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Word Problems

Several types of everyday problems expressed in language form can be solved by translating the language sentences into equations.

Problems on Numbers
 

Example :

One number is half the other. Sum of two numbers is equal to 6. Find the numbers.

 Solution :

 Let one number be

  x. The other number =

 Sum of two numbers = 6.

 Hence, x + = 6

 

Example :  

The numbers are 4 and 2.
Check : The sum of two numbers = 6. Hence the answer is correct.

 

Example :

The numerator of a fraction is 3 less than its denominator. If the numerator is increased by 1 and the denominator is increased by 3, the fraction becomes equal to . Find the original fraction.

 Solution :

 Let the denominator of the original fraction be x.

 If the numerator is 3 less than the denominator, then numerator =

  x – 3.

 Original fraction = .

 The denominator of the new fraction =

  x + 3

 The numerator of the new fraction = (

  x-3) + 1 = x-2.

 The new fraction = 
 The new fraction becomes.

 Hence , 
 

 
2 (x - 2) = 1 (x +3)
Apply cross multiplication
2 x - 4 = x + 3
Simplify LHS and RHS
2 x - x = 3 + 4
Transpose x to LHS and (- 4) to RHS
x = 7


... original fraction = 

Check : Original numerator + 1    
           Original denominator + 3  

         =  which is correct.
Simplify LHS and RHS

 

Example :

The sum of two numbers is 35. Their difference is 13. Find the numbers.

  Solution :

 Let one number be x. Then, the other number is (35 - x).

 The difference of the numbers is 13.

 (35 - x) - x = 13
 

(35 - x) - x = 13
 
35 - 2 x = 13.
Simplify LHS
(- 2) x = 13 - 35
Transpose 35 to RHS
(- 2) x = (- 22)
Simplify RHS
x = 
Transpose ( - 2) to RHS
x = 11
 


The two numbers are 11 and 24.

Check: Sum of numbers = 11+24= 35;
difference of numbers =  24 - 11= 13, hence the answer is correct.

 

 

Example :

Two number are in the ratio 3 : 4. sum of two numbers is equal to 56. Find the numbers?

Solution :

Let the numbers be 3x and 4x. 

Their sum is equal to 56.

7x = 56
x = 8
... the numbers are 3 x 8 = 24, and 4 x 8 = 32. 

 
Example :

The sum of three consecutive numbers is 243. Find the numbers.

Solution :

Method 1: Let the numbers be x, x+1, x+2.

The sum is x + (x +1) + (x +2) = 243

3x + 3 = 243

...3x = 240 or x = 80.

The numbers are 80, 81, 82.

Problems Involving Place Values


Remember that if the digits of a 2 digit number are a (units) and b (tens) then the number is 10 b + a.

 
Example :

In a 2-digit number, the tens digit is 4 times the units digit. When the digits are reversed, the new number formed is 54 less than the original number. Find the original number.

Solution :

Let the units digit of the number be x.

The tens digit = 4x
 

  Tens Units Number
Old number 4 x x (4 x ´ 10) + (x ´ 1) = 41 x
New number
Difference between the two = 41x - 14x = 54 (given)
x 4 x (x ´ 10) + (4 x ´ 1) = 14 x
 

 

Example : 

27x =54

x = 2
The old number = (8 × 10) + (2 × 1) = 82
The new number =(2 × 10) + (8 × 1) = 28
Check: Difference between the old and new number
          = 82 - 28 = 54, which is correct.

Age Problems

 

Example :

Anil is 9 years older than Ajith. In 10 years, Anil will be twice as old as Ajith was 10 years ago. Find their present ages.

Solution :

Let Ajith's present age be x. Look at the table given here. 
YearsAnil AjithPresentx+9x10 years ago
x - 1010 years hencex+9+10 = x + 19
 

The problem says 10 years hence, Anil will be twice as old as Ajith was 10 years ago.

Anil's age 10 years hence = x+19 ………(1)

Ajith's age 10 years ago = x-10 ……….(2)

Therefore (1) = 2 × (2)

x + 19 = 2(x - 10)

x - 2x = -20 - 19 or -x = -39
...   x = 39
Their present ages are: Anil = 48 years; Ajith = 39 years.
Check: 10 years hence Anil will be 58 years. Ajith was 29 years old, 10 years ago

Checking   58 = 2 × 29 Here L.H.S = R.H.S.





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