Modulus Operator – CS101 – Udacity
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https://mathbitsnotebook.com/Algebra1/LinearEquations/LEGraphiLines.html
https://mathbitsnotebook.com/Algebra1/FunctionGraphs/FNGTypeLinearQuad.html
http://www.mathbitsnotebook.com/Algebra1/LinearEquations/LEoutline.html
Problem 1
Answer: D
Solution:
We maximize the number of sandwiches Mike and Ike can buy by finding the lowest multiple of 4.5 that is less than 30 This number is 6.
Therefore, they can buy 6 sandwiches for 4.5*6=27. They spend the remaining money on soft drinks, so they buy30-27=3 soft drinks. Combining the items, Mike and Ike buy 6+3=9 items.
中文解析：
4.5*6=27. 因此买6个Sandwiches， 剩下的3元钱买3个Drinks.
Problem 2
Answer: E
Solution:
Using the diagram we find that the larger side of the small rectangle is 2 times the length of the smaller side. Therefore the longer side is 5*2=10. So the area of the identical rectangles is 5*10=50. We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is 50*3=150.
中文解析：
长方形的短边的长度是5， 则AD=5+5=10. CD=AD=10. 即长方形的长边是10. ABCD的面积是: AB *BC =(10+5)*10=150.
答案是E。
Problem 3
Answer: E
Solution:
Consider subtracting 1 from each of the fractions. Our new fractions would then be 4/11, 4/15, 4/13. Since 4/15 < 4/13 < 4/11, it follows that the answer is E.
中文解析：
15/11=1+4/11；
19/15=1+4/15；
17/13=1+4/13.
4/11 大于4/13， 大于4/15. 因此 15/11 > 17/13 > 19/15. 答案是E。
Problem 4
Answer: D
Solution:
Each side of the rhombus is 52/4=13. Let the diagonals BD and AC intersect at E. In a rhombus, diagonals are perpendicular and bisect each other, which means AE=12. Consider one of the right triangles ABE. AB=13, AE=12. Using Pythagorean theorem, we find that BE=5.
Thus the values of the two diagonals are 24, 10. The area of a rhombus is =d1*d2/2=24*10/2=120.
中文解析：
菱形的四条边长度相同，周长是52， 因此边长是52/4=13.
菱形的两条对角线互相垂直平分。在直角三角形中ABE中， AE=24/2=12. AB=13， 根据勾股定理， BE=5. 因此BD=5*2=10.
根据菱形面积公式， 菱形面积等于两条对角线相乘，除以2. 因此面积是：24*10/2=120.
答案是D。
Problem 5
Answer: B
Solution:
First, the tortoise walks at a constant rate, ruling out D Second, when the hare is resting, the distance will stay the same, ruling out E and C. Third, the tortoise wins the race, meaning that the non-constant one should go off the graph last, ruling out A. Therefore, the answer B is the only one left.
中文解析：
兔子中间休息了一段时间，因此随着时间的流逝，距离没有变化，因此兔子的曲线上应该有横着的一段直线，距离保持不变，只有选项A和B符合题意。乌龟比兔子先到达终点，选项A不符合这个特征（选项A中是兔子先到达终点）。因此答案只剩下了B。
Problem 6
Answer: C
Solution:
Lines of symmetry go through point P, and there are 8 directions the lines could go, and there are 4 dots at each direction.4*8/80=2/5.
中文解析：
如上图所示，红色的线是对称轴， 除了P点外，其上有4*8=32个点。32个点和总点数的比值是: 32/80,即 2/5.答案是C。
Problem 7
Answer: A
Solution:
We can compare each of the scores with the average of 81: 76->-5; 94->+13; 87->+6; 100->+19. So the last one has to be -33 (since all the differences have to sum to 0), which corresponds to 81-33=48.
中文解析 ：
前三次的平均分是76， 94， 87. 五次考试的平均分是81. 考试成绩的最高分是100分， 为了让未知的一个考试成绩是最低分， 则让另一个未知的考试成绩是100分。即四次考试成绩分别是76， 94， 87， 100， 五次考试的平均分是81，则第5次考试成绩是48分。答案是A。
Problem 8
Answer: E
Solution:
After Gilda gives 20% of the marbles to Pedro, she has 80% of the marbles left. If she then gives 10% of what’s left to Ebony, she has 0.8*0.9=72% of what she had at the beginning. Finally, she gives 25% of what’s left to her brother, so she has 0.75*0.72=54% of what she had in the beginning left.
中文解析：
Gilda给出20%后，剩余80%， 又给出剩余的10%后，剩余80%*90%。又给出剩余的25%后，剩余80%*90%*75%。即54%。答案是E。
Problem 9
Answer: B
Solution:
Using the formula for the volume of a cylinder, we get Alex, Pi*3*3*12=108Pi. Felicia, Pi*6*6*6=216Pi. The answer is 108/216=1/2.
中文解析：
Alex的圆柱，半径是3， 高是12， 体积是：Pi*3*3*12;
Felicia的圆柱，半径是6， 高是6， 体积是：Pi*6*6*6;
因此这两个圆柱的比值是: 1:2. 答案是B。
Problem 10
Answer: B
Solution:
20+26+16+22+16=100, so the mean is 20. The median is (16,16,20,22, 26) 20. . The coach figures out that actually 21 people come on Wednesday. The new mean is 21, while the new median is(16,20, 21,22,26) 21. The median and mean both increased by 1.
中文解析：
周三原来的数据是16， 修改后的数据是21. 增加了5. 会给整体平均值增加1.
这5个数字原来从小到大排列是：16，16，20，22， 26， 其中位数是20.
把周三的数据从16修改成21后，这5个数从小到大的排列是：16， 20， 21， 22， 26. 其中位数时21. 中位数也增加了1.
因此答案是B。
Problem 11
Answer: D
Solution:
Let x be the number of students taking both a math and a foreign language class.
we get 70+54-x=93. Solving gives us x=31. But we want the number of students taking only a math class. Which is 70-31=39.
中文解析：
总人数是93人，参加数学课的是70人， 参加外语课的是54人，因此两个课程都参加的是：（70+54）-93=31人。
参加数学课的共70人，只参加数学而不参加外语的有70-31=39人。答案是D。
Problem 12
Answer: A
Solution:
B is on the top, and R is on the side, and G is on the right side. That means that (image 2) W is on the left side. From the third image, you know that P must be on the bottom since G is sideways. That leaves us with the back, so the back must be A. The front is opposite of the back, so the answer is R.
中文解析：
以第一个图形的R为中心，其上面是B， 右边是G， 从第二个图可以看出，R的左边是W， 从第三个图可以看出R的下边是P，
因此， aqua和R颜色相对. 答案是A。
Problem 13
Answer: A
Solution:
Note that the only positive 2-digit palindromes are multiples of 11, namely 11, 22, ……99. . Since N is the sum of 2-digit palindromes, N is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so N=110 is a candidate answer.
中文解析：
2位数中的回文数分别是：11， 22， 33，……,99. 由这些数中的两个数相加，得到的最小的三位数是：11+99=110. 这就是题目所求的N。N的各位数字的和是：1+1+0=2. 答案是A。
Problem 14
Answer: C
Solution:
Let Sunday =0(mod 7); Monday=1(mod 7); Tuesday=2(mod 7); Wednesday=3(mod 7); Thursday=4(mod 7); Friday=5(mod 7); Saturday=6(mod 7).
10=3(mod 7); 20=6(mod 7); 30=2(mod 7); 40=5(mod 7); 50=1(mod 7) 60=4(mod 7).
Which clearly indicates if you start form a x=3(mod 7) you will not get a y=0(mod7).
Any other starting value may lead to a y=0(mod 7). Which means our answer is Wednesday.
中文解析：
Isabella选中的日期除以7之后的余数不能是7或0（周日）。Isabella选的6个日期之间的间隔都是10.
10除7的余数是3，
20除7的余数是6，
30除7的余数是2，
40除7的余数是5，
50除以7的余数是1，
60除以7的余数是4.
因此，从10除以7的余数是3这天开始。余数是3， 应该是星期三。
答案是C。
Problem 15
Answer: B
Solution:
The number of people wearing caps and sunglasses is 2/5 *35=14. So then 14 people out of the 50 people wearing sunglasses also have caps. 14/50=7/25.
中文解析：
由于：随机选一个戴帽子的人，他同时戴墨镜的概率是2/5， 另外戴帽子的人有35人，因此戴帽子同时戴墨镜的人有14人。
戴墨镜的人有50人，同时戴帽子和戴墨镜的有14人，因此随机选一个戴墨镜的人，他同时戴帽子的概率是：14/50， 即7/25. 答案是B。
Problem 16
Answer: D
Solution:
If he travels 15 miles at a speed of 30 miles per hour, he travels for 30 min. Average rate is total distance over total time so (15+d)/(0.5+t)=50. where d is the distance left to travel and t is the time to travel that distance. solve for d to get d=10+50t. you also know that he has to travel 55 miles per hour for some time, so d=55t plug that in for d to get 55t=10+50t =>t=2. Thus d=2*55=110. The answer is 110.
中文解析：
第一段路程的距离是15， 速度是30. 第二段路程的速度是55， 距离假设是d.
平均速度=总距离/总时间。两端路程的平均速度是: (15+d)/(15/30+d/55) 。
题目告知平均速度是50， 因此（15+d)/(15/30+d/55）=50. 求得：d=110.
答案是D。
Problem 17
Answer: B
Solution:
We rewrite: (1/2) * ( 3*2/ 2*3) *( 4*3/3*4)*…*(98*98/98*99)*(100/99)
The middle terms cancel, leaving us with (1*100) / (2*99)=50/99.
中文解析：
去掉括号，并且约分后得到：=1/2 * 100/99=50/99. 答案是B。
Problem 18
Answer: C
Solution:
We have a 2 die with 2 evens and 4 odds on both dies. For the sum to be even, the rolls must consist of 2 odds or 2 evens.
Ways to roll 2 odds (Case 1): The total number of ways to roll 2 odds is 4*4=16, as there are 4 choices for the first odd on the first roll and 4 choices for the second odd on the second roll.
Ways to roll 2 evens (Case 2): Similarly, we have 2*2=4 ways to roll 2 evens.
Totally, we have 6*6=36 ways to roll 2 dies.
Therefore the answer is (16+4)/36=5/9.
中文解析：
1，2，3，5，7，8 ，这几个数字中有4个奇数， 2个偶数。
两个数的和是偶数，要求这两个都是奇数或者都是偶数。
如果第一个色子的结果是奇数，第二个色子必须也是奇数，才能两数的和是偶数。相应的概率是: 4/6 * 4/6=4/9.
如果第二个色子的结果是偶数， 第二个色子必须也是偶数，才能两数的和是偶数。相应的概率是：2/6*2/6=1/9.
综合这两种情况，其概率是：4/9+1/9=5/9. 答案是C。
Problem 19
Answer: C
Solution:
Three top teams, say team A, team B , and team C , must beat the other three teams and each get 3*3*2=18 points. For games between A , B , C , : A team beats one team while losing to another which means (3+0)*2=6 points. This brings a total of 18+6=24 points.
中文解析：
前三名的分值一样，则后三名的分值也一样。
一个高分的team和后3名的team比赛时，均得3分。比赛2次，则得6分。和后三名的每个team比赛，共得：6+6+6=18分。
三个高分的team之间也需要比赛。如果平两场，得1+1=2分；如果赢一场，输一场，共得3+0=3分。这样分值更高。一个team和另一个team比赛，得3分， 和另一个team比赛又得3分，共6分。
合计起来，一个高分的team可得：18+6=24分。答案是C。
Problem 20
Answer: D
Solution:
If and only if x*x-5=+-4. If x*x-5=4, then x=9; => x=3 or x=-3.
If x*x-5=-4, then x*x=1 ; =>x=1 or x=-1. So, the answer is 4.
中文解析:
由于4*4=16, （-4）*（-4）=16，因此 x*x-5可以是4或者-4.
如果x*x-5=4，则x*x=9. x可以是3或者-3.
如果x*x-5=-4， 则x*x=1. x可以是1或-1.
x可以是-1，1， -3， 3 共4个可能的值。答案是D。
Problem 21
Answer: E
Solution:
Graphing the lines, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get 4*8/2=16.
中文解析：
对应的三条直线如上图所示：
围起来的三角形的底是8， 高是4 ，因此面积是16. 答案是E。
Problem 22
Answer: E
Solution:
Suppose the fraction of discount is x. That means (1+x)(1-x)=0.84. so 1-x*x=0.84. =>x*x=0.16, obtaining x=0.4. Therefore, the price was increased and decreased by 40%.
中文解析：
假设提高了x.
则提价后价格是1+x, 又降价x后，价格是：（1+x)*(1-x). 题目告知此时的价格是原来价格的84%。因此（1+x)(1-x)=0.84. 求解可得：x=0.4. 即40%。答案是E。
Problem 23
Answer: B
Solution:
We first start by setting the total number of points as 28, since LCM(4,7)=28. However, we see that this does not work since we surpass the number of points just with the information given (28/4+28*2/7+15>28). Next, we assume that the total number of points scored is 56. With this, we have that Alexa, Brittany, and Chelsea score: 56/4+56*2/7+15=45, and thus, the other seven players would have scored a total of 56-45=11.
中文解析：
总分的1/4 和2/7都必须是一个整数，因此我们通过求4和7的最小公倍数，28. 先假设总分是28. 那么Alex得了7分， Brittany得了8分， 题目告知Chelsea得了15分， 这三个人合起来已经是30分，超过总分28了，因此总分不能是28，我们假设总分是28*2，即56分。则A得了14分， B得了16分，题目告知C得了15分， 则，三人共得了45分，还剩下11分。即是题目所求的另外7个team的分数的总和。答案是B。
Problem 24
Answer: B
Solution:
Extend BD to G such that AG // BC. Then ADG similar with BDC; BEF similar with AEG. Since CD= 2 AD, CDB has 4 times the area of ADG. Since area of CBD is 240, then ADG=60. Area of ABD is 120, AED and ABE have the same height and BE=ED, thus area of AED=120/2=60. Since the area of AED=60, area of ADG=60, we have ED =DG . Thus triangle AEG has twice the side lengths and therefore four times the area of triangle EFF, giving BEF=120/4=30.
中文解析：
添加辅助线GD，使得GD//BC。
由于GD//BF， BE=ED， 因此三角形BEF和三角形GED全等。推出EG=EF。
因为AD: DC =1:2, 三角形ABD和三角形CBD等高， 因此三角形ABD的面积和三角形CBD的面积的比值也是1:2. 已知三角形ABC的面积是360， 因此三角形ABD的面积是360*1/3=120.
三角形ABE和三角形AED也是等高的， 又由于BE=ED，因此这两个三角形的面积一样大，都是三角形ABD的一半即60. 即，三角形AED的面积是60.
三角形AGD和三角形AFC， 由于GD//BC， AD：AC=1：3， 因此这两个三角形相似，且AG：AF=1：3. 则AG：GF=1：2.
又已知GF=EF， 因此AG=GE。
在三角形AGD和三角形GED，两个三角形等高， 且AG=GE ，因此三角形GED的面积是AED的一半，即30.
由于三角形BEF和三角形GED全等，因此三角形BEF的面积也是30. 答案是B。
Problem 25
Answer: C
Solution:
We use sticks and stones.
24-2-2-2=18. There are 18 stones and 2 sticks.
C(20, 2)=20!/(2!*18!)=190.
中文解析：
24个苹果分各3个小朋友， 每人至少2个。
先给每个人各分2个苹果，则剩下24-2*3=18个苹果。
剩下的18个苹果分给3个小朋友， 根据stick and stone 方法，从18+2个stick中，选出2个位置给stick。即C（20，2）。计算得到190. 答案是C。