Tom

[jainrahul1985]

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 deprives himself of snacks and dessert altogether. If his average calorie intake for a 10 day period was 2820, on how many days did he not splurge?
a)1 b)2 c)3 d)4 e)5
OA C

[VivianKerr]

normal day: 2400 + 200 = 2600 calories total

splurge day: 2400 + 3(200) = 3000 calories total

deprive day: 2400 calories total

2820 - 2400 = he ate 220 calories of snacks on average per day

We know that if he hadn't splurged at all, then his average would have been 200, so the 220 average shows he definitely splurged at least a few days. Just by estimation, we could eliminate A and E.

That's 120 excess calories total, from 10 days: so 1200 excess calories. The splurge days had 400 calories more per day, so 1200/400 = 3 days of splurge.

[sl750]

Total calorie intake = 28200

Let's work backward using the answer choices

Number of days he didn't splurge = 2
Number of days he splurges = 8

Total regular calorie intake 2400*2 = 4800
Splurge count = (2400+600)*8 = 24000
Total = 28800
This number is higher than total for 10 days. Let us pick a higher number

Number of days he didn't splurge = 3
Number of days he splurges = 7

Total regular calorie intake 2400*3 = 7200
Splurge count = (2400+600)*7 = 21000
Total = 28200. So n=3

[GMATGuruNY]

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 deprives himself of snacks and dessert altogether. If his average calorie intake for a 10 day period was 2820, on how many days did he not splurge?
a)1 b)2 c)3 d)4 e)5
OA C

This is a weighted average/mixture question.

Deprived days = 2400.
Splurge days = 2400 + 3*200 = 3000.
Mixture = 2820.

The deprived days are being combined with the splurge days to yield a mixture with an average of 2820.
We can use alligation:

The proportion needed of each ingredient in the mixture is equal to the distance between the OTHER 2 averages.

Proportion of deprived days = |3000-2820| = 180.
Proportion of splurge days = |2400-2820| = 420.
Ratio of deprived days to splurge days = 180:420 = 3:7.

Thus, there were 3 deprived days and 7 splurge days.

The correct answer is C.

[gmatclubmember]

Let on a days he takes a normal calorie intake (2600 cal)
on b days he splurges (3000 cal)
and on c days he deprives (2400 cal).
2600a+3000b+2400c=28200
=>13a+15b+12c=141
and we know that a+b+c=10=>13a+13b+13c=130
subtracting both equations:
2b-c=11.
if we take c=1, then b =6
c even would give fractional value of b so we discard c as even.
if c=3 then b=7 in this case a will be 0 which cannot be.
so C=1 and b=6 which gives a=3.So he ate normal for 3 days