COIN PROBLEM AND ALLEGATIONS & MIXTURES CONCEPT(2 problems)

[summer1350]

I need help with the following two problems which are in Allegations and mixtures concept.

 

1. If a man decides to travel 72 miles in 8 hours partly by foot and partly on a bicycle, his speed of foot being 8 mph and that on bicycle being 10 mph. What distance would he travel on foot?

(A) 24 miles

(B) 32 miles

© 48 miles

(D) 64 miles

(E) Cannot be determined

 

2. A sum of $80 is made up of 80 coins that are either dimes or quarters coins. Find out how many dimes or quarters coins are there in the total amount?

(A) 10

(B) 20

© 30

(D) 40

(E) 50

[achilles2040]

For 1st

let the man travel x miles on foot. he will then travel 72-x miles on bicycle. the total time taken is 8 hrs.

Therefore [ x/8 + (72-x)/10] = 8 since time= distance / speed .

solving the equation you will get x as 32miles.

 

for 2nd question:

the question seems to be flawed since even if u have 80 quarters it amounts to $20 . therefore the total amount given of $80 is flawed it should be something much less than that.

does somebody have a different explanation?

[zulkarnain]

1. Suppose he walks -

 

1 hour on foot (8 mile) and 7 hour on bicycle 70 mile, total 78 miles in 8 hours which doesn't fulfill the condition.

2 hour on foot (16 mile) and 6 hour on bicycle 60 mile, total 76 miles in 8 hours which doesn't fulfill the condition.

3 hour on foot (24 mile) and 5 hour on bicycle 50 mile, total 74 miles in 8 hours which doesn't fulfill the condition.

4 hour on foot (32 mile) and 4 hour on bicycle 40 mile, total 72 miles in 8 hours which fulfills the condition.

5 hour on foot (50 mile) and 3 hour on bicycle 30 mile, total 80 miles in 8 hours which doesn't fulfill the condition.

6 hour on foot (48 mile) and 2 hour on bicycle 20 mile, total 68 miles in 8 hours which doesn't fulfill the condition.

7 hour on foot (56 mile) and 1 hour on bicycle 10 mile, total 66 miles in 8 hours which doesn't fulfill the condition.

 

So the answer is B.

[summer1350]

Thanks Achilles and Zulkarnain. :)

 

@ Achilles: I found a solution to a similar coin problem posted above. But somehow, I can't relate this to that. Only dimes have been asked here. Please go through the following:

 

Laura has 20 coins consisting of quarters and dimes. If she has a total of $3.05, how many dimes does she have?

(A) 3 (B) 7 © 10 (D) 13 (E) 16

Let D stand for the number of dimes, and let Q stand for the number of quarters. Since the total number of coins in 20, we get D + Q = 20, or Q = 20 - D. Now, each dime is worth 10 cents, so the value of the dimes is 10D. Similarly, the value of the quarters is 25Q = 25(20 - D). Summarizing this information in a table yields

Dimes Quarters Total Number D 20 - D 20 Value 10D 25(20 - D) 305

Notice that the total value entry in the table was converted from $3.05 to 305 cents. Adding up the value of the dimes and the quarters yields the following equation:

10D + 25(20 - D) = 305

10D + 500 - 25D = 305

-15D = -195

D = 13

Hence, there are 13 dimes, and the answer is (D).

[achilles2040]

This is a typical denomination question. let there be D number of dimes and Q number of quarters. therefore D+Q= 20. [since total number of coins is 20]

now we know that 1 dime is equal to 10 cents therefore the amount of D dimes will be 10D cents . similarly 1 quarter equal to 25 cents therefore the amount Q quarter will be 25Q cents .

It is given that the total is $3.05 = 305 cents since [$1 = 100 cents].

therefore 10D + 25Q = 305

and D+Q = 20

by solving the two equations we get D= 13.

 

Now if we solve the first problem in the same way we will get two equations:

 

Q+D= 80

25Q+10D=8000

solving these we will get Q= 528. and D negative. therefore there is some problem with the question. also if we consider that there are only quarters that is 80 quarters it will amount to $20 and not $80.

[achilles2040]

This is a typical denomination question. let there be D number of dimes and Q number of quarters. therefore D+Q= 20. [since total number of coins is 20]

now we know that 1 dime is equal to 10 cents therefore the amount of D dimes will be 10D cents . similarly 1 quarter equal to 25 cents therefore the amount Q quarter will be 25Q cents .

It is given that the total is $3.05 = 305 cents since [$1 = 100 cents].

therefore 10D + 25Q = 305

and D+Q = 20

by solving the two equations we get D= 13.

 

Now if we solve the first problem in the same way we will get two equations:

 

Q+D= 80

25Q+10D=8000

solving these we will get Q= 528. and D negative. therefore there is some problem with the question. also if we consider that there are only quarters that is 80 quarters it will amount to $20 and not $80.