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Question Stats:
73% (01:12) correct
27% (01:27) wrong
based on 1726 sessions
If each term in the sum \(a_1+a_2+a_3+...+a_n\) is either 7 or 77 and the sum equals 350, which of the following could be equal to n? A. 38 B. 39 C. 40 D. 41 E. 42
Spoiler: OA
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7x + 77(n-x) = 350 --& where x is the number of times 7 repeats7x + 7(11)(n-x) = 350dividing both sides by 7x + 11 (n-x) = 50trying different optionsnow (n-x) has to be 1 because if its more than 1 (ex. 2) then 11(n-x) = 22 and x will be 37 or more which takes the total beyond 50.Therefore now trying options we get38 --& 37 + 11(1) = 4839 --& 38 + 11(1) = 4940 --& 39 + 11(1) = 50 ... this is the right answer
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If each term in the sum \(a_1+a_2+a_3+...+a_n\) is either 7 or 77 and the sum equals 350, which of the following could be equal to n? A. 38 B. 39 C. 40 D. 41 E. 42Number of approaches are possible. For example, approach #1: Since the units digit of 350 is zero then the number of terms must be multiple of 10. Only answer choice which is multiple of 10 is C (40).To illustrate consider adding:*7*7...7777----=350So, several 7's and several 77's, note that the # of rows equals to the # of terms. Now, to get 0 for the units digit of the sum the # of rows (# of terms) must be multiple of 10. Only answer choice which is multiple of 10 is C (40).Answer: C.Approach #2: \(7x+77y=350\), where \(x\) is # of 7's and \(y\) is # of 77's, so # of terms \(n\) equals to \(x+y\);\(7(x+11y)=350\) --& \(x+11y=50\) --& now, if \(x=39\) and \(y=1\) then \(n=x+y=40\) and we have this number in answer choices.Answer: C.Hope it helps.
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Thank you, Bunuel. Your first solution is a good one (the second one is rather non-deterministic for me).
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Thanks Bunuel for both the solutions. Very much appreciated.
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I used the answer choices to figure this one out ...Given that we have 77 and 7 , therefore out of all the options we can get exactly 40 options that will yield 350 if we use 7 , 39 times and 77 ones ... 7 x 39 = 273
, adding 77 we get 350..It took me about two and half minutes to do this by testing each answer out ..
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LM wrote:If each term in the sum a1+a2+a3+.....+an is either 7 or 77 and the sum equals 350, which of the following could be equal to n? A. 38 B. 39 C. 40 D. 41 E. 42This is as good as saying all the numbers are either 1 or 11 and the sum equals 50Let 1s be x and 11s be y, thus making n = x+yx + 11y = 50(x+y)+10y = 50 x+y = 10(5-y)Hence x+y must be a multiple of 10 i.e. n must be multiple of 10. Only choice CIn case if the question asks the possible values of n, we can conclude that n could be 10,20,30,40,50
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Bunuel wrote:nonameee wrote:Can I ask someone to take a look at it as I don't understand the solutions provided (or rather I don't understand how they came up with the solutions)? Thanks.If each term in the sum a1+a2+a3+.....+an is either 7 or 77 and the sum equals 350, which of the following could be equal to n? A. 38 B. 39 C. 40 D. 41 E. 42Number of approaches are possible. For example: as units digit of 350 is zero then # of terms must be multiple of 10. Only answer choice which is multiple of 10 is C (40).To illustrate consider adding:*7*7...7777----=350So, several 7's and several 77's, note that the # of rows equals to the # of terms. Now, to get 0 for the units digit of the sum the # of rows (# of terms) must be multiple of 10. Only answer choice which is multiple of 10 is C (40).Answer: C.Or: \(7x+77y=350\), where \(x\) is # of 7's and \(y\) is # of 77's, so # of terms \(n\) equals to \(x+y\);\(7(x+11y)=350\) --& \(x+11y=50\) --& now, if \(x=39\) and \(y=1\) then \(n=x+y=40\) and we have this number in answer choices.Answer: C.Hope it helps.Hi Bunuel,Thanks for all your help, as usual. I answered the question using the second approach but would love to understand the first approach better. I must be missing something simple, but could you further explain why the number off terms must be a multiple of 10 to get a &0& in the units digit? That statement is tripping me up.
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egiles wrote:Hi Bunuel,Thanks for all your help, as usual. I answered the question using the second approach but would love to understand the first approach better. I must be missing something simple, but could you further explain why the number off terms must be a multiple of 10 to get a &0& in the units digit? That statement is tripping me up.Consider this 7+7+7+7+7+7+7+7+7+7=10*7=70 (the sum of ten 7's equals to 10 times that term).Hope it helps.
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7(1+1 + ... 11+11+.. n terms)=3501+1 + ... 11+11+.. n terms=50try out various combinations.Like 11*4+6 (10 terms)The options are bigger numbers. So you decrease 11's and increase 1's.11*1+39 (40 terms) which is our answer.Hope it helped..
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LM wrote:If each term in the sum a1+a2+a3+.....+an is either 7 or 77 and the sum equals 350, which of the following could be equal to n? A. 38 B. 39 C. 40 D. 41 E. 42let us suppose there are all 7's then there must be 50 7's for the sum to be 350 . but it is not one of the answer choices . so let us suppose there is one 77 so there must be 39 7's =& total terms 40 which is answer choice c.
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77 is the equivalent of eleven 7s.Look at answer choices. Clearly, the number of 7s should be much more than 77s.Take choice 1: 38. If there are 37 7s and 1 77, we get 37*7 + 77 = 336.To get 350, we need 2 more 7s.So, answer choice C.
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\(\frac{350}{7} = 50\); As all the options are below 50, there should be atleast one 77 presentTo add one 77, we require to remove eleven 7's (7 x 11)50 - 11 + 1 = 40 = Answer = CIf 40 wasn't there, then again same method:40 - 11 + 1 = 30
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Thank you so much Brunel. This is such a simple question but my mind went blank when I was trying to attempt in the practice test! After the test got over, I used the same approach as approach1. It was so simple! Damn
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LM wrote:If each term in the sum a1+a2+a3+.....+an is either 7 or 77 and the sum equals 350, which of the following could be equal to n? A. 38 B. 39 C. 40 D. 41 E. 42We will start with the bigger numbers. There can be maximum of 50 Sevens and one 77 will replace 11 Sevens. For there to be 50 terms: there will be One 77 (11 sevens) and 39 sevens. Hence 39 + 11 = 50 sevens. Hence the answer is C.
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350= 7a + 77b (a= no of 7s & b = no of 77s)If b=0 then a=50If a=0 then b=4 (+some remainder)both these answers are not in the options.And looking at the options and above range we can conclude that the sum contains atleast 1 term as 77put b=1, 350= 7a + 77(1)350 - 77 = 7a50-11 = aa=39a+b = 39+1 = 40 which is in the options hence C
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I think we can frame another question out of this one.. If each term in the sum a1+a2+a3+.....+an is either 7 or 77 and the sum equals 350, which of the following could be the minimum
value of n?A)10
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Well, I tried it the following way:a series of 7's and 77's would add up to 350.To start with I maximize 7's and minimize 77's.So only 1 77 should be enough and rest 7's.350 -77 - 273273/9 = 39So 39 * 7 + 1 * 77 = 350Correct answer therefore in 40.
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Bunuel wrote:nonameee wrote:Can I ask someone to take a look at it as I don't understand the solutions provided (or rather I don't understand how they came up with the solutions)? Thanks.If each term in the sum a1+a2+a3+.....+an is either 7 or 77 and the sum equals 350, which of the following could be equal to n? A. 38 B. 39 C. 40 D. 41 E. 42Number of approaches are possible. For example: as units digit of 350 is zero then # of terms must be multiple of 10. Only answer choice which is multiple of 10 is C (40).To illustrate consider adding:*7*7...7777----=350So, several 7's and several 77's, note that the # of rows equals to the # of terms. Now, to get 0 for the units digit of the sum the # of rows (# of terms) must be multiple of 10. Only answer choice which is multiple of 10 is C (40).Answer: C.Or: \(7x+77y=350\), where \(x\) is # of 7's and \(y\) is # of 77's, so # of terms \(n\) equals to \(x+y\);\(7(x+11y)=350\) --& \(x+11y=50\) --& now, if \(x=39\) and \(y=1\) then \(n=x+y=40\) and we have this number in answer choices.Answer: C.Hope it helps.Hi Bunuel,I noticed that you tried to clarify method 1 below but still not connecting.How are you drawing the conclusion that it has to be a multiple of 10? Why can't it be 2*175 which is NOT a multiple of 10 or 1*350 etc?Thanks!
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russ9 wrote:Bunuel wrote:nonameee wrote:Can I ask someone to take a look at it as I don't understand the solutions provided (or rather I don't understand how they came up with the solutions)? Thanks.If each term in the sum a1+a2+a3+.....+an is either 7 or 77 and the sum equals 350, which of the following could be equal to n? A. 38 B. 39 C. 40 D. 41 E. 42Number of approaches are possible. For example: as units digit of 350 is zero then # of terms must be multiple of 10. Only answer choice which is multiple of 10 is C (40).To illustrate consider adding:*7*7...7777----=350So, several 7's and several 77's, note that the # of rows equals to the # of terms. Now, to get 0 for the units digit of the sum the # of rows (# of terms) must be multiple of 10. Only answer choice which is multiple of 10 is C (40).Answer: C.Or: \(7x+77y=350\), where \(x\) is # of 7's and \(y\) is # of 77's, so # of terms \(n\) equals to \(x+y\);\(7(x+11y)=350\) --& \(x+11y=50\) --& now, if \(x=39\) and \(y=1\) then \(n=x+y=40\) and we have this number in answer choices.Answer: C.Hope it helps.Hi Bunuel,I noticed that you tried to clarify method 1 below but still not connecting.How are you drawing the conclusion that it has to be a multiple of 10? Why can't it be 2*175 which is NOT a multiple of 10 or 1*350 etc?Thanks!Do we have 175's or 350's to sum? We have 7's and 77's. Try to sum those to get the sum with units digit of 0, and you'll see that the number of terms must be 10, 20, 30, ....
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＞＞＞已知直线l1：A1x+B1y=1和l2：A2x+B2y=1相交于点P（2，3），则过点P1（..
已知直线l1：A1x+B1y=1和l2：A2x+B2y=1相交于点P（2，3），则过点P1（A1，B1）、P2（A2，B2）的直线方程为
题型：填空题难度：中档来源：不详
∵直线l1和直线l2交于P（2，3），∴把P（2，3）代入两直线得：2A1+3B1=1；2A2+3B2=1；通过观察得到：过点P1（A1，B1）、P2（A2，B2）的直线方程为2x+3y=1即2x+3y-1=0故答案为2x+3y-1=0
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据魔方格专家权威分析，试题“已知直线l1：A1x+B1y=1和l2：A2x+B2y=1相交于点P（2，3），则过点P1（..”主要考查你对&&直线的方程，两条直线的交点坐标&&等考点的理解。关于这些考点的“档案”如下：
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直线的方程两条直线的交点坐标
直线方程的定义：
以一个方程的解为坐标的点都是某条直线上的点，这个方程就叫做这条直线的方程，这条直线叫做这个方程的直线。
基本的思想和方法：
求直线方程是解析几何常见的问题之一，恰当选择方程的形式是每一步，然后釆用待定系数法确定方程，在求直线方程时，要注意斜率是否存在，利用截距式时，不能忽视截距为0的情形，同时要区分“截距”和“距离”。
直线方程的几种形式：
1.点斜式方程：（1），（直线l过点，且斜率为k）。（2）当直线的斜率为0°时，k=0，直线的方程是y=y1。当直线的斜率为90°时，直线的斜率不存在，它的方程不能用点斜式表示，但因l上每一点的横坐标都等于x1，所以它的方程是x=x1。 2.斜截式方程：已知直线在y轴上的截距为b和斜率k，则直线的方程为：y=kx+b，它不包括垂直于x轴的直线。 3.两点式方程：已知直线经过（x1，y1），（x2，y2）两点，则直线方程为：4.截距式方程：已知直线在x轴和y轴上的截距为a，b，则直线方程为：（a、b≠0）。5.一般式方程：（1）定义：任何直线均可写成：Ax+By+C=0（A，B不同时为0）的形式。（2）特殊的方程如：平行于x轴的直线：y=b（b为常数）；平行于y轴的直线：x=a（a为常数）。 几种特殊位置的直线方程：
求直线方程的一般方法:
(1)直接法：根据已知条件，选择适当的直线方程形式，直接求出直线方程．应明确直线方程的几种形式及各自的特点，合理选择解决方法，一般地，已知一点通常选择点斜式；已知斜率选择斜截式或点斜式；已知在两坐标轴上的截距用截距式；已知两点用两点式，这时应特别注意斜率不存在的情况．(2)待定系数法：先设出直线的方程，再根据已知条件求出假设系数，最后代入直线方程，待定系数法常适用于斜截式，已知两点坐标等．利用待定系数法求直线方程的步骤：①设方程；②求系数；③代入方程得直线方程，如果已知直线过一个定点,可以利用直线的点斜式求方程，也可以利用斜截式、截距式等形式求解．两条直线的交点：
两直线：，，当它们相交时，方程组有唯一的解，以这个解为坐标的点就是两直线的交点。 若方程组无解，两直线平行；若方程组有无数个解，则两直线重合。 两条直线的交点特别提醒：
①若方程组无解，则直线平行；反之，亦成立；②若方程组有无穷多解，则直线重合；反之，也成立；③当有交点时，方程组的解就是交点坐标；④相交的条件是
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