11001 1 2 − 100 1 2 = 110011_2-1001_2= 1 1 0 0 1 1 2 − 1 0 0 1 2 =

10101 0 2 101010_2 1 0 1 0 1 0 2

101010 1 2 1010101_2 1 0 1 0 1 0 1 2

SPM Mathematics Paper 1- Collections (8)

Authored by fara irina

Mathematics

9th - 12th Grade

CCSS covered

Used 122+ times

SPM Mathematics Paper 1- Collections (8)

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40 questions

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MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Round off 0.05273 correct to three significant figures.

0.05

0.053

0.0527

0.0528

Tags

CCSS.5.NBT.A.4

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

$\frac{0.0048}{\left(2\times10^{-2}\right)^2}=$

$1.2\times10^{-7}$

$1.2\times10$

$2.4\times10^{-7}$

$2.4\times10$

Tags

CCSS.8.EE.A.4

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

$2.31\times10^4$

$2.31\times10^5$

$3.51\times10^7$

$3.51\times10^9$

Tags

CCSS.8.G.C.9

CCSS.HSG.GMD.A.3

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

$110011_2-1001_2=$

$1010_2$

$10101_2$

$101010_2$

$1010101_2$

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Given $105_8=\left(2\times5^2\right)+\left(3\times5\right)+p$ . Find the value of p

Tags

CCSS.1.NBT.B.2B

CCSS.1.NBT.B.2C

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

$1100_2+142_5+15_8=$

$110$

$110_8$

$110_5$

$110_2$

Tags

CCSS.7.NS.A.1C

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

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SPM Mathematics Paper 1- Collections (8)

Round off 0.05273 correct to three significant figures.

0.0048(2×10−2)2=\frac{0.0048}{\left(2\times10^{-2}\right)^2}=(2×10−2)20.0048​=

1100112−10012=110011_2-1001_2=1100112​−10012​=

Given 1058=(2×52)+(3×5)+p105_8=\left(2\times5^2\right)+\left(3\times5\right)+p1058​=(2×52)+(3×5)+p . Find the value of p

11002+1425+158=1100_2+142_5+15_8=11002​+1425​+158​=

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$\frac{0.0048}{\left(2\times10^{-2}\right)^2}=$

$110011_2-1001_2=$

Given $105_8=\left(2\times5^2\right)+\left(3\times5\right)+p$ . Find the value of p

$1100_2+142_5+15_8=$