New!
Get regular updates about newly solved problems
via algebra.com's RSS system.
Recent problems solved by 'ankor@dixie-net.com'
ankor@dixie-net.com answered: 15646 problems
Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 4740..4769 , 4770..4799 , 4800..4829 , 4830..4859 , 4860..4889 , 4890..4919 , 4920..4949 , 4950..4979 , 4980..5009 , 5010..5039 , 5040..5069 , 5070..5099 , 5100..5129 , 5130..5159 , 5160..5189 , 5190..5219 , 5220..5249 , 5250..5279 , 5280..5309 , 5310..5339 , 5340..5369 , 5370..5399 , 5400..5429 , 5430..5459 , 5460..5489 , 5490..5519 , 5520..5549 , 5550..5579 , 5580..5609 , 5610..5639 , 5640..5669 , 5670..5699 , 5700..5729 , 5730..5759 , 5760..5789 , 5790..5819 , 5820..5849 , 5850..5879 , 5880..5909 , 5910..5939 , 5940..5969 , 5970..5999 , 6000..6029 , 6030..6059 , 6060..6089 , 6090..6119 , 6120..6149 , 6150..6179 , 6180..6209 , 6210..6239 , 6240..6269 , 6270..6299 , 6300..6329 , 6330..6359 , 6360..6389 , 6390..6419 , 6420..6449 , 6450..6479 , 6480..6509 , 6510..6539 , 6540..6569 , 6570..6599 , 6600..6629 , 6630..6659 , 6660..6689 , 6690..6719 , 6720..6749 , 6750..6779 , 6780..6809 , 6810..6839 , 6840..6869 , 6870..6899 , 6900..6929 , 6930..6959 , 6960..6989 , 6990..7019 , 7020..7049 , 7050..7079 , 7080..7109 , 7110..7139 , 7140..7169 , 7170..7199 , 7200..7229 , 7230..7259 , 7260..7289 , 7290..7319 , 7320..7349 , 7350..7379 , 7380..7409 , 7410..7439 , 7440..7469 , 7470..7499 , 7500..7529 , 7530..7559 , 7560..7589 , 7590..7619 , 7620..7649 , 7650..7679 , 7680..7709 , 7710..7739 , 7740..7769 , 7770..7799 , 7800..7829 , 7830..7859 , 7860..7889 , 7890..7919 , 7920..7949 , 7950..7979 , 7980..8009 , 8010..8039 , 8040..8069 , 8070..8099 , 8100..8129 , 8130..8159 , 8160..8189 , 8190..8219 , 8220..8249 , 8250..8279 , 8280..8309 , 8310..8339 , 8340..8369 , 8370..8399 , 8400..8429 , 8430..8459 , 8460..8489 , 8490..8519 , 8520..8549 , 8550..8579 , 8580..8609 , 8610..8639 , 8640..8669 , 8670..8699 , 8700..8729 , 8730..8759 , 8760..8789 , 8790..8819 , 8820..8849 , 8850..8879 , 8880..8909 , 8910..8939 , 8940..8969 , 8970..8999 , 9000..9029 , 9030..9059 , 9060..9089 , 9090..9119 , 9120..9149 , 9150..9179 , 9180..9209 , 9210..9239 , 9240..9269 , 9270..9299 , 9300..9329 , 9330..9359 , 9360..9389 , 9390..9419 , 9420..9449 , 9450..9479 , 9480..9509 , 9510..9539 , 9540..9569 , 9570..9599 , 9600..9629 , 9630..9659 , 9660..9689 , 9690..9719 , 9720..9749 , 9750..9779 , 9780..9809 , 9810..9839 , 9840..9869 , 9870..9899 , 9900..9929 , 9930..9959 , 9960..9989 , 9990..10019 , 10020..10049 , 10050..10079 , 10080..10109 , 10110..10139 , 10140..10169 , 10170..10199 , 10200..10229 , 10230..10259 , 10260..10289 , 10290..10319 , 10320..10349 , 10350..10379 , 10380..10409 , 10410..10439 , 10440..10469 , 10470..10499 , 10500..10529 , 10530..10559 , 10560..10589 , 10590..10619 , 10620..10649 , 10650..10679 , 10680..10709 , 10710..10739 , 10740..10769 , 10770..10799 , 10800..10829 , 10830..10859 , 10860..10889 , 10890..10919 , 10920..10949 , 10950..10979 , 10980..11009 , 11010..11039 , 11040..11069 , 11070..11099 , 11100..11129 , 11130..11159 , 11160..11189 , 11190..11219 , 11220..11249 , 11250..11279 , 11280..11309 , 11310..11339 , 11340..11369 , 11370..11399 , 11400..11429 , 11430..11459 , 11460..11489 , 11490..11519 , 11520..11549 , 11550..11579 , 11580..11609 , 11610..11639 , 11640..11669 , 11670..11699 , 11700..11729 , 11730..11759 , 11760..11789 , 11790..11819 , 11820..11849 , 11850..11879 , 11880..11909 , 11910..11939 , 11940..11969 , 11970..11999 , 12000..12029 , 12030..12059 , 12060..12089 , 12090..12119 , 12120..12149 , 12150..12179 , 12180..12209 , 12210..12239 , 12240..12269 , 12270..12299 , 12300..12329 , 12330..12359 , 12360..12389 , 12390..12419 , 12420..12449 , 12450..12479 , 12480..12509 , 12510..12539 , 12540..12569 , 12570..12599 , 12600..12629 , 12630..12659 , 12660..12689 , 12690..12719 , 12720..12749 , 12750..12779 , 12780..12809 , 12810..12839 , 12840..12869 , 12870..12899 , 12900..12929 , 12930..12959 , 12960..12989 , 12990..13019 , 13020..13049 , 13050..13079 , 13080..13109 , 13110..13139 , 13140..13169 , 13170..13199 , 13200..13229 , 13230..13259 , 13260..13289 , 13290..13319 , 13320..13349 , 13350..13379 , 13380..13409 , 13410..13439 , 13440..13469 , 13470..13499 , 13500..13529 , 13530..13559 , 13560..13589 , 13590..13619 , 13620..13649 , 13650..13679 , 13680..13709 , 13710..13739 , 13740..13769 , 13770..13799 , 13800..13829 , 13830..13859 , 13860..13889 , 13890..13919 , 13920..13949 , 13950..13979 , 13980..14009 , 14010..14039 , 14040..14069 , 14070..14099 , 14100..14129 , 14130..14159 , 14160..14189 , 14190..14219 , 14220..14249 , 14250..14279 , 14280..14309 , 14310..14339 , 14340..14369 , 14370..14399 , 14400..14429 , 14430..14459 , 14460..14489 , 14490..14519 , 14520..14549 , 14550..14579 , 14580..14609 , 14610..14639 , 14640..14669 , 14670..14699 , 14700..14729 , 14730..14759 , 14760..14789 , 14790..14819 , 14820..14849 , 14850..14879 , 14880..14909 , 14910..14939 , 14940..14969 , 14970..14999 , 15000..15029 , 15030..15059 , 15060..15089 , 15090..15119 , 15120..15149 , 15150..15179 , 15180..15209 , 15210..15239 , 15240..15269 , 15270..15299 , 15300..15329 , 15330..15359 , 15360..15389 , 15390..15419 , 15420..15449 , 15450..15479 , 15480..15509 , 15510..15539 , 15540..15569 , 15570..15599 , 15600..15629 , 15630..15659, >>NextTravel_Word_Problems/638504: Kate runs twice as fast as she walks. On her way to school, she walks for twice the length of time as she runs. In this way she takes 20 minutes to get there. On her way home, she runs for twice the length of time as she walks. How long does she take to get home?
1 solutions
Answer 402407 by ankor@dixie-net.com(15661)   on 2012-08-19 16:04:17 (Show Source):
You can put this solution on YOUR website! Kate runs twice as fast as she walks.
On her way to school, she walks for twice the length of time as she runs.Le
In this way she takes 20 minutes to get there.
On her way home, she runs for twice the length of time as she walks.
How long does she take to get home?
:
Let s = the walking speed
and
2s = the running speed
:
From the information given, we know she walked  (20)min and ran  (20)min
Using decimals, walked 13.33 min and ran 6.67 min
Find the distance; dist = time * speed
13.33s + 6.67(2s) = 26.67s is the distance one way
:
"On her way home, she runs for twice the length of time as she walks."
let t = time she walked going home
then
2t = time she ran going home
Another distance equation
st + 2s(2t) = 26.67s
st + 4st = 26.67s
Divide thru by s
t + 4t = 26.67
5t = 26.67
t = 
t = 5 spent walking and 10 spend running
then total time to return
5 + 10 = 16 min to return home
|
Miscellaneous_Word_Problems/638374: A wire of length 16 inches is to be cut into two pieces; then each piece will be bent to form a square. Find the length of the two pieces assuming that the sum of the areas of the two squares is 10 square inches.
1 solutions
Answer 402279 by ankor@dixie-net.com(15661) on 2012-08-18 17:16:18 (Show Source):
You can put this solution on YOUR website!A wire of length 16 inches is to be cut into two pieces; then each piece will be
bent to form a square.
Find the length of the two pieces assuming that the sum of the areas of the two
squares is 10 square inches.
:
Let the length of the sides of the two square be x & y
:
4x + 4y = 16
simplify, divide by 4
x + y = 4
y = (4-x)
and
x^2 + y^2 = 10
:
Replace y with (4-x)
x^2 + (4-x)^2 = 10
FOIL (4-x)(4-x)
x^2 + 16 - 8x + x^2 = 10
Combine like terms on the left
x^2 + x^2 - 8x + 16 - 10 = 0
2x^2 - 8x + 6 = 0
simplify, divide by 2
x^2 - 4x + 3 = 0
Factors to
(x-1)(x-3) = 0
Two solutions
x = 1, then y = 3
and
x = 3, then y = 1
:
:
You can check if this is true in the two original equations
|
Travel_Word_Problems/638031: The distance between city A and B is 28 km. Two hikers left from each city (One from city A to city B and the other from city B to city A) heading towards each other at the same speed. The hiker from city A rested for an hour after having walked 9 km and then continued at a speed greater than 1 km/hour than his previous speed, then rested again 4 km away from city B. There he met the hiker from city B, who walked at a constant speed without resting until the two met. What were the speeds of the two hikers?
1 solutions
Answer 402276 by ankor@dixie-net.com(15661) on 2012-08-18 16:56:50 (Show Source):
You can put this solution on YOUR website!The distance between city A and B is 28 km.
Two hikers left from each city (One from city A to city B and the other from
city B to city A) heading towards each other at the same speed.
The hiker from city A rested for an hour after having walked 9 km and then
continued at a speed greater than 1 km/hour than his previous speed, then
rested again 4 km away from city B.
There he met the hiker from city B, who walked at a constant speed without
resting until the two met.
What were the speeds of the two hikers?
:
Let s = speed of the two hikers
then
(s+1) = speed of the A after his 1 hr rest
:
Find the distance hiker A traveled after his rest, to meet B, 4 km from City B
28-9-4 = 15 km
:
Write a time equation; time = dist/speed
 +  + 1 = 
:
+ = - 1
:
 = 
:
 =
:
 =
Multiply both sides by s
 = (4 - s)
24s + 9 = (s+1)(4-s)
FOIL the right side
24s + 9 = 4s - s^2 + 4 - s
24s + 9 = 3s - s^2 + 4
Combine on the left
s^2 + 24s - 3s + 9 - 4 = 0
s^2 + 21s + 5 = 0
Actually this equation has two negative solutions so the given scenario is not possible.
It seems unreasonable, to find a speed, where A travels 24 km (+ a rest),
while B travels 4 km
|
Travel_Word_Problems/638020: At the same time, two hikers took off, one from city B to city A and the other from city A to city B. They hiked the same distance and their speeds were constant. They met on the way, and it was revealed that the hiker from city B to city A traveled two km more than the other hiker. This hiker arrived to city A 40 minutes after the two met, and the hiker from city A arrived to city B an hour and a half after they met. What is the distance of the trail?
1 solutions
Answer 402254 by ankor@dixie-net.com(15661) on 2012-08-18 13:25:18 (Show Source):
You can put this solution on YOUR website!At the same time, two hikers took off, one from city B to city A and the other
from city A to city B.
They hiked the same distance and their speeds were constant.
They met on the way, and it was revealed that the hiker from city B to city A
traveled two km more than the other hiker.
This hiker arrived to city A 40 minutes after the two met, and the hiker from
city A arrived to city B an hour and a half after they met.
What is the distance
:
Do this using minutes for the time
let t = travel time to the point where they met
:
let x = dist from A to the meeting point
and
(x+2) = dist from B to the meeting point
then
x + (x+2) = 2x+2 km is the distance from A to B
:
A<-----40min------*---------t min-------< B(hiker B's path)
A------x km-------*-------(x+2)km-------B
A>-----t Min------*--------90min--------> B (hiker A's path)
:
Hiker B,
 = 
Cross multiply
tx = 40(x+2)
tx = 40x + 80
t = 
:
Hiker A
 = 
Cross multiply
t(x+2) = 90x
t = 
:
t=t, therefore:
=
Cross multiply
90x^2 = (x+2)(40x+80)
FOIL
90x^2 = 40x^2 + 80x + 80x + 160
90x^2 = 40x^2 + 160x + 160
Combine on the left
90x^2 - 40x^2 - 160x - 160 = 0
50x^2 - 160x - 160 = 0
Simplify, divide by 10
5x^2 - 16x - 16 = 0
Factors to
(5x + 4)(x - 4) =
The positive solution is all we want here
x = 4
"What is the distance?"
2(4) + 2 = 10 km is the distance from A to B
:
:
Confirm this, find t
t = 
t = 60 min
Find the speed of A & B
A's time in hrs (90+60)/60 = 2.5 hrs
A's speed 10/2.5 = 4 km/hr
and
B's time in hrs (40+60)/60 = 1 hrs
A's speed 10/1.67 ~ 6 km/hr
:
Find how far each traveled in 1 hr (t)
B: 6 km
A: 4 km
---------
dif: 2 km
|
Linear_Equations_And_Systems_Word_Problems/638354: A professor announces that course grades will be computed by taking 40% of a student's project score (0-100 points) and adding 60% of the students final exam score (0-100 points). If a student gets an 86 on the project, what scores can she get on the final exam to get a course grade of at least 90?
These are some conclusions I've made, which may or may not be correct.
40%(86)+60%(x) is greater than or equal to 90. I am confused by the word "And" in the problem. Does this let me know that this is a compound inequality? If so, what other numbers do I use to create another inequality...100?
1 solutions
Answer 402160 by ankor@dixie-net.com(15661) on 2012-08-17 18:55:42 (Show Source):
You can put this solution on YOUR website! professor announces that course grades will be computed by taking 40% of a
student's project score (0-100 points) and adding 60% of the students final
exam score (0-100 points).
If a student gets an 86 on the project, what scores can she get on the final exam to get a course grade of at least 90?
:
Here is one equation for this, the relationship between the project and final
exam is 4:6
:
x = students grade on the final exam
 => 90
 => 90
multiply both sides by 10
344 + 6x => 900
6x => 900 - 344
x => 556/6
x => 93 on the final exam to achieve an average of 90 for the course
:
:
You can confirm this: .4(86) + .6(93) = 90.2 (we rounded up 92.6667)
|
Surface-area/638263: How long is the piece of wood needed to frame a picture which measures 25 cm by 36 cm and the width of the wood is 3 cm?
1 solutions
Answer 402155 by ankor@dixie-net.com(15661) on 2012-08-17 18:11:09 (Show Source):
You can put this solution on YOUR website!How long is the piece of wood needed to frame a picture which measures
25 cm by 36 cm and the width of the wood is 3 cm?
:
A 3 cm wide frame will add 6 cm to each dimension
31 by 42 cm will be the dimensions which include the frame
:
Total amt of wood will be the perimeter
p = 2(31)+2(42)
p = 146 cm for 3 cm wide frame
|
Travel_Word_Problems/637688: x & y travel 4000m n finish race in tie.first x travel 50% than Y and Y travels 50% faster than x n end in tie.before y travelled 50% faster then X what distance had it covered?
1 solutions
Answer 401928 by ankor@dixie-net.com(15661) on 2012-08-15 22:01:30 (Show Source):
You can put this solution on YOUR website!Attempt to rewrite this so it makes sense.
:
x & y travel 4000m and finish the race in a tie.
At first x travels 50% faster than Y and then,
Y travels 50% faster than x, and the race ends in a tie.
Before y traveled 50% faster then X, what distance had he covered?
:
let s = the original speed of y
and
1.5s = the original speed of x
then
1.5(1.5s)= 2.25s = final speed of y
:
Let d = distance y traveled at speed s
then
(4000-d) = speed y traveled at speed 2.25s
:
Write a time equation:
y's total travel time = x's travel time
 +  = 
Multiply by least common multiple: 27s
27s* + 27s* = 27s*
Cancel out the denominators, results:
27d + 12(4000-d) = 18(4000)
27d + 48000 - 12d = 72000
27d - 12d = 72000 - 48000
15d = 24000
d = 
d = 1600 meters traveled before y changed his speed to 1.5 times x's speed
|
Age_Word_Problems/637410: 14. The present age of a father is equal to the sum of the ages of his 5 children. 12 years hence, the sum the ages of his children will be twice the ages of their father. Find the present age of the father.
1 solutions
Answer 401899 by ankor@dixie-net.com(15661) on 2012-08-15 19:31:40 (Show Source):
You can put this solution on YOUR website!The present age of a father is equal to the sum of the ages of his 5 children.
12 years hence, the sum the ages of his children will be twice the ages of their father.
Find the present age of the father.
:
let f = father's present age
let s = sum of the 5 children's now
then
f = s
"12 years hence, the sum the ages of his children will be twice the ages of their father."
(12 yrs hence, add 12 to each of the five children)
2(f + 12) = s + 5(12)
2f + 24 = s + 60)
2f = s + 60 - 24
2f = s + 36
replace s with f
2f = f + 36
2f - f = 36
f = 36 yrs is the father's present age
|
Travel_Word_Problems/637714: An ambulance left the hospital and drove south. 3 hours later a police car left driving 42 k/h faster attempting to catch up to the ambulance. The officer caught up after 2 hours. Find the average speed of the ambulance.
1 solutions
Answer 401869 by ankor@dixie-net.com(15661) on 2012-08-15 16:25:55 (Show Source):
You can put this solution on YOUR website!An ambulance left the hospital and drove south.
3 hours later a police car left driving 42 k/h faster attempting to catch up to the ambulance.
The officer caught up after 2 hours. Find the average speed of the ambulance.
:
Let a = the speed of the ambulance
then
(a+42) = the speed of the police car
:
From the information given, we know the travel time of the ambulance = 5 hrs
and the Cop = 2 hrs
:
When the cop catches the amb, they will have traveled the same distance
Write a distance equation
dist = speed * time
:
5a = 2(a+42)
5a = 2a + 84
5a - 2a = 84
3a = 84
a = 84/3
a = 28 mph speed of the ambulance
:
:
Confirm this by finding the distances, should be equal
5(28) = 140 km
2(28+42) = 140 km
|
Mixture_Word_Problems/637707: A metallurgist needs to make 12.4 lbs. of an alloy containing 50% gold. He is going to melt and combine one metal that is 60% gold with another metal that is 40% gold. How much of each should he use?
1 solutions
Answer 401851 by ankor@dixie-net.com(15661) on 2012-08-15 15:15:51 (Show Source):
You can put this solution on YOUR website!A metallurgist needs to make 12.4 lbs. of an alloy containing 50% gold.
He is going to melt and combine one metal that is 60% gold with another metal that is 40% gold.
How much of each should he use?
:
Common sense says that it would be half 50% alloy and half 40% alloy
But here is a the mixture equation
:
Let x = amt of 60% gold alloy required
the total is to be 12.4 lbs, therefore:
(12.4-x) = amt of 40% alloy required
:
Decimal equiv equation
.60x + .40(12.4-x) = .50(12.4)
.60x + 4.96 - .40x = 6.2
.60x - .40x = 6.2-4.96
.20x = 1.24
x = 1.24/.2
x = 6.2 lbs of the 60% alloy
then
12.4 - 6.2 = 6.2 lbs of 40% alloy
:
:
Check this
.6(6.2) + .4(6.2) = .5(12.4)
3.72 + 2.48 = 6.2
6.2 = 6.2
|
Quadratic_Equations/637693: A kicker starts a football game by "kicking off". The quadratic function y = -12⋅x2 + 36⋅x models the football's height after x seconds. How long is the football in the air?
1 solutions
Answer 401844 by ankor@dixie-net.com(15661) on 2012-08-15 14:41:30 (Show Source):
You can put this solution on YOUR website!A kicker starts a football game by "kicking off".
The quadratic function y = -12⋅x2 + 36⋅x models the football's height after x seconds.
How long is the football in the air?
:
y = -12x^2 + 36x
When the ball hits the ground, y = 0; therefore:
-12x^2 + 36x = 0
Factor out -12x
-12x(x - 3) = 0
Two solutions
x = 0
and
x = 3 seconds in the air
|
Travel_Word_Problems/637717: A car needs 1 minute 30 secondes to travel a distance of 1 mile.
At this rate how many miles will the car travel in 1 hour?
1 solutions
Answer 401839 by ankor@dixie-net.com(15661) on 2012-08-15 14:35:10 (Show Source):
You can put this solution on YOUR website!A car needs 1 minute 30 seconds to travel a distance of 1 mile.
At this rate how many miles will the car travel in 1 hour?:
:
Find the mph; speed = dist/time
Change 1.5 min to hrs, 1.5/60 = .025 hrs
:
s = 
s = 40 mph
|
Mixture_Word_Problems/637708: 9 quarts of a sugar solution was mixed 6 quarts of a 90% sugar solution to make an 84% sugar solution. What is the percent concentration of the first solution?
1 solutions
Answer 401836 by ankor@dixie-net.com(15661) on 2012-08-15 14:22:37 (Show Source):
You can put this solution on YOUR website!9 quarts of a sugar solution was mixed 6 quarts of a 90% sugar solution
to make an 84% sugar solution.
What is the percent concentration of the first solution?
:
Let x = the decimal equiv of the required percentage of the 1st solution
The total amt: 9 + 6 = 15 qts
:
9x + .90(6) = .84(15)
9x + 5.4 = 12.6
9x = 12.6-5.4
9x = 7.2
x = 7.2/9
x = .8 which is an 80% solution
:
|
Average/637704: 6-w=1 4/5
I know how to find the answer by using common sense. However, I'm used to problems like this:
w - 6=1 4/5
I would know to use the opposite operation and add 6 to both sides, etc...I just don't know the formula on figuring out the original problem. My textbook does not give me any guidance or example problems.
1 solutions
Answer 401834 by ankor@dixie-net.com(15661) on 2012-08-15 14:11:25 (Show Source):
You can put this solution on YOUR website!6-w=1 4/5
convert mixed number to an improper fraction; 5/5 + 4/5
6 - w = 
multiply both sides by 5 to get rid of the denominator
5(6-w) = 9
30 - 5w = 9
Subtract 9 from both sides, add 5w to both sides
30 - 9 = 5w
21 = 5w
Divide both sides by 5
w =  or 4  as a mixed number
:
:
Check this out in the original problem using improper fractions
6 - =
Convert 6 to an improper fraction with 5 as the denominator
 - =
|
Rate-of-work-word-problems/637716: Answer this school maths question - If it takes three men three hours to dig a hole, how long would it take one man to dig a hole of the same size? *
1 solutions
Answer 401830 by ankor@dixie-net.com(15661) on 2012-08-15 13:51:46 (Show Source):
You can put this solution on YOUR website!If it takes three men three hours to dig a hole, how long would it take
one man to dig a hole of the same size?
:
Find the number of man-hours required to dig a hole. 3*3 = 9 man-hrs
let h = no. of hrs for 1 man to do it
1h = 9
h = 9 hrs
|
Age_Word_Problems/637661: four times the sum of the digits of a two-digit number is equal to the number.if digits are reserved, the resulting number is 27 greater than the original number. what is the number?
1 solutions
Answer 401829 by ankor@dixie-net.com(15661) on 2012-08-15 13:47:10 (Show Source):
You can put this solution on YOUR website!Let a = the 10's digit
let b = the units
then
10a+b = the original number
:
Write an equation for each statement:
:
"four times the sum of the digits of a two-digit number is equal to the number."
4(a+b) = 10a + b
4a + 4b = 10a + b
4b - b = 10a - 4a
3b = 6a
divide both sides by 3
b = 2a
:
"if digits are reserved, the resulting number is 27 greater than the original number."
10b + a = 10a + b + 27
10b - b = 10a - a + 27
9b = 9a + 27
divide by 9
b = a + 3
replace b with 2a (from the 1st statement)
2a = a + 3
2a - a = 3
a = 3
you can find b, I'm sure
what is the number?
|
Radicals/637713: What is the best way to solve this equation in step by step ?
I'm profoundly grateful for help.
1 solutions
Answer 401827 by ankor@dixie-net.com(15661) on 2012-08-15 13:21:03 (Show Source):
You can put this solution on YOUR website!solve this equation in step by step ?
 = 
Square both sides
 = 
:
FOIL
 )*( 
:
Results
 -  - +  =
:
2x/2x = 1, square root of 1 is 1 therefore
- 1 - 1 + =
:
- 2 + =
:
Multiply each term by 2x, to clear the denominators, results
2(2) - 4x + x^2 = x
Combine on the left to form a quadratic equation
x^2 - 5x - x + 4 = 0
x^2 - 5x + 4 = 0
Factors to:
(x-4)(x-1) = 0
two solutions
x=4
x=1
|
Linear_Equations_And_Systems_Word_Problems/637513: During the class period,the number of girls is ten more than four times the number of boys.formulate two linear equations from the statement to determine the number of boys in the class if the total number of students in that class is one hundred.let the number of boys be represented by variable y and let the number of girls be represented by variable x.
1 solutions
Answer 401749 by ankor@dixie-net.com(15661) on 2012-08-14 21:02:10 (Show Source):
You can put this solution on YOUR website!During the class period,the number of girls is ten more than four times the number of boys.
formulate two linear equations from the statement to determine the number of
boys in the class if the total number of students in that class is one hundred.
let the number of boys be represented by variable y and
let the number of girls be represented by variable x.
:
Write an equation for each statement:
:
"the number of girls is ten more than four times the number of boys.'
x = 4y + 10
:
"the total number of students in that class is one hundred."
x + y = 100
|
Age_Word_Problems/637476: Jack is married to Jill. their son, junior, asked each of them to reveal their ages. Juniors parents decided to tell him but in the form of a puzzle. Jack told Junior, "If you reverse the digits in my age, you'll get your mothers age."
Jill told her son, "the sum of my age and your dad's age is equal to 11 times the difference in our ages"
"Wait a minuet," said junior, "I cant figure out your ages with just those two clues"
"You're right" said Jack "Remember that I am older than you're mother."
What are the ages of Jack and Jill?
1 solutions
Answer 401725 by ankor@dixie-net.com(15661) on 2012-08-14 19:41:16 (Show Source):
You can put this solution on YOUR website!Jack is married to Jill. their son, junior, asked each of them to reveal their ages.
Juniors parents decided to tell him but in the form of a puzzle.
Jack told Junior, "If you reverse the digits in my age, you'll get your mothers age."
:
let a = first digit in Jack's age
let b = the 2nd digit
Then
10a + b = Jack's age
and
10b + a = Jill's age
:
"the sum of my age and your dad's age is equal to 11 times the difference in our ages"
(10a+b)+(10b+a) = 11[(10a+b)-(10b+a)]
11a + 11b = 11[10a+b-10b-a]
11(a + b) = 11(9a-9b)
divide both sides by 11
a + b = 9a - 9b
b + 9b = 9a - a
10b = 8a
Divide both sides by 2
5b = 4a
b =  a
Only single digit integer solution
a = 5, b = 4
therefore
54 is Jack's age
and
45 is Jill's age
:
:
See if that checks out in the statement:
"the sum of my age and your dad's age is equal to 11 times the difference in our ages"
54 + 45 = 11(54-45)
99 = 11(9)
|
Mixture_Word_Problems/637157: Calculate how much 20% alcohol solution and 60% alcohol solution must be mixed to end up with exactly 12 gallons of a 50% alcohol solution. You'll need ____ gallons of the 60% solution.
1 solutions
Answer 401509 by ankor@dixie-net.com(15661) on 2012-08-13 20:13:07 (Show Source):
You can put this solution on YOUR website!Calculate how much 20% alcohol solution and 60% alcohol solution must be mixed
to end up with exactly 12 gallons of a 50% alcohol solution.
You'll need ____ gallons of the 60% solution.
:
Let x = amt of 60% solution required
the total amt will be 12 gals, therefore:
(12-x) = amt of 20% solution
:
.60x + .20(12-x) = .50(12)
.60x + 2.4 - .20x = 6
.60x - .20x = 6 - 2.4
.40x = 3.6
x = 3.6/.4
x = 9 gal of 60% solution required
:
:
Check
.6(9) + .2(3) = 6
|
Travel_Word_Problems/636846: At 8:00 AM a freight car left from city A to city B carrying citrus fruits. At the same time, a bike rider left from city B to city A. The freight car's speed was 32 km/hour greater than the bike rider.
They drove the same trail and their speeds were constant. The freight car arrived to city B, waited there for 1/4 of an hour, and then started driving back to city A. On the way back, the car met the bike rider 38 kilometers away from from city A. The distance between the two cities is 64 kilometers. What was the speed of the bike rider?
Thanks so much for your time! I really appreciate it!
1 solutions
Answer 401317 by ankor@dixie-net.com(15661) on 2012-08-12 22:06:43 (Show Source):
You can put this solution on YOUR website!At 8:00 AM a freight car left from city A to city B carrying citrus fruits.
At the same time, a bike rider left from city B to city A.
The freight car's speed was 32 km/hour greater than the bike rider.
They drove the same trail and their speeds were constant.
The freight car arrived to city B, waited there for 1/4 of an hour, and then
started driving back to city A.
On the way back, the car met the bike rider 38 kilometers away from from city A.
The distance between the two cities is 64 kilometers.
What was the speed of the bike?
:
Let s = speed of the bike
then
(s+32) = speed of the freight
:
A point 38 km from A would be 64-38 = 26 km from B
freight traveled 64 + 26 = 90 km when they met
bike traveled 26 km when they met
:
A time equation.
freight time = bike time
 +  = 
multiply by 4s(s+32) to clear the denominators, assemble a quadratic equation
:
solve with the quadratic formula, the positive solution
s = 11.125 km is the bike speed
|
Travel_Word_Problems/636832: The distance between city A and B is 90 kilometers. At the same time, a car and a motorcycle left from city A to city B. The motorcycle drove at a constant speed the entire drive and the car passed 1/3 of the drive 30 kilometers per hour faster than the motorcycle's speed, stopped for half an hour, and then continued the drive at a speed 20% slower than the speed driven in the first 1/3 of the drive. The car arrived 15 minutes before the motorcycle to city B (they drove the same distance). What was the speed of the motorcycle?
This is supposed to be easy, I just keep messing up somewhere.
I know Distance=Speed*Time
The motorcycle's speed is x, and the time is 90/x (because the distance is 90)
The car's speed is x+30, and the time is 30/x+30 (because 1/3 of 90 is 30)
Then the speed is .80(x+30) and the time is 60/.8x+24 (the remaining distance being 60 km)
but the car stopped for half an hour, and then arrived 15 minutes before the motorcycle, so I only add 1/4 of an hour to the time in the second equation..
After this step I keep getting stuck... Could you please help me finish?
Thanks so much! I don't know what I would do without your help!
1 solutions
Answer 401299 by ankor@dixie-net.com(15661) on 2012-08-12 20:24:47 (Show Source):
You can put this solution on YOUR website!The distance between city A and B is 90 kilometers.
At the same time, a car and a motorcycle left from city A to city B.
The motorcycle drove at a constant speed the entire drive and the car passed
1/3 of the drive 30 kilometers per hour faster than the motorcycle's speed,
stopped for half an hour, and then continued the drive at a speed 20% slower
than the speed driven in the first 1/3 of the drive.
The car arrived 15 minutes before the motorcycle to city B (they drove the same distance).
What was the speed of the motorcycle?
This is supposed to be easy, I just keep messing up somewhere.
I know Distance=Speed*Time
The motorcycle's speed is x, and the time is 90/x (because the distance is 90)
The car's speed is x+30, and the time is 30/x+30 (because 1/3 of 90 is 30)
Then the speed is .80(x+30) and the time is 60/.8x+24 (the remaining distance being 60 km)
--------------------------------------------
From what I see here, you have a time equation:
 +  + .5 =  - .25
we can simplify the 2nd fraction, divide .8 into 60 and we have
+  + .5 = - .25
subtract .5 from both sides, add fractions with same denominator and we have
 = - .75
Multiply by x(x+30) to clear denominators, results:
105x = 90(x+30) - .75x(x+30)
105x = 90x + 2700 - .75x^2 - 22.5x
Arrange as a quadratic equation on the left
.75x^2 + 22.5x - 90x + 105x - 2700 = 0
.75x^2 + 37.5x - 2700 = 0
Simplify, divide by .75. results
x^2 + 50x - 3600 = 0
This will factor to
(x+90)(x-40) = 0
The positive solution
x = 40 mph is the motorcycle speed
:
Check this out in the original time equation
|
Travel_Word_Problems/636835: A river boat's speed is 5 kilometers/hour on standing water when it is charged and 15 kilometers an hour on standing water when it is empty. The boat sets sail charged up the river (against the current) to a distance of 81 kilometers. It returns (against the current) empty. The total sailing time there and back, including 3 hours of docking for unloading the cargo, is 48 hours. What is the speed of the current?
I have no clue how to do this problem, and never did a problem like this before. Thanks for all the help, I really appreciate it!
1 solutions
Answer 401253 by ankor@dixie-net.com(15661) on 2012-08-12 14:59:06 (Show Source):
You can put this solution on YOUR website!A river boat's speed is 5 kilometers/hour on standing water when it is charged
and 15 kilometers an hour on standing water when it is empty.
The boat sets sail charged up the river (against the current) to a distance
of 81 kilometers.
It returns (against the current) empty.
The total sailing time there and back, including 3 hours of docking for
unloading the cargo, is 48 hours.
What is the speed of the current?
:
The way this reads, it is against the current up and back, but that doesn't
make sense. Assume it returns with the current
:
Let c = rate of the current
then
(5-c) = effective speed upstream, charged
and
(15+c) = effective speed downstream, empty
:
From the information given, we know the sailing time is 45 hrs (3hrs to unload)
:
Write a time equation; time = dist/speed
:
Time up + time down = 45 hrs
 +  = 45
:
multiply by (5-c)(15+c)*
(5-c)(15+c)* + (5-c)(15+c)* = 45(5-c)(15+c)
:
Cancel out the denominators, results
81(15+c) + 81(5-c) = 45(75+5c-15c-c^2)
1215 + 81c + 405 - 81c = 45(75-10c-c^2)
1620 = 3375 - 450c - 45c^2
:
combine like terms on the left
45c^2 + 450c - 3375 + 1620 = 0
45c^2 + 450c - 1755 = 0
:
Simplify, divide by 45, results
c^2 + 10c - 39 = 0
Factors to
(c+13)(c-3) = 0
the positive solution
c = 3 km/h is the current
:
:
Check the solution find the time each way
81/(5 - 3) = 40.5 hrs
81/(15 +3) = 4.5
----------------------
total time: 45 hrs; confirms our solution of current is 3 km/h
:
Did all this make sense to you here? C
|
Age_Word_Problems/636780: Two years hence, John's mother will be twice his age.Five years hence his father will be thrice his present age.What is the difference between the ages of his parents,if the sum of all the three ages is 105?
1 solutions
Answer 401245 by ankor@dixie-net.com(15661) on 2012-08-12 14:11:11 (Show Source):
You can put this solution on YOUR website! Two years hence, John's mother will be twice his age.
Five years hence his father will be thrice his present age.
What is the difference between the ages of his parents,if the sum of all the
three ages is 105?
:
let j = John's present age
let m = mother's present age
let f = father's present age
:
write an equation for each statement:
:
" Two years hence, John's mother will be twice his age."
m + 2 = 2(j+2)
m + 2 = 2j + 4
m = 2j + 4 - 2
m = 2j + 2
:
"Five years hence his father will be thrice his present age."
Note that it says "PRESENT AGE"
f + 5 = 3j
f = 3j - 5
:
"the sum of all the three ages is 105?"
j + m + f = 105
replace m with (2j+2); replace f with (3j-5)
j + (2j+2) + (3j-5) = 105
6j - 3 = 105
6j = 105 + 3
6j = 108
j = 108/6
j = 18 is John's present age
then
f = 3(18)-5
f = 49 yrs is Father's age
and
m = 2(18) + 2
m = 38 is Mother's age
:
"What is the difference between the ages of his parents,"
49 - 38 = 11 yrs, probably a story here.
:
:
Check this
18 + 38 + 49 = 105
|
Travel_Word_Problems/636806: Jeff starts driving at 55mph from the same point lauren starts driving at 70mph. They drive in the opposite directions and Lauren has a half hour head start. How long will they be able to talk on their cellphones that have a 280 mile range?
The answers we have to chose from are in form of a fraction with a whole number.
1 solutions
Answer 401240 by ankor@dixie-net.com(15661) on 2012-08-12 13:43:33 (Show Source):
You can put this solution on YOUR website!Jeff starts driving at 55mph from the same point Lauren starts driving at 70mph.
They drive in the opposite directions and Lauren has a half hour head start.
How long will they be able to talk on their cellphones that have a 280 mile range?
:
t = talking time as they travel in opposite direction
:
Write a distance equation; dist = speed * time
:
55t + 70(t+.5) = 280
55t + 70t + 35 = 280
125t = 280 - 35
125t = 245
t = 245/125 = 49/25
t = 1  hrs they can talk on the cell phone
|
Travel_Word_Problems/636755: Wendy took a trip from devenport to ohama, a distance of 300mi. she traveled part of the way by bus, which arrived at the train station just in time for Wendy to complete her journey by train. the bus averaged 40 mi/h and the train 60 mi/h. the entire trip took 5 1/2 h. how long did Wendy spend on the train?
1 solutions
Answer 401235 by ankor@dixie-net.com(15661) on 2012-08-12 13:16:44 (Show Source):
You can put this solution on YOUR website!Wendy took a trip from Davenport to Omaha, a distance of 300mi.
She traveled part of the way by bus, which arrived at the train station just
in time for Wendy to complete her journey by train.
The bus averaged 40 mi/h and the train 60 mi/h.
The entire trip took 5 1/2 h. how long did Wendy spend on the train?
:
Let t = time on the train
the entire trip took 5.5 hrs, therefore
(5.5-t) = time on the bus
:
Write a distance equation: dist = speed * time
:
Bus dist + Train dist = 300 mi
40(5.5-t) + 60t = 300
220 - 40t + 60t = 300
-40t + 60t = 300 - 220
20t = 80
t = 80/20
t = 4 hrs on the train
:
:
Check this out, bus time = 1.5 hrs, find dist on each
40*1.5 = 60 mi
60*4 = 240 mi
-----------------
total dist 300 mi
|
Graphs/636594: / = a fraction
1/5x + 1/3y = -8/15
1/5x + 4y = 34/5
solve the system by the substitution method.
1 solutions
Answer 401215 by ankor@dixie-net.com(15661) on 2012-08-12 11:03:27 (Show Source):
You can put this solution on YOUR website!x + y = 
x + 4y = 
Get rid of the fractions, multiply the first equation by 15, the 2nd eq by 5
3x + 5y = -8
x + 20y = 34
Use the 2nd equation for substitution in the 1st equation
x = -20y + 34
:
3(-20y+34) + 5y = -8
-60y + 102 + 5y = -8
-55y = -8 -102
-55y = -110
y = 
y = +2
:
Replace y with 2 in x = -20y + 34
x = -20(2) + 34
x = -40 + 34
x = -6
:
Solution x=-6, y=2
Check this in the equation: 3x + 5y = -8
3(-6) + 5(2) = -8
-18 + 10 = -8; confirms our solution
|
Equations/636508: If I am building a feature in the garden 4metresx 4 metres, to get it square what is the measurement across the diagonal. We worked it out to 5.6565 but it looks wrong.
1 solutions
Answer 401123 by ankor@dixie-net.com(15661) on 2012-08-11 16:18:56 (Show Source):
You can put this solution on YOUR website!If I am building a feature in the garden 4metres x 4 metres, to get it square what is the measurement across the diagonal.
d =  = 5.65685 ~ 5 m, 66 cm is close enough
|
|
