WORD PROBLEMS - NUMBER OF SOLVED EXAMPLES AND EXERCISES

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Method of Solving Word Problems

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Example 1 of Word Problems

A man is 35 years old and his son is 7 years old now.
In how many years will the son be half as old as his father.

solution to Example 1 of Word Problems :

Let x be the number of years after which the son will be
half as old as his father.
After x years,
the age of the father = present age + x = 35 + x.
the age of the son = present age + x = 7 + x.
But after x years, the age of the son = (1⁄2) the age of the father.
⇒ 7 + x = (1⁄2)(35 + x)

This is the Linear Equation formed by converting
the given word statements to the symbolic language.

Now we have to solve this equation.
Multiplying both sides with 2, we get
2(7 + x) = 2(1⁄2)(35 + x) ⇒ 2 x 7 + 2x = 35 + x
⇒ 14 + 2x = 35 + x ⇒ 2x - x = 35 - 14 ⇒ x = 21.
∴ 21 years is the required answer.

Check:
After 21 years, the father's age = 35 + 21 = 56.
After 21 years, the son's age = 7 + 21 = 28. = half of 56. (verified)

Example 2 of Word Problems

The present age of a man is three times that of his son. Six years ago
the age of the man was four times that of his son. Find the ratio
of their ages 6 years later.

solution to Example 2 of Word Problems :

Let x be the present age of the son.
As per data, present age of man = three times that of his son.
∴ present age of man = 3x.
six years ago, the age of the son = x - 6.
six years ago, the age of the father = 3x - 6.
As per data, age of man six years ago = four times that of his son.
⇒ 3x - 6 = 4(x - 6) = 4x - 24

This is the Linear Equation formed by converting
the given word statements to the symbolic language.

Now we have to solve this equation.
⇒ 3x - 4x = -24 + 6 = -18.
⇒ -x = -18 ⇒ x = 18.
∴ Present age of son = x = 18
and present age of man = 3x = 3 x 18 = 54.
6 years later, age of son = 18 + 6 = 24
and age of man = 54 + 6 = 60.
the ratio of their ages 6 years later = 60⁄24 = 5⁄2 = 2.5.
i.e. 6 years later, age of man = 2.5 times that of his son. Ans.

Check:
present age of man = 54 = 3 x 18 = three times that of his son. (verified.)
Six years ago the age of the man = 54 - 6 = 48 = 4(12) = 4 (18 - 6)
= four times that of his son. (verified.)

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Example 3 of Word Problems

There are some lotus flowers in a pond and some bees are hovering around.
If one bee lands on each flower, one bee will be left. If two bees land on
each flower, one flower will be left. Find the number of bees and the
number of flowers in the pond.

solution to Example 3 of Word Problems :

Let x be the number of flowers.
As per data, If one bee lands on each flower, one bee will be left.
That means Number of bees is one more than that of flowers.
∴ Number of bees = x + 1.
Also By data, If two bees land on each flower, one flower will be left.
That means Number of bees = 2 x (Number of flowers - 1)
x + 1 = 2(x - 1)

This is the Linear Equation formed by converting
the given word statements to the symbolic language.

Now we have to solve this equation.
x + 1 = 2x -2 ⇒ x - 2x = -2 -1
⇒ -x = -3 ⇒ x = 3.
Number of flowers = x = 3. Ans.
Number of bees = x + 1 = 3 + 1 = 4. Ans.

Check:
Out of 4 bees, if one bee lands on each of the 3 flowers,
then one bee is left which is same as per data. (verified.)
Out of 4 bees, if two bees land on each of the 2 flowers,
then one flower is left which is same as per data. (verified.)

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Exercise on Word Problems

Solve the following Word problems.

  1. Present ages of A and B are 20 years and 5 years respectively.
    After how many years will the age of A be twice that of B ?
  2. A person travelled 5⁄8 of the distance by train, 1⁄4 by bus
    and the remaining 15 km, by boat. Find the total distance travelled by him.
  3. A milk man has some buffaloes, cows and goats. He has goats equal to
    5⁄2 times the number of cows and cows equal to 3⁄2 times the
    number of buffaloes. If there are 150 animals with the milk man,
    how many of each category has he with him ?


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Answers to Exercise on Word Problems

The Answers to the Word Problems of the
exercise given above, are given below.

  1. 10
  2. 120 km
  3. 24 buffaloes, 36 cows, 90 goats.