Another Simple QR Problem...but

[Contach]

how would you do something like this in less than 1 minute?

4. The front tires on a formula car have circumferences of 3.52 feet and the rear tires have circumferences of 5.28 feet. On a 4200-mile cross-country race, how many more revolutions does a front tire make than a rear tire? (1 mile = 5280 feet)
A. 2100
B. 8400
C 2 100 00
D. 2 800 000
E. 4 200 000



mother@!#$@#$@ impossible to do in under a minute. I dare you to time yourself.

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[AdaAda87]

Is the answer C?
If so I did it in 45 seconds. haha. I probably did it wrong, but just keep practicing man! practice makes perfect!

 

[rdhdds1]

Is the answer C?
If so I did it in 45 seconds. haha. I probably did it wrong, but just keep practicing man! practice makes perfect!

The answer is "C". I do not remember how to do the problem but remember seeing it somewhere before.

 

[huk1985]

I'm really good with simple math and this took me almost 2 minutes. Some things take you less than ten seconds. I guess these problems you just skip and come back if you have time.

 

[slick90210]

how would you do something like this in less than 1 minute?

4. The front tires on a formula car have circumferences of 3.52 feet and the rear tires have circumferences of 5.28 feet. On a 4200-mile cross-country race, how many more revolutions does a front tire make than a rear tire? (1 mile = 5280 feet)
A. 2100
B. 8400
C 2 100 00
D. 2 800 000
E. 4 200 000

mother@!#$@#$@ impossible to do in under a minute. I dare you to time yourself.

Best way to go about it is starting with 1 mile.... 5280/5.28 = 1000 rotations per mile
5.28/3.52 = 1.5x (approx. in your head.... 2x 3.52 = 7.04.... 5.28 is the middle so 1.5) how many more rotation per mile on the front

1000 rotations on back.... 1000 x 1.5 = 1500 in the front.
500 more rotations per mile.... 500 x 4 200 miles = 2 100 000

 

[250rsavage]

the way I did it was once you find the rotations, 1500 and 1000, now just multiple 15*42 and 10 *42,since everything has the same power of 10. now just subtract and you will get 210, or 210000 if you want to add all the zeros back on.

hopefully this helps

 

[Contach]

Fair enough. good job.
Now try this: any good strategies?
18. Stella has 3 quarters and 4 dimes. How many different amounts can she form with these 7 coins, using at least one coin in each amount?

 

[AdaAda87]

no answer choice this time? is the answer 18?

 

[slick90210]

lol what are the options? Is the answer 12?

 

[yakuza]

(4200 x 5280)/(5.28-3.52)

Is this the right way to do this problem?

 

[Streetwolf]

Fair enough. good job.
Now try this: any good strategies?
18. Stella has 3 quarters and 4 dimes. How many different amounts can she form with these 7 coins, using at least one coin in each amount?

Since no two possible combos of coins add to the same amount, you don't have to worry about counting one amount twice with two different groups of coins. If you had 5 dimes for example, then 5 dimes is the same amount as 2 quarters and you'd have to only count one of them.

There are 3 quarters and 4 dimes which means you have 4 possibilities for the quarters (0, 1, 2, 3) and 5 possibilities for the dimes (0, 1, 2, 3, 4). So you could use any of the 4 for the quarters and any 5 for the dimes which is 4 x 5 = 20.

Since you have to use at least one coin, you can't use (0, 0). So 20 - 1 = 19, your answer.

 

[Contach]

the answer is 19.

Can you explain the logic behind 4 x 5?
Is this a permutation, combination? Neither I suspect.

And Yakuza: no thats not right.
It should be: (4200 x 5280)/5.28 - (4200 x 5280)/3.52 ... which is quite different.

 

[slick90210]

the answer is 19.

Can you explain the logic behind 4 x 5?
Is this a permutation, combination? Neither I suspect.

And Yakuza: yes that's right. Now try doing those calculations in less than a minute.

4 x 5 is the total amount of possibilities (including 0 nickels and 0 dimes)....
think about it this way.
You are going to have 4 groups each with 5 possibilities equalling 20 total. Now you cannot use 0 dimes and 0 nickles so the first possibility is scratched off.

0 nickles x 0 dime (X) 1 nickle x 0 dimes. 2 nicles x 0 dimes ~
0 nickles x 1 dime ~
0 nickles x 2 dimes ~
0 nickles x 3 dimes ~
0 nickles x 4 dimes ~
0 nickles x 5 dimes ~

 

[harrygt]

21 000 000 is the answer. I guess you missed one of the zeros in C.
It took me 95 seconds. NOt that bad. You can always skip these type of questions and do them at the end.

 

[Contach]

21 000 000 is the answer. I guess you missed one of the zeros in C.
It took me 95 seconds. NOt that bad. You can always skip these type of questions and do them at the end.

Yeah sorry, I was meant to type: C) 2 100 000
There is no option that is 21 million. sorry bud, you are wrong.
90 seconds for a wrong answer - sucks....

 

[Contach]

4 x 5 is the total amount of possibilities (including 0 nickels and 0 dimes)....
think about it this way.
You are going to have 4 groups each with 5 possibilities equalling 20 total. Now you cannot use 0 dimes and 0 nickles so the first possibility is scratched off.

0 nickles x 0 dime (X) 1 nickle x 0 dimes. 2 nicles x 0 dimes ~
0 nickles x 1 dime ~
0 nickles x 2 dimes ~
0 nickles x 3 dimes ~
0 nickles x 4 dimes ~
0 nickles x 5 dimes ~

Hrm.. Thanks! that chart helps.