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Sia ? 4 years, 5 months ago
Given : number should be possible to be constructed from any date of the year by adding the number of the month to the number of the day. with that logic, December 21st would become the number 33, since December is the twelfth month, and 21 + 12 = 33. March 23 would become 26, and so on.
To Find :how many different numbers can be made using dates of a regular calendar.
a. 28
b. 34
c. 42
d. 18
Solution:
January has 1 to 31 days and month 1 Hence
from 2 to 32
December has 1 to 31 days month 12 Hence
from from 12 to 43
Hence all possible numbers are from
2 to 43
Which counts to be 42
Hence 42 is correct answer
42 different numbers can be made using dates of a regular calendar.
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Balwant Kumar 4 years, 5 months ago
Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of a given triangle as shown in figure.
Now, (3,4) is the mid-point of AB, therefore,
{tex}3=\frac{x_1+x_2}{2}\ {/tex}and {tex}4=\frac{y_1+y_2}{2}{/tex}
x1 + x2 = 6 and y1 + y2 = 8 ..... (i)
(2,0) is the mid-point of BC, then,
{tex}2=\frac{x_2+x_3}{2}{/tex} and {tex} 0=\frac{y_2+y_3}{2}{/tex}
x2 + x3 = 4 and y2 + y3 = 0 ..........(ii)
(4,1) is the mid-point of AC, then,
{tex}4=\frac{x_1+x_3}{2}{/tex} and {tex}1=\frac{y_1+y_3}{2}{/tex}
x1 + x3 = 8 and y1 + y3 = 2 .........(iii)
Subtracting (ii) from (iii), we get,
x1 - x2 = 4 and y1 - y2 = 2 ........ (iv)
Adding (i) and (iv), we get,
2x1 = 10 and 2y1 = 10
x1 = 5 and y1 = 5
From (i), we have,
x2 = 6 - 5 = 1 and y2 = 8 - 5 = 3
From (ii), we have,
x3 = 4 - 1 = 3 and y3 = 0 - y2 = 0 - 3 = -3
Thus (5, 5), (1, 3) and (3, -3) are the vertices of triangle.
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Ayushi Porwal 4 years, 5 months ago
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