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Difference between revisions of "2005 CEMC Gauss (Grade 7) Problems"
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[[2005 CEMC Gauss (Grade 7) Problems/Problem 10|Solution]]
 
[[2005 CEMC Gauss (Grade 7) Problems/Problem 10|Solution]]
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= Part B: Each correct answer is worth 6 points =
  
 
== Problem 11 ==
 
== Problem 11 ==

Revision as of 20:44, 22 October 2014

Part A: Each correct answer is worth 5 points

Problem 1

The value of $\frac{3 \times 4}{6}$ is

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 6$

Solution

Problem 2

The value of $0.8 - 0.07$ is

$\text{(A)}\ 0.1 \qquad \text{(B)}\ 0.71 \qquad \text{(C)}\ 0.793 \qquad \text{(D)}\ 0.01 \qquad \text{(E)}\ 0.73$

Solution

Problem 3

[asy] //make the picture small and square size(80,80); //draw an arc from 0 to 75 degrees of radius 1 centered at the origin draw(arc((0,0),1,0,75)); //draw a line with an arrow on the end pointing to 25 degrees (scale it down by .95 so that it stays inside the arc) //dir(25) is a unit vector pointing 25 degrees draw((0,0)--scale(.95)*dir(25),Arrow); //put a label at the end of the 75 degree unit vector (and position it out 75 degrees) MarkPoint("9",dir(75),dir(75)); //put a label at the end of the 75 degree unit vector (rotate label 75 degrees to create a tick) MarkPoint(75,"-",scale(.85)*dir(75),dir(75)); MarkPoint("9.2",dir(60),dir(60)); MarkPoint(60,"-",scale(.85)*dir(60),dir(60)); MarkPoint("9.4",dir(45),dir(45)); //put a label at the end of the 45 degree unit vector - use NE to move it slightly NorthEast of this point (or use dir(45)) MarkPoint(45,"-",scale(.85)*dir(45),NE); MarkPoint("9.6",dir(30),dir(30)); MarkPoint(30,"-",scale(.85)*dir(30),dir(30)); MarkPoint("9.8",dir(15),dir(15)); MarkPoint(15,"-",scale(.85)*dir(15),dir(15)); //put a label at the end of the 0 degree unit vector - use E to move it slightly East of this point (or use dir(0)) MarkPoint("10",dir(0),E); MarkPoint(0,"-",scale(.85)*dir(0),dir(0)); [/asy]


Contestants on "Gauss Reality TV" are rated by an applause metre. In the diagram, the arrow for one of the contestants is pointing to a rating closest to:

$\text{(A)}\ 9.4 \qquad \text{(B)}\ 9.3 \qquad \text{(C)}\ 9.7 \qquad \text{(D)}\ 9.9 \qquad \text{(E)}\ 9.5$

Solution

Problem 4

Twelve million added to twelve thousand equals

$\text{(A)}\ 12,012,000 \qquad \text{(B)}\ 12,120,000 \qquad \text{(C)}\ 120,120,000 \qquad \text{(D)}\ 12,000,012,000 \qquad \text{(E)}\ 12,012,000,000$

Solution

Problem 5

The largest number in the set {$0.109, 0.2, 0.111, 0.114, 0.19$} is

$\text{(A)}\ 0.109 \qquad \text{(B)}\ 0.2 \qquad \text{(C)}\ 0.111 \qquad \text{(D)}\ 0.114 \qquad \text{(E)}\ 0.19$

Solution

Problem 6

At a class party, each student randomly selects a wrapped prize from a bag. The prizes include books and calculators. There are $27$ prizes in the bag. Meghan is the first to choose a prize. If the probability of Meghan choosing a book for her prize is $2/3$, how many books are in the bag?

$\text{(A)}\ 15 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 21 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 18$

Solution

Problem 7

Karen has just been chosen the new “Math Idol”. A total of $1,480,000$ votes were cast and Karen received $83\%$ of them. How many people voted for her?

$\text{(A)}\ 830,000 \qquad \text{(B)}\ 1,228,400 \qquad \text{(C)}\ 1,100,000 \qquad \text{(D)}\ 251,600 \qquad \text{(E)}\ 1,783,132$

Solution

Problem 8

In the diagram, what is the measure of $\angle ACB$ in degrees? [asy] size(300); draw((-60,0)--(0,0)); draw((0,0)--(64.3,76.6)--(166,0)--cycle); label("$A$",(64.3,76.6),N); label("$93^\circ$",(64.3,73),S); label("$130^\circ$",(0,0),NW); label("$B$",(0,0),S); label("$D$",(-60,0),S); label("$C$",(166,0),S); [/asy]

$\text{(A)}\ 57^\circ \qquad \text{(B)}\ 37^\circ \qquad \text{(C)}\ 47^\circ \qquad \text{(D)}\ 60^\circ \qquad \text{(E)}\ 17^\circ$

Solution

Problem 9

A movie theatre has eleven rows of seats. The rows are numbered from $1$ to $11$. Odd-numbered rows have $15$ seats and even-numbered rows have $16$ seats. How many seats are there in the theatre?

$\text{(A)}\ 176 \qquad \text{(B)}\ 186 \qquad \text{(C)}\ 165 \qquad \text{(D)}\ 170 \qquad \text{(E)}\ 171$

Solution

Problem 10

In relation to Smiths Falls, Ontario, the local time in St. John’s, Newfoundland, is $90$ minutes ahead, and the local time in Whitehorse, Yukon, is $3$ hours behind. When the local time in St. John’s is 5:36 p.m., the local time in Whitehorse is

$\text{(A)}$ 1:06 p.m. $\text{(B)}$ 2:36 p.m. $\text{(C)}$ 4:06 p.m. $\text{(D)}$ 12:06 p.m. $\text{(E)}$ 10:06 p.m.

Solution

Part B: Each correct answer is worth 6 points

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Solution

Problem 22

Solution

Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

See also

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