Matematika 3
Authored by Dewi Marsiana
Education
1st - 5th Grade
Used 2+ times
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15 questions
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1.
OPEN ENDED QUESTION
2 mins • 1 pt
(4+9)^2 – (4^2- 9^2) =⋯
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Answer explanation
(4+9)^2 – (4^2- 9^2)= (13)^2 - (16+81) = 169 - 97 = 72
2.
OPEN ENDED QUESTION
2 mins • 1 pt
Jika 3NN345 habis dibagi 9, tentukan nilai N.
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Answer explanation
N = 1, maka 9 x 1 = 9
N = 2, maka 9 x 2 = 18
3 + N + N + 3 + 4 + 5 = 9n
3+ 2N + 12 = 9n
15 + 2N = 9n, maka n yang paling mungkin adalah 3, sehingga 9n = 27
15 + 2N = 27
2N = 27 – 15
2B = 12
N = 6
3.
OPEN ENDED QUESTION
2 mins • 1 pt
Umur Amir lebih tua 3 tahun dari Budi. Budi usianya 4 tahun lebih muda dari Cipto. Ketika Cipto berusia 22 tahun, maka usia Amir adalah …
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Answer explanation
Umur Amir = Umur Budi + 3
Umur Budi = Umur Cipto – 4
Umur Cipto = 22 tahun
Umur Budi = 22 – 4 = 18 tahun
Umur Amir = 18 + 3 = 21 tahun
4.
OPEN ENDED QUESTION
2 mins • 1 pt
Nilai (2000+2002+2004+2006+2008)/(2000+2001+2002+2003+2004) adalah …
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Answer explanation
. (2000 + 2002+ 2004 + 2006 + 2008)/(2000 + 2001 + 2002 + 2003 + 2004) = 10.020/10.010= 1002/1001
5.
OPEN ENDED QUESTION
2 mins • 1 pt
Bilangan bulat yang mendekati adalah 7 (pembulatan ke atas)
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Answer explanation
5/2 + 2+4/7 ∶ 1/4 = (5/2 + 2) + (4/7 x 4/1) = 4 1/2 + 16/7 = 4 1/2 + 2 2/7 = 6 + (1/2 + 2/7)
6.
OPEN ENDED QUESTION
2 mins • 1 pt
Hari ini hari Senin, jatuh hari apakah 2005 hari yang akan datang?Senin = 2002, Selasa (2) = 2003, Rabu (20) = 2004, Kamis (3) = 2005
Jadi, 2005 hari yang akan datang adalah jatuh pada hari Kamis
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Answer explanation
Hari berulang setiap 7 hari. Senin akan bertemu Senin setelah 7 hari ke depan, sehingga 2005 : 7 = 286, maka sisa 3.
7.
OPEN ENDED QUESTION
2 mins • 1 pt
Nilai 1/(2+2/((3+3/5) ) ) adalah ...
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Answer explanation
1/(2+2/((3+3/5) ) ) = 1/(2+2/((15+3/5) ) ) = 1/(2+2/(18/5 ) )= 1/(2+(2 x 5/18) ) = 1/(2+10/18 ) = 1/(2+ 5/9 ) = 9/23
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