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One earned 5% profit while the other earned a 3% profit. If the investor's total profit was $86, how much was invested in each mutual fund? 1 solutions
Answer 448705 by mananth(12270) on 2013-04-05 02:07:17 (Show Source):
You can put this solution on YOUR website! Part I 3.00% per annum ------------- Amount invested =x
Part II 5.00% per annum ------------ Amount invested = y
2200
Interest----- 86.00
Part I 3.00% per annum ---x
Part II 5.00% per annum ---y
Total investment
x + 1 y= 2200 -------------1
Interest on both investments
3.00% x + 5.00% y= 86
Multiply by 100
3 x + 5 y= 8600.00 --------2
Multiply (1) by -3
we get
-3 x -3 y= -6600.00
Add this to (2)
0 x 2 y= 2000
divide by 2
y = 1000
Part I 3.00% $ 1200
Part II 5.00% $ 1000
CHECK
1200 --------- 3.00% ------- 36.00
1000 ------------- 5.00% ------- 50.00
Total -------------------- 86.00
m.ananth@hotmail.ca
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Quadratic_Equations/733129: We are learning this in class and my teacher goes way to fast. Could you please help me with the following question?
Solve each equation symbolically.
A. (X+3)^2=7
B. (X-2)^2-8=13
I honestly don't understand what "solve symbolically". If you would help me I would be extremely thankful.
Dakota 1 solutions
Answer 448300 by mananth(12270) on 2013-04-02 21:50:02 (Show Source):
You can put this solution on YOUR website!Solve each equation symbolically.
A. (X+3)^2=7
B. (X-2)^2-8=13
symbolically means using the quadratic formula
compare this equation with ax^2+bx+c
a=1, b=6, c=2
b^2-4ac= 36 -8
b^2-4ac= 28
x1=( -6 + 5.29 )/ 2
x1= -0.35
x2=( -6 -5.29 ) / 2
x2= -5.65
two values of x. -0.35 & -5.65
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Equations/732855: For some reason im really struggling with the equation (a-2)/3=(1-a)/2, could you possibly explain how i would work this out, thankyou
Emma 1 solutions
Answer 448082 by mananth(12270) on 2013-04-02 08:01:24 (Show Source):
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Money_Word_Problems/732810: An investor invested a total of 2800 in two mutual funds. One fund earned a 6% profit while the other earned a 4% profit. If the investors total profit was $152, how much was invested in each mutual fund? 1 solutions
Answer 448075 by mananth(12270) on 2013-04-02 06:35:30 (Show Source):
You can put this solution on YOUR website!Part I 6.00% per annum ------------- Amount invested =x
Part II 4.00% per annum ------------ Amount invested = y
2800
Interest----- 152.00
Part I 6.00% per annum ---x
Part II 4.00% per annum ---y
Total investment
x + 1 y= 2800 -------------1
Interest on both investments
6.00% x + 4.00% y= 152
Multiply by 100
6 x + 4 y= 15200.00 --------2
Multiply (1) by -6
we get
-6 x -6 y= -16800.00
Add this to (2)
0 x -2 y= -1600
divide by -2
y = 800
Part I 6.00% $ 2000
Part II 4.00% $ 800
CHECK
2000 --------- 6.00% ------- 120.00
800 ------------- 4.00% ------- 32.00
Total -------------------- 152.00
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Sequences-and-series/732799: The sum of the first 100 terms of an arithmetic progression is 10000; the first, second and fifth terms of this progression are three consecutive terms of a geometric progression. Find the first term, a, and the non-zero common difference, d. of the arithmetic progression 1 solutions
Answer 448074 by mananth(12270) on 2013-04-02 06:31:58 (Show Source):
You can put this solution on YOUR website!S 100 = 100/2[2a+99d]
S100 =50[2a+99d]
a, a+d, a+4d
(a+d)^2=a(a+4d)
a^2+2ad+d^2=a^2+4ad
d^2=2ad
d=2a
S100 =50[2a+99d]
S100= 50*100d
S100=5000d
10,000=5000d
d=2
Therefore a=1
Check
s100 = 100/2[2+198]
s100= 50*200
s100 = 10,000
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Rate-of-work-word-problems/732844: how to solve the sum of the digits of A 3-digit number is 16 if the digits are reversed and the resulting number added to the original number the sum is 1049 if the resulting number is subtracted from the original number the difference is 297 what is the number? 1 solutions
Answer 448072 by mananth(12270) on 2013-04-02 06:20:29 (Show Source):
You can put this solution on YOUR website!First condition
x+y+z=16................................(1)
Second condition
100x+10y+z+100z+10y+x=1049
101x+20y+101z=1049............................(2)
third condition
100x+10y+z-(100z+10y+x)=297
99x-99z=297
/99
x-z=3
x=z+3.......................................(3)
substitute x in (1) & (2)
z+3+y+z=16
y+2z=13...................(4)
101(z+3)+20y+101z=1049
101z+303+20y+101z=1049
202z+20y=1049-303
202z+20y=746..................(5)
1.00 y + 2.00 z = 13.00
Total value
20.00 y + 202.00 z = 746.00
Eliminate y
multiply (1)by 202.00
Multiply (2) by -2.00
202.00 y 404.00 z = 2626.00
-40.00 y -404.00 z = -1492.00
Add the two equations
162.00 y = 1134.00
/ 162.00
y = 7
plug value of y in (1)
1.00 y + 2.00 z = 13.00
7.00 + 2.00 z = 13.00
2.00 z = 13.00 -7.00
2.00 z = 6.00
z = 3
y=7,z=3,therefore x= 6
100*6+70+3
673 is the number
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Sequences-and-series/732802: An arithmetic series has first term 1000 and common difference -1.4. calculate the value of the first negative term of the series, and the sum of all the positive terms 1 solutions
Answer 448070 by mananth(12270) on 2013-04-02 04:22:34 (Show Source):
You can put this solution on YOUR website!a= 1000
d= -1.4
1000/1.4 = 714.29
there are 714.29 =ve numbers
add 1.4 you get 715.69
so 716th number will bwe negative
t716 = 1000+(716-1)-1.4
t716 = -1
-1 is the first negative number.
S715 will be the sum
S715 = 715/2[2*1000+714*-1.4]
S715 = 357643
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Miscellaneous_Word_Problems/732769: $6,000 dollars is invested in two different accounts earning 3% and 5% interest. At the end of one year, the two accounts earned $220 in interest. How much money was invested at 5%? 1 solutions
Answer 448034 by mananth(12270) on 2013-04-01 21:24:23 (Show Source):
You can put this solution on YOUR website!
Part I 5.00% per annum ------------- Amount invested =x
Part II 3.00% per annum ------------ Amount invested = y
6000
Interest----- 220.00
Part I 5.00% per annum ---x
Part II 3.00% per annum ---y
Total investment
x + 1 y= 6000 -------------1
Interest on both investments
5.00% x + 3.00% y= 220
Multiply by 100
5 x + 3 y= 22000.00 --------2
Multiply (1) by -5
we get
-5 x -5 y= -30000.00
Add this to (2)
0 x -2 y= -8000
divide by -2
y = 4000
Part I 5.00% $ 2000
Part II 3.00% $ 4000
CHECK
2000 --------- 5.00% ------- 100.00
4000 ------------- 3.00% ------- 120.00
Total -------------------- 220.00
m.ananth@hotmail.ca
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Travel_Word_Problems/732448: A certain aircraft can fly 630 miles with the wind in 3 hours and travel the same distance against the wind in 7 hours. What is the speed of the wind? 1 solutions
Answer 447850 by mananth(12270) on 2013-04-01 00:46:41 (Show Source):
You can put this solution on YOUR website!630 miles in 3 hours
with wind speed = 210 mph
630 in 7 hours
against wind speed = 630/7 = 90 mph
plane speed = x
wind speed = y
x+y = 210
x-y=90
add up the equations
2x=300
/2
x= 150 mph
so wind speed = 60 mph
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Travel_Word_Problems/732447: A private plain traveled from Seattle to the rugged wilderness, at an average speed of 156 mph.on the return trip the average speed was 182 mph if the total traveling time was 9 hours, how far was Seattle from the wilderness? 1 solutions
Answer 447849 by mananth(12270) on 2013-04-01 00:37:10 (Show Source):
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Coordinate-system/732433: what is the equation of the line that is parallel to the line that has equation y=2x-3 and with y-intercept of 5.How do you solve the inequality of 8(2-x)-9>2x+5?Thank you. 1 solutions
Answer 447848 by mananth(12270) on 2013-04-01 00:27:30 (Show Source):
You can put this solution on YOUR website!y= mx +b is the equation in slope intercept form
where b is the y intercept
y=2x-3
slope = 2
a parallel line will have same slope which is =2
m=2 , point(0,5)
y=2x+5
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Sequences-and-series/732158: The tickets in a raffle are numbered 1, 2, 3, and so on. The price of a ticket is the number of cents equal to the number of the ticket. If the raffled article cost $100, what is the least number of tickets that must be sold so that those conducting the raffle will not lose money? 1 solutions
Answer 447575 by mananth(12270) on 2013-03-31 07:00:48 (Show Source):
You can put this solution on YOUR website!It is an AP
a=1
d=1
Sn=$100= 10000 cents
Sn=n/2[2a+(n-1)d]
10000=n/2[2+n-1]
20000=n^2+n
n^2+n-20000=0
Find the roots of the equation by quadratic formula
a= 1 , b= 1 , c= -20000
b^2-4ac= 1 + 80000
b^2-4ac= 80001
x1=( -1 + 282.84 )/ 2
x1= 140.92
x2=( -1 -282.84 ) / 2
x2= -141.92
Ignore negative value Minimum tickets to be sold = 141 tickets
m.ananth@hotmail.ca
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Sequences-and-series/732150: a particle sliding down an inclined plane travels 2 feet the first second. In any second after the first it slides 4 feet farther than it did in the preceding second. how far will it slide in 7th second? how far will it slide in 7 seconds? how long will it take to slide 128 feet? 1 solutions
Answer 447574 by mananth(12270) on 2013-03-31 05:07:51 (Show Source):
You can put this solution on YOUR website!a particle sliding down an inclined plane travels 2 feet the first second. In any second after the first it slides 4 feet farther than it did in the preceding second. how far will it slide in 7th second? how far will it slide in 7 seconds? how long will it take to slide 128 feet?
t1=2 =a
t2=2+4=6
t3=6+4=10
a=2
d=4
tn= a+(n-1)d
t7=2+(7-1)4
t7= 26, 26 feet in the 7th. second
Sn=n/2[2a+(n-1)d]
S7=7/2[2*2+(7-1)4]
S7=7/2[28]
S7=98, 98 feet in 7 seconds
128=n/2[4+(n-1)2]
256=4n
256=4n +2n^2 -2n
2n^2+2n-256=0
/2
n^2+n-128=0
solve the quadratic for x
you will get n=10.82
In 10.82 sec it will go 128 feet
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Age_Word_Problems/732019: 1. If ram is 10 years younger the his brother shyam and after 7 years ram's age will be half of shaym's age. find actual age of ram and shyam.
2. mat and prateek started a business with initial contribution of 1000 and 12500
respectively, after a year profit earned is is Rs.100/-. what is the profit of mat and prateek.
3. If worker finishes a wall construction in 10 days and 2nd worker in 7 days, how many days it will take to finish if both of them work together. 1 solutions
Answer 447565 by mananth(12270) on 2013-03-31 01:01:47 (Show Source):
You can put this solution on YOUR website!1. If ram is 10 years younger the his brother shyam and after 7 years ram's age will be half of shaym's age. find actual age of ram and shyam.
shyam age =x
Ram = x-10
After 7 years
Ram age = x-10+7 = x-3
shyam age = x+7
1/2(x+7)= (x-3)
x+7 = 2(x-3)
x+7=2x-6
x=13
you find out Ram's age
2. mat and prateek started a business with initial contribution of 1000 and 1250
respectively, after a year profit earned is is Rs.100/-. what is the profit of mat and prateek.
The ratio of their investment = 1000/1250 = 4/5
so profit is shared in that ratio
3. If worker finishes a wall construction in 10 days and 2nd worker in 7 days, how many days it will take to finish if both of them work together.
they will do 1/10 + 1/7 = 17/70 of the job in 1 day
so they will take 70 /17 days together.
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Travel_Word_Problems/732123: John and Mary leave their house at the same time and drive in opposite directions. John drives at 90 mi/h and travels 35 mi farther than Mary, who drives at 40 mi/h. Mary's trip takes 15 min longer than John's. For what length of time does each of them drive? 1 solutions
Answer 447564 by mananth(12270) on 2013-03-31 00:50:14 (Show Source):
You can put this solution on YOUR website!john and Mary leave their house at the same time and drive in opposite directions. John drives at 90 mi/h and travels 35 mi farther than Mary, who drives at 40 mi/h. Mary's trip takes 15 min longer than John's. For what length of time does each of them drive?
let mary travel x miles
john travels x+35 miles
John speed = 90 mph
Mary 40 mph
Time Mary = x/40
Time john = (x+35)/90
Time Mary -1/4 = Time John
x/40 - 1/4 = (x+35)/90
x/40-(x+35)/90 = 1/4
90x-40(x+35)= 900
90x-40x-1400=900
90x-40x-1400=900
50x=2300
x=46 miles
John time = (46+35)/90 = 0.9 hours
Mary time = 46/40=1.15 hours
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Travel_Word_Problems/732118: a boat travels 2 miles upstream and 2 miles downstream in 5 hours. in still water, the boat travels 2 miles/hour.
1. what is speed of river's current?
2. how fast does it travel upstream?
3. how fast does it travel downstream?
4. how long does it take to travel downstream?
5. how long does it take to travel upstream?
Thank you. Gwen 1 solutions
Answer 447563 by mananth(12270) on 2013-03-31 00:29:36 (Show Source):
You can put this solution on YOUR website!let current speed = x
boat speed =2 mph
against current 2-x
with current x+2
2/(x+2)+2/(2-x) = 5
2(2-x)+2(x+2)=5(4-x^2)
4-2x+2x+4 = 20-5x^2
5x^2=12
x^2= 12/5
x= sqrt(12/5)
upstream speed = 2- 1.55= 0.45 mph
downstream speed = 2+sqrt(12/5)= 3.55 mph
downstream time = 2/3.55= 0.56 hours
upstream time = 2/0.45=4.44
Total time = 5 hours
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Travel_Word_Problems/732013: A plane is flying the 2553mile trip from los Angeles to Holohulu into a 60 mph headwind.if the speed of the plane in still air is 310mph.How far from Los Angeles is the plane's point of no retutn 1 solutions
Answer 447548 by mananth(12270) on 2013-03-30 21:15:40 (Show Source):
You can put this solution on YOUR website!given information
310 + 60 = 370 mph speed with the wind
310 - 60 - 250 mph speed against the wind
:
Let d = dist to LA from point of no return
Then
(2553-d) = dist to Honolulu from point of no return
:
: Time = dist/speed
:
Return to LA= Continue to Honolulu time
d/360 = 2553-d/250
Cross multiply:
250d = 360(2553-d)
:
250d = 919080 - 360d
:
610d = 919080
d = 1506.69
Check solution by finding the distance to LA and calculating the time to both:
2553-1506.69= = 1046.31 mi Honolulu
:
Time to L A: 1506.69/360 = 4.19 hrs
Time to Honolulu: 1046.31/250 = 4.19 hrs
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Money_Word_Problems/732113: an investor invested a total of 2700 dollars in two mutual funds. one funded 5% and the other at 2% profit.if the investors total profit was $93 how much was invested in each mutual fund 1 solutions
Answer 447534 by mananth(12270) on 2013-03-30 20:53:41 (Show Source):
You can put this solution on YOUR website!Part I 2.00% per annum ------------- Amount invested =x
Part II 5.00% per annum ------------ Amount invested = y
2700
Interest----- 93.00
Part I 2.00% per annum ---x
Part II 5.00% per annum ---y
Total investment
x + 1 y= 2700 -------------1
Interest on both investments
2.00% x + 5.00% y= 93
Multiply by 100
2 x + 5 y= 9300.00 --------2
Multiply (1) by -2
we get
-2 x -2 y= -5400.00
Add this to (2)
0 x 3 y= 3900
divide by 3
y = 1300
Part I 2.00% $ 1400
Part II 5.00% $ 1300
CHECK
1400 --------- 2.00% ------- 28.00
1300 ------------- 5.00% ------- 65.00
Total -------------------- 93.00
m.ananth@hotmail.ca
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Linear-equations/731931: graph the system of linear inequalities 4x -2 y < 0 and 2x + 4y <8
1 solutions
Answer 447395 by mananth(12270) on 2013-03-30 03:38:30 (Show Source):
You can put this solution on YOUR website!4x -2 y < 0
2y>4x
y>2x
2x + 4y <8
4y<8-2x
y<2-(1/2)x
The equality
y=2x
Generate points
(1,2),(2,4),(-1,-2)
y=2-(1/2)x
((2,1), (4,0) , (-2,3)
The inequality
since y < 2x(something) the region below is the solution
and y<2-(1/2)the region below is the solution
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Linear-equations/731936: find the slope-intercept form for the line satisfying the following conditions.
passing through (7, -6) and (6, -5)
My email is ..... cherrymcknight@att.net
1 solutions
Answer 447389 by mananth(12270) on 2013-03-30 01:50:26 (Show Source):
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Miscellaneous_Word_Problems/731906: 1. Suppose a refrigerator that sells for $700 costs $85 a year for electricity. Write an expression for the cost to buy and run this refrigerator for x years.
2. Suppose another refrigerator costs $1000 and $25 a year for electricity. Write an expression for the total cost for the refrigerator over x years.
3. Over 10 years which refrigerator costs the most? By how much?
4. In how many years will the total costs for the two refrigerators be equal? 1 solutions
Answer 447359 by mananth(12270) on 2013-03-29 22:05:54 (Show Source):
You can put this solution on YOUR website!first refrigerator cost
C(x1)=700+85x
C(10)= 700+850
C(10)= 1550
refrigerator II
C(x2) = 1000+25x
C10) =1000+250
C(10)=1250
Now you know it.
same cost
700+85x=1000+25x
60x=300
x= 5 years
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