NCTM December calendar problems.

[guilage]

I came here in november to ask for help when I got stuck in some of the problems and, well, I decided to come back cause I am stuck again and I need help!

Out of the 31 problems, there are 12 that are confusing me, some of them look simple but the answers I find seems senseless to me. Here they are:

- The increasing list of 5 integers {3, 4, 5, 8, 9} has a sum of 29. How many increasing lists of five different single-digit positive integers have a sum of 33?

- Different positive integers can be written in the eight empty circles so that the product of any three integers in a straight line is 3240. What is the largest possible sum of the eight numbers surrounding 45? Picture: http://i49.tinypic.com/ct65v.jpg

- If a+b/a=6 and b+c/c=7, find the ratio of a to c.

- Six dice are stacked on the floor, as shown. On each die, the 1 is opposite the 6, the 2 is opposite the 5, and the 3 is opposite the 4. What is the maximum possible sum of numbers on the 21 visible faces?

- A line with slope equal to 1 and a line with slope equal to 2 intersect at point P (1, 6). What is the area of PQR? Picture: http://i50.tinypic.com/2h7f605.jpg

- How many integers n have the property that the product of n digits is 0, where 5000 is smaller or equal to n, and n is bigger or equal to 6000?

- Angela and Barry share some land. The ratio of the area of Angela’s portion to Barry’s is 3:2. Both grow corn and peas on their portion. The entire piece of land is planted in corn and peas in the ratio of 7:3. On Angela’s land, the ratio of corn and peas is 4:1. Find the ratio of corn to peas on Barry’s land.

- For how many integers x is the value of -6/(x+1) an integer?

- Judi leans a 25 m ladder against a vertical wall, with the bottom of the ladder 7 m from the wall. As she pulls the bottom of the ladder away from the wall, the top of the ladder slides 4 m down the wall. How far did she pull the bottom of the ladder from its original position?

- The perimeter of a semicircular region is 20. Find the area of the region.

- On Monday, Hank drove to work at an average speed of 70 km/h and arrived 1 minute late. On Tuesday, he left at the same time and took the same route. This time he drove at an average speed of 75 km/h and arrived 1 minute early. How long is his route to work?

- If 2^x=15 and 15^y=32, find the value of xy.


Thanks for the help!

 

[chrisr]

The simple start to your first one is 4+5+7+8+9=33.
9+8+7=24, so the other 2 numbers must add to 9.
You get an alternative by changing 5 to 6 and 4 to 3 and that may be it as 8 and 7 are already included.
Are there any more?

Second one, 3240/45 = 72, so the other integers on opposite sides of the central 45 are factors of 72.
9 and 8, 18 and 4, 36 and 2, 72 and 1, 3 and 24, 6 and 12.
Now all you have to do is choose the four groups with the largest total.

Third one, you're missing parentheses, allowing for alternative interpretations of your equation.
Assuming this is (a+b)/a = 6 and (b+c)/c = 7,
then you want a/c = what?

a+b = 6a and b+c = 7c, therefore b=5a and also b=6c, so 5a=6c, so a=6c/5, so a/c = what?.

If no-one does the rest, I'll have a look after.

 

[Deleted member 4993]

I came here in november to ask for help when I got stuck in some of the problems and, well, I decided to come back cause I am stuck again and I need help!

Out of the 31 problems, there are 12 that are confusing me, some of them look simple but the answers I find seems senseless to me. Here they are:

- The increasing list of 5 integers {3, 4, 5, 8, 9} has a sum of 29. How many increasing lists of five different single-digit positive integers have a sum of 33?

- Different positive integers can be written in the eight empty circles so that the product of any three integers in a straight line is 3240. What is the largest possible sum of the eight numbers surrounding 45? Picture: http://i49.tinypic.com/ct65v.jpg

- If a+b/a=6 and b+c/c=7, find the ratio of a to c.

- Six dice are stacked on the floor, as shown. On each die, the 1 is opposite the 6, the 2 is opposite the 5, and the 3 is opposite the 4. What is the maximum possible sum of numbers on the 21 visible faces?

- A line with slope equal to 1 and a line with slope equal to 2 intersect at point P (1, 6). What is the area of PQR? Picture: http://i50.tinypic.com/2h7f605.jpg

- How many integers n have the property that the product of n digits is 0, where 5000 is smaller or equal to n, and n is bigger or equal to 6000?

- Angela and Barry share some land. The ratio of the area of Angela’s portion to Barry’s is 3:2. Both grow corn and peas on their portion. The entire piece of land is planted in corn and peas in the ratio of 7:3. On Angela’s land, the ratio of corn and peas is 4:1. Find the ratio of corn to peas on Barry’s land.

- For how many integers x is the value of -6/(x+1) an integer?

- Judi leans a 25 m ladder against a vertical wall, with the bottom of the ladder 7 m from the wall. As she pulls the bottom of the ladder away from the wall, the top of the ladder slides 4 m down the wall. How far did she pull the bottom of the ladder from its original position?

- The perimeter of a semicircular region is 20. Find the area of the region.

- On Monday, Hank drove to work at an average speed of 70 km/h and arrived 1 minute late. On Tuesday, he left at the same time and took the same route. This time he drove at an average speed of 75 km/h and arrived 1 minute early. How long is his route to work?

- If 2^x=15 and 15^y=32, find the value of xy.


Thanks for the help!

Are these problems for:

- .....................extra credit
-......................take-home or
-......................home-work

??

 

[soroban]

Hello, guilage!

\(\displaystyle \text{Given: }\:\begin{array}{ccccc}\dfrac{a+b}{a} &=& 6 & [1] \\ \\[-3mm] \dfrac{b+c}{c} &=&7 & [2] \end{array}\)

\(\displaystyle \text{Find the ratio of }a\text{ to }c.\)

\(\displaystyle \text{From [1], we have: }\:1 + \frac{b}{a} \:=\:6 \quad\Rightarrow\quad \frac{b}{a} \:=\:5 \;\;[3]\)

\(\displaystyle \text{From [2]. we have: }\:\frac{b}{c} + 1 \:=\:7 \quad\Rightarrow\quad \frac{b}{c} \:=\:6\;\;[4]\)

\(\displaystyle \text{Divide [4] by [3]: }\;\frac{\frac{b}{c}}{\frac{b}{a}} \:=\:\frac{6}{5} \quad\Rightarrow\quad\boxed{ \frac{a}{c} \:=\:\frac{6}{5}}\)

How many integers n have the property that the product of n digits is 0,
. . where: .\(\displaystyle 5000 \,\leq\, n \,<\,6000\) ?

From 5000 to 6000, there are 1001 integers.

If the product of its digits is zero, then at least one of its digit is zero.

How many of the numbers contain no zeros?

Since 6000 already contains zeros, we consider the numbers in the 5000s.

The first digit is 5.
. . There are 9 choices for the second digit.
. . There are 9 choices for the third digit.
. . There are 9 choices for the fourth digit.
Hence, there are: .\(\displaystyle 9^3 = 729\text{ numbers with no zeros.}\)

Then there are: .\(\displaystyle 1001 - 729 \:=\:272\text{ numbers with some zeros.}\)

\(\displaystyle \text{Therefore, there are }\boxed{272}\text{ numbers whose product-of-digits is zero.}\)

\(\displaystyle \text{For how many integers }x\text{ is the value of }\,\frac{-6}{x+1}\,\text{ an integer?}\)

\(\displaystyle \text{The denominator }(x+1)\text{ must be a factor of the numerator }(-6).\)

\(\displaystyle \text{This happens when the denominator is: }\:\text{-}6,\:\text{-}3,\:\text{-}2,\:\text{-}1,\;1,\;2,\;3,\;6\)

\(\displaystyle \text{Therefore, there are }\boxed{eight}\text{ values of }x\!:\;\;\text{-}7,\:\text{-}4,\:\text{-}3,\:\text{-}2,\:0,\;1,\;2,\;5\)

The perimeter of a semicircular region is 20. .Find the area of the region.

            * * *
        *           *
      *               *
     *                 *

    *                   *
    * . - - - * - - - - *
         r         r

The circumference of the semicircular arc is: \(\displaystyle \pi r\)
The diameter is: \(\displaystyle 2r\)
\(\displaystyle \text{Hence, the perimeter is: }\:\pi r + 2r \:=\:20 \quad\Rightarrow\quad r(\pi + 2) \:=\:20 \quad\Rightarrow\quad r \:=\:\frac{20}{\pi+2}\)

\(\displaystyle \text{Therefore, the area is: }\:A \;=\;\tfrac{1}{2}\pi r^2 \;=\;\tfrac{1}{2}\pi\left(\frac{20}{\pi+2}\right)^2 \quad\Rightarrow\quad \boxed{A \;=\;\frac{200\pi}{(\pi+2)^2}}\)

\(\displaystyle \text{If }2^x\,=\,15\,\text{ and }\,15^y\,=\,32,\,\text{ find the value of }xy.\)

\(\displaystyle \text{We have: }\;\begin{array}{cccc}2^x &=& 15 & [1] \\ 15^y &=& 32 & [2] \end{array}\)

\(\displaystyle \text{Raise [1] to the power }y\!:\quad (2^x)^{y} \:=\:15^y \quad\Rightarrow\quad 2^{xy} \:=\:15^y\;\;[3]\)

\(\displaystyle \text{Equate [3] and [2]: }\;2^{xy} \;=\;15^y \;=\;32 \quad\Rightarrow\quad 2^{xy} \:=\:32\)

\(\displaystyle \text{Therefore: }\;2^{xy} \:=\:2^5 \quad\Rightarrow\quad\boxed{ xy \:=\:5}\)

 

[guilage]

Subhotosh Khan

they are take home problems, we need to do them every month.

 

[Mrspi]

Subhotosh Khan

they are take home problems, we need to do them every month.

YOU need to do them every month.

WE don't have to.

And, unless you show some effort on the problems you're stuck on, SOME of us WON'T do them for you.

 

[chrisr]

For the 6th one, you surely meant integers less than or equal to 6000, but Soroban spotted that.

Number 4, the dice has no illustration.
I'm assuming that 3 faces are visible for 3 of these dices and the other 3 have 4 faces visible,
by whatever means.
Since 6, 5 and 4 are opposite the the small values, then conceivably, these digits are potentially visible on
all 6 dice, and therefore the digit 3 could also be visible on three dice unless there is a restriction
that has not been shown.
Hence, finding the maximum total visible is easy from there.

Number 5, if you sketch these lines, then the line with slope=1 goes through (1,6) and (0,5) and (-5,0).
The line with slope=2 goes through (1,6), (0,4) and (-2,0).
The triangle's base length is -2-(-5) = 3, it's perpendicular height is 6,
it has the same area as half the parallelogram that it fits inside and a parallelogram has the same area
as a rectangle with same height and base. The rest is straightforward.

Angela and Barry....
Angela has 3/5 of the land and 4/5 of that is planted with corn.
This 4/5 is (4/5)(3/5) of the total land, which is 12/25.
If corn covers 7/10 of the total land, then this is 35/50,
therefore Angela has corn on 24/50 of the entire land and Barry has (35-24)/50 of the entire land planted with corn.
As Barry has access to 20/50 of the entire land, then he has 20 parts in 50.
If 11 parts are corn, what is his ratio of corn to peas?

Judi and the ladder... use Pythagoras' theorem
h[sup:1chqj763]2[/sup:1chqj763]+7[sup:1chqj763]2[/sup:1chqj763]=25[sup:1chqj763]2[/sup:1chqj763], where h is the perpendicular height up the side of the wall of the ladder.
Having found h, use Pythagoras' theorem again,
(h-4)[sup:1chqj763]2[/sup:1chqj763]+(x+7)[sup:1chqj763]2[/sup:1chqj763]=25[sup:1chqj763]2[/sup:1chqj763] and solve for x.

Hank....
Av speed = distance travelled divided by time taken.
His speed is measured in km/hr, so convert minutes to hours.
The distance is the same, while speed and time varies, so write distance for both cases.
d=speed(time) = (x+1/60)70 = (x-1/60)75.
70x + 70/60 = 75x-75/60
70/60+75/60 = 75x-70x.
Now you solve for x.
x is the time it "should" take him.
Use that x to then solve for distance.

 

[soroban]

Hello, guilage!

A year late, but here's my approach to two more . . .

7. Angela and Barry share some land. The ratio of the area of Angela’s portion to Barry’s is 3:2.
Both grow corn and peas on their portion. The entire piece of land is planted in corn and peas in the ratio of 7:3.
On Angela’s land, the ratio of corn and peas is 4:1. Find the ratio of corn to peas on Barry’s land.

We have this information:

. . \(\displaystyle \begin{array}{c||c|c||c|} & \text{Corn} & \text{Peas} & \text{Total} \\ \hline \hline \text{Angela} & 4a & a & 5a \\ \hline \text{Barry} & C & P & C+P \\ \hline \hline \text{Total} & 4a+C & a+P & 5a+C+P \\ \hline \end{array}\)

\(\displaystyle \text{The ratio of their areas is }3:2 \qquad\frac{5a}{C+P} \:=\:\frac{3}{2} \quad\Rightarrow\quad 10a \:=\:3C + 3P\;\;[1]\)

\(\displaystyle \text{The ratio of corn to peas is }7:3 \qquad\frac{4a+C}{a+P} \:=\:\frac{7}{3} \quad\Rightarrow\quad a \:=\:\frac{7P-3C}{5}\;\;[2]\)


\(\displaystyle \text{Substitute [2] into [1]: }\;\;10\left(\frac{7P-3C}{5}\right) \:=\:3C + 3P \quad\Rightarrow\quad 14P - 6C \:=\:3C + 3P \quad\Rightarrow\quad 9C \:=\:11P\)

\(\displaystyle \text{Therefore: }\;\frac{C}{P} \:=\:\frac{11}{9}\)

11. On Monday, Hank drove to work at an average speed of 70 km/h and arrived 1 minute late.
On Tuesday, he drove at an average speed of 75 km/h and arrived 1 minute early. How long is his route to work?

\(\displaystyle \text{Hank drives }D\text{ km to work at his normal speed and takes }T\text{ hours.}\)

\(\displaystyle \text{At 70 kph, the drive took }\tfrac{1}{60}\text{ hour more: }\;\frac{D}{70} \;=\;T + \tfrac{1}{60} \quad\Rightarrow\quad D \;=\;70T + \tfrac{70}{60}\;\;[1]\)

\(\displaystyle \text{At 75 kph, the drive took }\tfrac{1}{60}\text{ hour less: }\;\frac{D}{75} \;=\;T - \tfrac{1}{60} \quad\Rightarrow\quad D \;=\;75T - \tfrac{1}{60}\;\;[2]\)

\(\displaystyle \text{Subtract [2] - [1]: }\;5T - \tfrac{145}{60} \;=\;0 \quad\Rightarrow\quad T \:=\:\tfrac{29}{60}\)

\(\displaystyle \text{Substitute into [1]: }\;D \:=\:70\left(\tfrac{29}{60}\right) + \frac{70}{60} \;=\;\tfrac{2100}{60} \;=\;35\)


\(\displaystyle \text{Therefore, he drives to work }35\text{ km.}\)