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## Question

Here are few elementary algebra problems from a collection of Russian math test problems for school students.

While they may appear rudimentary to math students, keep in mind that those present just an example of algebra test problems that every high school graduate was expected to solve within a short time limit. Compare that to some standard tests of today, like ACT and SAT in the United States. Also, the arrangement of the problems gives an interesting historical perspective. Enjoy.

1. Several years ago there were 10,000 trees in the forest. Now there are 15,600 trees in the forest. Over how many years the growth took place, if each year the number of trees increases by 0.1?

(From a High School Diploma test, 1875).

2. A dirigible first traveled 40 km going against a 30 km/hr wind and then returned, spending 2.5 hours total for the round trip. Find the dirigible’s speed in a still air.

(From a High School Diploma test, 1913)

3. A collective farm worker, set out from his village to the railroad station. Having walked 3 km after one hour, he observed that if he continued at the same speed, he would arrive 40 min late for his train. So he paced up and walked at 4 km/hr for the rest of the trip. Thus he arrived to the station 45 min before his train’s scheduled departure.

Find the distance from the village to the station.

4. A man walked every day along 1 km stretch of the streetcar rail. After 1 month he had counted 45 passing trams that moved in the same direction and 120 trams that moved toward him. Given that the man walked 1 km in 12 min, find the speed of tram.

(From Military School tests, 1947).

5. Two bodies simultaneously departed from point A on the circumference of a circle, moving along the circumference in the opposite directions. When the two bodies met, it turned out the first body had traveled the distance 7 cm greater than the second. After the meeting the bodies continued their motion and the first body arrived to the point A 4 seconds later, whereas the second body arrived to the point A 9 seconds later.

Find the circumference.

(From a university admission test, 1958)

## Recommended Posts

• 0 is question 1 meant to be a 0.1% increase every year? If so then...

10,000*(1.01^x)=15600

10,000*(1.01^44.7)=15601

44.7 years is close enough for me ##### Share on other sites
• 0
is question 1 meant to be a 0.1% increase every year? If so then...

10,000*(1.01^x)=15600

10,000*(1.01^44.7)=15601

44.7 years is close enough for me Sorry, the question #1 meant 0.1 annual rate of increase. Which is 10%.

While your solution method is correct, the answer is different. That is to say you'd get a partial credit for this one on the exam, something like 80%.

Edited by Prime
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• 0 Sorry, the question #1 meant 0.1 annual rate of increase. Which is 10%.

While your solution method is correct, the answer is different. That is to say you'd get a partial credit for this one on the exam, something like 80%.

Ahh i saw the 0.1 and thought you meant a 1% increase but just forgot to put the 1 infront of the decimals... ah well my bad

10,000*1.1^x=15600

10,000*1.1^4.67=15606

once again close enough ##### Share on other sites
• 0 2. A dirigible first traveled 40 km going against a 30 km/hr wind and then returned, spending 2.5 hours total for the round trip. Find the dirigible’s speed in a still air.

(From a High School Diploma test, 1913)

Headwind rate = x - 30 km/hr

Tailwind rate = x + 30 km/hr

Total Time = (40 km / headwind) + (40 km / tailwind)

2.5 = (40 / x - 30) + (40 / x + 30)

2.5 = (40(x + 30) + 40(x - 30)) / ((x - 30)(x + 30))

2.5 = 80x / (x^2 - 900)

2.5x^2 - 2250 = 80x

2.5x^2 - 80x - 2250 = 0

x = {-18, 50}

In still air, dirigible's speed is 50 km/hr

What's a dirigible?

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• 0
Ahh i saw the 0.1 and thought you meant a 1% increase but just forgot to put the 1 infront of the decimals... ah well my bad

10,000*1.1^x=15600

10,000*1.1^4.67=15606

once again close enough Interesting to note, school students didn't have calculators back in 1875.

What's a dirigible?

Headwind rate = x - 30 km/hr

Tailwind rate = x + 30 km/hr

Total Time = (40 km / headwind) + (40 km / tailwind)

2.5 = (40 / x - 30) + (40 / x + 30)

2.5 = (40(x + 30) + 40(x - 30)) / ((x - 30)(x + 30))

2.5 = 80x / (x^2 - 900)

2.5x^2 - 2250 = 80x

2.5x^2 - 80x - 2250 = 0

x = {-18, 50}

In still air, dirigible's speed is 50 km/hr

Correct solution.

Dirigible is a blimp.

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• 0 [spoiler='wouldn't #2 be

']assuming the same headwind on the way back (tailwind) then wouldn't it just be round trip distance 80km divided by the amount of time to travel 2.5 hours

80/2.5=32km/hr = average speed of trip.

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• 0 [spoiler='wouldn't #2 be

']assuming the same headwind on the way back (tailwind) then wouldn't it just be round trip distance 80km divided by the amount of time to travel 2.5 hours

80/2.5=32km/hr = average speed of trip.

If your airspeed was 32 km/hour against a 30 km/h headwind, your ground speed would be only 2km/hour. At that rate it would take 20 hours just to do the outbound 40 km; On the way back your ground speed would be 62 km/hour, and it would take just under 40 minutes to return. The total of 20.6 hours is far too long...

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• 0 is question 1 meant to be a 0.1% increase every year? If so then...

10,000*(1.01^x)=15600

10,000*(1.01^44.7)=15601

44.7 years is close enough for me This solution is for 1%, not 10% as posed by the question nor 0.1% as assumed in the intro.

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• 0 Interesting to note, school students didn't have calculators back in 1875.

Correct solution.

Dirigible is a blimp.

Interesting to note, school students didn't have calculators back in 1875.

Actually we are still not allowed use calculators until we graduate from highschool. We have to use tables to find logs, sines, cosines etc.

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• 0 Interesting to note, school students didn't have calculators back in 1875.

Sure they did. They just didn't run on batteries.

log(1.56)/log(1.1) would not have taken long for a high-school student in 1875, armed with a slide-rule, to compute to two or three digits. In fact, I'll wager they would have finished faster, because they wouldn't be wasting time copying all thirteen or so digits that their calculator display spewed at them onto the page. ##### Share on other sites
• 0
Actually we are still not allowed use calculators until we graduate from highschool. We have to use tables to find logs, sines, cosines etc.

Interesting! I guess, that depends on your location. In the US high school students do rely on calculators. Too much so, in my opinion.

The problems 3, 4, and 5 are still unsolved.

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• 0 Here are few elementary algebra problems from a collection of Russian math test problems for school students.

3. A collective farm worker, set out from his village to the railroad station. Having walked 3 km after one hour, he observed that if he continued at the same speed, he would arrive 40 min late for his train. So he paced up and walked at 4 km/hr for the rest of the trip. Thus he arrived to the station 45 min before his train’s scheduled departure.

Find the distance from the village to the station.

Let X = the distance between when he changed speeds and station

The difference in time between the 3km/hr and 4km/h speeds is 1.33 hours. Or,

Xkm/3km/hr = Xkm/4km/hr + 1.33 hours

solving for X yeilds

X = 16 km.

## Checking this work...

To walk 16 km @ 4km/hr takes 4 hrs.

To walk 16 km @ 3km/hr takes 5.33 hrs

The difference is 1.33, as desired

To get the total distance we have to add back the 3km he took to figure this out, so the total distance is 19 KM.

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• 0
Sure they did. They just didn't run on batteries.

log(1.56)/log(1.1) would not have taken long for a high-school student in 1875, armed with a slide-rule, to compute to two or three digits. In fact, I'll wager they would have finished faster, because they wouldn't be wasting time copying all thirteen or so digits that their calculator display spewed at them onto the page. When I went to school, we used printed tables. Although we did study slide rule at some point, high school students in Russia could not be expected to be able to find/afford their own slide rule.

The actual answer given in the book for this problem is ~4.7 years. Perhaps, there were some set requirements for rounding answers at those tests.

Let X = the distance between when he changed speeds and station

The difference in time between the 3km/hr and 4km/h speeds is 1.33 hours. Or,

Xkm/3km/hr = Xkm/4km/hr + 1.33 hours

solving for X yeilds

X = 16 km.

## Checking this work...

To walk 16 km @ 4km/hr takes 4 hrs.

To walk 16 km @ 3km/hr takes 5.33 hrs

The difference is 1.33, as desired

To get the total distance we have to add back the 3km he took to figure this out, so the total distance is 19 KM.

Sorry, that's incorrect. You overlooked that ...

The man walked only part of the way at 4 km/hr. The first part he walked at 3 km/hr.

Also, the difference in the time is not 1.33 hours, but rather 1 hour and 25 min.

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• 0 In the US high school students do rely on calculators. Too much so, in my opinion.

Hear, Hear!

I took far too much time coming up with this answer:

Let v1 be the speed of the faster-moving ...um, thingy, v2 be the speed of the other, and c is the circumference.

9v2 + 4v1 = c

9v2 - 4v1 = 7

c/v2 - c/v1 = 5

If you don't keep making algebra mistakes, you end up with c = 14.7 cm.

Maybe there's a more direct route, but I don't see it.

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• 0 Sorry, that's incorrect. You overlooked that ...

The first part is immaterial; the time delta is computed from the point at which he switched velocity.

Of course, confusing 80 minutes with 85 minutes was a mistake...

The man walked only part of the way at 4 km/hr. The first part he walked at 3 km/hr.

Also, the difference in the time is not 1.33 hours, but rather 1 hour and 25 min.

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• 0 Let X = the distance between when he changed speeds and station

The difference in time between the 3km/hr and 4km/h speeds is 1.4166 hours (Doh!). Or,

Xkm/3km/hr = Xkm/4km/hr + 1.4166 hours

solving for X yeilds

X = 17 km.

## Checking this work...

To walk 17 km @ 4km/hr takes 4.25 hrs.

To walk 17 km @ 3km/hr takes 5.667 hrs

The difference is 1.4166, as desired

To get the total distance we have to add back the 3km he took to figure this out, so the total distance is 20 KM.

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• 0 When I went to school, we used printed tables. Although we did study slide rule at some point, high school students in Russia could not be expected to be able to find/afford their own slide rule.

Really? I was always under the impression that public education was very well funded in Soviet times, and even going as far back as Catherine the Great's reign.

I think when I was in high-school, calculators were banned from math class, but were allowed in science classes. I have memories of trig and log tables printed on the back pages of our exams, but I don't remember using them much. Angles were always pi/3 or something equally easy to remember, and most other things were easy enough to compute in your head or on scrap paper. Also, when giving an answer, things like ln(13.8) were perfectly acceptable as well.

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• 0 4. A man walked every day along 1 km stretch of the streetcar rail. After 1 month he had counted 45 passing trams that moved in the same direction and 120 trams that moved toward him. Given that the man walked 1 km in 12 min, find the speed of tram.

(From Military School tests, 1947).

Assuming he left at sufficiently random times (relative to the train schedule) and that a month was long enough to smooth any bias...

Let Vt = (true) Tram Velocity

Vf = (apparent) Velocity of oncoming trams

Vr = (apparent) Velocity of trams going same direction

# From the problem, he walks 1 km in 12 minutes, or 5km/hour

Vt = Vf - 5 = Vr + 5

~ or ~

Vf = Vt + 5

Vr = Vt - 5

# The ratio of Vf/Vt should equal the number of trams seen in each direction, so

Vf/Vr = 120/45 = 2.6667

(Vt + 5)/(Vt - 5) = 2.66

Vt = 11 km/hour = Speed of Tram

To check:

(11 + 5)/(11 - 5 ) = 16/6 = 2.6667

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• 0
Hear, Hear!

I took far too much time coming up with this answer:

Let v1 be the speed of the faster-moving ...um, thingy, v2 be the speed of the other, and c is the circumference.

9v2 + 4v1 = c

9v2 - 4v1 = 7

c/v2 - c/v1 = 5

If you don't keep making algebra mistakes, you end up with c = 14.7 cm.

Maybe there's a more direct route, but I don't see it.

Sorry, that's not the answer I got. (And mine matched the one in the book.)

Interesting way of constructing equations though. Perhaps, still some algebra mistakes in resolving the system.

To be fair, the time limit for a problem like that is not like time limit on ACT, where you get 60 min to solve 60 problems. A typical Russian university admission math test would be something like 4 or 5 problems with 4 hours to solve them. So if you can solve this under one hour you'd be OK, as long as you also could simplify multistory, three line long trigonometrical and algebraic equations in the other part of the test fast enough.

Depending on university, those tests were designed to sieve the competition of 3 - 15 applicants per one available place. ##### Share on other sites
• 0 let c = circumference of circle

let t = time first and second body meet

c/2-3.5 / 4 = speed of first body

c/2+3.5 / 9 = speed of second body

(c/2-3.5 / 4 )*t = c/2 + 3.5

(c/2+3.5 / 9 )*t = c/2 - 3.5

then I just solved let x = distance (km)

x/3 = t + 40/60

x-3 / 4 = t - 45/60 - 1

Then I just solved again (too lazy to type out everything <_< )

Umm...I don't get question 4 about the trams...care to elaborate? ##### Share on other sites
• 0
Hear, Hear!

I took far too much time coming up with this answer:

Let v1 be the speed of the faster-moving ...um, thingy, v2 be the speed of the other, and c is the circumference.

9v2 + 4v1 = c

9v2 - 4v1 = 7

c/v2 - c/v1 = 5

If you don't keep making algebra mistakes, you end up with c = 14.7 cm.

Maybe there's a more direct route, but I don't see it.

After solving your system of equations, I arrived to the same answer as in the book. So you must have made an error in calculations somewhere. You'd get a partial credit for constructing a correct system of equations. Whether it would be enough to pass, depended on competition. (And if you didn't pass -- 2 years of military service it was.)

I was more lucky in constructing my equations. They yielded smaller intermediate results.

Let X = the distance between when he changed speeds and station

The difference in time between the 3km/hr and 4km/h speeds is 1.4166 hours (Doh!). Or,

Xkm/3km/hr = Xkm/4km/hr + 1.4166 hours

solving for X yeilds

X = 17 km.

## Checking this work...

To walk 17 km @ 4km/hr takes 4.25 hrs.

To walk 17 km @ 3km/hr takes 5.667 hrs

The difference is 1.4166, as desired

To get the total distance we have to add back the 3km he took to figure this out, so the total distance is 20 KM.

Assuming he left at sufficiently random times (relative to the train schedule) and that a month was long enough to smooth any bias...

Let Vt = (true) Tram Velocity

Vf = (apparent) Velocity of oncoming trams

Vr = (apparent) Velocity of trams going same direction

# From the problem, he walks 1 km in 12 minutes, or 5km/hour

Vt = Vf - 5 = Vr + 5

~ or ~

Vf = Vt + 5

Vr = Vt - 5

# The ratio of Vf/Vt should equal the number of trams seen in each direction, so

Vf/Vr = 120/45 = 2.6667

(Vt + 5)/(Vt - 5) = 2.66

Vt = 11 km/hour = Speed of Tram

To check:

(11 + 5)/(11 - 5 ) = 16/6 = 2.6667

Problems 3 and 4 are now officially sooved.

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• 0

let x = distance (km)

x/3 = t + 40/60

x-3 / 4 = t - 45/60 - 1

Then I just solved again (too lazy to type out everything <_< )

Umm...I don't get question 4 about the trams...care to elaborate? let c = circumference of circle

let t = time first and second body meet

c/2-3.5 / 4 = speed of first body

c/2+3.5 / 9 = speed of second body

(c/2-3.5 / 4 )*t = c/2 + 3.5

(c/2+3.5 / 9 )*t = c/2 - 3.5

then I just solved Also correct answers. Except, I didn't get, what did 1.4 mean for the problem 5?

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