Word problems (1) - answers
(1)Daniel bought a calculator on-line. The calculator was advertised at $24.99 before the tax. When the tax was included, Daniel paid $26.86. What percent of his money did he spend on tax? Price before tax = $24.99
Price after tax (including tax) = $26.86
Tax = after tax – before tax = 26.86 – 24.99 = $ 1.87
Percentage tax based on price before tax value = )(100%) tax value
before tax
( = )(100%)
24.99 1.87 ( = 7.48% nearest to two decimal place
(2)In a shipment of 400 parts, 14 are found to be defective. How many defective parts should be expected in a shipment of 1000?
14 defective parts in 400 parts Using percentage concept
Percentage of defective parts = )(100%) parts
of no. total
parts defective of
no.
( = )(100%)
400 14
( = 3.5%
For total of 1000 parts
Assume that the 1000 parts have the same percentage (3.5%) of defective parts. %)
100 )( parts of no. total
parts defective of
no.
( = 3.5%
%) 100 )( 1000
parts defective of
no.
( = 3.5%
No. of defective parts = )(1000) 100
5 . 3
( = 35 <- - this is expected no. of defective parts in 1000
Using proportion concept 14 defective parts in 400 parts
parts of no. total
parts defective of
no. =
400 14
1000
parts defective of
no.
= 400
14
<- - plug in total number No. of defective parts = (
400
(3)A shop buys food items at a wholesale and then marks them up for retail sale. They recently sold a food item at a 20% discount off of their marked-up price. What percentage mark-up did they originally apply to the food item if they broke even on that sale?
Break even ==> purchase price = Sale price, or no profit, no loss Pretend that
x = purchase price
y% = their mark-up percentage
Marked-up selling price = )(purchaseprice) 100
y 100
( = ) x
100 y 100
(
Sell one item at 20% discount off of their marked-up selling price
Selling price of that item = 20% less of marked-up price = 80% of marked-up price = )(marked-upsellingprice)
100 80
( = (0.8) ) x
100 y 100
(
They break even on that item purchase price = Selling price x = (0.8) ) x
100 y 100
(
1 = (0.8) )
100 y 100
( <- - - - divide both sides by ‘x’ 100 = (0.8)(100 + y) <- - - - multiply both sides by 100 125 = 100 + y <- - - - divide both sides by 0.8 25 = y
Mark up percentage = 25%
(4)A can of spray costs $3.99 for 4 ounces. Another brand costs $5.99 for 6 ounces. Which can is the better buy?
Spray A
$ 3.99 for 4 ounces = ) once 4
$ 3.99
( = 0.9975 <- - - cost of one ounce ($/ounce) Spray B
$5.99 for 6 ounces = ) once 6
$ 5.99
( = 0.9983 <- - - cost of one ounce ($/ounce) 0.9975 < 0.9983
A is cheaper than B. A is better buy.
(5)A pound of rye grass seed covers 120 square feet of lawn. How many pounds are needed to seed a lawn measuring 60 feet by 50 feet?
One pound of see covers 120 square feet = 1 lb per 120 sq.ft area
of Size
seeds of
amount =
sq.ft 120
Size of your lawn = 60 feet by 50 feet = 60 x 50 = 3000 square feet area
of Size
seeds of
amount =
sq.ft 120
lb 1
Amount of seeds = (
sq.ft 120
lb 1
) (Size of area) = (
sq.ft 120
lb 1
) (3000 sq.ft) = 25 pounds
(6)If a study revealed that 2 out of 5 families run into trouble with a computer and it is known that 300 families had trouble with the computer, how many families were surveyed?
2 out of 5 families have computer problem families
of no. total
problem computer
have which families of
no. =
5 2
300 families have computer problem, families
of no. total
300
= 5 2
300 = 5 2
(total no. of families) <- - - - multiply both sides by (total no. families) 750 = total no. of families surveyed
(7)A worker in an assembly line takes 3 hours to produce 26 parts. At that rate how many parts can she produce in 9 hours?
Speed of one worker = 3 hours to produce 26 parts produced
parts of no.
hours work of
no. =
26 3
The same worker, the same speed, but she works 9 hours -produced
parts of no.
9 =
26 3
9 = ( 26
3 ) (no. of parts produced) <- - - - multiply both sides by no. of parts 78 = no. of parts produced
* OR *
In 3 hours, she finishes 26 parts 9 hours is 3 times longer than 3 hours
So she could finish (26)(3) = 78 parts in 9 hours
3 in 1500 children have cancer risk
no. total
cancer of risk with
no. =
1500 3
632 540 12
cancer of risk with no.
= 1500
3
No. with risk of cancer = ( 1500
3 )(12 540 632) = 25081.264 = 25081
(9)For a complete circle with radius r, the area A = r2 and the circumference S = 2r.
a) Using the fact that area and arc length are proportional to the central angle, find the formulas for the area and the arc length of a circular sector in terms of radius R and central angle .
b) In the above diagram, if R1= 30 m, 1= 60 degree, R2= 15 m, 2= 45 degree, find the sum of the areas of the two sectors, and the perimeter of each sector.
Ans (a)
Area of a circular sector
Area ‘A’ is proportional to central angle ‘’
If the central angle = 360 degree => it is one full circle => area = r2
= 360 degree will make the A = r2
angle Central
Area =
θ A
= 360
r2
<- - - we will multiple both sides by
A = ( 360
r2
) <- - - formula to find A if you know ‘’ in degand radius ‘r’
Arc length of a circular sector
Arc length ‘S’ is proportional to central angle ‘’
If the central angle = 360 degree => it is one full circle => arc length S = 2r
= 360 degree will make the S = 2r angle
Central length
Arc =
θ S=
360 r 2 =
180 r
S = ( 180
r
) <- - - formula to find S if you know ‘’ in degand radius ‘r’
Ans (b)
R1= 30 m, 1= 60 degree R2= 15 m, 2= 45 degree
Sum of the areas = A1+ A2= ( 360
R2 1
)1+ ( 360
R 2 2
)2
= )(R θ R θ )
360
( 2 2
2 1 2 1
= )[(30 )(60) (15 )(45)]
360
( 2 2
= 178.125 squared meter unit
= about 559.6 square-meter to one decimal place
Perimeter of sector one = (2)(radius) + (arc length) = 2 R1+ S1= 2 R1+ ( 180
R1
)
1
= (2)(30) + ( 180
(30)
) (60) = about 91.4 meters to one decimal place
Perimeter of sector two = (2)(radius) + (arc length) = 2 R2+ S2= 2 R2 + ( 180
R2
) 2 = (2)(15) + (
180 (15)
) (45) = about 41.8 meters to one decimal place
(10)The (x,y) coordinates of the points on a parabola are related to each other by the equation: y = a x2+ bx + c
Where a, b, and c are constants for each parabola graph.
a) write down the formula to compute x values of its x-intercepts in terms of a, b, and c b) write down the conditions so that the whole parabola will be above the x-axis c) write down the conditions so that the whole parabola will be below the x-axis d) find the x- and y- intercepts of y = x2+ 2x + 1 and graph the parabola
(a)
x-values of the x-intercepts of a graph => x-coordinates of the points where the graph passes x-axis Those points are on the graph and also on the x-axis… Their y-values must be zero.
y = a x2+ bx + c 0 = a x2+ bx + c
Solve for x, the answer is x =
a 2
ac b b
(b) & (c)
The parabola must not pass x-axis (no x-intercepts). There is no solution to 0 = a x2+ bx + c
a 2
ac b b
- 2
must be undefined. b2– 4ac must be negative.
b) The conditions so that the whole parabola will be above the x-axis It must be open up, and there are no x-intercepts
The conditions are: a > 0 and b2– 4ac < 0
(c) The conditions so that the whole parabola will be below the x-axis It must be open down, and there is no x-intercepts
The conditions are: a < 0 and b2– 4ac < 0
d) Find the x- and y- intercepts of y = x2+ 2x + 1 and graph the parabola y = x2+ 2x + 1
y = a x2+ bx + c
a = 1 (bigger than zero) <- - - Open up b = 2
c = 1
b2– 4ac = (2)2- 4 (1)(1) = 4 – 4 = 0 <- - - One x-intercept (it is the vertex) y-intercepts (x = 0)
y = x2+ 2x + 1 y = (0)2+ 2(0) + 1 = 1 (0, 1)
x-intercepts (y = 0) y = x2+ 2x + 1 0 = x2+ 2x + 1 0 = (x + 1)(x + 1) -1 = x
The parabola graph opens up
At x = -1, y = 0 - - - - it is x-intercept and it is the lowest point as well as vertex (-1,0) We will find y-values for about two points on the left of -1 and two points on the right of -1
x y = x2+ 2x + 1 (x, y)
-3 (-3)2+ 2(-3) + 1 = 4 (-3, 4) -2 (-2)2+ 2(-2) + 1 = 1 (-2, 1)
-1 0 (-1, 0) Vertex
0 (0)2+ 2(0) + 1 = 1 (0, 1) 1 (1)2+ 2(1) + 1 = 4 (1, 4)
Notes:
For M to be a real number M 0
If M is given, we always take M 0
If x2= N is given, then x = N and N must be greater than or equal to zero
