and 7 for normal day respectively:

2*26=52 + 8*30 = ...too big
7*26=182 + 3*30 = ...272

Answer is A

Without splurging, maximum intake over 10 days = 10 * (2400 + 200) = 26000
With splurging, intake over 10 days = 10 * average = 27200
Difference = 27200 - 26000 = 1200
Triple dessert calories = 200*3 = 600 which is 400 more than 200
1200/400 = 3 days
Answer = A