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# A clock store sold a certain clock to a collector for 20 per

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A clock store sold a certain clock to a collector for 20 per  [#permalink]

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Updated on: 06 May 2014, 09:27
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A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270
B. $250 C.$240
D. $220 E.$200

Originally posted by hsk on 11 Jun 2007, 21:02.
Last edited by Bunuel on 06 May 2014, 09:27, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: A clock store sold a certain clock to a collector for 20 per  [#permalink]

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06 May 2014, 20:22
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hsk wrote:
A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270
B. $250 C.$240
D. $220 E.$200

Assume numbers and then use ratios to fit them in with the actual numbers.

Say the store bought the clock for $100 and sold it to the collector for$120. The store bought back from the collector for $60 and sold it back at$60*18/10 = $108 Here, difference between original cost ($100) and buy back price ($60) is$40. But actually it is given to be $100 which is 2.5 times 40. So the shop sold the clock a second time for$108*2.5 = $270 _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >

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11 Jun 2007, 21:42
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$270 Original Price = x Sold to collector at 1.2x re-buy price = 1.2x * 0.5 = 0.6x therefore, x - 0.6x = 100; x = 250 second selling price = 1.8 * 0.6 * 250 = 270 Hope, this is not too confusing ##### General Discussion Senior Manager Joined: 04 Mar 2007 Posts: 419 ### Show Tags 12 Jun 2007, 10:52 1 100 % - original price 120 % - sell 1 60 % - bought 2 180*60/100 = 108 % - sell 2 100$ - 40%
x - 108 %
x= 100*108/40 = 270

A
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Re: A clock store sold a certain clock to a collector for 20 per  [#permalink]

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02 Jun 2015, 00:44
7
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution

Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$

Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$

Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$

Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$

Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$

Now, we are given that difference between clock's original price and clock's buy back price = $$100$$

$$x - 0.6x = 100$$ i.e. $$x = 250$$

We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$

Hope this helps

Regards
Harsh
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Re: A clock store sold a certain clock to a collector for 20 per  [#permalink]

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11 Jul 2015, 13:39
EgmatQuantExpert wrote:
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution

Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$

Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$

Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$

Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$

Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$

Now, we are given that difference between clock's original price and clock's buy back price = $$100$$

$$x - 0.6x = 100$$ i.e. $$x = 250$$

We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$

Hope this helps

Regards
Harsh

Dear,

I have a basic question.

The author states: "The shop then sold the clock again at a profit of 80 percent on its buy-back price". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8.

Where my confuse is?

Kind regards...
Gonzalo
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Re: A clock store sold a certain clock to a collector for 20 per  [#permalink]

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11 Jul 2015, 13:54
1
1
gbascurs wrote:
EgmatQuantExpert wrote:
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution

Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$

Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$

Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$

Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$

Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$

Now, we are given that difference between clock's original price and clock's buy back price = $$100$$

$$x - 0.6x = 100$$ i.e. $$x = 250$$

We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$

Hope this helps

Regards
Harsh

Dear,

I have a basic question.

The author states: "The shop then sold the clock again at a profit of 80 percent on its buy-back price". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8.

Where my confuse is?

Kind regards...
Gonzalo

Gonzalo, Profit is calculated = Selling price - Cost price. Profit % is always calculated on the cost price of the purchase and not on the selling price. You are calculating profit % on the selling price. This is where you are going wrong. I am assuming that in your example, you are buying something at 80 while you are selling it at 100, giving you an absolute profit of 20$while your profit % will be 20/80 = 25% and not 20/100 = 20%. Now, in the question above, lets say the original cost of the clock to store was C$ and then it sold the same to the collector at 20% profit.
This means the clocks' selling price was C (1.2) and this becomes cost price for the collector.
Now, when the collector tries to sell the same clock to the store, the store buys it for 50% the price at which the collector bought it.

Thus, you get = 1.2*0.5*C = 0.6 C

Furthermore, the store sells the clock for the second time for 80% profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy-back * 1.8 = 1.8 * 0.6 C

Finally given that C - 0.6 C = 100 ----> C = 250$Thus, the cost of the clock the second time around = 1.8*0.6 C = 1.8 * 0.6 * 250 = 270$. Hence A is the correct answer.
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Joined: 25 Dec 2012
Posts: 17
Re: A clock store sold a certain clock to a collector for 20 per  [#permalink]

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11 Jul 2015, 14:20
Engr2012 wrote:
gbascurs wrote:
EgmatQuantExpert wrote:
The question is a bit wordy and gives us a series of percentages on which we need to work on. Going step wise is the key to not miss out on any of the information. Presenting the detailed step wise solution

Step-I- Store buys the clock
Let's assume the original price of clock paid by the store to be $$x$$

Step-II- Collectors buys the clock from the store
Extra amount paid by collector to buy the clock = $$20$$% of $$x$$

Therefore price at which collector buys the clock = $$x + 20$$% of $$x = 1.2x$$

Step-III- Store buys back the clock from collector
Price at which the store buys the clock = $$50$$% of price collector paid $$= 50$$% of $$1.2x = 0.6x$$

Step-IV- Store resells the clock
Price at which store resells the clock = $$0.6x + 80$$% of $$0.6x = 0.6x * 1.8$$

Now, we are given that difference between clock's original price and clock's buy back price = $$100$$

$$x - 0.6x = 100$$ i.e. $$x = 250$$

We are asked to find the price at which the store resells the clock = $$0.6x * 1.8 = 0.6 * 250 * 1.8 = 270$$

Hope this helps

Regards
Harsh

Dear,

I have a basic question.

The author states: "The shop then sold the clock again at a profit of 80 percent on its buy-back price". Why is it suppose to multiply 0.6x*1.8? I think my confuse is related to the profit concept. For me, if I buy something at 100, and the cost of it is 80. I have a profit of 20/100 = 20%. Using the same logic on the question, if I buy something at 100 and I get a profit of 80%, the sales price must be 100/0.2. Thus, it is 500. Profit is 400/500=80%. It is a totally different aproach than to say 100*1,8.

Where my confuse is?

Kind regards...
Gonzalo

Gonzalo, Profit is calculated = Selling price - Cost price. Profit % is always calculated on the cost price of the purchase and not on the selling price. You are calculating profit % on the selling price. This is where you are going wrong. I am assuming that in your example, you are buying something at 80 while you are selling it at 100, giving you an absolute profit of 20$while your profit % will be 20/80 = 25% and not 20/100 = 20%. Now, in the question above, lets say the original cost of the clock to store was C$ and then it sold the same to the collector at 20% profit.
This means the clocks' selling price was C (1.2) and this becomes cost price for the collector.
Now, when the collector tries to sell the same clock to the store, the store buys it for 50% the price at which the collector bought it.

Thus, you get = 1.2*0.5*C = 0.6 C

Furthermore, the store sells the clock for the second time for 80% profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy-back * 1.8 = 1.8 * 0.6 C

Finally given that C - 0.6 C = 100 ----> C = 250$Thus, the cost of the clock the second time around = 1.8*0.6 C = 1.8 * 0.6 * 250 = 270$. Hence A is the correct answer.

Thank you a lot.

Here is the key of my misunderstanding: "Profit % is always calculated on the cost price of the purchase". I assumed it was on the sales price.
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A clock store sold a certain clock to a collector for 20 per  [#permalink]

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10 Nov 2015, 05:00
1
x=starting price
therefore
x +20% = $$x +1/5x$$ = $$6x/5$$
50% = 3x/5 BB Price
as per Q
x - $$3x/5$$=100
x=250
BB =150
80 %increment of 150 = 270

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A clock store sold a certain clock to a collector for 20 per  [#permalink]

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10 Nov 2015, 05:20
1
hsk wrote:
A clock store sold a certain clock to a collector for 20 percent more than the store had originally paid for the clock. When the collector tried to resell the clock to the store, the store bought it back at 50 percent of what the collector had paid. The shop then sold the clock again at a profit of 80 percent on its buy-back price. If the difference between the clock's original cost to the shop and the clock's buy-back price was $100, for how much did the shop sell the clock the second time? A.$270
B. $250 C.$240
D. $220 E.$200

Let us assume the initial price (CP) of the clock to be 100x
We assume 100x to avoid the usage of decimals in case of x and avoid the usage of unitary method in case of 100

Transaction 1:
Store sold the clock to collector.
Selling Price = 20% more than CP = 120x

Transaction 2:
The collector sold it back to the store.
The new CP for the store = 60x (Store bought it back for 50% of the Selling Price)

Transaction 3:
Store sold it back at a profit of 80% on 60x = 108x

Now we are told that 100x - 60x = 100
This means x = 100/40

Hence the second selling price = 108*100/40 = 270
Option A
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# In the xy-plane, the straight-line graphs of the three equations above

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In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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23 Jul 2015, 10:47
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y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?

(1) a = 2
(2) r = 17
[Reveal] Spoiler: OA

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Last edited by Bunuel on 13 Nov 2017, 21:24, edited 4 times in total.
Edited the question.

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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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23 Jul 2015, 11:30
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mcelroytutoring wrote:
DS 136 from OFG 2016 (new question)

y = ax - 5
y = x + 6
y = 3x + b

In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?

1) a = 2

2) r = 17

Solution provided by : mcelroytutoring

Let's start by substituting the point (p,r) into all equations in place of (x,y) which will make step #2 a bit easier to comprehend but is not necessary to solve the question. Then, let's consider the number of variables left in each equation.

#1: r = ap - 5 (3 variables R,A,P)
#2: r = p + 6 (2 variables R,P)
#3: r = 3p + b (3 variables R,B,P)

1) a = 2 allows us to reduce equation #1 to the variables r and p, which are the same two variables as equation #2. Thus we have simultaneous equations. As soon as we verify that the equations are different, we know that we can solve for both variables. Once we know r and p, we can substitute in equation #3 to solve for b. Sufficient.

2) r = 17 allows us to do the same thing, more or less. It reduces equation #2 to only one variable, allowing us to solve for p. Once we have p (and r), we can use equation #3 to solve for b. Sufficient.
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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24 Sep 2015, 01:53
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if each of the 3 equations contains points (p,r) this means that they intersect in that point
1. a=2
Find the intercept Intercept for three simultaneous equations
y=2x-5
y=x+6
y=3x+b
Let's use the first 2 equations: plug y=x+6 in the secod equation
x+6=2x-5 -> x=11, y=17 we can use the values to calulate b in the 3rd equation
17=33+b -> b=-16 SUFFICIENT

2. Here we have directly the value for Y, let's plug it in the 2nd equation
y=x+6 -> 17=x+6 -> x=11, y=17; We can plug these values in the 3rd equation and find b as we did above
17=33+b -> b=-16 SUFFICIENT

Answer (D) Most important point is here to catch the hint about intersection of 3 lines at one point
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In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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26 Oct 2015, 10:02
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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26 Oct 2015, 10:50
2
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I think it's D.

Keeping point (p,r) in all the equations we get :

p = ar -5 -----(1)
p = r + 6 ------(2)
p = 3r + b ------(3)

Now consider (1) if a = 2 from (1) and (2) we get
r = 11 , p=17 and putting in (3) we can get b.

Similarly for (2) we can get the values for r and p and hence can get the value for b.

So both statements individually are correct to answer the question.

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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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26 Oct 2015, 19:14
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m2k wrote:
I think it's D.

Keeping point (p,r) in all the equations we get :

p = ar -5 -----(1)
p = r + 6 ------(2)
p = 3r + b ------(3)

Now consider (1) if a = 2 from (1) and (2) we get
r = 11 , p=17 and putting in (3) we can get b.

Similarly for (2) we can get the values for r and p and hence can get the value for b.

So both statements individually are correct to answer the question.

Approach if right but the values you derived are wrong. According to me r = 17 and that is what stmt b also states.

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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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10 Jun 2016, 11:23
2
KUDOS
y = ax - 5 ... eq 1
y = x + 6 ... eq 2
y = 3x + b ...eq 3

Total of 4 variables are present.

Statement 1 : a = 2

Insert in eq 1

We have y = 2x -5 and y = x+6

Solving we get x = 11 and y = 17

Substitute in eq 3 and we get value of b

Statement 2: r=17

This means the y co- ordinate is 17
Substitute in eq 2 we get x as 11

Again can find value of b from equation 3

Hence D
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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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18 Jan 2017, 15:18
we used information from both 1 and 2 then how can the answer be D... should it not be C... some one kindly clarify.......... clearly am a zero in ds and that too a big one

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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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18 Jan 2017, 20:18
1
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y = ax - 5---------1
y = x + 6---------2
y = 3x + b--------3
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?

(1) a = 2
(2) r = 17

All three line intersect each other at common point (p,r).
1. given a = 2
putting in equation 1 .= y=2x-5
equating 1(after replacing value of a) and 2 we will get value of (p,r)
putting (p,r) in equation 3 we will get value for b---suff..

2 given r = 17.
putting in equation 2 we will get value of x.i'e p.
Now as we know common point of intersection ,putting p,r in equation 3 , we will get value of b

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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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22 Jan 2017, 15:51
thanks sobby for your response... highly appreciated...

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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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16 Mar 2017, 08:58
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(eq1) $$y = ax - 5$$
(eq2) $$y = x + 6$$
(eq3) $$y = 3x + b$$

In the xy-plane, the straight-line graphs of the three equations above each contain the point (p, r). If a and b are constants, what is the value of b?

1) $$a = 2$$
2) $$r = 17$$

Solution:

1) $$a = 2$$
- Putting the value of a in eq1, we get: $$y = 2x - 5$$
- At this point you can solve for (x, y), plug (x, y) in (eq3) and solve for (b) [though this approach might take few seconds]
(alternatively, faster method)
- you can skip solving for (x, y) and deduce that given 3 equations and 3 unknowns (since a is given in statement 1) we can solve for all of them (including b), since the lines are have different slopes i.e. different lines. Hence, we can get single value of 'b', proving the condition SUFFICIENT.
NOTE: 3 equations and 3 unknowns does not ALWAYS mean that we can find 3 unknown. We have to make sure that 2 of them or all of them are not the same line.

2) $$r = 17$$
- Since, point (p, r) lie on all the line, we can plugin the point in above equation
$$r = ap - 5 => 17 = ap - 5$$
$$r = p + 6 => 17 = p + 6$$
$$r = 3p + b => 17 = 3p + b$$
- Again, we do not need to solve for all the variables and just recognize that the above equations will lead to single value of b. Hence, SUFFICIENT.

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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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30 Apr 2017, 23:51
y = ax - 5
y = x + 6
y = 3x + b

In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?

1) a = 2

2) r = 17

My 2 cents.

It is important to realize from the Question stem that the 3 equations intersect as (p,r).

For 1), as we know a =2, we can equate the first and second equation to get the value of x, and then use that value of x to find value of y and the find value of b.

For 2), similarly, use r = 17 (which is value of y) to find value of x using the second equation. And then plug it back to the third equation.

So D.

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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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10 Sep 2017, 19:04
Would it be correct to simply say that we have 4 variables with 3 equations so eliminating any one variable gets us to three equations and three variables and is therefore sufficient? Is that logic sound?

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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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13 Nov 2017, 20:21
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",

i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables

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Re: In the xy-plane, the straight-line graphs of the three equations above [#permalink]

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13 Nov 2017, 22:31
Cheryn wrote:
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",

i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables

The highlighted statement in effect says that all these 3 lines meet each other at one point and so there is a single value of (x,y) that satisfies these 3 equations. It is only because of this highlighted statement you can solve this set of equations for a unique value of x,y, a and b.

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Re: In the xy-plane, the straight-line graphs of the three equations above   [#permalink] 13 Nov 2017, 22:31
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If r and s are positive integers, can the fraction r/s be expressed as

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If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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20 Oct 2012, 08:21
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If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?

(1) s is a factor of 100
(2) r is a factor of 100
[Reveal] Spoiler: OA
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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20 Oct 2012, 10:02
IMO it is A, if the denominator is a factor of 100 then it could be 1; 2; 5; 10; 20.. if you divide all the positive integer by these number you will have a finite decimal result.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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Finite decimals are decimals which end. eg 0.5,0.25, etc. Non finite are numbers like 1/3,1/6 etc ie 0.33333333333333333333333333333333333333333333333333333.......... or 0.66666666666666666666666666666666666666666666666666666..........

1)S is a factor of 100. So S cannot have more than two 2s and two 5s. Any number divisible be 2 or 5 gives a finite decimal. Since R and S are both positive integers, there can be no 0s in the decimal places either. So Sufficient

2)R is a factor of 100. Cant say anything form this. R can be 1. 1/10 is finite. 1/3 is not. Insufficient.

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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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23 Oct 2012, 07:59
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asveaass wrote:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?

1) s is a factor of 100
2) r is a factor of 100

I don't understand the answer explanation in the OG, could someone please explain in detail?

THEORY:

Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)

Questions testing this concept:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html

BACK TO THE ORIGINAL QUESTION:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?

(1) s is a factor of 100. Factors of 100 are: 1, 2, 4, 5, 10, 20, 25, 50 and 100. All these numbers are of the form $$2^n5^m$$ (for example 1=2^0*5^0, 2=2^1*5^0, ...), therefore no matter what is the value of r, r/s will always will be terminating decimal. Sufficient.

(2) r is a factor of 100. We need to know about the denominator. Not sufficient.

Hope it's clear.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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02 Jan 2013, 01:25
you rock bunuel, your explanations are awesome.. thanks a lot for this one!
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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24 Dec 2013, 11:12
But it has been said the 'decimal with finite number of non-zero digits'. Now if we take 2/50 then it will be 0.04, which means its a finite decimal but definitely it does not have all the non-zero digits after decimal point. So, can anybody explain?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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04 May 2014, 11:31
Hi Bunuel,

Speaking from a prime perspective, I could factor "100" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?

If the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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05 May 2014, 02:31
halloanupam wrote:
But it has been said the 'decimal with finite number of non-zero digits'. Now if we take 2/50 then it will be 0.04, which means its a finite decimal but definitely it does not have all the non-zero digits after decimal point. So, can anybody explain?

0.04 has finite number of non-zero digits: 4 is not followed by any non-zero digit.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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05 May 2014, 02:35
russ9 wrote:
Hi Bunuel,

Speaking from a prime perspective, I could factor "100" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?

If the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?

Not entirely. The denominator can have some other primes as well but if those primes can be reduced the fraction still would be terminating. For example, consider fraction 3/6. The denominator has 3 in it, but it ca be reduced to get 3/6=1/2=0.5.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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26 Sep 2014, 23:28
Bunuel wrote:
asveaass wrote:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?

1) s is a factor of 100
2) r is a factor of 100

I don't understand the answer explanation in the OG, could someone please explain in detail?

THEORY:

Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)

Why is it then 130/13 or 121/11 would give finite ... infact they properly divide...
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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27 Sep 2014, 01:28
ani781 wrote:
Bunuel wrote:
asveaass wrote:
If r and s are positive integers, can the fraction r/s be expressed as a decimal with only a finite number of nonzero digits?

1) s is a factor of 100
2) r is a factor of 100

I don't understand the answer explanation in the OG, could someone please explain in detail?

THEORY:

Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction $$\frac{6}{15}$$ has 3 as prime in denominator and we need to know if it can be reduced.)

Why is it then 130/13 or 121/11 would give finite ... infact they properly divide...

The rule above is for reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term). When you reduce 130/13 to the lowest term you get 10 and when you reduce 121/11 you get 11: 10/(2^0*5^0) and 11/(2^0*5^0) respectively.

Check the links in my post above to practice more on this type of questions.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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27 Sep 2014, 02:08
A, of course. I had got it as soon as I saw it. This is a very simple problem probably fetched from RS
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If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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Bunuel wrote:

THEORY:

Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example $$\frac{x}{2^n5^m}$$, (where x, n and m are integers) will always be terminating decimal.

Hi Bunuel,

Just looking for some clarification, in the highlighted text above, please correct me if I am wrong, by including the word "and" did you mean that if the denominator contains only 2's or only 5's then too the fraction is a terminating decimal? In other words, the denominator doesn't need to contain, both - 2's & 5's!

Thank you!
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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17 Nov 2014, 23:36
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Yes, examples such as $$\frac{17}{2^7}$$, $$\frac{31}{5^{18}}$$, and $$\frac{21}{[(2^5)(5^7)]}$$ are all terminating decimals.

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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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18 Nov 2014, 12:44
dabral wrote:
dmmk

Yes, examples such as $$\frac{17}{2^7}$$, $$\frac{31}{5^{18}}$$, and $$\frac{21}{[(2^5)(5^7)]}$$ are all terminating decimals.

Dabral

Thank you dabral
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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03 Dec 2014, 11:12
Bunuel wrote:
russ9 wrote:
Hi Bunuel,

Speaking from a prime perspective, I could factor "100" into prime factors and since I get 2^2 and 5^2, that should suffice in coming up with the right answer, correct? I don't need to pull up every single factors - right?

If the denominator had ANY other primes then it would NOT be a terminating decimal. Is that correct?

Not entirely. The denominator can have some other primes as well but if those primes can be reduced the fraction still would be terminating. For example, consider fraction 3/6. The denominator has 3 in it, but it ca be reduced to get 3/6=1/2=0.5.

Bunuel I did not understand from your post whether we can have other primes in denominator? Can you pls. repeat?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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Ergenekon

For a fraction to be a terminating decimal the only primes that can be present in the denominator are 2 and 5, and this only applies to reduced form of the fraction. If there is any other prime in the denominator, then the fraction will be non-terminating. For example, $$\frac{21}{[(2^4)(5^3)(11^2)]}$$, $$\frac{11}{[(5^6)(7^3)]}$$, and $$\frac{22}{(7^4)}$$ are all non-terminating decimals.

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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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04 Dec 2014, 00:35
dabral wrote:
Ergenekon

For a fraction to be a terminating decimal the only primes that can be present in the denominator are 2 and 5, and this only applies to reduced form of the fraction. If there is any other prime in the denominator, then the fraction will be non-terminating. For example, $$\frac{21}{[(2^4)(5^3)(11^2)]}$$, $$\frac{11}{[(5^6)(7^3)]}$$, and $$\frac{22}{(7^4)}$$ are all non-terminating decimals.

Dabral

Thanks Dabral. Got it now.
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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Hello, what if r is the same as s? Then we don't have a finite decimal but an integer. Then A should not be sufficient as the answer can be a decimal or an integer. Am I wrong in my thinking?
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Re: If r and s are positive integers, can the fraction r/s be expressed as [#permalink]

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Hello, what if r is the same as s? Then we don't have a finite decimal but an integer. Then A should not be sufficient as the answer can be a decimal or an integer. Am I wrong in my thinking?

An integer IS a decimal with a finite number of nonzero digits. For example, integer 51 has 2 (finite) number of digits.
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Re: If r and s are positive integers, can the fraction r/s be expressed as   [#permalink] 12 Mar 2015, 03:15

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# 2,600 has how many positive divisors?

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2,600 has how many positive divisors? [#permalink]

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2,600 has how many positive divisors?

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D. 24
E. 48
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By factorization, you can write 2600 as 2600=2^3*5^2*13^1. Number of factors = (3+1)(2+1)(1+1)=24
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29 Oct 2010, 02:47
vgan4 wrote:
Number of factors = (3+1)(2+1)(1+1)=24

can' t understand how we came to this. What does this multiplication mean?
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ulm wrote:
vgan4 wrote:
Number of factors = (3+1)(2+1)(1+1)=24

can' t understand how we came to this. What does this multiplication mean?

Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

Back to the original question:
If p and q are prime numbers, how many divisors does the product $$p^3*q^6$$ have?

According to above the number of distinct factors of $$2,600=2^3*5^2*13$$ would be $$(3+1)(2+1)(1+1)=24$$.

For more on this check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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29 Oct 2010, 03:27
Yes, thanks,
i forgot
"The number of factors of n will be expressed by the formula $$(p+1)(q+1)(r+1)$$"
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29 Oct 2010, 23:57
Thanx Bunuel
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Re: 2,600 has how many positive divisors? [#permalink]

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20 Mar 2017, 01:27
shrive555 wrote:
2,600 has how many positive divisors?

A. 6
B. 12
C. 18
D. 24
E. 48

2600 = 13 x 2^3 x 5^2

no of possible divisor = (1+1)(2+1)(3+1) = 4.3.2 = 24

Option D
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Re: 2,600 has how many positive divisors? [#permalink]

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03 Apr 2017, 00:16
2600= 26*100= 2*13*2^2*5^2
=2^3*5^2*13

Total factors= (3+1)(2+1)(1+1)= 4*3*2=24
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Re: 2,600 has how many positive divisors? [#permalink]

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05 Jan 2018, 11:24
shrive555 wrote:
2,600 has how many positive divisors?

A. 6
B. 12
C. 18
D. 24
E. 48

To determine the number of positive divisors, we break 2,600 into primes, add 1 to the exponent of each unique prime, and then multiply those values together.

We see that 2,600 = 26 x 100 = 2 x 13 x 4 x 25 = 2^3 x 5^2 x 13^1.

Now we add 1 to each exponent and multiply those results:

(3 + 1)(2 + 1)(1 +1) = 24

Thus, 2,600 has 24 positive divisors.

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Re: 2,600 has how many positive divisors? [#permalink]

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05 Jan 2018, 11:56
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BOOKMARKED
shrive555 wrote:
2,600 has how many positive divisors?

A. 6
B. 12
C. 18
D. 24
E. 48

Experts, is there a fast way to find all the divisors? Thanks.[/quote]

If the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) =(5)(4)(2) = 40
----------ONTO THE QUESTION-------------------------

2600 = (2)(2)(2)(5)(5)(13)
= (2^3)(5^2)(13^1)
So, the number of positive divisors of 2600 = (3+1)(2+1)(1+1) =(4)(3)(2) = 24

Cheers,
Brent
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Re: 2,600 has how many positive divisors?   [#permalink] 05 Jan 2018, 11:56
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# General Solution for $\cos(\frac{x}{2}-1) =\cos^2(1-\frac{x}{2})$

I'm looking for an algebraic solution to : $$\cos(\frac{x}{2}-1) = \cos^2(1-\frac{x}{2})$$. So I simplified the equation: first off, $$\cos(\frac{x}{2}-1) = \cos(1-\frac{x}{2})$$. Then I divided both sides by that. and so I'm left with two things to solve:

$$\cos(\frac{x}{2}-1) = 0$$ (because I divided both sides by that expression, I have to also include the $$0$$ solution too). and $$\cos(\frac{x}{2}-1) = 1$$. And the general solution would be, I think, the union of those.

However, I'm kinda lost at this point. I've attempted to solve each equation. First off, I know that $$\cos(x) = 0$$ at $$\frac{\pi}{2}$$ and $$\frac{3\pi}{2}$$. So, the general solution for $$\cos(x) = 0$$ would be $$x=\frac{\pi}{2} +2\pi k, \cup \ \frac{3\pi}{2}+2\pi k, k\in Z.$$ I got up to this point, but don't know how to proceed.

The thing that's most confusing to me is I don't know how the $$-1$$ in the argument plays into the solution. Does it just change the graph to the right? Playing with desmos shows that graph is shifting by 2, but I thought that it'd shift by 1. More importantly: does it also affect the period of the function?

Additional question: In my book the answers are given in a different form. For example, the union I wrote would be written as: $$x= (-1)^k\frac{\pi}{2} + \pi k, k \in Z.$$ And in every case the period is "reduced" to $$\pi k$$. Why is that?

• So if the solution for $\cos(x) =0$ is $(x) =\pi/2+2\pi k$ then the solution to $\cos(x/2-1)=0$ is $(x/2-1)=\pi/2+2\pi k$ similarly for the other solution. – kingW3 Jun 23 '20 at 23:49
• well, $\cos (-u) = \cos u$ – Will Jagy Jun 23 '20 at 23:51

Okay so, you noticed that $${\cos\left(\frac{x}{2}-1\right)=\cos\left(1-\frac{x}{2}\right)}$$. And as you say, you end up with

$${\Leftrightarrow \cos^2\left(\frac{x}{2}-1\right)=\cos\left(\frac{x}{2}-1\right)}$$

And this implies

$${\cos\left(\frac{x}{2}-1\right)\left(\cos\left(\frac{x}{2}-1\right)-1\right)=0}$$

(as was pointed out by someone else - indeed it's probably bad practice to divide through here by $${\cos}$$. Not that it's incorrect, since you did take into account the fact we would then miss the $$0$$ solution (which was awesome!!!) - but unnecessary).

For simplicity, we can replace $${\frac{x}{2}-1}$$ with $${u}$$ and just rearrange for $${x}$$ at the end. So

$${\Rightarrow \cos(u)\left(\cos(u)-1\right)=0}$$

Now.... we want to solve $${\cos(u)=0}$$. As you said, one solution is $${\frac{\pi}{2}}$$... and if you take a look at the graph, you will notice that every other $$0$$ to the cosine function can be "reached" by hopping foward and backwards by multiplies of $${\pi}$$... you are right in thinking indeed, the period of the cosine function is $${2\pi}$$, but in fact $${0}$$ reoccurs every $${\pi}$$ radians. There is no problem with this. In order to have a full period ($${2\pi}$$ in this case), every value the function takes on must reoccur - only $$0$$ has. This actually means that

$${\cos(u)=0\Leftrightarrow u=\frac{\pi}{2}+n\pi, n \in \mathbb{Z}}$$

Now we can solve $${\cos(u)=1}$$ (the other solution we need). Of course we have a $${1}$$ at the point $${x=0}$$, and you may notice that we can reach all the other 1's by jumping foward and backwards by multiples of $${2\pi}$$ this time. Hence the solution is

$${\cos(u)=1\Leftrightarrow u=0 + 2n\pi =2n\pi, n \in \mathbb{Z}}$$

Now, you can simply plug back in the definition of $${u}$$ in terms of $${x}$$, adding $${1}$$ and multiplying both sides by two and you will end up with the solutions being

$${x=\pi + 2n\pi + 2=(2n+1)\pi + 2, n \in \mathbb{Z}}$$

$${x=4n\pi + 2, n \in \mathbb{Z}}$$

• This explained things very clearly and helped me solved another, similar problem too!. Thanks!!. – Ebrin Jun 24 '20 at 0:33
• No problem at all! :) – Riemann'sPointyNose Jun 24 '20 at 0:35

You're making things more complicated than they are

First, cosine is an even function, so the equation can be written $$\cos\Bigl(\frac{x}{2}-1\Bigr) =\cos^2\Bigl(\frac{x}{2}-1\Bigr)$$ as well. Note the equation implies $$\cos\bigl(\frac{x}{2}-1\bigr)\ge 0$$.

Also, if $$0, note that $$c Therefore the equation is equivalent to $$\cos\Bigl(\frac{x}{2}-1\Bigr)=0\:\text{ or }\:1.$$ Now this is easy to solve in terms of congruences: $$\begin{cases} \cos\Bigl(\frac{x}{2}-1\Bigr)=0\iff \frac{x}{2}-1\equiv \frac\pi 2 \mod\pi \iff\frac x2\equiv 1+\frac\pi 2\mod\pi \\ \cos\Bigl(\frac{x}{2}-1\Bigr)=1\iff \frac{x}{2}-1\equiv 0\mod2\pi \iff\frac x2\equiv 1\mod2\pi \end{cases}$$ and ultimately $$x\equiv2+\pi\bmod 2\pi \quad\text{ or }\quad x\equiv2\bmod 4\pi.$$

• Oh, the first part explains why the answer only contained the positive answers. So, the -1 in the argument just corresponds to the graph shifting to the right by two, right? It has no impact on the period. – Ebrin Jun 24 '20 at 0:07
• Absolutely no impact. – Bernard Jun 24 '20 at 0:09
• @kingW3: You're right. I shouldn't calculate directly on screen. Thank you for pointing it! – Bernard Jun 24 '20 at 0:17

Approach to the problem I would like to discuss an approach and leave solving the equations to you. $$\cos(y) = 0$$ implies that y is an odd multiple of $$\frac{\pi}{2}$$ or $$(2n + 1)\frac{\pi}{2}$$, where n is an integer. You can even write $$(2n - 1)$$, which is fine too as it gives odd vales. Similarly, $$\cos(\frac{x}{2} - 1) = 0$$ implies that $$\frac{x}{2} - 1 = (2n + 1)\frac{\pi}{2}.$$ You can simplify it further. Then solve the next one which is $$\cos(\frac{x}{2} - 1) = 1$$which implies that $$\frac{x}{2} - 1 = 2n\pi$$ (even multiple of pi), where $$n$$ is an integer. The $$'n'$$ in both solutions can be same because the first solution will always give an odd value of $$\frac{\pi}{2}$$ and the second solution will always give an even value of $$\frac{\pi}{2},$$ so we can use the same symbol 'n'. Now, to unify both solutions: Observe if there is any similarity between first and second solution. Is there any value(s) or range for which both solutions are same? If yes, we can get an even compact expression. If no, then write both solutions joined together with union symbol.

Regarding graph To interpret a function $$f(x - 1)$$interms of $$f(x)$$: The value which you get on putting $$x$$ in a function $$f(x)$$ will now [in $$f(x - 1)]$$ be obtained by setting $$x$$ as $$(x+1)$$ in $$f(x - 1)$$. Put $$x+1$$ in place of $$x$$ in $$f(x-1)$$, you will get $$f(x)$$ again. What does it mean? It means that you will have to shift your graph of $$f(x)$$ towards right by $$1$$ unit. Practice by drawing the graph of $$f(x) = x^2.$$ Then draw $$f(x) = (x - 1)^2.$$ But your problem has $$\cos(\frac{x}{2} - 1)$$ which is similar to $$f(ax + b)$$. We can now interpret $$'b'$$ as we did in last para. What about $$'a'$$?

Think like this. Every value of function $$f(x)$$ is now obtained at $$\frac{x}{a}$$ when the function is transformed to $$f(ax)$$. To show this, set $$x$$ as $$\frac{x}{a}$$ in $$f(ax)$$, you will get f(x) again. It also means that the frequency of the function $$f(ax)$$ is now decreased by $$a$$ factor of $$'a'$$.

In case of $$f(\frac{x}{a})$$, it can be said that every old value of $$f(x)$$ is obtained at $$'ax'$$ (show). Here the frequency is increased by a factor of $$'a'$$. Every old value of $$f(x)$$ is obtained quickly with $$f(\frac{x}{a}).$$ You can relate this with period since frequency and period are related. Or, you could interpret it in terms of period in the first place itself. If the frequency has increased, it means the value of the function will be repeated quickly, thereby decreasing the period. Period of $$\cos(x)$$ is $$2\pi$$. Period of $$\cos(ax+b)$$ will be $$\frac{2\pi}{a}.$$ The constant b only shifts the values, it does not account for frequency or period.

Thus, your interpretation of graph of $$\cos(\frac{x}{2} - 1)$$ is incomplete. To obtain its graph, we have to expand the graph of $$\cos(x)$$ by a factor of $$2$$, then shift the graph towards right by $$1$$ unit.

Regarding your last doubt In academics it is completely fine to represent the solution as simple as possible according to you. Generally, we want to make it more compact. The solution which you have wrote there. Put integral values of $$k$$. You will find that you have actually written odd multiples of $$\frac{\pi}{2}$$ in any case. Try to put several odd and even (positive integers for ease) integral values for $$k$$ and interpret the result yourself.

I thank @Ebrin for his help in formatting this answer.

• Sorry for the late reply, I've read your answer thoroughly and found it very helpful. Thanks for taking the time and explaining things! – Ebrin Jun 28 '20 at 15:59

So if $$\cos W = 0$$ then $$W = \pm \frac \pi 2 + 2k\pi$$. Notice $$-\frac \pi 2 + 2k \pi = \frac \pi 2 + (2k-1)\pi$$. And $$\frac \pi 2 + 2k \pi = -\frac \pi 2 + (2k+1) \pi$$. So to simplify $$W = \frac \pi 2 + m\pi$$ would be the simplest way to state this.

(Alternativily as $$0 = -0$$ and $$\cos(W \pm \pi) = -\cos W$$ we'd note that for an $$W = \pm \frac \pi 2 +2k \pi$$ then $$\mp \frac \pi 2+ (2k+1)\pi$$ is a solution).

So $$\frac x2 -1 = \frac \pi 2 + m\pi$$ then $$x = (2m+1)\pi + 2$$

And if $$\cos W = 1$$ then .... well... do I need to point out that solution is $$W=m\pi$$?

So $$\frac x2 -1= m\pi$$ so $$x=2m\pi +1$$.

So the solutions are $$x = k \pi + 2$$. If $$k$$ is odd then $$\cos(\frac x2 -1)=\cos(\frac k2 \pi) = 1$$. And if $$k$$ is odd then $$\cos(\frac x2 -1)=\cos(\frac {k-1}2\pi + \frac \pi 2) = 0$$.


# scal2frq

Scale to frequency

## Description

example

frq = scal2frq(A,wname,delta) returns the pseudo-frequencies corresponding to the scales given by A and the wavelet specified by wname and the sampling period delta. The output frq is real-valued and has the same dimensions as A.

frq = scal2frq(A,wname) is equivalent to frq = scal2frq(A,wname,1).

## Examples

collapse all

This example shows how the pseudo-frequency changes as you double the scale.

Construct a vector of scales with 10 voices per octave over five octaves.

vpo = 10;
no = 5;
a0 = 2^(1/vpo);
ind = 0:vpo*no;
sc = a0.^ind;

Verify that the range of scales covers five octaves.

log2(max(sc)/min(sc))
ans = 5.0000

If you plot the scales, you can use a data cursor to confirm that the scale at index $n+10$ is twice the scale at index $n$. Set the y-ticks to mark each octave.

plot(ind,sc)
title('Scales')
xlabel('Index')
ylabel('Scale')
grid on
set(gca,'YTick',2.^(0:5))

Convert the scales to pseudo-frequencies for the real-valued Morlet wavelet. First, assume the sampling period is 1.

pf = scal2frq(sc,"morl");
T = [sc(:) pf(:)];
T = array2table(T,'VariableNames',{'Scale','Pseudo-Frequency'});
disp(T)
Scale     Pseudo-Frequency
______    ________________

1          0.8125
1.0718         0.75809
1.1487         0.70732
1.2311         0.65996
1.3195         0.61576
1.4142         0.57452
1.5157         0.53605
1.6245         0.50015
1.7411         0.46666
1.8661         0.43541
2         0.40625
2.1435         0.37904
2.2974         0.35366
2.4623         0.32998
2.639         0.30788
2.8284         0.28726
3.0314         0.26803
3.249         0.25008
3.4822         0.23333
3.7321          0.2177
4         0.20313
4.2871         0.18952
4.5948         0.17683
4.9246         0.16499
5.278         0.15394
5.6569         0.14363
6.0629         0.13401
6.498         0.12504
6.9644         0.11666
7.4643         0.10885
8         0.10156
8.5742        0.094761
9.1896        0.088415
9.8492        0.082494
10.556         0.07697
11.314        0.071816
12.126        0.067006
12.996        0.062519
13.929        0.058332
14.929        0.054426
16        0.050781
17.148        0.047381
18.379        0.044208
19.698        0.041247
21.112        0.038485
22.627        0.035908
24.251        0.033503
25.992         0.03126
27.858        0.029166
29.857        0.027213
32        0.025391

Assume that data is sampled at 100 Hz. Construct a table with the scales, the corresponding pseudo-frequencies, and periods. Since there are 10 voices per octave, display every tenth row in the table. Observe that for each doubling of the scale, the pseudo-frequency is cut in half.

Fs = 100;
DT = 1/Fs;
pf = scal2frq(sc,"morl",DT);
T = [sc(:)/Fs pf(:) 1./pf(:)];
T = array2table(T,'VariableNames',{'Scale','Pseudo-Frequency','Period'});
T(1:vpo:end,:)
ans=6×3 table
Scale    Pseudo-Frequency     Period
_____    ________________    ________

0.01           81.25         0.012308
0.02          40.625         0.024615
0.04          20.313         0.049231
0.08          10.156         0.098462
0.16          5.0781          0.19692
0.32          2.5391          0.39385

Note the presence of the $\Delta t=\frac{1}{Fs}$ factor in scal2frq. This is necessary in order to achieve the proper scale-to-frequency conversion. The $\Delta t$ is needed to adjust the raw scales properly. For example, with:

f = scal2frq(1,'morl',0.01);

You are really asking what happens to the center frequency of the mother Morlet wavelet, if you dilate the wavelet by 0.01. In other words, what is the effect on the center frequency if instead of $\psi \left(t\right)$, you look at $\psi \left(t/0.01\right)$. The $\Delta t$ provides the correct adjustment factor on the scales.

You could have obtained the same results by first converting the scales to their adjusted sizes and then using scal2frq without specifying $\Delta t$.

max(pf-pf2)
ans = 0

The example shows how to create a contour plot of the CWT using approximate frequencies in Hz.

Create a signal consisting of two sine waves with disjoint support in additive noise. Assume the signal is sampled at 1 kHz.

Fs = 1000;
t = 0:1/Fs:1-1/Fs;
x = 1.5*cos(2*pi*100*t).*(t<0.25)+1.5*cos(2*pi*50*t).*(t>0.5 & t<=0.75);
x = x+0.05*randn(size(t));

Obtain the CWT of the input signal and plot the result.

[cfs,f] = cwt(x,Fs);
contour(t,f,abs(cfs).^2);
axis tight;
grid on;
xlabel('Time');
ylabel('Approximate Frequency (Hz)');
title('CWT with Time vs Frequency');

## Input Arguments

collapse all

Scales, specified as a positive real-valued vector.

Wavelet, specified as a character vector or string scalar. See wavefun for more information.

Sampling period, specified as a real-valued scalar.

Example: pf = scal2frq([1:5],"db4",0.01)

collapse all

### Pseudo-Frequencies

There is only an approximate answer for the relationship between scale and frequency.

In wavelet analysis, the way to relate scale to frequency is to determine the center frequency of the wavelet, Fc, and use the following relationship:

${F}_{a}=\frac{{F}_{c}}{a}$

where

• a is a scale.

• Fc is the center frequency of the wavelet in Hz.

• Fa is the pseudo-frequency corresponding to the scale a, in Hz.

The idea is to associate with a given wavelet a purely periodic signal of frequency Fc. The frequency maximizing the Fourier transform of the wavelet modulus is Fc. The centfrq function computes the center frequency for a specified wavelet. From the above relationship, it can be seen that scale is inversely proportional to pseudo-frequency. For example, if the scale increases, the wavelet becomes more spread out, resulting in a lower pseudo-frequency.

Some examples of the correspondence between the center frequency and the wavelet are shown in the following figure.

Center Frequencies for Real and Complex Wavelets

As you can see, the center frequency-based approximation (red) captures the main wavelet oscillations (blue). The center frequency is a convenient and simple characterization of the dominant frequency of the wavelet.

## References

[1] Abry, P. Ondelettes et turbulence. Multirésolutions, algorithmes de décomposition, invariance d'échelles et signaux de pression. Diderot, Editeurs des sciences et des arts, Paris, 1997.

## Version History

Introduced before R2006a