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polygon, Point A and B , Circles & Fruit Dealer

by goyalsau » Sat Dec 25, 2010 5:11 am
If each interior angle of a regular polygon is 150 degrees, then it is

Octagon
Decagon
Quadrilateral
Dodecagon
Tetrahedron


A man starts from point A and walks northwards at the speed of 6 km/hr. He then turns right and walks eastwards at the speed of 4 km/hr, turns right again and walks southwards at the speed of 8 km/hr and turns left and walks eastwards at the rate of 6 km/hr. The time he walks in various directions is inversely proportional to the speed with which he walks. If the total distance that he covers is equal to 100 km then find how far he is from the starting point?

25 km
25√2 km
50 km
50√2 km
100 km


Hundred concentric circles are drawn with radius 1, 2, 3....100 cm respectively. The region bounded by the inner most circle is labelled 1. Every region bounded by any two consecutive circles is labeled 2, 3, 4...in the ascending order of their areas. Then, what is the sum of the areas of all the regions that are labelled even as a fraction of the total area of all the regions?

49/100
51/100
1/2
101/200
99/200


A fruit dealer fixes selling price of watermelon including 10% tax on the selling price and 30% profit. He sells it further at a surcharge of 15% to a juice centre. The owner of the juice centre mixes 25% water to the pure juice and offers a discount of 4% to his customer. If all the juice of the watermelon costs Rs. 161 to the customer and there is no tax on juice, then what is the cost price of the watermelon for the fruit dealer?

Rs. 83.30
Rs. 87.50
Rs. 76.50
Rs. 81.50
Rs. 70
Saurabh Goyal
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by N:Dure » Sat Dec 25, 2010 6:49 am
1)

To know the total interior angles of any polygon, use this formula: (n-2) * 180, where n is the number of polygon sides

So in this case: (n-2) * 180 = 150 n

180n - 360 = 150n

n= 12

A dodecagaon is a polygon with 12 sides.
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by anshumishra » Sat Dec 25, 2010 7:56 am
If each interior angle of a regular polygon is 150 degrees, then it is

Octagon
Decagon
Quadrilateral
Dodecagon
Tetrahedron


A man starts from point A and walks northwards at the speed of 6 km/hr. He then turns right and walks eastwards at the speed of 4 km/hr, turns right again and walks southwards at the speed of 8 km/hr and turns left and walks eastwards at the rate of 6 km/hr. The time he walks in various directions is inversely proportional to the speed with which he walks. If the total distance that he covers is equal to 100 km then find how far he is from the starting point?

25 km
25√2 km
50 km
50√2 km
100 km
Saurabh,

Please post one question per thread. That way we can get lots of discussions around it, and learn more.

1.As N:dure has already solved it, knowing that (for triangle the angle is 60degree, for square it is 90 degree, for pentagon it is 108 degree and hexagon it is 120 degree helps you to deduce it, in case you have forgotten).

150 = (n-2)*180/n
=> n = 12.

2.
The time he walks in various directions is inversely proportional to the speed with which he walks. This means that distance travelled in all the directions are equal :

If you draw it ( can't show the point B and C without attaching an image, but you can draw it yourself in direction North, East and then south ), this is how it looks.

A............D...........E

AD = DE = 25 =>
AE = 50 km.
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by anshumishra » Sat Dec 25, 2010 8:10 am
Hundred concentric circles are drawn with radius 1, 2, 3....100 cm respectively. The region bounded by the inner most circle is labelled 1. Every region bounded by any two consecutive circles is labeled 2, 3, 4...in the ascending order of their areas. Then, what is the sum of the areas of all the regions that are labelled even as a fraction of the total area of all the regions?

49/100
51/100
1/2
101/200
99/200
Area for the series for the region labelled as even :
(3,7,11,.....,199)pi

Area for all :
(1,3,5,7,.....,199) pi

So the required fraction = (3+7+11+.....+199)/(1+3+5+......+199) = 101/200 (If i didn't make any calculation mistake).
Thanks
Anshu

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by goyalsau » Sat Dec 25, 2010 9:26 am
N:Dure wrote:1)

To know the total interior angles of any polygon, use this formula: (n-2) * 180, where n is the number of polygon sides

So in this case: (n-2) * 180 = 150 n

180n - 360 = 150n

n= 12

A dodecagaon is a polygon with 12 sides.
Thanks N'Dure, I posted this problem for just one simple reason. I already know till octagon but it is really compulsory to remember after that...... like this dodecagaon, Hendecagon, Nonagon ...
I already know till Hexagon, Heptagon is kind of half missing and half remember. I know octagon But this Dodecagon - 12 sides, Hendecagon - 11 sides, & Nanogon - 9 sides . are all new for me.........

Do i really need to remember them , or this is just a weired question. That we will never see in Real Test........
Saurabh Goyal
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by anshumishra » Sat Dec 25, 2010 11:17 am
A fruit dealer fixes selling price of watermelon including 10% tax on the selling price and 30% profit. He sells it further at a surcharge of 15% to a juice centre. The owner of the juice centre mixes 25% water to the pure juice and offers a discount of 4% to his customer. If all the juice of the watermelon costs Rs. 161 to the customer and there is no tax on juice, then what is the cost price of the watermelon for the fruit dealer?

Rs. 83.30
Rs. 87.50
Rs. 76.50
Rs. 81.50
Rs. 70
Let x be the cost price of watermelon for the fruit dealer, and s be the selling price fixed by dealer :

x ---30%----> S ----10%---> 1.1S -----15%-------> 1.1 *1.15S ------25%----> 1.1*1.15*1.25S ------(-4%)-----> 1.1*1.15*1.25*0.96S

Given;
1.3x = S
1.1*1.15*1.25*0.96S = 161
=> x = 161/(1.3*1.1*1.15*1.25*0.96) ~= 81.5 Rs.
Thanks
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by prachich1987 » Sat Dec 25, 2010 12:47 pm
anshumishra wrote:
Hundred concentric circles are drawn with radius 1, 2, 3....100 cm respectively. The region bounded by the inner most circle is labelled 1. Every region bounded by any two consecutive circles is labeled 2, 3, 4...in the ascending order of their areas. Then, what is the sum of the areas of all the regions that are labelled even as a fraction of the total area of all the regions?

49/100
51/100
1/2
101/200
99/200
Area for the series for the region labelled as even :
(3,7,11,.....,199)pi

Area for all :
(1,3,5,7,.....,199) pi

So the required fraction = (3+7+11+.....+199)/(1+3+5+......+199) = 101/200 (If i didn't make any calculation mistake).
I didn't understand how you calculated the area of circle here
I can't see any square terms
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by anshumishra » Sat Dec 25, 2010 12:52 pm
prachich1987 wrote:
anshumishra wrote:
Hundred concentric circles are drawn with radius 1, 2, 3....100 cm respectively. The region bounded by the inner most circle is labelled 1. Every region bounded by any two consecutive circles is labeled 2, 3, 4...in the ascending order of their areas. Then, what is the sum of the areas of all the regions that are labelled even as a fraction of the total area of all the regions?

49/100
51/100
1/2
101/200
99/200
Area for the series for the region labelled as even :
(3,7,11,.....,199)pi

Area for all :
(1,3,5,7,.....,199) pi

So the required fraction = (3+7+11+.....+199)/(1+3+5+......+199) = 101/200 (If i didn't make any calculation mistake).
I didn't understand how you calculated the area of circle here
I can't see any square terms
The area was calculated like
pi*(2^2-1^2) = pi*(2+1)(2-1) = 3pi
pi*(3^2-2^2) = pi*(3+2)(3-2) = 5pi
...................

Area for even region was calculated like :
pi(2^2-1^2) = 3*pi
pi(4^2-3^2) = 7*pi
....................
Thanks
Anshu

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by prachich1987 » Sat Dec 25, 2010 12:55 pm
anshumishra wrote:
prachich1987 wrote:
anshumishra wrote:
Hundred concentric circles are drawn with radius 1, 2, 3....100 cm respectively. The region bounded by the inner most circle is labelled 1. Every region bounded by any two consecutive circles is labeled 2, 3, 4...in the ascending order of their areas. Then, what is the sum of the areas of all the regions that are labelled even as a fraction of the total area of all the regions?

49/100
51/100
1/2
101/200
99/200
Area for the series for the region labelled as even :
(3,7,11,.....,199)pi

Area for all :
(1,3,5,7,.....,199) pi

So the required fraction = (3+7+11+.....+199)/(1+3+5+......+199) = 101/200 (If i didn't make any calculation mistake).
I didn't understand how you calculated the area of circle here
I can't see any square terms
The area was calculated like
pi*(2^2-1^2) = pi*(2+1)(2-1) = 3pi
pi*(3^2-2^2) = pi*(3+2)(3-2) = 5pi
...................

Area for even region was calculated like :
pi(2^2-1^2) = 3*pi
pi(4^2-3^2) = 7*pi
....................
thnx Anshu
hi goyalsau.. all 4 were very nice problems
may I know the source?
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by goyalsau » Sun Dec 26, 2010 6:56 am
prachich1987 wrote: hi goyalsau.. all 4 were very nice problems
may I know the source?
@ Anshu....Thanks Anshu , Great Work, I must say you stick to the basics and that is very important. Specially in that Circle and Fruit Dear problem, After all the search about Gmat , It is said that its very rare that somebody score above 51 in quant , 51 in quant is itself 99% and above that, but i must say, You have all the potential to go beyond That....... I wish you Good Luck.. :D :D :D :D

@prachi..... These questions are from www.wmprep.com I have joined that institute sometime back But my acess is about to expire on 31th of December. So i thought Till than as much i can explore from that is good.
Wish you good luck too, I can see that you are putting a lot of hard work, that is great...... Keep the good work going and hope to see you in other post. Bye.
Saurabh Goyal
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by HPengineer » Sun Dec 26, 2010 11:14 am
in that 4th question where it says Rs. what does that actually stand for?

also why is the 10% tax calculated on S and not X?
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by anshumishra » Sun Dec 26, 2010 11:20 am
HPengineer wrote:in that 4th question where it says Rs. what does that actually stand for?

also why is the 10% tax calculated on S and not X?
Rs. stands for Rupees (It is Indian currency).

As per question, there is 10% tax on the selling price. S is the selling price set by fruit dealer and X is the cost price paid by fruit dealer.
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by anshumishra » Sun Dec 26, 2010 11:22 am
goyalsau wrote:
prachich1987 wrote: hi goyalsau.. all 4 were very nice problems
may I know the source?
@ Anshu....Thanks Anshu , Great Work, I must say you stick to the basics and that is very important. Specially in that Circle and Fruit Dear problem, After all the search about Gmat , It is said that its very rare that somebody score above 51 in quant , 51 in quant is itself 99% and above that, but i must say, You have all the potential to go beyond That....... I wish you Good Luck.. :D :D :D :D
Thanks Saurabh ! Wish you too the same !!!
Thanks
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by prachich1987 » Sun Dec 26, 2010 9:54 pm
goyalsau wrote:
prachich1987 wrote: hi goyalsau.. all 4 were very nice problems
may I know the source?
@ Anshu....Thanks Anshu , Great Work, I must say you stick to the basics and that is very important. Specially in that Circle and Fruit Dear problem, After all the search about Gmat , It is said that its very rare that somebody score above 51 in quant , 51 in quant is itself 99% and above that, but i must say, You have all the potential to go beyond That....... I wish you Good Luck.. :D :D :D :D

@prachi..... These questions are from www.wmprep.com I have joined that institute sometime back But my acess is about to expire on 31th of December. So i thought Till than as much i can explore from that is good.
Wish you good luck too, I can see that you are putting a lot of hard work, that is great...... Keep the good work going and hope to see you in other post. Bye.
Thanks Saurabh for wishes!!!
All the best!!
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