One lily pad, doubling in size every day, covers a pond in 30 days. How long would it take eight lily pads to...
A lily pad sits on a pond. It doubles in size every day. It takes 30
days for it to cover the pond. If you start with 8 lily pads instead,
how many days does it take to cover the pond?
I think that the answer is $27$, but I don't really think that makes sense intuitively. I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.
sequences-and-series algebra-precalculus
add a comment |
A lily pad sits on a pond. It doubles in size every day. It takes 30
days for it to cover the pond. If you start with 8 lily pads instead,
how many days does it take to cover the pond?
I think that the answer is $27$, but I don't really think that makes sense intuitively. I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.
sequences-and-series algebra-precalculus
1
Hint: Set up an algebraic equation relating to what the word problem is saying. You should see your answer then fairly easily.
– John Omielan
22 hours ago
3
It'll take $27$ days only. Think of it like this that (for the first case) on the second day no of lily pads will become $2$, on the third day $4$ and on the fourth $8$. In your second case you start your first day at this point
– Sauhard Sharma
22 hours ago
6
Just out of curiosity, where does 30/4 come from Joseph?
– Peter
18 hours ago
add a comment |
A lily pad sits on a pond. It doubles in size every day. It takes 30
days for it to cover the pond. If you start with 8 lily pads instead,
how many days does it take to cover the pond?
I think that the answer is $27$, but I don't really think that makes sense intuitively. I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.
sequences-and-series algebra-precalculus
A lily pad sits on a pond. It doubles in size every day. It takes 30
days for it to cover the pond. If you start with 8 lily pads instead,
how many days does it take to cover the pond?
I think that the answer is $27$, but I don't really think that makes sense intuitively. I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.
sequences-and-series algebra-precalculus
sequences-and-series algebra-precalculus
edited 12 mins ago
Blue
47.7k870151
47.7k870151
asked 22 hours ago
josephjoseph
46710
46710
1
Hint: Set up an algebraic equation relating to what the word problem is saying. You should see your answer then fairly easily.
– John Omielan
22 hours ago
3
It'll take $27$ days only. Think of it like this that (for the first case) on the second day no of lily pads will become $2$, on the third day $4$ and on the fourth $8$. In your second case you start your first day at this point
– Sauhard Sharma
22 hours ago
6
Just out of curiosity, where does 30/4 come from Joseph?
– Peter
18 hours ago
add a comment |
1
Hint: Set up an algebraic equation relating to what the word problem is saying. You should see your answer then fairly easily.
– John Omielan
22 hours ago
3
It'll take $27$ days only. Think of it like this that (for the first case) on the second day no of lily pads will become $2$, on the third day $4$ and on the fourth $8$. In your second case you start your first day at this point
– Sauhard Sharma
22 hours ago
6
Just out of curiosity, where does 30/4 come from Joseph?
– Peter
18 hours ago
1
1
Hint: Set up an algebraic equation relating to what the word problem is saying. You should see your answer then fairly easily.
– John Omielan
22 hours ago
Hint: Set up an algebraic equation relating to what the word problem is saying. You should see your answer then fairly easily.
– John Omielan
22 hours ago
3
3
It'll take $27$ days only. Think of it like this that (for the first case) on the second day no of lily pads will become $2$, on the third day $4$ and on the fourth $8$. In your second case you start your first day at this point
– Sauhard Sharma
22 hours ago
It'll take $27$ days only. Think of it like this that (for the first case) on the second day no of lily pads will become $2$, on the third day $4$ and on the fourth $8$. In your second case you start your first day at this point
– Sauhard Sharma
22 hours ago
6
6
Just out of curiosity, where does 30/4 come from Joseph?
– Peter
18 hours ago
Just out of curiosity, where does 30/4 come from Joseph?
– Peter
18 hours ago
add a comment |
5 Answers
5
active
oldest
votes
Starting with 8 pcs. is like 3 days have gone by. So 27 days remain.
add a comment |
Hint $#1$:
At the end of the $30$ days with one lilypad, the doubling means that the lilypad now encompasses the area of $2^{30}$ of the original lilypads.
In that light, starting with $2^3 = 8$ lilypads and each one doubling in size per day, how many doublings will it take for you to get to $2^{30}$?
(I know you've already solved this, I just think rewording/reframing the question might make it a bit easier to grasp on the intuitive level.)
Hint $#2$:
If that doesn't help ease your intuition behind your answer (which to my understanding is correct), keep in mind that starting with $8$ lilypads is basically no different than your first scenario after $3$ days. Sure, you have more lilypads, but since each doubles in size, it's no different than one lilypad of the same size as those $8$ put together then doubling. The number of lilypads differs, but we're focused on the total area encompassed.
add a comment |
Your answer is correct
If we denote the size of the lily pad by $x$, then after $1$ day, the coverage becomes $2x$ , . . . Also let's denote the size of pond by $y$. The assumptions imposes that:$$2^{30}xge y$$and $$2^{29}x<y$$Now starting with $8$ lily pads we obtain $$2^{27}cdot 8xge y$$and$$2^{26}cdot 8x< y$$which shows that the completion of the process takes long as much as $27$ days.
The intuition is that:
If we start with $1$ lily pad, after $3$ days we will have $8$ of them since the size of lily pads doubles each day. Starting with $8$ lily pads in first place and observing their growth is equivalent to observe the growth of $1$ lily pad during the days skipping the first $3$ days. If whole the process takes long $30$ days, then or modified process takes long $30-3=27$ days which is totaly intuitive.
add a comment |
$27$ is the right answer. Consider at which point the original lily pad is 8 times its original size (after three days), and how long it takes it to cover the lake from there.
Also, this relative "indifference" to a seemingly large disparity in starting point ("$x$ times more to start with means it takes $30-y$ days rather than $30/y$") is exactly what exponential growth means.
add a comment |
I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.
Lily pad doubles in size every day, so it is increasing as a geometric progression.
$$begin{array}{c|c|c|c|c|c|c|c|c}
text{n-th day}&1&2&3&4&cdots&26&27&28&29&30\
hline
text{size of $1$ lily pad}&1&2&2^2&2^3&cdots&2^{25}&2^{26}&2^{27}&2^{28}&color{red}{2^{29}} end{array}$$
If you start with $8$ lily pads, each doubling on its own, then:
$$begin{array}{c|c|c|c|c|c|c|c|c}
text{n-th day}&1&2&3&4&cdots&26&27&28&29&30\
hline
text{size of $8$ lily pads}&2^3&2^4&2^5&2^6&cdots&2^{28}&color{red}{2^{29}}&2^{30}&2^{31}&2^{32} end{array}$$
Because when each of $8$ lily pads keeps doubling per day, the $8$ lily pads increase $8$ times faster in size altogether than that of one lily pad. So, you must multiply the size of one lily pad on any day by $8=2^3$ to find the total size of $8$ lily pads.
As this source informs, the Giant Water Lily may grow as large as $8$ to $9$ feet ($2.4-2.7$m) in diameter.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067148%2fone-lily-pad-doubling-in-size-every-day-covers-a-pond-in-30-days-how-long-wou%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Starting with 8 pcs. is like 3 days have gone by. So 27 days remain.
add a comment |
Starting with 8 pcs. is like 3 days have gone by. So 27 days remain.
add a comment |
Starting with 8 pcs. is like 3 days have gone by. So 27 days remain.
Starting with 8 pcs. is like 3 days have gone by. So 27 days remain.
answered 21 hours ago
zolizoli
16.7k41845
16.7k41845
add a comment |
add a comment |
Hint $#1$:
At the end of the $30$ days with one lilypad, the doubling means that the lilypad now encompasses the area of $2^{30}$ of the original lilypads.
In that light, starting with $2^3 = 8$ lilypads and each one doubling in size per day, how many doublings will it take for you to get to $2^{30}$?
(I know you've already solved this, I just think rewording/reframing the question might make it a bit easier to grasp on the intuitive level.)
Hint $#2$:
If that doesn't help ease your intuition behind your answer (which to my understanding is correct), keep in mind that starting with $8$ lilypads is basically no different than your first scenario after $3$ days. Sure, you have more lilypads, but since each doubles in size, it's no different than one lilypad of the same size as those $8$ put together then doubling. The number of lilypads differs, but we're focused on the total area encompassed.
add a comment |
Hint $#1$:
At the end of the $30$ days with one lilypad, the doubling means that the lilypad now encompasses the area of $2^{30}$ of the original lilypads.
In that light, starting with $2^3 = 8$ lilypads and each one doubling in size per day, how many doublings will it take for you to get to $2^{30}$?
(I know you've already solved this, I just think rewording/reframing the question might make it a bit easier to grasp on the intuitive level.)
Hint $#2$:
If that doesn't help ease your intuition behind your answer (which to my understanding is correct), keep in mind that starting with $8$ lilypads is basically no different than your first scenario after $3$ days. Sure, you have more lilypads, but since each doubles in size, it's no different than one lilypad of the same size as those $8$ put together then doubling. The number of lilypads differs, but we're focused on the total area encompassed.
add a comment |
Hint $#1$:
At the end of the $30$ days with one lilypad, the doubling means that the lilypad now encompasses the area of $2^{30}$ of the original lilypads.
In that light, starting with $2^3 = 8$ lilypads and each one doubling in size per day, how many doublings will it take for you to get to $2^{30}$?
(I know you've already solved this, I just think rewording/reframing the question might make it a bit easier to grasp on the intuitive level.)
Hint $#2$:
If that doesn't help ease your intuition behind your answer (which to my understanding is correct), keep in mind that starting with $8$ lilypads is basically no different than your first scenario after $3$ days. Sure, you have more lilypads, but since each doubles in size, it's no different than one lilypad of the same size as those $8$ put together then doubling. The number of lilypads differs, but we're focused on the total area encompassed.
Hint $#1$:
At the end of the $30$ days with one lilypad, the doubling means that the lilypad now encompasses the area of $2^{30}$ of the original lilypads.
In that light, starting with $2^3 = 8$ lilypads and each one doubling in size per day, how many doublings will it take for you to get to $2^{30}$?
(I know you've already solved this, I just think rewording/reframing the question might make it a bit easier to grasp on the intuitive level.)
Hint $#2$:
If that doesn't help ease your intuition behind your answer (which to my understanding is correct), keep in mind that starting with $8$ lilypads is basically no different than your first scenario after $3$ days. Sure, you have more lilypads, but since each doubles in size, it's no different than one lilypad of the same size as those $8$ put together then doubling. The number of lilypads differs, but we're focused on the total area encompassed.
answered 22 hours ago
Eevee TrainerEevee Trainer
5,2091834
5,2091834
add a comment |
add a comment |
Your answer is correct
If we denote the size of the lily pad by $x$, then after $1$ day, the coverage becomes $2x$ , . . . Also let's denote the size of pond by $y$. The assumptions imposes that:$$2^{30}xge y$$and $$2^{29}x<y$$Now starting with $8$ lily pads we obtain $$2^{27}cdot 8xge y$$and$$2^{26}cdot 8x< y$$which shows that the completion of the process takes long as much as $27$ days.
The intuition is that:
If we start with $1$ lily pad, after $3$ days we will have $8$ of them since the size of lily pads doubles each day. Starting with $8$ lily pads in first place and observing their growth is equivalent to observe the growth of $1$ lily pad during the days skipping the first $3$ days. If whole the process takes long $30$ days, then or modified process takes long $30-3=27$ days which is totaly intuitive.
add a comment |
Your answer is correct
If we denote the size of the lily pad by $x$, then after $1$ day, the coverage becomes $2x$ , . . . Also let's denote the size of pond by $y$. The assumptions imposes that:$$2^{30}xge y$$and $$2^{29}x<y$$Now starting with $8$ lily pads we obtain $$2^{27}cdot 8xge y$$and$$2^{26}cdot 8x< y$$which shows that the completion of the process takes long as much as $27$ days.
The intuition is that:
If we start with $1$ lily pad, after $3$ days we will have $8$ of them since the size of lily pads doubles each day. Starting with $8$ lily pads in first place and observing their growth is equivalent to observe the growth of $1$ lily pad during the days skipping the first $3$ days. If whole the process takes long $30$ days, then or modified process takes long $30-3=27$ days which is totaly intuitive.
add a comment |
Your answer is correct
If we denote the size of the lily pad by $x$, then after $1$ day, the coverage becomes $2x$ , . . . Also let's denote the size of pond by $y$. The assumptions imposes that:$$2^{30}xge y$$and $$2^{29}x<y$$Now starting with $8$ lily pads we obtain $$2^{27}cdot 8xge y$$and$$2^{26}cdot 8x< y$$which shows that the completion of the process takes long as much as $27$ days.
The intuition is that:
If we start with $1$ lily pad, after $3$ days we will have $8$ of them since the size of lily pads doubles each day. Starting with $8$ lily pads in first place and observing their growth is equivalent to observe the growth of $1$ lily pad during the days skipping the first $3$ days. If whole the process takes long $30$ days, then or modified process takes long $30-3=27$ days which is totaly intuitive.
Your answer is correct
If we denote the size of the lily pad by $x$, then after $1$ day, the coverage becomes $2x$ , . . . Also let's denote the size of pond by $y$. The assumptions imposes that:$$2^{30}xge y$$and $$2^{29}x<y$$Now starting with $8$ lily pads we obtain $$2^{27}cdot 8xge y$$and$$2^{26}cdot 8x< y$$which shows that the completion of the process takes long as much as $27$ days.
The intuition is that:
If we start with $1$ lily pad, after $3$ days we will have $8$ of them since the size of lily pads doubles each day. Starting with $8$ lily pads in first place and observing their growth is equivalent to observe the growth of $1$ lily pad during the days skipping the first $3$ days. If whole the process takes long $30$ days, then or modified process takes long $30-3=27$ days which is totaly intuitive.
edited 21 hours ago
answered 21 hours ago
Mostafa AyazMostafa Ayaz
14.5k3937
14.5k3937
add a comment |
add a comment |
$27$ is the right answer. Consider at which point the original lily pad is 8 times its original size (after three days), and how long it takes it to cover the lake from there.
Also, this relative "indifference" to a seemingly large disparity in starting point ("$x$ times more to start with means it takes $30-y$ days rather than $30/y$") is exactly what exponential growth means.
add a comment |
$27$ is the right answer. Consider at which point the original lily pad is 8 times its original size (after three days), and how long it takes it to cover the lake from there.
Also, this relative "indifference" to a seemingly large disparity in starting point ("$x$ times more to start with means it takes $30-y$ days rather than $30/y$") is exactly what exponential growth means.
add a comment |
$27$ is the right answer. Consider at which point the original lily pad is 8 times its original size (after three days), and how long it takes it to cover the lake from there.
Also, this relative "indifference" to a seemingly large disparity in starting point ("$x$ times more to start with means it takes $30-y$ days rather than $30/y$") is exactly what exponential growth means.
$27$ is the right answer. Consider at which point the original lily pad is 8 times its original size (after three days), and how long it takes it to cover the lake from there.
Also, this relative "indifference" to a seemingly large disparity in starting point ("$x$ times more to start with means it takes $30-y$ days rather than $30/y$") is exactly what exponential growth means.
edited 21 hours ago
answered 22 hours ago
ArthurArthur
111k7107189
111k7107189
add a comment |
add a comment |
I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.
Lily pad doubles in size every day, so it is increasing as a geometric progression.
$$begin{array}{c|c|c|c|c|c|c|c|c}
text{n-th day}&1&2&3&4&cdots&26&27&28&29&30\
hline
text{size of $1$ lily pad}&1&2&2^2&2^3&cdots&2^{25}&2^{26}&2^{27}&2^{28}&color{red}{2^{29}} end{array}$$
If you start with $8$ lily pads, each doubling on its own, then:
$$begin{array}{c|c|c|c|c|c|c|c|c}
text{n-th day}&1&2&3&4&cdots&26&27&28&29&30\
hline
text{size of $8$ lily pads}&2^3&2^4&2^5&2^6&cdots&2^{28}&color{red}{2^{29}}&2^{30}&2^{31}&2^{32} end{array}$$
Because when each of $8$ lily pads keeps doubling per day, the $8$ lily pads increase $8$ times faster in size altogether than that of one lily pad. So, you must multiply the size of one lily pad on any day by $8=2^3$ to find the total size of $8$ lily pads.
As this source informs, the Giant Water Lily may grow as large as $8$ to $9$ feet ($2.4-2.7$m) in diameter.
add a comment |
I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.
Lily pad doubles in size every day, so it is increasing as a geometric progression.
$$begin{array}{c|c|c|c|c|c|c|c|c}
text{n-th day}&1&2&3&4&cdots&26&27&28&29&30\
hline
text{size of $1$ lily pad}&1&2&2^2&2^3&cdots&2^{25}&2^{26}&2^{27}&2^{28}&color{red}{2^{29}} end{array}$$
If you start with $8$ lily pads, each doubling on its own, then:
$$begin{array}{c|c|c|c|c|c|c|c|c}
text{n-th day}&1&2&3&4&cdots&26&27&28&29&30\
hline
text{size of $8$ lily pads}&2^3&2^4&2^5&2^6&cdots&2^{28}&color{red}{2^{29}}&2^{30}&2^{31}&2^{32} end{array}$$
Because when each of $8$ lily pads keeps doubling per day, the $8$ lily pads increase $8$ times faster in size altogether than that of one lily pad. So, you must multiply the size of one lily pad on any day by $8=2^3$ to find the total size of $8$ lily pads.
As this source informs, the Giant Water Lily may grow as large as $8$ to $9$ feet ($2.4-2.7$m) in diameter.
add a comment |
I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.
Lily pad doubles in size every day, so it is increasing as a geometric progression.
$$begin{array}{c|c|c|c|c|c|c|c|c}
text{n-th day}&1&2&3&4&cdots&26&27&28&29&30\
hline
text{size of $1$ lily pad}&1&2&2^2&2^3&cdots&2^{25}&2^{26}&2^{27}&2^{28}&color{red}{2^{29}} end{array}$$
If you start with $8$ lily pads, each doubling on its own, then:
$$begin{array}{c|c|c|c|c|c|c|c|c}
text{n-th day}&1&2&3&4&cdots&26&27&28&29&30\
hline
text{size of $8$ lily pads}&2^3&2^4&2^5&2^6&cdots&2^{28}&color{red}{2^{29}}&2^{30}&2^{31}&2^{32} end{array}$$
Because when each of $8$ lily pads keeps doubling per day, the $8$ lily pads increase $8$ times faster in size altogether than that of one lily pad. So, you must multiply the size of one lily pad on any day by $8=2^3$ to find the total size of $8$ lily pads.
As this source informs, the Giant Water Lily may grow as large as $8$ to $9$ feet ($2.4-2.7$m) in diameter.
I think that, intuitively, the answer should be less than $30/4$ since it is increasing at an exponential rate.
Lily pad doubles in size every day, so it is increasing as a geometric progression.
$$begin{array}{c|c|c|c|c|c|c|c|c}
text{n-th day}&1&2&3&4&cdots&26&27&28&29&30\
hline
text{size of $1$ lily pad}&1&2&2^2&2^3&cdots&2^{25}&2^{26}&2^{27}&2^{28}&color{red}{2^{29}} end{array}$$
If you start with $8$ lily pads, each doubling on its own, then:
$$begin{array}{c|c|c|c|c|c|c|c|c}
text{n-th day}&1&2&3&4&cdots&26&27&28&29&30\
hline
text{size of $8$ lily pads}&2^3&2^4&2^5&2^6&cdots&2^{28}&color{red}{2^{29}}&2^{30}&2^{31}&2^{32} end{array}$$
Because when each of $8$ lily pads keeps doubling per day, the $8$ lily pads increase $8$ times faster in size altogether than that of one lily pad. So, you must multiply the size of one lily pad on any day by $8=2^3$ to find the total size of $8$ lily pads.
As this source informs, the Giant Water Lily may grow as large as $8$ to $9$ feet ($2.4-2.7$m) in diameter.
answered 19 hours ago
farruhotafarruhota
19.5k2737
19.5k2737
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067148%2fone-lily-pad-doubling-in-size-every-day-covers-a-pond-in-30-days-how-long-wou%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Hint: Set up an algebraic equation relating to what the word problem is saying. You should see your answer then fairly easily.
– John Omielan
22 hours ago
3
It'll take $27$ days only. Think of it like this that (for the first case) on the second day no of lily pads will become $2$, on the third day $4$ and on the fourth $8$. In your second case you start your first day at this point
– Sauhard Sharma
22 hours ago
6
Just out of curiosity, where does 30/4 come from Joseph?
– Peter
18 hours ago