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Terkunci (paket kadaluarsa)
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Lampu Merah, Lampu HijauPro
Bebas
SMP 7
Matematika
BODMAS (Fraction, Decimal, Percentage)
Ms. Madya
45
Pertanyaan yang ditambahkan (10/ 20)
Izinkan jawaban yang salah
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# 1

Pertanyaan jawaban singkat

2.9+1.1400%200%=​\frac{2.9+1.1}{400\%-200\%}=​

  • 2

Petunjuk

Change the percentage to a whole number.

# 2

Pilihan ganda

25%÷120.3=​25\%\div\frac12-0.3=​

  • 0.4

  • 0.1

  • 0.2

  • 0.3

# 3

Pertanyaan jawaban singkat

(65+0.75)×(60%÷30%)=​\left(\frac65+0.75\right)\times\left(60\%\div30\%\right)=​

  • 39/10

  • 3.9

  • 390%

Petunjuk

The answer can be a fraction, a decimal, or a percentage. You may choose any of them.

# 4

Pertanyaan jawaban singkat

(120%÷425)(0.8750.75)×2=​\left(120\%\div\frac{4}{25}\right)-\left(0.875-0.75\right)\times2=​

  • 7.25

  • 29/4

  • 725%

  • 7.25 or 29/4 or 725%

Petunjuk

The answer can be a fraction, a decimal, or a percentage. You may choose any of them.

# 5

OX

0.90.35+56÷50%=2​0.9-0.35+\frac56\div50\%=2​1360​\frac{13}{60}​

# 6

OX

(25+0.75)×20%=20%​\left(\frac25+0.75\right)\times20\%=20\%​

# 7

Pilihan ganda

2.4÷(0.6+0.2)310+25%​2.4\div\left(0.6+0.2\right)-\frac{3}{10}+25\%​

=2.4÷0.8310+25%​=2.4\div0.8-\frac{3}{10}+25\%​

=n0.3+m​=n-0.3+m​

=2.95​=2.95​

  • n=3​n=3​ and m=2.5​m=2.5​

  • n=3​n=3​ and m=0.25​m=0.25​

  • n=0.3​n=0.3​ and m=0.25​m=0.25​

  • n=0.3​n=0.3​ and m=2.5​m=2.5​

# 8

Pilihan ganda

4.5+92×16%=​\sqrt{4.5+\frac92}\times16\%=​

  • 66%

  • 48%

  • 58%

  • 72%

# 9

Pilihan ganda

Find the value of n.


0.25+20%×n=94​0.25+20\%\times n=\frac94​

  • 10

  • 100

  • 50

  • 0

# 10

Pertanyaan jawaban singkat

40%×1510+810÷0.2=​40\%\times\frac{15}{10}+\frac{8}{10}\div0.2=​

  • 23/5

  • 4.6

  • 460%

Petunjuk

The answer can be a fraction, a decimal, or a percentage. You may choose any of them.

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