The sum of the first 2005 terms of the sequence 1, 2, 3, 4, 1, 2, 3, 4, $\ldots$ is$5011$$5110$$5020$$5010$$501$The sequence repeats every 4 terms. How many times will the pattern $1,2,3,4$ occur in the first 2005 terms? Since 2005 divided by 4 gives a quotient of 501 and a remainder of 1, then the first 2005 terms contain the pattern $1, 2, 3, 4$ a total of 501 times (ending at the 2004th term). Also, the 2005th term is a 1. Therefore, the sum of the first 2005 terms is $501(1+2+3+4)+1=501(10)+1=5011$.Singers in a singing competition are rated by an applause metre. In the diagram, the arrow for one of the contestants is pointing to a rating that is closest to $9.4$$9.3$$9.7$$9.9$$9.5$Since the arrow is pointing between 9.6 and 9.8, it is pointing to a rating closest to 9.7.If $\sqrt{5+n}=7$, the value of $n$ is$4$$9$$24$$44$$74$Since $\sqrt{5+n}=7$ and $7=\sqrt{49}$, then $5+n=49$, so $n=44$.A piece of string fits exactly once around the perimeter of a square whose area is 144. Rounded to the nearest whole number, the area of the largest circle that can be formed from the piece of string is$144$$733$$113$$452$$183$Since the area of the square is 144, each side has length $\sqrt{144}=12$. The length of the string equals the perimeter of the square which is $4 \times 12=48$. The largest circle that can be formed from this string has a circumference of 48 or $2\pi r=48$. Solving for the radius $r$, we get $r=\frac{48}{2\pi}\approx7.64$. The maximum area of a circle that can be formed using the string is $\pi(\frac{48}{2\pi})^2 \approx\pi(7.64)^2\approx 183$.Suppose that $k>0$ and that the line with equation $y = 3kx + 4k^2$ intersects the parabola with equation $y = x^2$ at points $P$ and $Q$, as shown. If $O$ is the origin and the area of $\triangle OPQ$ is 80, then the slope of the line is$4$$3$$\frac{15}{4}$$6$$\frac{21}{4}$First, we find the coordinates of the points \(P\) and \(Q\) in terms of \(k\) by finding the points of intersection of the graphs with equations \(y=x^2\) and \(y=3kx+4k^2\). Equating values of \(y\), we obtain \(x^2 = 3kx+4k^2\) or \(x^2 - 3kx-4k^2 = 0\). We rewrite the left side as \(x^2 - 4kx + kx + (-4k)(k) = 0\) which allows us to factor and obtain \((x-4k)(x+k)=0\) and so \(x=4k\) or \(x=-k\). Since \(k>0\), \(P\) is in the second quadrant and \(Q\) is in the first quadrant, then \(P\) has \(x\)-coordinate \(-k\) (which is negative). Since \(P\) lies on \(y=x^2\), then its \(y\)-coordinate is \((-k)^2 = k^2\) and so the coordinates of \(P\) are \((-k,k^2)\). Since \(Q\) lies on \(y=x^2\) and has \(x\)-coordinate \(4k\), then its \(y\)-coordinate is \((4k)^2 = 16k^2\) and so the coordinates of \(Q\) are \((4k,16k^2)\). Our next step is to determine the area of \(\triangle OPQ\) in terms of \(k\). Since the area of \(\triangle OPQ\) is numerically equal to 80, this will give us an equation for \(k\) which will allow us to find the slope of the line. To find the area of \(\triangle OPQ\) in terms of \(k\), we drop perpendiculars from \(P\) and \(Q\) to \(S\) and \(T\), respectively, on the \(x\)-axis. The area of \(\triangle OPQ\) is equal to the area of trapezoid \(PSTQ\) minus the areas of \(\triangle PSO\) and \(\triangle QTO\). Trapezoid \(PSTQ\) has parallel bases \(SP\) and \(TQ\) and perpendicular height \(ST\). Since the coordinates of \(P\) are \((-k,k^2)\), then \(SP=k^2\). Since the coordinates of \(Q\) are \((4k,16k^2)\), then \(TQ=16k^2\). Also, \(ST=4k-(-k)=5k\). Thus, the area of trapezoid \(PSTQ\) is \(\frac{1}{2}(SP+TQ)(ST) = \frac{1}{2}(k^2+16k^2)(5k) = \frac{85}{2}k^3\). \(\triangle PSO\) is right-angled at \(S\) and so has area \(\frac{1}{2}(SP)(SO) = \frac{1}{2}(k^2)(0-(-k)) = \frac{1}{2}k^3\). \(\triangle QTO\) is right-angled at \(T\) and so has area \(\frac{1}{2}(TQ)(TO) = \frac{1}{2}(16k^2)(4k-0) = 32k^3\). Combining these, the area of \(\triangle POQ\) equals \(\frac{85}{2}k^3 - \frac{1}{2}k^3 - 32k^3 = 10k^3\). Since this area equals 80, then \(10k^3 = 80\) or \(k^3 = 8\) and so \(k=2\). This means that the slope of the line is \(3k\) which equals 6. There are six identical red balls and three identical green balls in a pail. Four of these balls are selected at random and then these four balls are arranged in a line in some order. How many different-looking arrangements are possible?1516101112Since 4 balls are chosen from 6 red balls and 3 green balls, then the 4 balls could include: 4 red balls, or 3 red balls and 1 green ball, or 2 red balls and 2 green balls, or 1 red ball and 3 green balls. There is only 1 different-looking way to arrange 4 red balls. There are 4 different-looking ways to arrange 3 red balls and 1 green ball: the green ball can be in the 1st, 2nd, 3rd, or 4th position. There are 6 different-looking ways to arrange 2 red balls and 2 green balls: the red balls can be in the 1st/2nd, 1st/3rd, 1st/4th, 2nd/3rd, 2nd/4th, or 3rd/4th positions. There are 4 different-looking ways to arrange 1 red ball and 3 green balls: the red ball can be in the 1st, 2nd, 3rd, or 4th position. In total, there are \(1+4+6+4=15\) different-looking arrangements. 4 red balls, or3 red balls and 1 green ball, or2 red balls and 2 green balls, or1 red ball and 3 green balls.Shuxin begins with \(10\) red candies, \(7\) yellow candies, and \(3\) blue candies. After eating some of the candies, there are equal numbers of red, yellow, and blue candies remaining. What is the smallest possible number of candies that Shuxin ate?\(17\)\(7\)\(11\)\(20\)\(14\)For there to be equal numbers of each colour of candy, there must be at most \(3\) red candies and at most \(3\) yellow candies, since there are \(3\) blue candies to start. Thus, Shuxin ate at least \(7\) red candies and at least \(4\) yellow candies. This means that Shuxin ate at least \(7 + 4 = 11\) candies. We note that if Shuxin eats \(7\) red candies, \(4\) yellow candies, and \(0\) blue candies, there will indeed be equal numbers of each colour.A circular spinner is divided into three sections. An arrow is attached to the centre of the spinner. The arrow is spun once. The probability that the arrow stops on the largest section is \(50\%\). The probability it stops on the next largest section is 1 in 3. The probability it stops on the smallest section is\(\frac14\)\(\frac25\)\(\frac16\)\(\frac27\)\(\frac{3}{10}\)The probability that the arrow stops on the largest section is 50% or \(\frac12\). The probability that it stops on the next largest section is 1 in 3 or \(\frac13\). Thus, the probability that the arrow stops on the smallest section is \(1-\frac12-\frac13=\frac66-\frac36-\frac26=\frac16\).In the diagram, each letter from \(A\) to \(H\) is equal to a different integer from \(1\) to \(8\). Also, \(H\) is a perfect square and is \(1\) more than \(D\) \(5\) and \(8\) are in the same row \(C\) is a multiple of both \(G\) and \(D\) \(B\) is the largest prime number in the set The value of \(B+G\) is even What is the value of \(F\)?\(2\)\(6\)\(1\)\(7\)\(8\)\(H\) is a perfect square and is \(1\) more than \(D\).The perfect squares from \(1\) to \(8\) inclusive are \(1\) and \(4\). Since \(H\) is one more than \(D\), \(H\) cannot be equal to \(1\) (since \(D\neq0\)), and so \(H=4\) and \(D=3\).\(B\) is the largest prime number in the set, and so \(B=7\).\(C\) is a multiple of both \(G\) and \(D\). Since \(D=3\), then \(C=6\) and \(G\) is equal to either \(1\) or \(2\) (since \(6\) is a multiple of each of these).The value of \(B+G\) is even, and since \(B=7\), then \(G=1\).The letters which have not been assigned values are \(A\), \(E\) and \(F\), and the integers which have not been assigned are \(2\), \(5\) and \(8\).Since \(5\) and \(8\) are in the same row and \(A\) and \(E\) are the two remaining unassigned letters that are in the same row, then \(A\) and \(E\) are \(5\) and \(8\) in some order, which leaves \(F=2\).