I have the following word problems (finding ages, finding distance)

[Kulla_9289]

1) Rose is one and a half times older than her daughter. Seven years ago, she was four times as old as her daughter. Find their present ages.
My working:
Let x = daughter. Rose = (3/2)x. Seven years ago, (x-7) is daughter and (3/2)x - 7 is Rose. So equating them: (3/2)x - 7 = 4(x-7). On solving, x is is 8.4. So, daughter is 8.4 years old and Rose is 12.6.

2) Nathan travels at an average speed of 48 km/h from his home to his office. If he travels at an average speed of 60 km/h instead, he would arrive at his office 6 minutes earlier. Find the distance between his home and his office.
My working:
Distance = speed × time. So, D = 48 × t and D = 60 × (t-6). On equating: 48t = 60(t-6) ==> 4t = 5(t-6) ==> 5t-4t = 30. So, t = 30. On substituting into D = 48 × t, D = 48 × 30, which is 1440. So distance between home and office is 1440 km.

Are the answers correct? They seem a bit off.

 

[blamocur]

(3/2)x - 7 = 4(x-7). On solving, x is is 8.4.

Have you checked your solution by plugging it into the equation?

 

[Kulla_9289]

(3/2)(8.4) - 7 = 4(8.4-7)
12.6 - 7 = 5.6
5.6 = 5.6

 

[Mr. Bland]

1) Rose is one and a half times older than her daughter. Seven years ago, she was four times as old as her daughter. Find their present ages.
My working:
Let x = daughter. Rose = (3/2)x. Seven years ago, (x-7) is daughter and (3/2)x - 7 is Rose. So equating them: (3/2)x - 7 = 4(x-7). On solving, x is is 8.4. So, daughter is 8.4 years old and Rose is 12.6.

One and a half times older [imath]\left(* (1 + \frac{3}{2})\right)[/imath] or one and a half times as old [imath]\left(*\frac{3}{2}\right)[/imath]? Your math is correct, but using [imath]\frac{5}{2}[/imath] as the ratio gives integer answers, which I suspect is what was intended.

2) Nathan travels at an average speed of 48 km/h from his home to his office. If he travels at an average speed of 60 km/h instead, he would arrive at his office 6 minutes earlier. Find the distance between his home and his office.
My working:
Distance = speed × time. So, D = 48 × t and D = 60 × (t-6). On equating: 48t = 60(t-6) ==> 4t = 5(t-6) ==> 5t-4t = 30. So, t = 30. On substituting into D = 48 × t, D = 48 × 30, which is 1440. So distance between home and office is 1440 km.

While it is true that if Nathan drives at 48 km/h for 30 hours or at 60 km/h for (30 - 6 = 24) hours he'll travel 1440 km, the question isn't asking the distance at which he'll arrive 6 hours earlier.

 

[blamocur]

(3/2)(8.4) - 7 = 4(8.4-7)
12.6 - 7 = 5.6
5.6 = 5.6

Looks correct to me. I've made a mistake solving it myself, but assumed that there was a mistake in your solution. Since most 4.2 year olds cannot have children I suspected that your solution is wrong. Now I suspect that the wording of the problem might be incorrect.

 

[Kulla_9289]

One and a half times older (∗(1+32))\left(* (1 + \frac{3}{2})\right)(∗(1+23)) or one and a half times as old (∗32)\left(*\frac{3}{2}\right)(∗23)? Your math is correct, but using 52\frac{5}{2}25 as the ratio gives integer answers, which I suspect is what was intended.

English is confusing. I am here to solve mathematics not figure out the pithy words.

While it is true that if Nathan drives at 48 km/h for 30 hours or at 60 km/h for (30 - 6 = 24) hours he'll travel 1440 km, the question isn't asking the distance at which he'll arrive 6 hours earlier.

I have made a fool out of myself, and I genuinely apologise.
My working:
Distance = speed × time. So, D = 48 × t and D = 60 × (t-0.1). On equating: 48t = 60(t-0.1) ==> 4t = 5(t-0.1) ==> 5t-4t = 0.5. So, t = 0.5. On substituting into D = 48 × t, D = 48 × 0.5, which is 24. So distance between home and office is 24 km.

 

[Mr. Bland]

I am here to solve mathematics not figure out the pithy words.

Like @blamocur pointed out, there's an additional context clue regarding the ages at which people could have children. It seems entirely reasonable for a 28-year-old woman to have a 7-year-old daughter.

 

[Kulla_9289]

The present ages are 14 and 35

 

[blamocur]

One and a half times older (∗(1+32))\left(* (1 + \frac{3}{2})\right)(∗(1+23)) or one and a half times as old (∗32)\left(*\frac{3}{2}\right)(∗23)?

As can be seen from my posts, English isn't my first languange, but: if "One and a half times older" means [imath]\left(* (1 + \frac{3}{2})\right)[/imath], then what does "two times older" mean?

Still, I have to admit that your interpretation yields much more meaningful answer.

 

[blamocur]

English is confusing.

Couldn't agree more

I am here to solve mathematics not figure out the pithy words.

Many serious errors can be attributed to ambiguous problem statements.

 

[Mr. Bland]

As can be seen from my posts, English isn't my first languange, but: if "One and a half times older" means [imath]\left(* (1 + \frac{3}{2})\right)[/imath], then what does "two times older" mean?

"Older" a as a word means greater than some reference age. In this case, the reference age is that of Rose's daughter. For Rose to be older than her daughter, her age must necessarily be greater than that of her daughter. As a multiplier in terms of the daughter's age, the figure must be greater than [imath]1[/imath].

Consider the case where Rose and her daughter are the same age (due to time travel, presumably). How much older than her daughter is Rose? Rose is "not at all older", or in other words, "zero times older". The multiplier, to be applied to the daughter's age, takes the form of [imath]*(1 + 0)[/imath].

For Rose to be "two times older" than her daughter means is that in addition to the age of her daughter, she is another two times her daughter's age. The multiplier takes the form of [imath]*(1 + 2)[/imath], or three times her daughter's age in total.

Contrast with "as old", which implies a direct ratio. For Rose to be "two times as old" as her daughter means the multiplier is taken verbatim as [imath]*2[/imath].

To make matters even more confusing, English speakers will occasionally use the phrase "half again", which unto itself means [imath]*1\frac{1}{2}[/imath]. The words in isolation have no apparent meaning.

To make matters even more even more confusing, English speakers will also often suggest relationships using language like "three times younger", or perhaps that some object is "three times smaller" than another object. This is usually meant to express [imath]*\frac{1}{3}[/imath], but syntactically that's not what the words mean. Because "younger" and "smaller" are relative to some reference measurement, the multipler should take the form of [imath]*(1-x)[/imath], but people use those words to describe multipliers in the form of [imath]*\frac{1}{x}[/imath]. I can't correct people's behavior--no matter how wrong they are--so the best I can do is warn others about it.

 

[Steven G]

2) Nathan travels at an average speed of 48 km/h from his home to his office. If he travels at an average speed of 60 km/h instead, he would arrive at his office 6 minutes earlier. Find the distance between his home and his office.
My working:
Distance = speed × time. So, D = 48 × t and D = 60 × (t-6). On equating: 48t = 60(t-6) ==> 4t = 5(t-6) ==> 5t-4t = 30. So, t = 30. On substituting into D = 48 × t, D = 48 × 30, which is 1440. So distance between home and office is 1440 km.

Are the answers correct? They seem a bit off.

48t = 60(t-6) is NOT correct
Either your t is in hours or minutes (or something else?)

Suppose your t is in minutes.
Then D = 48*t which really is (48km/hr)(6min) = 288km* min/hr which is NOT just a distance as D should be in.

Now suppose t is in hours.
Then D = 48*t= (48km/hr)*(6 hr) = 288km which has the unit km which is a distance that D should be in. The problem is that t is not 6 hours, it is 6 minutes.

If you let the units help you by simply including then you would have seen that something is wrong.

From the very beginning you were told that all distances were in km but that time was in hours and minutes. You need to convert all the times to a single unit-whether it be hours, minutes, seconds or milli seconds.

For the record 6 minutes = 6 minutes * 1 = 6minutes * 1hr/60 minutes = 6*(1/60) hr = 0.1hr.

 

[Kulla_9289]

@Steven G I did. I fixed it.
My working:
Distance = speed × time. So, D = 48 × t and D = 60 × (t-0.1). On equating: 48t = 60(t-0.1) ==> 4t = 5(t-0.1) ==> 5t-4t = 0.5. So, t = 0.5. On substituting into D = 48 × t, D = 48 × 0.5, which is 24. So distance between home and office is 24 km.

 

[Steven G]

@Steven G I did. I fixed it.
My working:
Distance = speed × time. So, D = 48 × t and D = 60 × (t-0.1). On equating: 48t = 60(t-0.1) ==> 4t = 5(t-0.1) ==> 5t-4t = 0.5. So, t = 0.5. On substituting into D = 48 × t, D = 48 × 0.5, which is 24. So distance between home and office is 24 km.

I will not even look at your work since there are no units shown. Units help you know if your answer even makes sense. For example, if you conclude that D = 6ft*3ec= 18 ft*sec you know that your wrong since ft*sec is not a unit of distance.