Marmite or Kangaroo meat, anyone?

Probability Level 1

A local grocery store in the outback newly opened. They were offering 1 free bottle Marmite to every $11^\text{th}$ customer and 1 free pound of kangaroo meat for every $13^\text{th}$ customer. If there were 1000 customers that visited them on opening day, how many customers walked away with free goodies?

160 166 172 1000

3 solutions

Caleb Townsend
Mar 13, 2015

The number of people who received free Marmite is $m = \lfloor\frac{1000}{11}\rfloor = 90.$

The number of people who received free kangaroo meat is $k = \lfloor\frac{1000}{13}\rfloor = 76.$

But the question asks for how many customers received free goodies; we can't double count those who received both Marmite and kangaroo meat. This number of customers is $d = \lfloor\frac{1000}{11\times 13}\rfloor = 6.$

So the number of customers with free goodies is $f = 90 + 76 - 6 = \boxed{160}$

I got this problem wrong because I misread it as "how many goodies were given away?", but here's my simulation:

Python 2.7:

# outback grocery store simulator v1.0
winners = 0
for customer in xrange(1,1001):
    if customer % 11 == 0:
        # give away a bottle of marmite
        winners += 1
    elif customer % 13 == 0:
        # give away some kangaroo meat
        winners += 1
print "Answer:", winners

Thanks for the laugh.

Brock Brown - 6 years, 2 months ago

$\text{CORRECT !!}$

Vaibhav Prasad - 6 years, 3 months ago

Why we use floor value here ??

janki patel - 1 year, 1 month ago

Otto Bretscher
Mar 30, 2015

Since I dislike messy arithmetic, I will assume that there were 1001 customers; recall that $1001=7\times11\times13$ . Now $\frac{1001}{11}=7\times13=91$ got the Marmite, $7\times11=77$ got the kangaroo meat, and 7 got both, so that $91+77-7=161$ walked away with at least one gift. But there was no 1001th customer, so the answer is 160.

Oh, that's a nice approach!

Chung Kevin - 6 years, 2 months ago

I love the way you think, Otto !!

Bob Kadylo - 4 years, 8 months ago

Where did you get the 7 from?

Balaji Vengatesh - 1 year, 1 month ago

Nandni Jotwani
Jul 4, 2016

This is a simple question, First we can find how many multiples of 11 & 13 are there less than 1000 by A.P The largest multiple of 11 less than 1000 is 990 , and the smallest is 11 so by applying a $n$ = a + (n-1)d we can find n = 90 . By the same method we can find for 13 as well which will come out to be n= 76 Now we find the multiples common to 11 & 13 that is multiples of 143 , by using a= 143 and L = 858 Which comes out to be n=6 Now putting all in the formula we get, 90+ 76 -6 = 160

Marmite or Kangaroo meat, anyone?

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