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ALGEBRA WORD PROBLEM - NUMBER OF SOLVED EXAMPLES AND EXERCISES

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Method of Solving Algebra Word Problem

if you have not already done so.

That knowledge is a prerequisite here.

















Example 1 of Algebra Word Problem

In a two-digit number, the unit's digit is 7 more than the ten's digit.
Sum of the digits is half of the whole number. Find the digits and number.

solution to Example 1 of Algebra Word Problem :

Let x be the ten's digit of the two-digit number.
Then its unit's digit = x + 7.
We know
the value of a two-digit number = its unit's digit + 10(its ten's digit).
∴ the number = (x + 7) + 10(x) = 11x + 7.
Sum of the digits = (x + 7) + (x) = 2x + 7.
By data,
Sum of the digits is half of the whole number
∴ 2x + 7 = (1⁄2)(11x + 7).

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(2x + 7) = 2(1⁄2)(11x + 7) ⇒ 2 x 2x + 2 x 7 = 11x + 7
⇒ 4x + 14 = 11x + 7
⇒ 4x - 11x = 7 - 14 ⇒ -7x = -7 ⇒ x = (-7)⁄(-7)
x = 1.
∴ten's digit of the two-digit number = x = 1. Ans
and unit's digit of the two-digit number = x + 7 = 1 + 7 = 8.Ans.
∴ the required number is 18. Ans.

Check:
the number = 18
Sum of the digits = 1 + 8 = 9. = (1⁄2)18 = (1⁄2)the number (verified.)

















Example 2 of Algebra Word Problem

In a two-digit number, unit's digit is 3 more than the ten's digit.
The number formed by interchanging the digits and the original number
are in the ratio 7:4. Find the number.


solution to Example 2 of Algebra Word Problem :

Let x be the ten's digit of the two-digit number.
Then its unit's digit = x + 3.
We know
the value of a two-digit number = its unit's digit + 10(its ten's digit).
∴ the original number = (x + 3) + 10(x) = 11x + 3.
The number formed by interchanging the digits = (x) + 10(x + 3)
= x + 10x + 10 x 3 = 11x + 30.
By data the ratio of the number formed by interchanging the digits
and the original number is 7:4.
⇒ (11x + 30) : (11x + 3) = 7 : 4 ⇒ (11x + 30)⁄(11x + 3) = 7⁄4
cross multiplying, we get
4(11x + 30) = 7(11x + 3) ⇒ 4 x 11x + 4 x 30 = 7 x 11x + 7 x 3
⇒ 44x + 120 = 77x + 21 ⇒ 44x - 77x = 21 - 120
⇒ -33x = -99 ⇒ x = (-99)⁄(-33) = 3.
∴ten's digit of the two-digit number = x = 3.
and unit's digit of the two-digit number = x + 3 = 3 + 3 = 6.
∴ the required number is 36. Ans.

Check:
The original number = 36
The number formed by interchanging the digits = 63
Their ratio = 63 : 36 = 7 x 9 : 4 x 9 = 7 : 4 which is the given ratio. (verified.)













Example 3 of Algebra Word Problem

A person walks from one place to another place at a speed of 4km/hr.
If he walks at a speed of 5km/hr, he reaches the place 7 minutes earlier.
Find the distance between the two places.

solution to Example 3 of Algebra Word Problem :

Let x km be the distance between the two places.
We know
time = (distance)⁄(speed)
∴ the time taken to reach the place at a speed of 4km/hr
= (xkm)⁄(4km/hr) = x⁄4.hrs. = (x⁄4) x 60 minutes. = 15x minutes.
and the time taken to reach the place at a speed of 5km/hr
= (xkm)⁄(5km/hr) = x⁄5.hrs. = (x⁄5) x 60 minutes. = 12x minutes.
By data the time taken at 5km/hr is 7 minutes less than that at 4km/hr.
That means 12x = 15x - 7

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

Now we have to solve this equation.
⇒ 12x - 15x = -7 ⇒ -3x = -7
x = (-7)⁄(-3) = 7⁄3
∴ The distance between the two places = x km = (7⁄3)km. Ans.

Check:
The time taken to reach the place at a speed of 4km/hr
= 15x = 15(7⁄3) = 5(7) = 35 minutes.
The time taken to reach the place at a speed of 5km/hr = 12x = 12(7⁄3)
= 4(7) = 28 minutes. = 35 minutes - 7 minutes which is as per data.(verified.)









Exercise on Algebra Word Problem

Solve each of the following Algebra Word problem.

  1. In a two-digit number, the unit's digit is two more than ten's digit.
    Sum of the digits is equal to 1⁄4 of the whole number.
    Find the number.
  2. In a two-digit number, the unit's digit is twice the ten's digit. If sum
    of the digits is added to the whole number, the result is equal to 30.
    Find the number.






Answers to Exercise on Algebra Word Problem

The Answers to each of the Algebra Word Problem
of the exercise given above, are given below.

  1. 24
  2. 24















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