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KUIZ MATEMATIK TINGKATAN 3

Authored by Abu Umar

Mathematics

3rd Grade

Used 125+ times

KUIZ MATEMATIK TINGKATAN 3
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20 questions

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1.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

(−1 12)3=\left(-1\ \frac{1}{2}\right)^3=

−3 38-3\ \frac{3}{8}

−2 78-2\ \frac{7}{8}

2 782\ \frac{7}{8}

3 383\ \frac{3}{8}

2.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Permudahkan: 2p4÷6p9q2÷pq3Permudahkan:\ 2p^4\div6p^9q^2\div pq^3

13pq5\frac{1}{3}pq^5

13p6q5\frac{1}{3p^6q^5}

q3p4\frac{q}{3p^4}

3p4q\frac{3p^4}{q}

3.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Bundarkan 0.004055 betul kepada 3 angka bererti

0.004

0.0040

0.00405

0.00406

4.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Ungkapkan 0.000452 dalam bentuk piawai.

4.52×10−44.52\times10^{-4}

4.52×1044.52\times10^4

45.2×10−345.2\times10^{-3}

45.2×10345.2\times10^3

5.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Jenis akaun manakah yang menawarkan kadar faedah paling tinggi?

Akaun simpanan

Akaun simpanan tetap

Akaun semasa

Akaun wadiah

6.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Enciik Tan mendapat pinjaman kereta sebanyak RM125 000 dan tempoh bayaran ialah 5 tahun. Kadar faedah sama rata yang ditawarkan ialah 2.8% setahun. Hitung jumlah faedah yang perlu dibayarnya.

RM3 500

RM1 750

RM7 500

RM17 500

7.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Puan Liza menyimpan RM10 000 dalam satu akaun simpanan tetap selama satu tahun dengan faedah pengkompaunan 4 bulan sekali. Tentukan jenis faedah yang diterima oleh Puan Liza.

Faedah kompaun

Faedah mudah

Faedah sama rata

Faedah atas baki

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