8.2.1 Basic Measurements, PT3 Practice

Posted on April 1, 2017 by user

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Question 1:
Karen has 18 m of ribbon to tie 16 bars of chocolate and 28 boxes of sweets. A bar of chocolate needs 24 cm of ribbon and a box of sweets needs 38 cm of ribbon.
Calculate the length, in m, of the remaining ribbon.

Solution:
Length of ribbon needed
= (16 × 24 cm) + (28 × 38 cm)
= 384 cm + 1064 cm
= 1448 cm
= 14.48 m

Length of remaining ribbon
= 18 m – 14.48 m
= 3.52 m

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Question 2:
The length of a red string is 4 m 80 cm. The length of a brown string is ⅔ of the length of the red string. The length of a black string is twice the length of the brown string.
Calculate the total length, in m, of the brown and the black strings.

Solution:

\begin{array}{l} Length of brown string \\ = \frac{2}{3} \times 4 m 80 cm \\ = \frac{2}{3} \times 4.8 \\ = 3.2 m \\ Length of black string \\ = 2 \times 3.2 m \\ = 6.4 m \\ Total length = 3.2 m + 6.4 m \\ = 9.6 m \end{array}

[adinserter name="Block 3"]

Question 3:
Kelly has 13 m of cloth. She uses it to make 6 curtains and 4 tablecloths. Each curtain requires 1.25 m of cloth and each tablecloth requires 90 cm of cloth.
What is the length, in m, of the remaining cloth?

Solution:
Length of cloth used to make curtains
= 6 × 1.25 m
= 7.5 m

Length of cloth used to make tablecloth
= 4 × 90 cm
= 360 cm
= 3.6 m

Length of the remaining cloth
= 13 – 7.5 – 3.6
= 1.9 m

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Question 4:
A rope measuring 6 m 30 cm is cut into 1.6 m, 96 cm and the remainder into two equal parts. Find the length of each of the remaining parts, in cm.

Solution:
(6 m 30 cm – 1.6 m – 96 cm) ÷ 2
= (630 cm – 160 cm – 96 cm) ÷ 2
= 374 cm ÷ 2
= 187 cm

6.2.1 Integers, PT3 Practice

Posted on March 18, 2017 by user

Question 1:
Evaluate 5 + (–7)

Solution:

Therefore,
5 + (–7) = 5 – 7
= –2

Question 2:
Evaluate –6 – (–5)

Solution:

Therefore,
–6 – (–5) = –6 + 5
= –1

Question 3:
Simplify –3 + 7 + (–2)

Solution:

Therefore,
–3 + 7 + (–2) = 4 + (–2)
= 4 – 2
= 2

Question 4:
Simplify 5 + (–4) + (–3)

Solution:

Therefore,
5 + (–4) + (–3)
= 5 – 4 – 3
= –2

6.2 Integers

Posted on March 15, 2017 by user

6.2 Addition and Subtraction of Integers

6.2.1 Addition of Integers
1. Addition of integers can be done on a number line.
(a) To add a positive integer:
Move to the right (positive direction) on the number line.

(b) To add a negative integer:
Move to the left (negative direction) on the number line.

2. Integers with the same signs are called integers with like signs.
Example:
2 and 7, –25 and –5.

3. Integers with the different signs are called integers with unlike signs.
Example:
+2 and –7, –25 and 5.

Example 1:
Solve the following.
(a) 3 + (+4)
(b) 2 + (–5)

Solution:
(a)

Therefore,
3 + (+4) = 3 + 4
= 7

(b)

Therefore,
2 + (–5)
= –3

6.2.3 Multiplication and Division of Integers
1. Multiplication and division of like signs gives (+)

\begin{array}{l} (+) \times (+) = + (+) \div (+) = + \\ (-) \times (-) = + (-) \div (-) = + \end{array}

2. Multiplication and division of unlike signs gives (–)

\begin{array}{l} (+) \times (-) = - (+) \div (-) = - \\ (-) \times (+) = - (-) \div (+) = - \end{array}

Example:

(a) –25 ÷ 5 = –5

(b) 8 × (–5) = –40

5.2.2 Percentages, PT3 Practice

Posted on March 7, 2017 by user

Question 6:
Liza had 360 apples. She sold ⅝ of the apples. Then she gave 27 of the remaining apples to her neighbour.
Calculate the percentage of apples that Liza has left.

Solution:

\begin{array}{l} Number of apples sold \\ = \frac{5}{8} \times 360 \\ = 225 \\ Number of apples left \\ = 360 - 225 - 27 \\ = 108 \\ Percentage of apples left \\ = \frac{108}{360} \times 100 % \\ = 30 % \end{array}

Question 7:
Mei Ling took part in a science quiz competition. She answered 12 questions correctly. She answered 25% of the questions incorrectly.
Find the total number of questions in the quiz.

Solution:

\begin{array}{l} 75 % \to 12 correct answers \\ 25 % \to ? incorrect answers \\ Number of incorrect answers \\ = \frac{25}{75^{3}} \times 12 \\ = 4 \\ Total number of questions \\ = 12 + 4 \\ = 16 \end{array}

Question 8:
Diagram below shows the prices of two items.

Kenny buys a pair of track pants and 2 pieces of sweaters.
Calculate the total amount Kenny has to pay.

Solution:

\begin{array}{l} Price of a pair of track pants after 25% discount \\ = RM 150.00 - (\frac{25}{100} \times RM 150.00) \\ = RM 150.00 - RM37 .50 \\ = RM 112.50 \\ Total amount Kenny has to pay \\ = (RM 100.00 \times 2) + RM 112.50 \\ = RM3 12.50 \end{array}

Question 9:
Table below shows the prices and the discounts for the same brand of school bag sold at four shops, P, Q, R and S.

Which shop offers the cheapest price?

Solution:

\begin{array}{l} Shop P, \\ Price per item after discount \\ = RM 100 - (\frac{20}{100} \times RM 100) \\ = RM 80 \\ Shop Q, \\ Price per item after discount \\ = RM 110 - (\frac{30}{100} \times RM 110) \\ = RM77 \\ Shop R, \\ Price per item with no discount \\ = RM 170 \div 2 \\ = RM85 \\ Shop S, \\ Price per item with no discount = RM90 \\ Shop Q offers the cheapest price . \end{array}

Question 10:
Table below shows dividend received by Janice from her investment.

Janice reinvests RM204 from the total dividends she received into ASB investment.
What is the remaining percentage that she did not use in her reinvestment?

Solution:

\begin{array}{l} Dividends from ASB \\ = 3 % \times RM 2 000 \\ = \frac{3}{100} \times RM 2 000 \\ = RM 60 \\ Dividends from ASN \\ = 4 % \times RM 7 000 \\ = \frac{4}{100} \times RM 7 000 \\ = RM 280 \\ Total dividends \\ = RM 60 + RM 280 \\ = RM 340 \\ Percentage of dividends not reinvested \\ = \frac{340 - 204}{340} \times 100 \\ = \frac{136}{340} \times 100 \\ = 40 % \end{array}

5.2.1 Percentages, PT3 Practice

Posted on March 5, 2017 by user

Question 1:
Rahman bought a house at RM180 000. The market price of the house increases by 15 % per year. After a year, Rahman sold his house at the price of RM 205 000.
Is the selling price higher compared to the market price? Show your working.

Solution:

\begin{array}{l} Market price \\ = 115 % \times RM180 000 \\ = \frac{115}{100} \times RM180 000 \\ = RM 207 000 \\ RM 207 000 > RM 205 000 \\ The market price of the house is higher than the selling price of the house . \end{array}

Question 2:
A company makes a profit of RM 50 000 in year 2016. The projection of profit increase is fixed at 20 % from the previous year.
Predict the company profit in year 2018.

Solution:

\begin{array}{l} Profit in year 2017 \\ = RM 50 000 \times \frac{120}{100} \\ = RM6 0 000 \\ Profit in year 2018 \\ = RM6 0 000 \times \frac{120}{100} \\ = RM72 000 \end{array}

Question 3:
Sharon wants to buy a new car with the price of RM 78 500. Her father gives her some money, which is 30 % of the car price. The deposit for the car is

\frac{1}{5}

of the price.
Would her father’s contribution be enough for her deposit? Give your reason with calculation.

Solution:

\begin{array}{l} 30 % of the car price given by Sharon's father \\ = \frac{30}{100} \times RM 78500 \\ = RM 235 50 \\ Deposit of the car price needed \\ = \frac{1}{5} \times RM 78500 \\ = RM 15700 \\ RM 235 50 > RM 15700, therefore her father 's contribution is \\ enough for the deposit . \end{array}

Question 4:
The profit of Company P and Company Q in March are RM38 000 and RM48 000 respectively. In April, the profit of Company P has increased by 20%.
Given the profit for both companies are equal in April, find the percentages decrease in profit of Company Q.

Solution:

\begin{array}{l} Profit of Company P in April \\ = \frac{120}{100} \times 38000 \\ = RM45600 \\ Drop in profit of Company Q \\ = RM48000 - RM45600 \\ = RM24 00 \\ Percentage decrease in profit \\ = \frac{2400}{48000} \times 100 % \\ = 5 % \end{array}

Question 5:
A biscuit making company needs 200 tonnes of flour every month. Factory K supplies 80 tonnes, while the rest is supplied by Factory M. In a particular month, Factory K faces a problem which forces it to reduce its output by 15%.
Find the total percentage of the flour output of Factory M in order to accommodate the requirement of the biscuit making company.

Solution:

\begin{array}{l} New flour supply from Factory K \\ = \frac{85}{100} \times 80 \\ = 68 tonnes \\ New flour supply from Factory M \\ = 200 - 68 \\ = 132 tonnes \\ Total percentage of flour output of Factory M \\ = \frac{132}{120} \times 100 % \\ = 110 % \end{array}

4.2.2 Decimals, PT3 Practice

Posted on February 22, 2017 by user

Question 5:
Calculate the value of

\frac{4}{5} + 15.4 \div 4

and express the answer correct to one decimal place.

Solution:

\begin{array}{l} \frac{4}{5} + 15.4 \div 4 \\ = \frac{4}{5} + 3.85 \\ = 0.8 + 3.85 \\ = 4.65 \\ = 4.7 (one decimal place) \end{array}

Question 6:
Calculate the value of 84 – 3(14 + 48 ÷ 6) =

Solution:

\begin{array}{l} 84 - 3 (14 + 48 \div 6) \\ = 84 - 3 [14 + (48 \div 6)] \\ = 84 - 3 (14 + 8) \\ = 84 - 3 (22) \\ = 84 - 66 \\ = 18 \end{array}

Question 7:

Calculate the value of \frac{4 \times 0.18}{0.9} .

Solution:

\begin{array}{l} \frac{4 \times 0.18}{0.9} = \frac{0.72}{0.9} \\ = \frac{0.72 \times 10}{0.9 \times 10} \\ = \frac{7.2}{9} \\ = 0.8 \end{array}

Question 8:

\begin{array}{l} Calculate the value of (8.4 - 2.64) \div \frac{4}{3} and express the answer correct \\ to two decimal places . \end{array}

Solution:

\begin{array}{l} (8.4 - 2.64) \div \frac{4}{3} \\ = {5.76}^{1.44} \times \frac{3}{4} \\ = 4.32 \end{array}

4.2.1 Decimals, PT3 Practice

Posted on February 20, 2017 by user

Question 1:
62.4 – 36.02 + 12.6 =

Solution:

Question 2:
4.76 × 7.2 ÷ 0.3 =

Solution:

\begin{array}{l} 4.76 \times 7.2 \div 0.3 \\ = 34.272 \div 0.3 \\ = \frac{34.272}{0.3} \\ = \frac{34.272 \times 10}{0.3 \times 10} \\ = \frac{342.72}{3} \\ = 114.24 \end{array}

Question 3:
0.12 × (5.9 – 1.33) ÷ 0.8 =

Solution:

\begin{array}{l} 0.12 \times (5.9 - 1.33) \div 0.8 \\ = 0.12 \times 4.57 \div 0.8 \\ = 0.5484 \div 0.8 \\ = \frac{0.5484}{0.8} \\ = \frac{0.5484 \times 10}{0.8 \times 10} \\ = \frac{5.484}{8} \\ = 0.6855 \end{array}

Question 4:
14.83 + (20.04 – 10.68) ÷ 1.8 =

Solution:

\begin{array}{l} 14.83 + (20.04 - 10.68) \div 1.8 \\ = 14.83 + 9.36 \div 1.8 \\ = 14.83 + \frac{9.36 \times 10}{1.8 \times 10} \\ = 14.83 + \frac{93.6}{18} \\ = 14.83 + 5.2 \\ = 20.03 \end{array}

3.2.3 Fractions, PT3 Practice

Posted on February 9, 2017 by user

Question 11:
Table below shows the types of fruit tree planted by Azmi in his orchard.

(a) Calculate the number of guava trees.
(b) 100 mangosteen trees died due to fungus.
Find the total number of the remaining trees.

Solution:
(a)

\begin{array}{l} Number of Guava trees \\ = \frac{1}{3} \times 180 \\ = 60 \end{array}

(b)

\begin{array}{l} Number of mangosteen and guava trees \\ = 180 + 60 \\ = 240 \\ Fraction of mangosteen and guava trees = \frac{3}{5} \\ Total number of trees \\ = \frac{5}{3} \times 240 \\ = 400 \\ Total number of remaining trees \\ = 400 - 100 \\ = 300 \end{array}

Question 12:
A palm oil processing factory has a number of workers.

\frac{4}{7}

of the workers are skilled workers. If

\frac{3}{5}

of the general workers are 18 females, find the total number of workers in the factory.

Solution:

\begin{array}{l} \frac{3}{5} of the general workers are 18 females . \\ Let the number of general workers = w \\ \frac{3}{5} \times w = 18 \\ w = 18 \times \frac{5}{3} \\ = 30 \\ Hence the total number of workers in the factory \\ = 30 \div (\frac{7}{7} - \frac{4}{7}) \\ = 30 \div \frac{3}{7} \\ = 30 \times \frac{7}{3} \\ = 70 \end{array}

Question 13:
In a fruit basket,

\frac{2}{5}

are mangoes and the rest are kiwis.
If

\frac{2}{3}

of the mangoes or 12 of them are in good condition, calculate the number of kiwis in the basket.

Solution:

\begin{array}{l} Let the number of mangos \\ in the basket = x \\ \frac{2}{3} \times x = 12 \\ x = 12 \times \frac{3}{2} \\ x = 18 mangoes \\ Total number of mangoes and \\ kiwis in the basket \\ = 18 \times \frac{5}{2} \\ = 45 \\ Hence, the number of kiwis \\ in the basket \\ = (1 - \frac{2}{5}) \times 45 \\ = (\frac{5}{5} - \frac{2}{5}) \times 45 \\ = \frac{3}{5} \times 45 \\ = 27 \end{array}

3.2.2 Fractions, PT3 Practice

Posted on February 7, 2017 by user

Question 6:

1 \frac{1}{2} \div \frac{1}{4} \times \frac{4}{5} =

Solution:

\begin{array}{l} 1 \frac{1}{2} \div \frac{1}{4} \times \frac{4}{5} \\ = \frac{3}{2} \times \frac{4^{2}}{1} \times \frac{4}{5} \\ = \frac{24}{5} \\ = 4 \frac{4}{5} \end{array}

Question 7:

4 \frac{1}{5} \div 1 \frac{1}{6} \times 2 \frac{2}{9} =

Solution:

\begin{array}{l} 4 \frac{1}{5} \div 1 \frac{1}{6} \times 2 \frac{2}{9} \\ = \frac{21^{3}}{5} \times \frac{6^{2}}{7} \times \frac{20^{4}}{9^{3}} \\ = 8 \end{array}

Question 8:

2 \frac{3}{4} \times 1 \frac{1}{3} \div 7 \frac{1}{3} =

Solution:

\begin{array}{l} 2 \frac{3}{4} \times 1 \frac{1}{3} \div 7 \frac{1}{3} \\ = \frac{11}{4} \times \frac{4}{3} \div \frac{22}{3} \\ = \frac{11}{4} \times \frac{4}{3} \times \frac{3}{22^{2}} \\ = \frac{1}{2} \end{array}

Question 9:

4 \frac{5}{6} - 2 \frac{1}{3} \div \frac{5}{9} =

Solution:

\begin{array}{l} 4 \frac{5}{6} - 2 \frac{1}{3} \div \frac{5}{9} \\ = 4 \frac{5}{6} - (\frac{7}{3} \div \frac{5}{9}) \\ = 4 \frac{5}{6} - (\frac{7}{3} \times \frac{9^{3}}{5}) \\ = 4 \frac{5}{6} - \frac{21}{5} \\ = 4 \frac{5}{6} - 4 \frac{1}{5} \\ = \frac{25}{30} - \frac{6}{30} \\ = \frac{19}{30} \end{array}

Question 10:

(\frac{7}{9} + \frac{11}{15}) \div (7 \frac{8}{15} - 4 \frac{1}{5}) =

Solution:

\begin{array}{l} (\frac{7}{9} + \frac{11}{15}) \div (4 \frac{8}{15} - 3 \frac{1}{5}) \\ = (\frac{35}{45} + \frac{33}{45}) \div (4 \frac{8}{15} - 3 \frac{3}{15}) \\ = \frac{68}{45} \div 1 \frac{5}{15} \\ = \frac{68}{45} \div 1 \frac{1}{3} \\ = \frac{68}{45} \div \frac{4}{3} \\ = \frac{68^{17}}{45^{15}} \times \frac{3}{4} \\ = \frac{17}{15} \\ = 1 \frac{2}{15} \end{array}

3.2.1 Fractions, PT3 Practice

Posted on February 5, 2017 by user

Question 1:

3 \frac{2}{5} + 1 \frac{5}{6} - \frac{1}{5} =

Solution:

\begin{array}{l} 3 \frac{2}{5} + 1 \frac{5}{6} - \frac{1}{5} \\ = 3 \frac{12}{30} + 1 \frac{25}{30} - \frac{6}{30} \\ = 4 \frac{37}{30} - \frac{6}{30} \\ = 4 \frac{31}{30} \\ = 5 \frac{1}{30} \end{array}

Question 2:

6 \frac{1}{12} - 3 \frac{2}{3} - 1 \frac{5}{6} =

Solution:

\begin{array}{l} 6 \frac{1}{12} - 3 \frac{2}{3} - 1 \frac{5}{6} \\ = 6 \frac{1}{12} - 3 \frac{8}{12} - 1 \frac{10}{12} \\ = (5 \frac{13}{12} - 3 \frac{8}{12}) - 1 \frac{10}{12} \\ = 2 \frac{5}{12} - 1 \frac{10}{12} \\ = 1 \frac{17}{12} - 1 \frac{10}{12} \\ = \frac{7}{12} \end{array}

Question 3:

30 - (\frac{2}{7} \div \frac{2}{21}) =

Solution:

\begin{array}{l} 30 - (\frac{2}{7} \div \frac{2}{21}) \\ = 30 - (\frac{2}{7} \times \frac{21^{3}}{2}) \\ = 30 - 3 \\ = 27 \end{array}

Question 4:

(7 \frac{1}{2} - 3 \frac{2}{3}) \times \frac{12}{69} =

Solution:

\begin{array}{l} (7 \frac{1}{2} - 3 \frac{2}{3}) \times \frac{12}{69} \\ = (7 \frac{3}{6} - 3 \frac{4}{6}) \times \frac{12}{69} \\ = (6 \frac{9}{6} - 3 \frac{4}{6}) \times \frac{12}{69} \\ = 3 \frac{5}{6} \times \frac{12}{69} \\ = \frac{23}{6} \times \frac{12^{2}}{69^{3}} \\ = \frac{2}{3} \end{array}

Question 5:

1 \frac{1}{5} \div (4 - 2 \frac{1}{2}) =

Solution:

\begin{array}{l} 1 \frac{1}{5} \div (4 - 2 \frac{1}{2}) \\ = 1 \frac{1}{5} \div (3 \frac{2}{2} - 2 \frac{1}{2}) \\ = \frac{6}{5} \div (1 \frac{1}{2}) \\ = \frac{6}{5} \div \frac{3}{2} \\ = \frac{6^{2}}{5} \times \frac{2}{3} \\ = \frac{4}{5} \end{array}