VIVA Voce Chapter 9 Exercise 9 Question 6






6. (a) Simplify each of the following ratios.
    1. a : b = 1½ : 2⅖                               
    1. b : c =  0.105 : 0.350
    1. Find the ratio a : b : c using the data in (a).
  (c) Alan, Bob and Cathy share $500 in the ratio a : b : c found in (b). Find Alan’s   
                Share correct to 2 decimal places. 
  1. i) 
  A : b
=1 1/2 : 2 2/5
=1 5/10 : 2 4/10
=15/10 : 24/10
=15 : 24
=5 : 8
    ii)
  B:c
=0.105 : 0.350
=21/100 : 7/20
=21/100 : 35/100
=21 : 35
=3 : 5


(b)
  A : b : c
=5: 8
=     3 : 5
_________
=15: 24: 40


(c)
15u+24u+40u=79u
79u-->$500
1u-->$500÷79=$6.33 (2d.p.)
15u-->$6.33 x 15=$94.95 



(Viva Voce) Chap 9 Exercise 9 Question {8}

My Video :



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http://www.youtube.com/watch?v=aGcYHNh1FW0

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Solution:

Average speed = Total Distance ÷ Total Time
                        =360km ÷4 1/2 h
                        = 80km/h


Petrol Consumption rate = Total Distance ÷ Total petrol
                                        =  360 km ÷ 37litres
                                        =  9.6 km/l



Average speed of remaining part of journey = Remaining Distance ÷ remaining time

Distance of highway = 110 km/h × 2h
                                 = 220 km

Remaining distance = 360 km − 220 km
                                 = 140 km
Remaining time  = 4 1/2 h − 2h
                           =2 1/2h

Average speed of remaining part of journey = 140km ÷ 2 1/2 h 
                                                                       = 56 km/h

Viva Voce: Question 4




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(Viva Voce) Chap 9 Exercise 9 Question 8





Youtube Video Link:
http://www.youtube.com/watch?v=acErK4gMLe4

I screwed up the last part...

(Viva Voce) Chap 9 Exercise 9 Question { 8 }

Solution for Q8

Q8a) Find the average speed.

Average speed = Total Distance Travelled/Total Time Taken
                         = 360km/4 1/2h
                         = 360km/4.5h
                         = 80km/h
Ans: 80km/h

Q8b)Find the petrol consumption rate.

Petrol Consumption Rate = 360km/37.5 l
                                         = 9.6km/l
Ans: 9.6km/l

Q8c) Distance Travelled on the highway= speed X time
                                                                = 110km/h x 2h
                                                                = 220km

Remaining Distance = 360km-220km
                                 = 140km 

Remaining Time = 4 1/2h - 2h
                           = 2 1/2h

Average sped for the remaining part of journey = Total Distance Travelled/Total Time Taken 
                                                                           = 140km/2 1/2h
                                                                           = 56km/h 

Ans: 56km/h

Mavis Cheng S1-01 

Math Viva Question #8




Sorry if late... and also i din see the examples before i did this viva and i was at my grandmother house tonight so i decided to use the white board immediately when i reached home. I only posted once i reach home after my grandmother house trip ._.

Owen here btwfyi...

(Viva Voce) Chap 9 Exercise 9 Question {8}




8. (a) Average speed : 360 km ÷ 4 1/2 h = 80 km/h
(b) Petrol consumption rate : 360 km ÷ 37.5 l = 9.6 km/l
(c) Distance covered at highway : 110 km/h * 2 h = 220 km
Remaining distance : 360 km - 220 km = 140 km
Remaining time : 4 1/2 h - 2 h = 2 1/2 h
Average speed for remaining distance of trip : 140 km ÷ 2 1/2 h = 56 km/h

(Viva Voce) Chap 9 Exercise 9 Question 2



Average Speed = Distance/Time
1hour=60mins

(a) Average speed of Car X=60km/45mins=1⅓km/min
=80km/h
Ans: 80km/h

(b) Average speed of Car Y = (60/80*72)km/h
= 54km/h
Ans: 54km/h

(c) Average Speed of Car X : Average speed of Car Y
80 : 54
40 : 27

Ans: 40:27

(Viva Voce) Chap 9 Exercise 9 Question 2



a) average speed of car X=60km/45min
                                        =1⅓km/min
1⅓km/minx60=80km
ANS:80km/hour


b)average speed of car Y=72km/80min(1h20min=80min)
                                    =9/10km/min
9/10km x60=54km
ANS:54km/hour


c)ratio of average speed of car X to car Y
80:54
40:27
ANS:40:27

Viva Question 4





(a) no. of parts -> 1 + 4
                          = 5
Volume of alcohol (1 part) -> 600 ÷ 5
                                            = 120
Volume of alcohol is 120cm3


(b) Volume of water (4 parts) -> 600 ÷ 5 x 4
                                                = 480 
Volume of water is 480cm3 


(c) 3 parts -> 480
      1 part (new volume of alcohol) -> 480  ÷ 3
                                                         = 160
Difference -> 160 - 120
                   = 40
The amount needed is 40cm3 

Viva Qn 10


 10 a) Distance already travelled = 54 km/h x 2/3 h = 36 km
               Remaining distance = 120 km - 36 km = 84 km
               Remaining time = 2 h - 2/3 hr = 1 1/3 h
               Average speed = 84 km ÷ 1 1/3h = 63 km/h 
      b) i) average speed = 120 km ÷ 2 1/6 h = 55 1/3 km/h
             ii) Total distance = 120 km + 120 km = 240 km
                 Total time = 2h + 2 1/6 h = 4 1/6 h
                 Average speed = 240 km ÷ 4 1/6h = 57.6 km/h

(Viva Voce) Chap 9 Exercise 9 Question {6}




   a     :     b
1 1/2 : 2 2/5
  1.5  :  2.4
   30  :   38
    5   :    8

   a     :     b     :     c
   5     :     8     :     c
   a     : 0.105  : 0.350
   a     :   105   :   350
   a     :     3      :   10
  15    :    24     :    80

15+24+80 = 119

alan : 15/119 * $500 $63.03 (2dp)

Viva Question 6



Question 6 :
a) i) a:b = 1 1/2 : 2 2/5
= 1.5 : 2.4
-changed to nearest whole number-
= 30 : 48
-simplified-
= 15 : 24
= 5 : 8

ii) b:c = 0.105 : 0.350
-changed to whole number-
= 105 : 350
-simplified-
= 21 : 70
= 3 : 10

b) a:b = 5:8
b:c = 3:10
Since 'b' is the repeated value in both ratios , the ratios have to be multiplied so that the value of 'b' will be the same in both ratios .
a:b = 15:24
b:c = 24:80
a:b:c = 15:24:80

c) a+b+c = 119 units
$500 ÷ 119 * 15 ( Alan's share in units )
≈ $63.03 (2 d.p)









Viva Question 4




1+4=5
600/5=120
Alcohol:120cm3 
120*4=480
Water:480cm3 
480/3=160
160-120=40
Ans:40cm3 



(Viva Voce) Chap 9 Exercise 9 Question 10





Qn 10 a) Distance already travelled = 54km/h x 2/3h = 36km
               Remaining distance = 120km - 36km = 84km
               Remaining time = 2h - 2/3hr = 1 1/3h
               Average speed = 84km ÷ 1 1/3h = 63km/h

          b) i) average speed = 120km ÷ 2 1/6h = 55 1/3km/h
             ii) Total distance = 120km + 120km = 240km
                 Total time = 2h + 2 1/6h = 4 1/6h
                 Average speed = 240km ÷ 4 1/6h = 57 3/5km/h OR 57.6km/h

(Viva Voce) Chapter 9 Question 10

a) 40 minutes= 2/3 hours
1 hour, he can travel 54km (The speed he used for the first 40 minutes)

40 minutes, he can travel 36km (distance already travelled)

120km -36km = 84km (remaining distance)

2 hours- 40 minutes= 1hour 20 minutes
                               = 1 1/3 hours

84km ÷ 1 1/3 hours = 63km/h


b) [i]
2 hours 10 min= 2 1/6 hours (Total Time taken)
120km ÷ 2 1/6 hours = 55 5/13km per hour

[ii]
2 hours 10 minutes + 2 hours = 2 1/6 hours (Total Time Taken)
240km ÷ 2 1/6 hours = 57 3/5 (Average Speed)

Wang Yi Chieh

(Viva Voce) Chap 9 Exercise 9 Question 6





6a)(i) a:b= 1. 5 : 2. 4
                     to
          a:b= 30 : 48
                     to
          a:b= 5 : 8

(ii)
         b : c = 0.105 : 0.350
                    to
         b : c = 105 : 350
                   to
         b : c = 3 : 10

6b)   a : b : c
      30 : 48
              3 : 10
                              
      30 : 48 : 160
      15 : 24 :80

6c) $500 ÷ 119 x 15 ≈ 63.03 (2 decimals places)

(Viva Voce) Chap 9 Exercise 9 Question 4


(a) 600cm^3 = 5 units
Alcohol = 1 unit
= 600cm^3 ÷ 5
= 120cm^3

(b)Water = 4 units
= 4 x 120cm^3
= 480cm^3

(c) 4 units(old) = 3 units(new)
= 480cm^3
1 unit(new) = Alcohol
= 160cm^3
Amount of alcohol added = 160cm^3 - 120cm^3
= 40cm^3


Title: (Viva Voce) Chap 9 Exercise 9 Question 2

average speed= total distance/total time

(a) distance=60Km
     time=45 min
            =3/4h
    average speed=60÷3/4
                          =80Km/h


(b)distance= 72 Km
    time= 1h 20min
           = 1/1/3
   average speed= 72÷1/1/3
                         = 54Km/h


(c) ratio
    X:Y
   80:54 ÷ 2
   40:27

RE: Reminder

Please remember to post your Viva's by today!!

(Viva Voce) Chap 9 Exercise 9 Question 2





The solution:


a) 60km ÷ 3/4 h
  =80km/h


b)72km ÷ 1 1/3h
 =54km/h


c) Speed of Car X: Speed of Car Y
  =40:27




I failed. Epicly.

(Follow-up) June Vacation Chap 9.2 RATE (L5) Can he catch up with her?

What is the 'slowest speed' that Shamus has to ride the bicycle so that he could catch up with Grace?

Note: The units we are dealing here are: Time in minutes & Distance in metres.
Speed therefore is metres per minute (m/min). We'll have to convert it to something that is more common (i.e. m/s or km/h)

By answering to the pointers in the original question...
  • What information do you know from the problem?
  • What else do you need to know to solve the problem?
  • Pick a reasonable number for the information you need. 
1. Information we know:
  • The ratio of speed (one is 3 times the other)
  • Maximum distance covered: 800 m
  • One started the journey 10 minutes later
2. To solve the problem, we need to know the speed or time taken by Grace or Shamus. Since we do not know these information, we could only provide answers for certain scenario to 'happen', for example, Shamus catches up with Grace at the school gate, when both have travelled 800m. We can also test another extreme whereby Shamus to catch up with Grace along the way to school.

3. By using the info given, we can solve the problem using Algebra:
Let Grace's speed be x m/min
Shamus' speed is therefore 3x m/min

Since distance travelled = 800m (imagine that Shamus catches up with Grace at the school gate)
Time taken by Grace = 800/x min
Time taken by Shamus = 800/(3x) min

Since the time difference is 10 min (as Grace started earlier)
Time taken by Grace - Time taken by Shamus = 10
800/x - 800/(3x) = 10
by solving the algebraic equation, we have x = 53 1/3 m/min
To convert it to m/s, we have x = 8/9 m/min which is the same as 0.889 m/s (correct to 3SF)

Shamus takes 5 min to catch up with Grace
Therefore, time they meet = 6.45 am

6AM Quiz (2011, 18 June): It's about Cats and Dogs

A dog takes 3 steps to walk the same distance for which the cat takes 4 steps. Suppose 1 step of the dog covers 30 cm. How many centimeters would the cat cover in taking 12 steps?

There are at least 2 methods to solve the above problem.
Describe any 2 methods that leads to the answer.

6 AM Quiz (11 June 2011)... There are more than one way!

What number multiplied by itself is equal to the product of 32 and 162?

There are 2 or more ways to solve this problem. Describe at least 2 methods :)

June Vacation Learning Activities

Dear Students

We have selected the topic, Rate (Textbook 1B, Chapter 9.2) as a self-directed lesson where you would re-visit a topic that you came across in Primary School.

A couple of new concepts are introduced, for example, what is the difference between Constant Rate and Average Rate?

Take time to attempt the 5 lessons. You may stagger the learning activities (e.g. one lesson a day).

Complete the activities before school re-opens as we'll be using your responses for discssions when school reopens.

June Vacation Chap 9.2 RATE (L1) Rate in Everyday Situation

What's defined in Wikipedia: Rate (mathematics)
In mathematics, a rate is a ratio between two measurements, often with different units.[1]. If the unit or quantity in respect of which something is changing is not specified, usually the rate is per unit time. However, a rate of change can be specified per unit time, or per unit of length or mass or another quantity. The most common type of rate is "per unit time", such as speed, heart rate and flux. Rates that have a non-time denominator include exchange rates, literacy rates and electric flux.

When we describe the units of a rate, the word "per" is used to separate the units of the two measurements used to calculate the rate (for example a heart rate is expressed "beats per minute"). A rate defined using two numbers of the same units (such as tax rates) or counts (such as literacy rate) will result in a dimensionless quantity, which can be expressed as a percentage (for example, the global literacy rate in 1998 was 80%) or fraction or as a multiple.

~~~~~~~~~~~~~~~~~~~~~
Rate is commonly used in our daily life. Here are some examples:

40 km/h - 40 kilometres per hour
30 steps/min - 30 steps in per minute
2 l/hour - 2 litres per hour
67 words/min - 60 words per 1 min
80 m/week - 80 metres per week
25km/l - 25 kilometres per 1 litre
$90/m³ - $90 per cubic metre

Think of 2 real life situations in which we use the concept of rate to describe useful information.

Here is an example:
The Singapore Flyer rotates at the rate of 0.24 m/s or 0.76 km/h
(http://www.singaporeflyer.com/en/about-us/fun-facts-about-singapore-flyer.html)

June Vacation Chap 9.2 RATE (L2) A Twist to the Red Riding Hood Adventure

Read the following short story... you will be able to find descriptions related to RATE...

Once upon a time, there was a little girl who lived in a village near the forest. Whenever she went out, the little girl wore a red riding cloak, so everyone in the village called her Little Red Riding Hood.

One morning, Little Red Riding Hood asked her mother if she could go to visit her grandmother as it had been a while since they'd seen each other.

"That's a good idea," her mother said. "We usually visit grandma twice a month. However, because of common tests, you have not visited Grandma this month! She would be upset if she doesn't see you soon!"
["Rate" in the story: 2 visits in ONE month]

Based on what you understand about the term "rate", you will now continue to build the story from where the last person has commented... Everyone will contribute to one part of this Marathon Story. Your part of the story should have at least one description related to to RATE.

Complete this in the Facebook Group "Mathematics In Real Life"
Title of Post: S1-01 Story Marathon (L2) A Twist to the Red Riding Hood Adventure
.

June Vacation Chap 9.2 RATE (L3) CONSTANT Rate vs AVERAGE Rate

Here are 2 scenarios:
  • The tap in the bathroom leaks - 2 drops of water droplets for every minute. 2 drops of water/minute is considered as "CONSTANT RATE".
  • Benoît Lecomte, the French long distance swimmer was the first man to swim across the Atlantic Ocean at the rate of 43.6 miles in a day. 43.6 miles/day is considered as "AVERAGE RATE".
Are you able to tell the difference between CONSTANT Rate and AVERAGE Rate?
When do we use CONSTANT rate and when to use AVERAGE rate?

Attempt the following Quiz... It may help you to sharpen your thoughts...

June Vacation Chap 9.2 RATE (L4) Self-Paced Learning

1. Mathematics Textbook 1B
(a) Go through Chapter 9 (p9 to p11)
(b) Attempt Exercise 9.2 Q1 to Q10.
Enter your answers to Exercise 9.2 under Comments. Label your Answers with Question numbers.

2. ACELearning Portal: eLearning Portal
> Subject: Secondary 1 Express
> Additional Arithmetic
> Ratio, Rate and Speed: Choose Rate

(a) Go through video lessons: Rate, Average Rate, Example 1, Example 2
(b) Try Practice Drills on your own to check your understanding
Note: An email has been sent to all students via the AceLearning

June Vacation Chap 9.2 RATE (L5) Can he catch up with her?

Grace left home at 6.30 am to walk to school. 10 minutes later, her brother Shamus saw her wallet on the dining table. He grabbed the wallet and hopped onto his bicycle and rode after Grace.

Shamus rode his bicycle 3 times as fast as Grace walked.
Q1: What time was it when Shamus caught up with Grace?
Q2: If the school is 800m from home, did Shamus reach Grace before she arrived at school?

Use the following to guide you...
  • What information do you know from the problem?
  • What else do you need to know to solve the problem?
  • Pick a reasonable number for the information you need.