Problem solving the following problems

[grexam]

1) n=7.19^3
Col A:the number of distinct positive factors of n
Col B:10


2)At 9:am train T left train station and teo hours later train S left the same station on parallel track.If train T
averaged 60km/hr and train S averaged 75km/hr until S passed T,at what time did S pass T?
a)2pm
b)5pm
c)6pm
d)7pm
e)9pm


3)If the perimeter of a traingle is 18,then the length of one sides cannot be
1
3
6
8
9


4)In a certain country a person is born every 3 seconds and a person dies every 10 seconds,Therefore the birth and
death rates account for a population growth rate of one person every

3(1/3)sec
4(2/7)sec
7sec
11(2/3)sec
13sec


5)If r and s are positive integers,each greater than 1,and
if 11(s-1)=13(r-1),what is the least possible value of r+s?
2
11
22
24
26

[chestnut.cc]

1) n=7.19^3
Col A:the number of distinct positive factors of n
Col B:10

Is that 7.19^3, 7*19^3 or (7*19)^3... though i think its A in all cases...

2)At 9:am train T left train station and teo hours later train S left the same station on parallel track.If train T
averaged 60km/hr and train S averaged 75km/hr until S passed T,at what time did S pass T?
a)2pm
b)5pm
c)6pm
d)7pm
e)9pm

When train S passes train T they will have travelled the same distance... 60t = 75(t-2)... 15t = 150... t=10... 7 pm... B

3)If the perimeter of a traingle is 18,then the length of one sides cannot be
1
3
6
8
9

E

4)In a certain country a person is born every 3 seconds and a person dies every 10 seconds,Therefore the birth and
death rates account for a population growth rate of one person every

3(1/3)sec
4(2/7)sec
7sec
11(2/3)sec
13sec

1 per 3 seconds means 1/3 per second... 1 per 10 seconds means 1/10 per second... growth rate = birth rate - death rate... 1/3 -1/10... 7/30... B

5)If r and s are positive integers,each greater than 1,and
if 11(s-1)=13(r-1),what is the least possible value of r+s?
2
11
22
24
26

(s-1)/(r-1) = 13/11... obviously neither (s-1) nor (r-1) can be less than 13 or 11... in which case the ratio would reduce to lower terms and the inequality wouldn't be satisfied... so the first approximations should be 14 and 12... ie r +s =26... E

[eo352971]

hey chestnut, how come for the 2nd problem it is 75 (t-2)? I understand why it works out, but my natural instinct is to make it (t+2) as the second train left 2 hours later...?

Thanks!

[chestnut.cc]

t is the time for which the first train has been travelling... since the second train left 2 hours later... it travels 2 hours less as compared to the first... hence t-2... additionally if you put t+2 into the eqn you will see that you get a -ve time... which isnt possible... of course you could always say t is the time taken by the second train... t+2 is then the time taken by the first train...

[eo352971]

Ahh, got it. I was foolishly thinking of t as 9:00 am and not the actual number of hours travelled. Thanks a lot!

[scaredOfQuant]

Hey chestnut , how did u solve the first problem ? I will appreciate if u can elaborate on that.

[chestnut.cc]

1) n=7.19^3
Col A:the number of distinct positive factors of n
Col B:10

Is that 7.19^3, 7*19^3 or (7*19)^3... though i think its A in all cases...

They haven't said integral positive factors... which means you can have an infinite no of factors for any of the above mentioned... A

[grexam]

@chestnut
Can u please explain how u solved 3 rd problem

[chestnut.cc]

3)If the perimeter of a traingle is 18,then the length of one sides cannot be
1
3
6
8
9
The sum of two sides of a triangle is always greater than the third side... so if the perimeter is 18... plug in A... the sum of the other two sides is 17... let the other 2 sides be 8.5 and 8.5... the sum of any two sides exceeds the third... do this with all the others and you will see that for 9 the sum of the other 2 sides is 9... which isnt possible... E

[grexam]

thanks chestnut