Download Warm-up Problems (with solutions)

Warm-up problemsx
March 6, 2010
1. A soft-drink company wanted to see which of two new drinks its consumers would prefer. To find
out, the company surveyed 400 people who had tried both new drinks. 254 of the people surveyed
liked drink A, 136 liked both drinks A and B, and 42 people liked neither. Which of the two drinks
was preferred by the larger number of people? Show how you found the answer.
2. Flora has an average of 56% on her first 7 exams. What should she have to make on her eighth
exam to obtain an average of 60% on the 8 exams?
3. A sum is formed by alternatively adding and subtracting consecutive odd integers starting with 1
and ending with 1413 as indicated. What is the sum
1 − 3 + 5 − 7 + 9 − 11 + . . . + 1409 − 1411 + 1413
(Do not add all these numbers... please!)
4. We have 6 hippos and we weigh them in pairs. After weighing all possible pairs, we obtain the
following list of the weights of all pairs. Here is the list:
{361, 364, 376, 377, 380, 389, 392, 393, 396, 398, 408, 411, 414, 426, 430}
What is the sum of the weights of all the 6 hippos?
5. A group of students at Math Circle in the Triangle consists of 8 students from the 6th, 7th and
8th grades. This means that their ages are between 10 and 14 . If the product of their ages is
388,694,592 . What is the sum of their ages?
6. In a sequence, every term after the second term is twice the sum of the two preceding terms. The
seventh term of the sequence is 8, and the ninth term is 24. What is the eleventh term of the
sequence?
7. In a sequence, each term is obtained by calculating the sum of the preceding two terms starting
from the third term. The 10th term is 199, and the 7th term is 47. What is the 8th term?
Solutions
1. (400 people)-(people that liked A)-(people that liked neither)=(people that like only B) , so 400254-42=104 people that like only B. We know that there are 136 that like both A and B, so it’s
easy to see that 104+136=240 like drink B. In conclusion, 254 like drink A, while only 240 like B,
so 14 more people like A.
2. Having an average of 56% is like getting 56% on each of the seven exams. Let’s say that the score
on the eight exam is T . Then we need to find it so the new average is 60%:
56% + 56% + 56% + 56% + 56% + 56% + 56% + T
8
(7)(56%) + T
8
(7)(56%) + T
= 60%
= 69%
= (60%)(8)
T
= (60%)(8) − (7)(56%)
T
= 88%
So Flora needs a score of 88% on the last test to get a final average of 60%.
3. Let’s group the second and the third term, then fourth and fifth, and so on...
1 − 3 + 5 − 7 + 9 − . . . + 1409 − 1411 + 1413 = 1 + (−3 + 5) + (−7 + 9) + . . . + (−1411 + 1413)
= 1 + 2 + 2 + 2 + ... + 2
1412
How many two’s do we have in this sum? First we notice that the sum has
+ 1 = 707 terms
2
since we must count all the even numbers from 1 to 1412, and 1413. Secondly, since when we
grouped them we skipped the first term, 1, we have 706/2=353 pairs that add up to 2. Therefore,
1 − 3 + 5 − 7 + 9 − 11 + . . . + 1409 − 1411 + 1413 = 1 + (2)(353) = 1 + 706 = 707
4. Each hippo is in 5 pairs, so he is weighted five times. The sum of all the weight-ins is: 5w1 + 5w2 +
5w3 + 5w4 + 5w5 + 5w6 = 361 + 364 + 376 + 377 + 380 + 389 + 392 + 393 + 396 + 398 + 408 + 411 +
414 + 426 + 430 = 5915
so then w1 + w2 + w3 + w4 + w5 + w6 = 5915/5 = 1183.
5. Let’s say a is the number of students from 6th grade (i.e. 11 year-old), b is the number of students
from 7th grade (12 year old), and c is the number of students from 8th grade (13 year old). Then:
(11 · . . . · 11)(12 · . . . · 12)(13 · . . . · 13) = 388694592 = 113 · 123 · 132 , so a = 3, b = 3, and c = 2. Thus
the sum of their ages is:
(3)(11) + (3)(12) + (2)(13) = 95
6. Let’s name the terms of the sequence a1 , a2 , . . . so that we can identify its position. Then we have
that
a7 , a8 , a9 , a10 , a11
8
24 ,
From 2(a7 + a8 ) = a9 we get 8 + a8 = 24/2 = 12, so a8 = 12 − 8 = 4. Then
a10 = 2(a8 + a9 ) = 2(4 + 24) = 56 and a11 = 2(a9 + a10 ) = 2(24 + 56) = 160
Thus the eleventh term equals 160.
7. The sequence is:
a7 , a8 , a9 , a10
47
199
We have: a10 = a8 +a9 = a8 +(a7 +a8 ) = a7 +2a8 , so from 199 = 47+2a8 we get a8 = (199−47)/2 =
76.