Page 22 - PBD Plus Matematik T1 (EG)

Matematik Tingkatan 1 Bab 1 Nombor Nisbah
(ii) Ramesh bergerak 8.7 m ke arah timur, Jawapan / Answer :
kemudian 12.9 m ke arah barat. Pecahan bahagian Diana
Seterusnya, dia bergerak sejauh 5.8 m Fraction of Diana’s part
1
ke arah timur. Perihalkan pergerakan = 10
Ramesh yang seterusnya supaya dia
boleh berada di kedudukan asalnya.
Ramesh moves 8.7 m to the east, then 12.9 m to Pecahan bahagian Alyenda Pecahan bahagian Julia
the west. Next, he moves 5.8 m to the east. Describe Fraction of Alyenda’s part Fraction of Julia’s part
5
4
Ramesh’s next movement so that he will be at his = 2 5 = 10 = 10
original position. SP 1.4.4
[3 markah / 3 marks] 5 bahagian / parts 75

Penerbitan Pelangi Sdn. Bhd.
Jawapan / Answer : 1 bahagian / part 75 ÷ 5 = 15

8.7 m – 12.9 m + 5.8 m 10 bahagian / parts 15 × 10 = 150

= –4.2 m + 5.8 m Julia mempunyai 150 lilin aroma.
= 1.6 m (ke barat dari titik asal) Julia has 150 scented candles.
(to the west from original position)
Mula 10. (a) Nita mempunyai RM235 di dalam bank dan
Start 8.7 m
telah berbelanja RM321, berapakah hutang
12.9 m
5.8 m yang perlu dilangsaikan Nita? SP 1.2.6
1.6 m Berhenti Nita had RM235 in the bank and spent RM321, how
Stop much is the debt that she has to settle?
Ramesh perlu bergerak 1.6 m ke timur [2 markah / 2 marks]
untuk kembali ke kedudukan asal. Jawapan / Answer :
Ramesh needs to move 1.6 m to the east so that he Hutang Nita / Nita’s debt = 235 – 321
will return to original position.
= –86
(b) Rajah di bawah menunjukkan satu peta Hutang Nita ialah RM86.
buih. Nita’s debt is RM86.
The diagram below shows a bubble map.
(b) Encik Wong menyimpan RM1.3 ribu setiap
K + 3 –4 – L bulan daripada hasil keuntungan syarikatnya
selama lima tahun. Dalam tempoh itu, dia
–20 telah membuat pengeluaran berjumlah
4
–60 ÷ N M × (–2) RM ribu sebanyak 7 kali. Hitung purata
5
simpanan Encik Wong dalam tempoh lima

Hitung nilai bagi K, L, M dan N. SP 1.2.1 SP 1.2.2 tahun. SP 1.5.3
Mr. Wong saved RM1.3 thousand each month from
State the values of K, L, M and N. his company profits for five years. Within that time, he
[2 markah / 2 marks] withdrew RM thousand for 7 times. Calculate Mr.
4
Jawapan / Answer : 5
K + 3 = –20 –4 – L = –20 Wong's average savings in five years.
K = –20 – 3 L = –4 + 20 [3 markah / 3 marks]
= –23 = 16 Jawapan / Answer :
Purata simpanan / Average savings
–60 ÷ N = –20 M × (–2) = –20 4

2
N = –60 ÷ (–20) M = –20 ÷ (–2) = 1.3 × 12 × 5 – × 7 ÷ 5
5
= 3 = 10 = (78 – 5.6) ÷ 5
= 72.4 ÷ 5
2
(c) Julia memberikan daripada lilin aromanya = RM14.48 ribu / thousand
5
kepada Alyenda dan 1 daripadanya kepada
10 (c) (i) Rajah di bawah menunjukkan satu garis
Diana. Bilangan lilin aroma yang dimiliki oleh nombor.
Julia sekarang ialah 75 unit. Cari bilangan asal The diagram below shows a number line.
lilin aroma yang dimiliki oleh Julia. SP 1.3.4
1
2
Julia gives of his scented candles to Alyenda and X –16 Y –8 Z
5 10
to Diana. The number of scented candles Julia has now Cari nilai X, Y dan Z. SP 1.1.3 SP 1.2.3
is 75 units. Find the number of scented candles Julia has Find the values of X, Y and Z.
initially. [3 markah / 3 marks]
[4 markah / 4 marks]

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