Questions on Word Problems: Numbers, consecutive odd/even, digits answered by real tutors!

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The tens digit of a number is 3 less than the units digit. If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3. What is the original number?
:
Let x = the 10's digit; Let y = units digit
Then
10x+y = "the number"
:
"The tens digit of a number is 3 less than the units digit.",therefore
x = (y-3)
:
" If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3.
:
Subtract the remainder from the number and you have:
= 4
:
multiply both sides by (x+y)
10x + y - 3 = 4(x+y)
:
10x + y - 3 = 4x + 4y
Arrange x on the left and y on the right with the remainder:
10x - 4x = 4y - y + 3
:
6x = 3y + 3
Substitute (y-3) for x:
6(y-3) = 3y + 3
:
6y - 18 = 3y + 3
:
6y - 3y = 3 + 18
:
3y = 21
y =
y = 7
then
x = 7 -3
x = 4
:
47 = "the number"
:
Check solution in the statement:
"If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3."
= 4, remainder of 3

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5(4x)-10x=20
20x-10x=20
10x=20
x=20/10
x=2 failed tests.
4*2=8 passed tests.
Proof:
5*8-10*2=20
40-20=20
20=20

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x+1.2x=157498
2.2x=157498
x=157498/2.2
x=71590 is the January profit
71590*1.2=85908 is the February profit.
Proof:
71590+85908=157498
157498=157498

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Let the first integer be x, the next consecutive integer is (x+1).
Expand this.
Simplify.
Subtract 217 from both sides.
Factor the trinomial.



or
The two integers are 8 and 9
Check:
=

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(x+1)^3-x^3=217
x^3+3x^2+3x+1-x^3=217
3x^2+3x-216=0
(3x-24)(x+9)=0
3x-24=0
3x=24
x=24/3
x=8 for the amallernteger.
8+1=9 for the larger integer.
proof;
9^3-8^3=217
729-512=217
217=217

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Integers=x, x+2, x+4
(x+x+2)/2=(x+4-x)^2
(2x+2)/2=4^2
2(x+1)/2=16
x+1=16
x=16-1
x=15 the smallest integer.
15+2=17 the next one.
15+4=19 the largest.
proof:
(15+17)/2=(19-15)^2
32/2=4^2
16=16

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I have 51 handle bars and 116 wheels. Using all the parts how many tricycles and bicycles can I assemble?
-----------------------------------------
Let # of tricycles be "t".
Let # of bicycles by "b".
---------------------------
Handle bar equation: t + b = 51
Wheel equation.....: 3t + 2b = 116
-------------------------------------
Rewrite the equations:
3t + 3b = 153
3t + 2b = 116
------------------
Subtract 2nd from 1st to solve for "b":
b = 37 (# of bicycles)
------
Substitute into t+b = 51 to solve for "t"
t + 37 = 51
t = 14 (# of tricycles)
===========================
Cheers,
Stan H.

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what is 3252 more than 892334?
892334+3252=895586 answer.

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From the problem, we know two things:
62,094 people were at the stadium
AND
this represented 85% of the stadium
.
If x = total capacity of the stadium
then mathematically:
.85x = 62094
x = 62094/.85
x = 73052 (this is the capacity)

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R+(R-8)+(R-8)/2=68
R+R-8+.5R-4=68
2.5R=68+12
2.5R=80
R=80/2.5
R=32 REPUBLICANS.
32-8=24 DEMOCRATS.
24.2=12 INDEPENDENTS.
PROOF:
32+24+12=68
68=68

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x(x+2)>=35
x^2+2x-35>=0
(x+7)(x-5)=0
x=5 for theshorter side.
5+2=7 for the longer side.

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There are 40 cows and chicken in the farmyard. One quiet afternoon, Jack counted and found that there were 100 legs in all. How many cows and chickens are there? Solve this problem by writing and solving an equation.
--------------
C = # of cows
K = # of chickens
4C + 2K = 100 (4 legs per cow, 2 per chicken)
C + K = 40
------------
Sub K = 40-C into eqn 1
4C + 2*(40-C) = 100
4C + 80 -2C = 100
2C = 20
C = 10
K = 40-C = 30
-----------
10 cows and 30 chickens

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There are probably more than one set of combinations, but here's one set: 7, 11 13, 19


7+11+13+19=50

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421 exceeds nine times a number by 61. What is the number?
"exceeds a number by 61" means "is 61 more than a number". So 421 is 61 more than 9 times a number".
a number=x

Equation:
421=9x+61

Now to solve it. Subtract 61 from both sides
360=9x
40=x

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"421 exceeds nine times a number by 61" translates to



Start with the given equation.


Subtract from both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


Reduce.


----------------------------------------------------------------------

Answer:

So the answer is which means that the number is 40

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Let the two consecutive even integers be n,n+2
As the larger, added to eight times the smaller, equals 110, we have
n + 2 + 8n = 110
Solving for n, we obtain:
9n + 2 = 110
9n = 108
n = 12
So the two consecutive even integers are 12 and 14.

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Let the three consecutive odd integers be n-2,n,n+2.
Then their sum is n-2+n+n+2.
As the sum is 279, we have
n-2+n+n+2=279
Solving for n, we have
3n=279
n= 279/3
n= 93
So the three consecutive odd integers are 91,93,95.

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Let the two consecutive numbers be n, n+ 1.
The sum of the first and 3 times the second can be expressed as
n + 3(n + 1)
Setting it equal to 55, we have
n + 3(n + 1) = 55
Solving for x, we have
n + 3n + 3 = 55
4n + 3 = 55
4n = 52
n = 13
So the two consecutive numbers are 13 and 14.

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At 11 am Kate left clinton tollbooth travelling west at 809 km/h. At 12 pm, Juan left the tollbooth traveling west at 72 km/h. At what time will they be 98 km apart
----------
Kate DATA:
rate = 809 km/h ; time = x + 1 hrs ; distance = 809(x+1) km
------------------------------
Juan DATA:
rate = 72 km/h ; time = x hrs ; distance = 72x km
--------------------
EQUATION:
distance + distance = 98 km
809(x+1) + 72x = 98
809x + 809 + 72x = 98
---------------
These numbers give a negative answer.
Most probably the 809 km/h rate for Kate is wrong.
------------------------------
Cheers,
Stan H.

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The trick here is to total of the 'fractional parts'.
So add


The total fraction when everyone is counted must be 1. We are short
So 2 two goalies must be worth of the group.


Now check your answers.
1/2 would be 12
1/6 would be 4
1/12 would be 2
Does 12+4+4+2+2 = 24?

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How many 9 digit numbers can be made from the set 1-9. No repeat digits so that all even digits remain adjacent?
------------------------------------
The adjacent even digits can be arranged in 4! ways.
Considering the 4-even as one clump, you have 5 odd and a clump to arrange.
-----------------
Total # of arrangements is 4! * 6! = 17280
=====================
Cheers,
Stan H.

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What is a two digit number equal to three times the product of its digits?

Solve for u







Do long division using only integers:

         3 
3t-1)10t+0
      9t-3
       t+3



So we can tell from that that

u must be larger than 3.  So try u = 4











So 24 is a solution

The product of its digits is 8, and 24 is 3 times 8.

Try u=5









 

So 15 is another solution

The product of its digits is 5, and 15 is 3 times 5.

Those are the only two solutions.

Edwin

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You need to find 2/3 of 69
Cancel the 3's as indicated.

46 of the handkerchiefs are blue.

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i) Problem: The moon travels an elliptical path with Earth as one focus. The maximum distance from the moon to Earth is 405,500 km and the minimum distance is 363,300 km.
a = 405,000 ; b = 363,300
Then c^2 = a^2-b^2 = 3.2256x10^10
So c = 179599.55
-------------------------------
(1) What is the eccentricity of the orbit?
e = c/a = 179599.55/405000 = 0.443456...
--------------------------------------------
(2) For a planet or satellite in an elliptical orbit around a focus of the ellipse, perigee (P) is defined to be its closest distance to the focus and apogee (A) is defined to be its greatest distance from the focus. Show that is equal to the eccentricity of the orbit.
Comment: This question is confused and meaningless.
---------------------------------
(3) Find the Apogee. Find the Perigee.
Focus is c = 179599.55 from the center
Apogee:distance from focus to furthest point = 179599.55+405000 = 584599.55 km
Perigee: distance from focus to closest point = 405000-179599.55 = 225400.45 km
================================
Cheers,
Stan H.

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There are pigs and chicken in the farm. The total no. of heads and legs of both animals is 100. How many pigs and chicken are there in the farm?
:
Let p = no. of pigs
Let c = no. of chicks
:
Legs expression:
4p + 2c
;
Head expression:
p + c
:
no. of legs = 100 - no. of heads
:
(4p + 2c) = 100 - (p+c)
4p + 2c = 100 - p - c
4p + p + 2c + c = 100
5p + 3c = 100
5p = 100 - 3c
p = 100/5 - (3/5)c
p = 20 - .6c
;
p has to be an integer, what integer value for c will provide this?
:
Turns out c = multiples of 5 from 5 thru 30 will satisfy
c p
5 17
10 14
15 11
20 8
25 5
30 2
:
Examples
5 chicks have 10 legs + 5 heads total = 15
17 pigs have 68 legs + 17 heads total = 85
and
30 chicks have 60 legs + 30 heads total = 90
2 pigs have 8 legs + 2 heads total = 10

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Let n be the measure of the smaller angle, then the larger one is n + 2.
As their sum is 180, we can set up an equation:
n + n + 2 = 180
Solving for n, we have
2n + 2 = 180
2n = 178
n = 89
So one angle is 89 degrees, another one is n + 2 = 89 + 2=91 degrees.

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The moon travels an elliptical path with Earth as one focus. The maximum distance from the moon to Earth is 405,500 km and the minimum distance is 363,300 km.
(1) What is the eccentricity of the orbit?

Eccentricity = c/a 
where c is the distance from the center to the focus of the ellipse 
a is the distance from the center to a vertex 

Here is a sketch:



 The ellipse represents the orbit of the moon. 



So the coordinates of M are (4.055,0)
and the coordinates of N are (-3.633,0)  

The ellipse has vertices are at M and N. The center of the ellipse,
R, is the midpoint between M and N, so we use the midpoint formula

midpoint = (, ),

midpoint = (,) = (0.211,0)

and find that the center of the ellipse is R(0.211,0)

Since the focus of the ellipse is the earth at (0,0) 
and the center of the ellipse is at R(0.211,0), the value of c
is c=0.211 units (distance from center to focus).  That is c = 211 km.

the value of a is a=  (the distance from the ellipse's center
(.211,0) to vertex M(4.055,0) is 4.055-.211 or 3.844 units, or a = 384400 km.

Therefore the eccentricity =  

-------------------------------------------------

(2) For a planet or satellite in an elliptical orbit around a focus
of the ellipse, perigee (P) is defined to be its closest distance to
the focus and apogee (A) is defined to be its greatest distance from 
the focus. Show that  is equal to the eccentricity 
of the orbit.
 


Now don't confuse the small  for the semi-major axis
with the capital  for the apogee.








------------------------------

(3) Find the Apogee and the Perigee of problem (1)

Apogee = 405,500 km
Perigee = 363,300 km 

Checking the eccentricity using  =  

Edwin

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B=M+1,500 OR M=B-1,500
B=T+1,900 OR T=B-1,900
B+M=5T
B+(B-1.500)=5(B-1,900)
2B-1,500=5B-9,500
5B-2B=-1,500+9,500
3B=8,000
B=8,000/3
B=2,666.67 LBS. IS THE WEIGHT OF THE BOAT.

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$1625 for tickets for its X members
We divide the total purchased price to the number of members to get the price for one:
= $_____/member
Unless "X" is a Roman numeral number = 10. Then,
=$162.50 for 1 ticket for 1 member
Thank you,
Jojo

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Given
--->
--> {{2g = s}}}
-->
Use substitution







wow, cheap goats !

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"A farmer sold 7 goats, 11 chickens, 5 sheeps and 3 cows for U$980"
__ 7g+11c+5s+3c=980

"One goat costs 3 times as much as one chicken" __ 1g=3c __ g/3=c

"half as much as one sheep" __ g=s/2 __ 2g=s

"a quarter of the cost of one cow" __ g=c/4 __ 4g=c

substituting __ 7g+11(g/3)+5(2g)+3(4g)=980 __ 32 2/3 g=980 __ dividing by 32 2/3 __ g=30

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Any two-digit number can be written as the sum of its respective 10's and units components.Assuming that the units place in the particular number is 'x' and the tens place is 'y';the number may be written as
10y+x.
Now given that
10y+x=2xy
Solving for unknowns:
x=6 and y=3
So the number is 36.

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Let's label cloudy and clear mornings and afternoons.
Cloudy mornings: M_cloudy
Cloudy afternoons: A_cloudy
Clear mornings: M_clear
Clear afternoons: A_clear

We know that there are only 13 cloudy days. Since it was never cloudy for an entire day, we know that the sum of the cloudy afternoons and cloudy mornings equal the number of cloudy days:
M_cloudy + A_cloudy = 13

We also know the number of clear mornings and clear afternoons:
M_clear = 11
A_clear = 12
M_clear + A_clear = 23

Since there is one morning and one afternoon per day, the total number of mornings plus the total number of afternoons will equal twice the total number of days:
(M_cloudy + A_cloudy + M_clear + A_clear)/2
= (13 + 23)/2
= 36/2
= 18

There were 18 days during Forburg's vacation.

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$81.25

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I'll do the first two to get you going in the right direction

a)

Start with the given function.


Plug in .


Square to get .


Combine like terms.

---

b)

Start with the given equation.


Plug in .


Square to get .


Multiply by


Multiply by


Combine the fractions


Combine like terms.

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This is a problem of statistics. Instead of get bogged down in statistics language and symbols, let's take a straightforward approach:
First, let's make a few observations: The MOST pennies there will be will be 65+12=77. The LEAST pennies there will be will be 42-12=30. So we know that 30 Now,
|x+a|=1
|x+b|=4
|x+c|=6
|x+d|=9
|x+e|=11
|x+f|=12
gives a set of numbers {a,b,c,d,e,f} such that the magnitude of the difference (+ or -) from x is our set of incorrect guesses.
Most interesting here, we can note the last two equations. We have to have two numbers that are either precisely 1 away from each other, or precisely 23 away from each other. We see that this can only happen with 65 and 42 from our number set. 65-11 or 65-12 must be our number. But we need 42+11 or 42+12 to be that same number. So 53 or 54 is our number.
And right away we can notice that if a number is 1 away, that we must be dealing with 53 due to the presence of 52 in our original number list. (54-1=53, not on the list)
So our x is 53.
And to check that we can say:
53-1=52
53-4=49
53+6=59
53+9=62
53-11=42
53+12=65
This all follows.

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Let the number be n, then the next higher number is n + 1
The sum of the number and its square can be written as n + n^2.
Twelve times the next higher number can be expressed as 12(n + 1).
As the sum of the number and its square is twelve times the next higher number, we have:
n + n^2 = 12(n + 1)
Solving the quadratic equation for x, we have
n + n^2 = 12n + 12
n^2 - 11n - 12 = 0
(n - 12)(n + 1) = 0
So n = 12 or n = -1
Therefore the number is 12 or -1.

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Let the number = x
x+x^2=12(x+1)
x+x^2=12x+12
x^2-11x-12=0
(x-12)(x+1)=0
x=12 or x=-1
.
Check:
12+144=12*13 True
-1+(-1)^2=12*0 True
.
Ed

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Any problem which has two variables can only be solved if there are at least two equations.

Two numbers whose sum is 23:
x + y = 23

The difference of whose squares is 207: (Let's make x represent the larger of the two):
x² - y² = 207

Let's solve the first equation for x:
x + y = 23
x = 23 - y

Now we can plug it into the second equation:
x² - y² = 207
(23-y)² - y² = 207
23² - 46y + y² - y² = 207
529 - 46y = 207
529 - 207 - 46y = 207 - 207
322 - 46 y = 0
322 = 46y
322/46 = 46y/46
y = 7

Plugging into the first equation:
x = 23 - y
x = 16

Therefore, y = 7 and x = 16

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The two consecutive odd number; x, (x+2)
:
(x+2)^2 - x^2 = 64
FOIL
x^2 + 4x + 4 - x^2 = 64
:
4x + 4 = 64
:
4x = 64 - 4
x =
x = 15 and 17 are the numbers

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4/5k+7/10=13/15k-3/5
Re-writing:





k=19-1/2=19.5
Thank you,
Jojo

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The sum of the digits of a two-digit number is 8. If 16 is added to the original number, the result is 3 times the original number with its digits reversed. Find the original number.
;
Let x = 10's digit
Let y = units digit
;
two digit number = 10x + y
:
Sum of digits:
x + y = 8
or
y = (8-x)
:
Write an equation for the statement:
"If 16 is added to the original number, the result is 3 times the original number with its digits reversed."
10x + y + 16 = 3(10y + x)
:
10x + y + 16 = 30y + 3x
:
10x - 3x + 16 = 30y - y
:
7x = 29y - 16
:
Find the original number.
:
substitute (8-x) for y in the above equation
7x = 29(8-x) - 16
:
7x = 232 - 29x - 16
:
7x + 29x = 232 - 16
:
36x = 216
x =
x = 6, then, of course, y = 2
:
Original number = 62
;
:
Check solution in the statement:
If 16 is added to the original number, the result is 3 times the original number with its digits reversed."
62 + 16 = 3(26)
78 = 78

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The answer is right, or (shown in Answer#111569)
But we'll try to show the person who addressed the question how we arrived to that answer if you don't mind.
Let, ----------> 1st digit
------------> 2nd digit
1st condition: the sum of its digit is 8,
-----------> eqn 1
2nd condition: the squares of its digits is 50,
-------> eqn 2
IN EQN 1 WE GET, and substitute in eqn 2:
------------> working eqn

--> divide the whole eqn by 2,
-------> it's perfect square that equals to:
, got 2 values:
,
Go back eqn 1,
-------> --->

AND,
-------> ---->

SO THERE YOU GO, THE ANSWERS EITHER OR
To check, this time try eqn 2: use values of or

------------>
Thank you,
Jojo

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It is an odd two-digit number,the sum of its digits is 8,the sum of the squares of its digits is 50. What is it?
ANS :my ans is either 71 or 17

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X+7