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Practice each question, then open the answer or worked solution.
Find three repeater machines that will do the same work as a (×64) machine. Draw them, or describe them using exponents.
Ans: We know that, the possible factors of 64 are 2, 4, 8. :
If 26 =64, 43 =64 and 82 =64
Hence, three repeater machines that would work as a (x 64) will be (x 26), (x 43) and (x 82). The diagram of (x 26), (x 43) and (x 82) is given below.
What will the following machine do to a 2 cm long piece of chalk?
Ans: The machine produce x 1100 = 1
So, if we insert a 2 cm long piece of chalk in that machine, the piece of chalk remains the same.
In a repeater machine with 0 as an exponent, the base machine is applied 0 times.
(a) What do these machines do to a piece of chalk?
(b) What do you think the value of 60 is?
Ans: Use the law of exponent a0 = 1.
Since, 30 = 1, 130 = 1, 290 = 1
So, machine (x 30), (x 130) and (x 290) produce nothing on not changing the piece 7 chalk.
Ans: Use law of exponent a0 = 1,
Similarly, (x 60) machine does not change the piece.
Shrinking Machine
In a shrinking machine, a piece of stick is compressed to reduce its length. If 9 cm long sandwich is put into the shrinking machine below, how many cm long will it be when it emerges?
Ans: According to the question, in a shrinking machine, a piece of stick is compressed to reduce its length. If a 9 cm long sandwich is put into the shrinking machine, then the length of the sandwich will be $9 \times \dfrac{1}{{{3^{ - 1}}}}$= 9 x 3 = 27 cm.
What happens when 1 cm worms are sent through these hook-ups?
(i)
(ii)
Ans: If you put 1 cm worms through ( x21) and (x2-1)
machine, the result will be = 1 x 2 x $\dfrac{1}{2}$ = 1cm.
Ans: If we send 1 cm worms through hooked machines (x 2-1) and (x 2)-2, result will be-
$1 \times \dfrac{1}{2} \times \dfrac{1}{{2 \times 2}} = \dfrac{1}{{2 \times 4}} = \dfrac{1}{8}{\text{ cm = 0}}{\text{.125 cm}}$
Sanchay put a 1cm stick of gum through a (1 × 3–2) machine. How long was the stick when it came out?
Ans: Length of the stick of gum that Sanchay put through the (1 x 3-2) machine = 1 cm
Negation (-) sign in power states that the given machine is a
shrinking machine.
Hence $1 \times \dfrac{1}{{{3^2}}} = 1 \times \dfrac{1}{9} = \dfrac{1}{9}{\text{ cm}}$
Therefore, the length of the stick when it came out is $\dfrac{1}{9}{\text{ cm}}$.
Ajay had a 1cm piece of gum. He put it through the repeater machine given below and it came out $\dfrac{1}{{{\mathbf{100}},{\mathbf{000}}}}$cm long. What is the missing value?
Ans: The 1 cm piece of gum that Ajay put came out as $\dfrac{1}{{100000}}$.
From this we can conclude that the machine is a shrinking
machine.
Therefore, the type of shrinking machine is ${\left( { \times \dfrac{{{1}}}{{10}}} \right)^5}$.
Hence, the value missing is 5.
Find a single machine that will do the same job as the given hook-up.
(a) a (× 23) machine followed by (× 2–2) machine.
(b) a (× 24) machine followed by $\left( { \times {{\left( {\dfrac{1}{2}} \right)}^2}}\right)$ machine.
(c) a (× 599) machine followed by a (5–100) machine.
Ans: If (x23) machine is followed by a (x2-2) machine,
the result will be- $2 \times 2 \times 2 \times \dfrac{1}{2} \times \dfrac{1}{2} = {2^1}$
Thus, a single machine (x 21) will also work as the given hook-up.
Ans: A (x24) machine is followed by a $\left[ {{{\left( {\dfrac{1}{2}} \right)}^2}} \right]$ machine.
Therefore, the result will be $2 \times 2 \times 2 \times 2 \times \dfrac{1}{2} \times \dfrac{1}{2} = 4$
Ans: A (×599) machine is followed by a (x5–100) machine.
Therefore, the result will be ${5^{99}} \times \dfrac{1}{{{5^{100}}}} = \dfrac{1}{5}$
Thus, a $\left( { \times \dfrac{1}{5}} \right)$ machine will work as the given hook-up.
Find a single repeater machine that will do the same work as each hook-up.
(a)
(b)
(c)
(d)
Ans: The work that can be performed by the repeater machine is
equal to 22 x 23 x 24 = 29.
Hence, (x 29) single machine can perform the same amount of work.
Ans: The work that can be performed by the repeater machine is equal to 1002 x 10010.
Hence, (x 10012) single machine can perform the same amount of work.
Ans: The work that can be performed by the repeater machine is equal to 710 x 750 x 71 = 761
Hence, (x761) single machine can perform the same amount of work.
Ans: The work that can be performed by the repeater machine is equal to 3y x 3y = 32y
Hence, (32y) single machine can perform the same amount of work.
(e)
Ans: The work that can be performed by the repeater machine is equal to ${2^2} \times {\left( {\dfrac{1}{2}} \right)^3} \times {2^4} = {2^6} \times \dfrac{1}{{{2^3}}} = {2^3}$
Hence, (x23) single machines can perform the same amount of work.
(f)
Ans: The work that can be performed by the repeater machine is equal to ${\left( {\dfrac{1}{2}} \right)^2} \times {\left( {\dfrac{1}{3}} \right)^2} = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{{2 \times 3 \times 2 \times 3}} = \dfrac{1}{{{6^2}}}$
Hence, $\left[ {{{\left( { \times \dfrac{1}{6}} \right)}^2}} \right]$ single machine can perform the same amount of work.
For each hook-up, determine whether there is a single repeater machine that will do the same work. If so, describe or draw it.
(a)
(b)
(c)
(d)
Ans: The work that can be performed by the hook-up machine = 73 × 72 = 75
So, (x 75) single machine can perform the same amount of work.
The diagram of a (x 75) single machine can be shown as below:
Ans: The work that can be performed by the hook-up machine = 23 × 33 = 8 × 9 = 72
Hence, it is not possible that a single machine to perform the same amount of work as (x 82) = 64 and (x 92) = 81.
Ans: The work that can be performed by the hook-up machine = ${2^2} \times {\left( {\dfrac{1}{3}} \right)^3} \times {5^4} = 4 \times \dfrac{1}{{27}} \times 625$ = $\dfrac{{2500}}{{27}} = 92.59$
Hence, it is not possible that a single machine to perform the same amount of work.
Ans: The work that can be performed by the hook-up machine = (0.5)2 x (0.5)3 = (0.5)5
So, (x (0.5)5) a single machine can perform the same amount of work.
The diagram of a (x 0.5)5 single machine can be shown as below:
(e)
Ans: The work that can be performed by the hook-up machine = 122 x 123 = 125
So, (x 125) single machine can perform the same amount of work.
The diagram of a (x 125) single machine can be shown as below:
