ALGEBRA WORD PROBLEMS - ANSWERS

1.   Jim has 16 coins in his pocket. Some are 10p coins
      and some are 5p coins. The total value of the coins
       is £1.25  how many of each coin does Jim have?

Jim has an unknown amount of 10p coins (call this unknown amount x) and an unknown amount of 5p coins (call this unknown amount y).

We know that the total of Jim's coins = £1.25.

At this point convert the £'s to pence to make the calculation easier - £1.25 = 125 pence.

So, 10x + 5y = 125

We are also told that Jim has 16 coins in his pocket, so:

x + y = 16

All you need to do now is solve the simultaneous equations:

10x + 5y = 125      (1)
    x +   y = 16        (2)

Multiply (2) by 5 and subtract (2) from (1) to cancel the y's:

10x + 5y = 125
  5x + 5y = 80

Subtracting we get: 5x = 45 hence x = 9
Substitute into (2) we get y = 16 - 9 = 7

Hence Jim has nine 10p coins and seven 5p coins

2.  When an object is dropped from a building which is
      x metres tall, the distance it travels is directly
      proportional to the square of the time taken during
      its fall.

    When t = 2 seconds, x = 20 metres.
    Calculate the value of x when the time taken is 5     seconds.

Any question with the word "proportional" in will expect you to do the following:

1) State the proportionality relationship:     x a t2

2) Create an equation:   x = kt2    (where k = constant)

3) Find k: When t = 2, x = 20
                 Hence: 20 = k(2)2
                             20 = 4k
                               k = 20/4 = 5.

4) Use the equation to solve another problem:

      x = 5(5)2

      x = 5 times 25 = 125m
3.  A ladder is 4m long and is resting against a wall at
     the top and is 1m from the wall at the bottom.
     How high up the wall does the ladder reach?

Ladders resting against walls always mean trigonometry or Pythagoras' Theorem.

Here, we're looking at Pythagoras as there are no angles given.

Draw a picture and you'll see that the hypotenuse is 4m long.

Hence 42 = 12 + x2

Leading to the solution x = Ö7 m or 2.65m (to 2 d.p.)

4.  James cycled from home to his friend's house for a
    distance of 26 km and later on cycled back the same     distance. James' average speed on the way to his     friend's house was x km/h.

    On his return journey James cycled 2 km/h quicker
    as it would get dark soon and his lights were not     working, so he managed to get back 10 minutes
    quicker than it took to reach his friends' house
    originally.
     a) Write in terms of x, the time in hours taken for:
     i) The outward journey
     ii) The return journey

     b) Show that x2 + 2x - 312 = 0
     c) Calculate James' average speed on the return
          journey.

Remember that:    Time = Distance / Average Speed
a)
(i) Outward Time = 26 / x
(ii) For the return journey, James' average speed is 2km/h quicker this can be represented as x + 2.

Hence Return Time = 26 / x + 2

b) Outward Time - Return Time = 10 mins

Convert the minutes to hours to be consistent = 1/6 hours
Hence,

c) Solve the quadratic to get a non-negative value for x, then remember to add 2 to get James' average speed on the return journey:

x = 16.69 hence James' average speed = 18.69km/h

5. A cuboid has a volume of 60 cm3. It has a square     base of length x cm and a height of h cm.

    Find an equation to represent x.
    Hence find x when h = 6 and find h when x = 3.

A Cuboid is a cube which is not quite a cube (like a 3d rectangle).

We are told that the base is square and has a length of x cm.
As the base is square all the lengths will be the same, hence the area of the base = x2.

We are given the volume as 60cm3 which we also know will be the area of the base times the height. Hence,

60 = x2 + h

To find x, rearrange the formula: x =  Ö (60 - h)

To find h, rearrage the formula: h = 60 - x2.

Hence x =  Ö54cm    (or 7.35cm to 2 d.p.)
and   h = 51cm
6.  I subtract 5 from a number and multiply the result
     by 6. The answer is the same as when I multiply the      number by 5 and subtract 10 from the result.
     What is the number?

Let the number = x

6(x - 5) = 5x - 10
6x - 30 = 5x - 10
x -30 = -10

x = 20
7.  Mrs Young is 7 times older than her daughter, but in      25 years' time she will only be twice as old as her      daughter. How old is Mrs Young?

It's weird that ages work in this way....

Let Mrs Young's age = m and her daughter's age = d

Today:
m = 7d

In 25 year's time:
Mrs Young's age = m + 25
Her daughter's age = d + 25

As in 25 years Mrs. Young is twice as old as her daughter, we can say:

m = 2(d + 25) = 2d + 50

We know also that m = 7d, so by substitution,

7d = 2d + 50
5d = 25
d = 5

If the daughter's age is 5 then as Mrs. Young's age = 7d,
she is 35 years old.
8.  The width of a rectangular area of a field is 4 metres      less than its length. The area enclosed is 60m2. Find
     a) The length of the rectangle and hence find
     b) the width of the rectangle.

a) Let the length = x metres, then the width = x - 4 metres.

Area = Length x Width
= x(x -4)
= x2 - 4x

We know that the area = 60m2, so we can say:

x2 - 4x = 60

Rearrange this as a quadratic:
x2 - 4x - 60 = 0

Factorising we get:
(x + 6) (x -10) = 0

x = -6 is not possible hence x must = 10 metres

b) If the length = 10m we know that the width is 4 metres less, hence width = 10 - 4 = 6 metres.