<p>Here it is: </p>
<p>If j, k, n re consecutive integers such that 0<j<k<n and the units ( ones) digit of the product jn is 9, what is the digit of k ?
A. 0
B. 1
C. 2
D. 3
E. 4</p>
<p>Edit : Another one need ur help</p>
<li>There are 25 trays on a table in the cafeteria. Each contain a cup only, a plate only, or both a cup and a plate. If 15 of the trays cotain cups, 21 contain plates, how many contain both a cup and a plate ?</li>
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<li>the answer is 11. it would be easy to make a venn diagram, so label each section "cups", "plates", and "cups and plates". you know that 15 contain cups, which means that 10 dont, so there would be only 10 trays that only have plates. also, you know that 21 contain plates, which means that only 4 only have cups. which means that there are 11 w/ both (25- (10 + 4) = 11)</li>
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<li>the answer is 0. i just guess the 3 consecutive numbers until i got to 9-10-11. same would be true for 19-20-21, 29-30-31, etc</li>
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<p>a slightly faster (but not much) way to solve number 1 is to realize that j and n must be different factors of 9. the factors of 9 are 1, 3, and 9; since j and n can't both be equal to 3, one must be 9 and one must be 1.</p>
<p>another approach to the cups</p>
<p>Let x be the number of trays with plates and cups</p>
<p>So 21-x is the number of just plates
15-x is the number of just cups
x is the number of both</p>
<p>All of these have to add up to 25</p>
<p>(21-x)+(15-x)+x = 25</p>
<p>21-x+15 = 25</p>
<p>36-x = 25</p>
<p>-x = -11</p>
<p>x = 11</p>
<p>j, k, n
As unscientific as backsolving method is, it's often the fastest.
Answer A. If the unit digits (t.u.d) of k is 0, then t.u.d. of j is 9 and t.u.d. of n is 1.
t.u.d. of jn is the t.u.d. of the the product of t.u.d.'s of j and n:</p>
<h1>9x1 = 9</h1>
<p>Cups'n'plates.
If you add 15 trays that contain cups (some also contain plates) and 21 trays that contain plates (some also contain cups), you'll count twice those trays that contain both a cup and a plate. The total number 25 includes these trays only once.
15 + 21 - 25 = 11</p>
<p>Thankx guys, your all genious :D BTW, I just bought a Barron Critical Workbook for increase my CR score, is it really helpful or I took a wrong step to buy that</p>
<p>Bought one too since CC guys said it's helpful. Just started working on the book myself :)</p>
<p>yeah the CR book by barron's is pretty good, helped my score go up by 100 pts. in CR section =D (600->700) now I'm trying to get a 750</p>
<p>If j, k, n are consecutive integers such that 0<j<k<n and the units ( ones) digit of the product jn is 9, what is the digit of k ?</p>
<p>So we know these numbers j,k,n are consecutive, so k = j+1 and n = j+2. When we're talking about the units digit of a product, we only have to look at the units digit of the numbers being multiplied. All the possibilities (a,b) where a and b are units digits and b = a + 2 (on a units digit ring) are the following:</p>
<p>(0,2), (1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9), (8, 0),(<--not 10 because we're only concerned with units digits), (9,1). </p>
<p>Out of these ordered pairs (a,b), the only product ab which equals 9 is the pair (9,1). </p>
<p>We know then that:
j = 10x + 9 for some x
k = 10x + 10 (11x)
n = 10x + 11 (11x + 1) </p>
<p>Therefore k has units digits 0. Of course, this problem can be solved intuitively in a matter of seconds by using actual numbers, like 9,10, and 11.</p>
<ol>
<li>There are 25 trays on a table in the cafeteria. Each contain a cup only, a plate only, or both a cup and a plate. If 15 of the trays cotain cups, 21 contain plates, how many contain both a cup and a plate ?</li>
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<p>Let x = number of trays with only a cup, y = number of trays with only a plate, and z = number of trays with both a cup and a plate.</p>
<p>We know: </p>
<p>x+y+z = 25
x+z = 15
y+z = 21</p>
<p>We can easily see:
y = 10
x = 4
z = 11.</p>
<p>The answer is 11.</p>