Some easy ones

[look_hd]

1. A dog is tethered to the corner of the outside of a ten foot by ten foot building by a leash which is 14 feet long as shown in the figure. How much area (outside of the building) can the dog roam?

 

 

http://img266.imageshack.us/img266/3362/p01prob28ap.gif

 

 

 

 

 

 

 

 

 

2. Three cubes of volume 1, 8, and 64 are glued together. What is the smallest possible surface area of the resulting configuration?

A) 112

B) 114

C) 116

D) 118

E) 120

 

 

3.What is [1(1!)+2(2!)+...+n(n!)]?

A) n! + n + 1

B) (n + 1)!n!

C) (n + 1)!

D) (n + 1)! + 1

E) (n + 1)! - 1

 

 

4. “Chicken Snaps” come in boxes of 5, 9, and 12. What is largest number of Chicken Snaps you cannot buy?

A) 11

B) 16

C) 19

D) 26

E) There is no largest number.

[stickler_nishat]

EBEE

 

easy ones!!!!!!:blush: what r the real answers???

[Daniel Sadoc]

1. A dog is tethered to the corner of the outside of a ten foot by ten foot building by a leash which is 14 feet long as shown in the figure. How much area (outside of the building) can the dog roam?

 

 

http://img266.imageshack.us/img266/3362/p01prob28ap.gif

 

 

 

 

155 pi = 3 pi 14 . 14 / 4 + pi 16 / 4 + pi 16 / 4

 

 

2. Three cubes of volume 1, 8, and 64 are glued together. What is the smallest possible surface area of the resulting configuration?

A) 112

B) 114

C) 116

D) 118

E) 120

 

 

126 - 2 - 4 = 120

 

 

3.What is [1(1!)+2(2!)+...+n(n!)]?

A) n! + n + 1

B) (n + 1)!n!

C) (n + 1)!

D) (n + 1)! + 1

E) (n + 1)! - 1

 

 

(E)

 

(2-1) 1! + (3-1) 2! + (4-1) 3! + ... + (n+1-1) n!

 

 

 

4. “Chicken Snaps” come in boxes of 5, 9, and 12. What is largest number of Chicken Snaps you cannot buy?

A) 11

B) 16

C) 19

D) 26

E) There is no largest number.

 

16 = 12 + 4 = 5 + 11 = 9 + 7

 

19 = 12 + 5 + 2 = 10 + 9 = 13 + 6 =15 + 4 = 18 + 1 = 12 + 7

 

26 = 5 + 9 + 5 + 5 + 3 = 24 + 2 = 27 - 1 = 20 + 6 = 18 + 8 = 12 + 9 + 5

 

So, is the answer (b) or (e) ?

 

Let

 

x = p (mod q)

 

if the remainder of x divided by q is equal to p.

 

Then, if

 

x = 0 (mod 5) or x = 2 (mod 5) or x = 4 (mod 5) and x > 16,

 

then x can be written as

 

x = 5 m + 9 n + 12 o

 

But what about the cases x = 1 (mod 5) or x = 3 (mod 5) ?

 

Ok. The answer is (b).

 

Consider the remainder of x divided by 12.

 

x = 0 (mod 12) ok

x = 1 (mod 12) then x - 25 = 0 (mod 12)

x = 2 (mod 12) then x - 14 = 0 (mod 12)

x = 3 (mod 12) then x - 15 = 0 (mod 12)

x = 4 (mod 12) then x - 28 = 0 (mod 12)

x = 5 (mod 12) then x - 5 = 0 (mod 12)

x = 6 (mod 12) then x - 18 = 0 (mod 12)

x = 7 (mod 12) then x - 7 - 12 = 0 (mod 12)

x = 8 (mod 12) then x - 8 - 12 = 0 (mod 12)

x = 9 (mod 12) then x - 9 = 0 (mod 12)

x = 10 (mod 12) then x - 10 = 0 (mod 12)

x = 11 (mod 12) then x - 35 = 0 (mod 12)

 

Is there an easiest way to prove the desired result?

 

Answer: (b)

[stickler_nishat]

2. Three cubes of volume 1, 8, and 64 are glued together. What is the smallest possible surface area of the resulting configuration?

A) 112

B) 114

C) 116

D) 118

E) 120

 

 

Side of three cubes respectively 1,2,4

 

if u place smaller two over the large one side by side, u will get the minimum surface area.

 

16*6+4*4+1*2=114

 

so, answer should be B

[priyankagre2005]

I am getting 120.

[denizen06]

2. Three cubes ....

Side of three cubes respectively 1,2,4

if u place smaller two over the large one side by side, u will get the minimum surface area.

 

16*6 +4*4+1*2=114

 

so, answer should be B

 

16*6 ---------16*5

+4*4----------4*4

+1*2..........1*4

+(something more for the common surface of 3 cube )......11

 

+something more for the common surface of 2nd and 3rd big surface )...3

=114

 

 

 

priyanka is wrong

great stickler

[stickler_nishat]

Dont Be sorry Denizen,

I am still getting 114. and i m sure abt it.

 

[N.R] Denizen u have changed ur opinion. Thats so fast........

[priyankagre2005]

Let me try again...

[denizen06]

I am still confused. Getting two answers correct. wait abit. pzled

Yap Priyanka is wrong sure.

 

126 -(4+2)*2 =114

 

the common surface must be subtracted twice.

that's wrong deniel and prianka

[priyankagre2005]

agree guys 114.

[denizen06]

4. “Chicken Snaps” come in boxes of 5, 9, and 12. What is largest number of Chicken Snaps you cannot buy?

A) 11

B) 16

C) 19

D) 26

E) There is no largest number.

 

17- 30 are possible

so all numbers > 21 possible

11 , 16 not possible.

hence 16 is the answer.

[stickler_nishat]

yeah........i understand.......U both r right.

Nice explanations.......Thanks to both D.

[denizen06]

U are welcome s_n.

Did you agreed with the same neighbor problems ?

No response down there.

come on dude. Let's make that one clean.

Btw is t_g your friend? A good one?

[look_hd]

I am still confused. Getting two answers correct. wait abit. pzled

Yap Priyanka is wrong sure.

 

126 -(4+2)*2 =114

 

the common surface must be subtracted twice.

that's wrong deniel and prianka

 

right on

 

 

OA:

DBEB

[riteshn]

q.1: In 155 pi = 3 pi 14 . 14 / 4 + pi 16 / 4 + pi 16 / 4, I got 3 pi 14 . 14 / 4 but from where pi 16 / 4 + pi 16 / 4 came? Whats the idea?

 

q2: if u place smaller two over the large one side by side, u will get the minimum surface area.

 

 

Is that given somewhere or we found out by trial and error?

 

Can somebody explain question 4? Accordingly to me you cannot buy any of the numbers as whole coz you cannot get any sum with the numbers of Chicken Snap given or did I miss something?

[riteshn]

Further light on my queries?

[askrach]

r u there online

I am getting 120.