New!
Get regular updates about newly solved problems
via algebra.com's RSS system.
Recent problems solved by 'mananth'
mananth answered: 12270 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, >>NextAge_Word_Problems/323957: Mary is presently twice as old as Maggy. Ten years ago, Maggy's age was one-third of Mary's then. Ten years from now, Maggy's age will be three-fifths of Mary's age then. How old is each?
1 solutions
Answer 231875 by mananth(12270)   on 2010-07-20 10:33:53 (Show Source):
You can put this solution on YOUR website!Let Maggy's age = x
Mary's age = 2x
..
ten years later their ages will be
Maggie= x+10
Mary = 2x+10
..
x+10 = 3/5 (2x+10)
5(x+10)=3(2x+10)
5x+50=6x+30
6x-5x=50-30
x=20 years Maggie's age
Mary's age = 2x = 40 years
|
test/323986: The shortest leg of a right triangle is 2 less than the hypotenuse and the hypotenuse is one longer than the longest leg right in quadratic equation form
1 solutions
Answer 231869 by mananth(12270) on 2010-07-20 10:21:00 (Show Source):
|
Equations/323975: Which quadratic equation has roots 7 and -2/3?
a) 2x^2 - 11x - 21=0
b) 3x^2 - 19x - 14=0
c) 3x^2 + 23x + 14=0
d) 2x^2 + 11x - 21=0
1 solutions
Answer 231856 by mananth(12270) on 2010-07-20 09:34:55 (Show Source):
|
Miscellaneous_Word_Problems/323966: A certain cake recipe states that the cake should be baked in a pan 8 inches in diameter. if jules wants to use the recipe to make a cake of the same depth but 12 inches in diameter, by what factor should he multiply the recipe ingredients ?
A 2 1/2 B 2 1/4 C 1 1/2 D 1 4/9 E 1 1/3
1 solutions
Answer 231848 by mananth(12270) on 2010-07-20 09:22:50 (Show Source):
You can put this solution on YOUR website!you have to compare the ratios of the area OR the ratio of the squares of the radius.
12" dia = 6" radius
8" diameter = 4" radius
..
r^2= 36
r^2=16
..
Ratio = 36/16
=2.25 times
OPTION B
|
Average/323945: A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
solution:
Speed= 60X5/18 m/sec = 50/3m/sec
length of the train = S X D = 50/3 X 9m = 150m
when do we use 5/18 ?
1 solutions
Answer 231825 by mananth(12270) on 2010-07-20 07:36:13 (Show Source):
|
Problems-with-consecutive-odd-even-integers/323940: using the five-step problem-solving process
A rancher outside Houston decides to build a silo inside his corral. The rectangular corral measures 23 m by 17 m and the silo's base is circular with a diameter of x meters. Find a polynomial for the remaining area of the corral (in square meters) after the silo is built.
1 solutions
Answer 231823 by mananth(12270) on 2010-07-20 07:28:09 (Show Source):
You can put this solution on YOUR website!Length of corral = 23 m
width of corral = 17m
..
Area of Corral = 17*23 = 391 m^2
..
diameter of silo = x
radius = x/2
..
Area of circle = pi*r^2
Area = pi*(x/2)^2
..
Area of corral - area of silo = area of remaining area
391- pi*(x/2)^2
|
test/323938:
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
1 solutions
Answer 231822 by mananth(12270) on 2010-07-20 07:23:00 (Show Source):
You can put this solution on YOUR website!60 km/hour = 60*1000 meters/ 3600 seconds
= 16.6 meters/second
..
speed = 16.6 meters/second
time = 9 seconds
distance traveled = length of the train
16.6 *9= 149.5 meters
|
Linear-equations/323934: Write the equation of the line which passes through (0, -4) and is parallel to the line 3x - y = 6.
1 solutions
Answer 231820 by mananth(12270) on 2010-07-20 07:10:47 (Show Source):
You can put this solution on YOUR website!line equation 3x-y=6
put it in the form y = mx +b
y=3x-6
the slope is 3
and b = -6
the line parallel to this will have the same slope.
m=3
plug the value (0,-4) in the equation y= mx+b
b=-4
y=3x-4 is the line passing through (0,-4)
|
Graphs/323881: Find the x- and y-intercepts apply elimination or substitution to find the coordinates of the point of intersection.
15x + 11y = -17
8x - 6y = 74
1 solutions
Answer 231812 by mananth(12270) on 2010-07-20 04:31:50 (Show Source):
You can put this solution on YOUR website!15x + 11y = -17................1
8x - 6y = 74...................2
multiply equation 1 by 6 and equation 2 by 11 and add them. to eliminate y
..
90x+66y=-102
88x-66y=814
178x=712
x= 712/178
x= 4
plug the value of x in any equation
90*4 +66y=-102
360+66y=-102
66y=-102-360
66y=-462
y=-7
The intercepts are (4, -7)
|
Equations/323891: Ray and Amir each bought a used car for the same price. Each needed repairs. The repairs on Ray's car cost an additional $800 and on Amir's, $200. Amir noted, "Ray, my total cost was really only 5/6 of your total cost." How much did each pay for his used car?
1 solutions
Answer 231811 by mananth(12270) on 2010-07-20 04:23:15 (Show Source):
You can put this solution on YOUR website!let the price of their cars be $X
Ray cost = x+800
Amir's cost = x+200
x+200= 5/6 ( x+800)
6(x+200)= 5(x+800)
6x+1200 = 5x+4000
x= 4000-1200
x= $2800 price of car
|
Word_Problems_With_Coins/323893: In a collection of coins consisting of only nickels and pennies, the total value is $1.10 and there are twice as many nickels as pennies. How many nickels are in the collection?
1 solutions
Answer 231810 by mananth(12270) on 2010-07-20 04:13:52 (Show Source):
|
Triangles/323922: The perimeter of a right triangle is 12 ft, and one of its legs measures 3 ft. Find the length of the other leg and the hypotenuse.
1 solutions
Answer 231808 by mananth(12270) on 2010-07-20 04:00:20 (Show Source):
You can put this solution on YOUR website!let x be the hypotenuse
and y be the other leg
one leg = 3
..
x^2-y^2= 9
(x+y)(x-y)=9
(x+y)=9
9(x-y)=9
x-y =1
x+y+3=12
..
Add the 2 equations
x-y+x+y+3=12+1
2x+3=13
2x=10
x=5 feet the hypotenuse
the second leg will be 4 feet
|
Length-and-distance/323902: What is the equation of the line perpendicular to the line joining (4, 2) and (3, -5) and passing through (4, 2)?
1 solutions
Answer 231807 by mananth(12270) on 2010-07-20 03:44:16 (Show Source):
You can put this solution on YOUR website!(4,2 ) (3, -5)
slope of the line joining them is
m= y2-y1/x2-x1
m=-5-2 / 3-4
=-7/-1
m=7
so the general equation is y=7x+b
plug the value of any one poin to get the value of b
2=28+b
b=-26
The line perpendicular to it will have slope o -1/7
The equation for that lineis y= -1/7 x -26
|
Mixture_Word_Problems/323909: chandra has 3 liters of a 14% solution of sodium hydroxide in a container. What is the amount and concentration of sodium hydroxide solution she must add to this in order to end up with 7 liters of 34% solution?
1 solutions
Answer 231806 by mananth(12270) on 2010-07-20 03:28:03 (Show Source):
You can put this solution on YOUR website!chandra has 3 liters of a 14% solution of sodium hydroxide in a container. What is the amount and concentration of sodium hydroxide solution she must add to this in order to end up with 7 liters of 34% solution?
3 liters 14%
total is 7 liters
so 4 liters x%
..
Total 7 liters 34%
..
Equation for concentration
3*0.14 +4*0.01x = 7*0.34
0.42+0.04x=2.38
0.04x=2.38-0.42
0.04x= 1.96
x= 49% and 4 liters is to be added
|
Miscellaneous_Word_Problems/323913: When 358, 528 and 698 are in turn divided by a certain number, the remainder is 1 in each case. Find the number.-
1 solutions
Answer 231802 by mananth(12270) on 2010-07-20 02:20:18 (Show Source):
You can put this solution on YOUR website!When 358, 528 and 698 are in turn divided by a certain number, the remainder is 1 in each case. Find the number.-
..
the difference between the numbers is 170
528-358=170
698-528=170
..
170= 17*10
!7 is the number
|
Miscellaneous_Word_Problems/323915: To lap a running track, Tom takes 65 seconds, Dick takes 70 seconds and Harry takes 75 seconds. If they start together, when will they next be together at the starting point?
1 solutions
Answer 231800 by mananth(12270) on 2010-07-20 02:02:17 (Show Source):
You can put this solution on YOUR website!In such problems you have to take the LCM of the numbers. That is when they are together
Factors of 60= 2*2*3*5
Factors of 70 are 2* 5*7
Factors of 75 are 3*5*5
The LCM will be 2100
After 2100 seconds they will be together after 2100 seconds
|
Triangles/323639: The area of a triangle with height 5 inches is equal to the area of a certain rectangle of length 10 inches and width 7 inches. What is the base of the triangle?
1 solutions
Answer 231596 by mananth(12270) on 2010-07-19 10:54:05 (Show Source):
You can put this solution on YOUR website!length of rectangle = 10
width of rectangle =7
Area = 70 sq.in
..
Area of triangle = 1/2 * base * h
h = 5 in
Area =70
base = area *2 / h
base = 70*2/5
=14*2
= 28 inches height
|
Human-and-algebraic-language/323635: I have to make an equation for this problem: A teacher goes to Company B to buy supplies and for every dollar amount spent over $20 there is a 30% discount given. I thought the equation would be y=.30x+20 or y=x-20(.30). Are either of these correct?
1 solutions
Answer 231595 by mananth(12270) on 2010-07-19 10:50:10 (Show Source):
You can put this solution on YOUR website!let him buy $x worth material above $20
So for every dollar he spends above $20 he pays $0.7
so for $x he will pay 0.7x
Totally he will pay 20 + 0.7x dollars
|
Money_Word_Problems/323543: Of the $50,000 that Natasha pocketed on her last real estate deal, $20,000 went to charity. She invested part of the remainder in Dreyfus New Leaders Fund with an annual yield of 16% and the rest in Templeton Growth Fund with an annual yield of 25%. If she made $6060 on these investments in 1 year, then how much did she invest in each fund?
1 solutions
Answer 231584 by mananth(12270) on 2010-07-19 10:17:50 (Show Source):
You can put this solution on YOUR website!$ 50,000 was the amount received.
20000 went to charity
So she invested $30,000
Let her invest x in New Leader - 16%
she will invest 30,000-x in Templeton 25%
..
Total she earned =
$9090
.
0.16x+0.25(30,000-x)=6060
0.16x+7500-0.25x=6060
-0.09x=-1440
x= -1440 / -0.09
x= $ 16,000 with Dreyfus New leaders
$14,000 with Templeton
|
Average/323608: The average of 6 numbers is 10. If the average of 4 numbers is 12, then what is the average of the remaining 2 numbers?
1 solutions
Answer 231571 by mananth(12270) on 2010-07-19 08:58:54 (Show Source):
You can put this solution on YOUR website!average of 6 numbers = 10
total of thses 6 numbers = 60.
..
Average of 4 numbers is 12
total of 4 numbers = 40
..
difference of these totals is the total of balance 2 numbers
60-48 = 12
12 is the total of 2 numbers
so the average = 12/2 = 6
..
Average of 2 numbers is 6
|
Average/323609: The average age of boys & girls in a class is 10.5 yrs. That of the boys is 10.6 & that of the girls is 10.1. If there are 60 boys in the class how many girls are there?
1 solutions
Answer 231556 by mananth(12270) on 2010-07-19 08:32:26 (Show Source):
You can put this solution on YOUR website!let the number of girls be x
Number of boys = 60
Total students = x+60
Average age of class = 10.5
The total age of the class = 10.5(x+60)
..
Average age of girls = 10.1
total age of girls = 10.1x
..
Average age of boys = 10.6
Number of boys = 60
Total age of boys = 10.6*60= 636 years
..
Total age of girls + total age of boys = Total age of class
10.1x+636=10.5(x+60)
10.1x +636=10.5x + 630
10.5x-10.1x = 636-630
0.4x= 6
x= 6/0.4
Multiply by 10 and divide by 10
x=60/4
x=15 girls
|
|
