TEST CORRECTIONS Test #03
Standard Index Form
a) Convert each of these three numbers from Standard Index Form to ordinary numbers (Full Numerical Form):i) \(2.236\times 10^3\)ii) \(4.3\times 10^4\)iii) \(3.1\times 10^{-4}\)
Hints Part (i): 2 Χ 103 can be written as: ┌──────────┤Chucking in loads of meaningless zeros after the decimal-point ╒═╤═╪═╤═╕ 2.0 0 0 0 0 Χ 103 ╔═════════════╗ ▼ ╟╖ 2.0 0 0 0 0 Χ 103 ║ Every time we move the decimal right, we knock 1 off the power ║ ║ So, we have to move the decimal right by 3 places ║ ║ to get this power down to zero ║ So, the number becomes: ╙╜ 2.0 0 0.0 0 Χ 100 = 2000.00 = 2000 │ ▲ ▲ ▲ └─┘─┘─┘ moving the decimal right by 3 places Hints Part (ii): 4.3 Χ 104 can be written as: ┌──────────┤ Bung in loads of meaningless zeros after the decimal-point ╒═╤═╪═╤═╕ 4.3 0 0 0 0 0 Χ 104 ╔═══════════╗ ▼ ╟╖ 4.3 0 0 0 0 0 Χ 104 ║ Every time we move the decimal right, we knock 1 off the power ║ ║ So, we have to move the decimal right by 4 places ║ ║ to get this power down to zero ║ So, the number becomes: ╙╜ 4.3 0 0 0.0 0 Χ 100 = 43000.00 = ...... │ ▲ ▲ ▲ ▲ └─┘─┘─┘─┘ moving the decimal right by 4 places Hints Part (iii): 3.1 Χ 10-4 can be written as: ┌──────────┤Putting in loads of meaningless zeros before any digits ╒═╤═╪═╤═╕ 0 0 0 0 0 3.1 Χ 10-4 ╔═══════════╗ ▼ ╟╖ 0 0 0 0 0 3.1 Χ 10-4 ║ Every time we move the decimal left, we add 1 to the power ║ ║ So, we have to move the decimal left by 4 places ║ ║ to get this power up to zero ║ So, the number becomes: ╙╜ 0 0.0 0 0 3.1 Χ 100 = 0.00031 ▲ ▲ ▲ ▲ │ └─└─└─└─┘ moving the decimal left by 4 places
b) Convert each of these numbers from Standard Index Form to ordinary numbers (Full Digital Form):i) \(2.12\times 10^4\)ii) \(6\times 10^3\)iii) \(1.4\times 10^{-3}\)
Hints Part (i): Write the number as: 2.1 2 0 0 0 0 0 Χ 104 Then, knock 1 off the power and └─┐ ⭭ move the decimal RIGHT by 1 place: 2 1.2 0 0 0 0 0 Χ 103 └─┐ ⭭ 2 1 2.0 0 0 0 0 Χ 102 └─┐ ⭭ 2 1 2 0.0 0 0 0 Χ 101 └─┐ ⭭ 2 1 2 0 0.0 0 0 Hints Part (ii): Write the number as: 6.00000 Χ 103, then move the decimal RIGHT by �places Hints Part (iii): Write the number as: 000001.4 Χ 10-3, then move the decimal LEFT by �places
c) Convert these numbers from Standard Index Form to ordinary numbers (Full Digital Form):i) \(8.23\times 10^3\)ii) \(5\times 10^5\)iii) \(2.1\times 10^{-4}\)
Hints Part (i): You need to manage this all by yourself Hints Part (ii): Sorry: No hints for this Hints Part (iii): Sorry! 🤭 a) Re-write the following numbers in Standard Index Form:i) \(1234.56\)ii) \(12,000\)iii) \(0.00123\)
Hints Part (i): Start by writing the number as: 1 2 3 4.5 6 Χ 100 Now move the decimal until there is just 1 digit in-front of it (this digit cannot be a zero): ╔═════════════╗ ▼ ╟╖ 1 2 3 4.5 6 Χ 100 ║ Every place we move the decimal left, we add 1 to the power ║ ┌─┘ ⭭ ╙╜ 1 2 3.4 5 6 Χ 101 ┌─┘ ⭭ 1 2.3 4 5 6 Χ 102 ┌─┘ ⭭ 1.2 3 4 5 6 Χ 103 = 1.23456 Χ 103 ▲ │ ┌──────────────────────┐ └─┤ Until there is only │ │ 1-digit in-front │ │ of the decimal-point │ └──────────────────────┘ Hints Part (ii): Start by writing the number as: 1 2 0 0 0.0 Χ 100 Now move the decimal until there is just 1 digit in-front of it (this digit cannot be a zero): ╔═════════════╗ ▼ ╟╖ 1 2 0 0 0.0 Χ 100 ║ Every place we move the decimal left, we add 1 to the power ║ ┌─┘ ⭭ ╙╜ 1 2 0 0.0 0 Χ 101 ┌─┘ ⭭ 1 2 0.0 0 0 Χ 102 ┌─┘ ⭭ 1 2.0 0 0 0 Χ 103 ┌─┘ ⭭ 1.2 0 0 0 0 Χ 104 = 1.20000 Χ 104 = 1.2 Χ 104 ▲ │ ┌──────────────────────┐ └─┤ Until there is only │ │ 1-digit in-front │ │ of the decimal-point │ └──────────────────────┘ Hints iii) We can only add the powers if the bases are the same! Start by writing the number as: 0.0 0 1 2 3 Χ 100 Now move the decimal until there is just 1 digit in-front of it (this digit cannot be a zero): ╔═════════════╗ ▼ ╟╖ 0.0 0 1 2 3 Χ 100 ║ Every place we move the decimal right, we knock 1 off the power ║ └─┐ ⭭ ╙╜ 0 0.0 1 2 3 Χ 10-1 └─┐ ⭭ 0 0 0.1 2 3 Χ 10-2 └─┐ ⭭ 0 0 0 1.2 3 Χ 10-3 = 0001.23 Χ 10-3 = 1.23 Χ 10-3 ⭫ ⭫ ⭫ ▲ These │ ┌──────────────────────────┐ zeros┐└─┤ Until there is only 1 │ don't └──┼(non-zero)-digit in-front│ count │ of the decimal-point │ └──────────────────────────┘
b) Re-write the following numbers in Standard Index Form:i) \(123.456\)ii) \(1,200,000\)iii) \(0.0123\)
Hints Part (i): Write the number as: 1 2 3.4 5 6 Χ 100 Then, move the decimal left 1 place ┌─┘ ⭭ and add '1' to the power: 1 2.3 4 5 6 Χ 101 ┌─┘ ⭭ 1.2 3 4 5 6 Χ 102 Hints Part (ii): Part (ii): Write as: 1,200,000.0 Χ 100, then move the decimal LEFT by � places and add �to the power Hints Part (iii): Part (iii): Write as: 0.0123 Χ 100, then move the decimal RIGHT by � places and knock � off the power
c) Convert the numbers below into Standard Index Form:i) \(12.3456\)ii) \(120\)iii) \(0.00000123\)
Hints You need to answer these three without my help but they are worked out in exactly the same way as Part (b) a) If \(a=1.25\times 10^5\) and \(b=4.5\times 10^3\), find, in Standard Index Form:i) \(a+b\)ii) \(a-b\)iii) \(a+3b\)
Hints We can only add numbers in standard form, if the exponents are the same We need to convert the one with the lower power: └───────────────────────┬─────┘ ˅ 1.25 Χ 105 + 4.5 Χ 103 ╘════╤════╛ ┌────────┴────────┐ ╒══════════════════════════════════════╕ │So convert this ═╪═══╪═► move the decimal left (increasing │ └────────┬────────┘ │ the power by 1 with each move) until │ ┌────────┴────────┐ │ it matches the power of the other № │ │ now add them ◄══╪═══╪═0.045 Χ 105 ⇐ 0.45 Χ 104 ⇐ 4.5 Χ 103 │ └────────┬────────┘ ╘══════════════════════════════════════╛ ╒════╧══════╕ 1.25 Χ 105 + 0.045 Χ 105 └──┬─┘ └─┬───┘ └─────ADD─────┬─┘ ┌──┴──┐ 1.295 Χ 105 Hints Part (ii): Start by working out 3b: a = 1.25 Χ 105 b = 4.5 Χ 103 3b = 3Χ4.5 Χ 103 └──┬──┘ ┌──┴─┐ 3b= 13.5 Χ 103 NOW make the powers the same: 1.25 Χ 105 - 13.5 Χ 103 ╘═════╤════╛ ┌────────┴────────┐ convert this one: └────────┬────────┘ ╒════════╧═══════════════════════════════╕ move the decimal to the left (increasing the power by 1 with each move) until the power is the same as the other number: 13.5 Χ 103 = 1.35 Χ 104 = 0.135 Χ 105 ╘════════╤══════════════════════════════╛ ┌────────┴─────────┐ now subtract them: └────────┬─────────┘ ╒═════╧═════╕ 1.25 Χ 105 - 0.135 Χ 105 = ... Χ 105 BUT, this ain't is Standard Index Form You need to shift the decimal until there is only 1 (non-zero) digit to the left of it... (and adjust the exponent accordingly!) Hints Part (iii): We can only subtract numbers in standard form, if the exponents are the same We need to convert the one with the lower power: └───────────────────────┬─────┘ ˅ 1.25 Χ 105 - 4.5 Χ 103 ╘════╤════╛ ┌────────┴────────┐ ╒══════════════════════════════════════╕ │So convert this ═╪═══╪═► move the decimal left (increasing │ └────────┬────────┘ │ the power by 1 with each move) until │ ┌────────┴────────┐ │ it matches the power of the other № │ │ now add them ◄══╪═══╪═0.045 Χ 105 ⇐ 0.45 Χ 104 ⇐ 4.5 Χ 103 │ └────────┬────────┘ ╘══════════════════════════════════════╛ ╒════╧══════╕ 1.25 Χ 105 + 0.045 Χ 105 └──┬─┘ └─┬───┘ └───SUBTRACT──┬─┘ ┌──┴──┐ 1.205 Χ 105
b) If \(p=4.5\times 10^6\) and \(q=5\times 10^5\), find, in Standard Form:i) \(p+q\)ii) \(p-q\)iii) \(3p-\frac{3}{2}q\)
Hints Part (i): ALLs wez gotz'ta do is ADD EM UP: Start by re-writing the one with the lower power: └─────────────────────┬───────┘ ˅ 4.5 Χ 106 + 5 Χ 105 ╘════╤════╛ ┌────────┴────────┐ ╒══════════════════════════════════════╕ │So convert this ═╪═══╪═► move the decimal left (increasing │ └────────┬────────┘ │ the power by 1 with each move) until │ ┌────────┴────────┐ │ it matches the power of the other № │ │ now add them ◄══╪═══╪═ 0.50 Χ 106 ⇐ 5.0 Χ 105 │ └────────┬────────┘ ╘══════════════════════════════════════╛ ╒════╧════╕ 4.5 Χ 106 + 0.5 Χ 106 = ... Χ 106 Hints Part (ii): We juz needz to SUBTRACTZ 'em: Start by re-writing the one with the smaller exponent: 4.5 Χ 106 - 5 Χ 105 ╘═══╤═══╛ ┌────────┴────────┐ convert this one: └────────┬────────┘ ╒════════╧═══════════════════════════════╕ move the decimal to the left (increasing the power by 1 with each move) until the power is the same as the other number: 5.0 Χ 105 = 0.50 Χ 106 ╘════════╤══════════════════════════════╛ ┌─────────┴─────────┐ now we can add them: └─────────┬─────────┘ ╒═══╧═════╕ 4.5 Χ 106 - 0.5 Χ 106 = ... Χ 106 Hints Part (iii): Start by working out 3p: and ³⁄₂q: ------------------------------------------------ p = 4.5 Χ 106 q = 5 Χ 105 3p = 3Χ4.5 Χ 106 ³⁄₂q = ³⁄₂Χ5 Χ 105 └──┬──┘ └──┬──┘ ┌─┴──┐ ┌─┴─┐ 3p = 13.5 Χ 106 ³⁄₂q = 7.5 Χ 105 NOW make the powers the same: 13.5 Χ 106 - 7.5 Χ 105 ╘════╤═══╛ ┌────────┴────────┐ convert this one: └────────┬────────┘ ╒════════╧═══════════════════════════════╕ move the decimal to the left (increasing the power by 1 with each move) until the power is the same as the other number: 7.5 Χ 105 = 0.75 Χ 106 ╘════════╤═══════════════════════════════╛ ┌────────┴─────────┐ now subtract them: └────────┬─────────┘ ╒════╧═════╕ 13.5 Χ 106 - 0.75 Χ 106 = ... Χ 106 BUT, this ain't is Standard Index Form You need to shift the decimal until there is only 1 (non-zero) digit to the left of it (and adjust the exponent accordingly!)
c) If \(x=7.2\times 10^8\) and \(y=5\times 10^6\), find, in Standard Form:i) \(x+2y\)ii) \(2x-3y\)iii) \(\frac{1}{3}x-\frac{3}{4}y\)
Hints Clue: Sorry, noooo mooore help a) If \(a=1.5\times 10^9\) and \(b=3\times 10^6\), find in Standard Form, the values of:i) \(ab\)ii) \(\frac{a}{b}\)iii) \(b^3\)
Hints Part (i): We want to find: 1.5 Χ 109 Χ 3 Χ 106 ╔═══════════╦════╗ ▼ ▼ ║ 1.5 Χ 109 Χ 3 Χ 106 add └─┬─┘ └┬┘ ╔═══╝ └─multiply─┬─┘ ║ ┌─┴─┐ ▼ 4.5 Χ 1015 Hints Part (ii): Part (ii): We want to find: 1.5 Χ 109 χ 3 Χ 106 ╔═══════════╦══════╗ ▼ ▼ ║ 1.5 Χ 109 ? 3 Χ 106 subtract └─┬─┘ └┬┘ ╔═══╝ └──divide──┬─┘ ║ ┌─┴──┐ ▼ 0.5 Χ 103 ╘═════╤══════╛ ┌────────────┴────────────┐ ╒═══════════════════════════════════════╕ │this isn't standard form:╪═══╪► move the decimal 1 place to the RIGHT│ └────────────┬────────────┘ │so there's just ONE (non-zero) digit to│ ┌────────────┴────────────┐ │the left of it. Knock 1 off the power│ │so the final answer is ◄═╪═══╪═ 5.0 Χ 10? ⇐ 0.50 Χ 103 │ └────────────┬────────────┘ ╘═══════════════════════════════════════╛ ╒════╧════╕ 5 Χ 10? Hints Part (iii): 3 We want to find: ⎧ 3 Χ 10₆ ⎫ ⎩ ⎭ We can DISTRIBUTE THE POWER (into the bracket) 3 ⎧ 3 Χ 10₆ ⎫ = 33 Χ (10₆)3 ⎩ ⎭ └┬┘ └──┬──┘ │ └────────────── This is just: 1018 this is 27 ────────────────┘ (i.e. multiply the powers) BUT, this ain't is Standard Index Form You need to shift the decimal until there is only 1 (non-zero) digit to the left of it (and adjust the exponent accordingly!)
b)If \(p=4.5\times 10^4\) and \(q=3\times 10^2\), find:i) \(pq\)ii) \(p\div q\)iii) \(\frac{p}{q^2}\)
Hints Part (i): We want to find: (4.5 Χ 104) Χ (3 Χ 102) => ... Χ 10? ╔═══════════╦════╗ ▼ ▼ ║ 4.5 Χ 104 Χ 3 Χ 102 add └─┬─┘ └┬┘ ╔═══╝ └─multiply─┬─┘ ║ ┌─┴──┐ ▼ 13.5 Χ 106 ╘═════╤══════╛ ┌────────────┴────────────┐ ╒═══════════════════════════════════════╕ │this isn't standard form:╪═══╪► move the decimal 1 place to the LEFT │ └────────────┬────────────┘ │ so there's just ONE (non-zero) digit │ ┌────────────┴────────────┐ │ in-front of it. Add 1 to the power │ │so the final answer is ◄═╪═══╪═ 1.35 Χ 10? ⇐ 13.5 Χ 106 │ └────────────┬────────────┘ ╘═══════════════════════════════════════╛ Hints Part (ii): We want to find: (4.5 Χ 104) ?χ(3 Χ 102) => ... Χ 10? Dividing the numbers gives: => ... Χ 10? SUBTRACTING the exponents gives: => ... Χ 102 Which is the answer! Hints Part (iii): Sorry matey; you get nothing from me!
c) Given \(u=4\times 10^6\) and \(v=8\times 10^5\), calculate the following, giving your answers in Standard Formi) \(\sqrt{u}\)ii) \(\frac{uv^2}{100}\)iii) \(\frac{u+v}{\sqrt[3]{2u}}\)
Hints Part (i): We want to find √u, which is actually the same as: u½ (although we don't need to know that to answer this) √u = √4 Χ 10⁶ = √4 Χ √10⁶ = √4 Χ √1,000,000 └─┬─┘ └─────┬─────┘ Everyone │ └──────────────── I'm pretty sure you can guess this! knows this ────┘ Don't forget to put your answer BACK into Standard Index Form Hints Part (ii): I would if I could, but I can't so I shan't... Hints Part (iii): I'm sorry, but you are all alone now. a) A litre mineral water contains \(1.25\times 10^5\) mineral particles.A litre of tap water contains \(4.5\times 10^3\) particles of tap water.If these two are mixed, how many mineral particles will be in the two litres of mixed water?
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