TOPICS
Unit-2(Examples)

Example-1 :-  Find the solution of 2x – 3 = 7.

Solution :-
  2x - 3 = 7
  2x = 7 + 3
  2x = 10
  x = 10/2
  x = 5
  Verification : 
  L.H.S = 2x - 3 = 2 x 5 - 3 = 10 - 3 = 7 = R.H.S 
    

Example-2 :-  Solve 2y + 9 = 4.

Solution :-
  2y + 9 = 4
  2y = 4 - 9
  2y = -5
   y = -5/2
  Verification : 
  L.H.S = 2y + 9 = 2 x (-5/2) + 9 = -5 + 9 = 4 = R.H.S 
    

Example-3 :- Solve x/3 + 5/2 = -3/2.

Solution :-
  x/3 + 5/2 = -3/2
  x/3 = -3/2 - 5/2
  x/3 = (-3 - 5)/2
  x/3 = -8/2
  x/3 = -4
    x = -4 x 3
    x = -12
  Verification : 
  L.H.S = x/3 + 5/2 = -12/3 + 5/2 = -4 + 5/2 = (-8 + 5)/2 = -3/2 = R.H.S 
    

Example-4 :-  Solve 15/4 - 7x = 9.

Solution :-
  15/4 - 7x = 9
  15/4 - 9 = 7x
  7x = (15 - 36)/4 
  7x = -21/4
  x = -21/(4 x 7)
  x = -3/4
  Verification : 
  L.H.S = 15/4 - 7x 
        = 15/4 - 7 x (-3/4) 
        = 15/4 + 21/4 
        = (15 + 21)/4 
        = 36/4
        = 9 = R.H.S 
    

Example-5 :-  What should be added to twice the rational number -7/3 to get 3/7.

Solution :-
  Let the number added is X in given rational number.
  According to question :
  X + 2 x (-7/3) = 3/7
  X + (-14/3) = 3/7
  X = 3/7 + 14/3
  X = (9 + 98)/21
  X = 107/21
  So, 107/21 be added to twice the rational number -7/3 to get 3/7.
    

Example-6 :- The perimeter of a rectangle is 13 cm and its width is 2 34 cm. Find its length.

Solution :-
  Let the length of the rectangle to be X cm. 
  The perimeter of the rectangle = 2 × (length + width) = 13 cm
  width = 2 34 cm = 11/4 cm
  Then, 2 x (X + 11/4) = 13
  2x + 22/4 = 13
  2x = 13 - 11/2
  2x = (26 - 11)/2
   x = 15/4
  Length of rectangle is 15/4 cm.
    

Example-7 :-  The present age of Sahil’s mother is three times the present age of Sahil. After 5 years their ages will add to 66 years. Find their present ages.

Solution :-
  Let Sahil’s present age be x years.
  So, mother's age is 3x years.
  According to question :
  (x + 5) + (3x + 5) = 66
  4x + 10 = 66
  4x = 66 - 10
  4x = 56
   x = 56/4
   x = 14
  Then, Sahil's present age be x = 14 years.
  And mother's age is 3x = 3 x 14 = 42 years.
    

Example-8 :-  Bansi has 3 times as many two-rupee coins as he has five-rupee coins. If he has in all a sum of ₹ 77, how many coins of each denomination does he have?

Solution :-
  Let the number of five-rupee coins that Bansi has be X. 
  Then the number of two-rupee coins he has is 3 times X or 3X. 
  The amount Bansi has: 
(i) from 5 rupee coins, ₹ 5 × X = ₹ 5X 
(ii) from 2 rupee coins, ₹ 2 × 3X = ₹ 6X 
  Hence the total money he has = ₹ 11X 
  But this is given to be ₹ 77; 
  therefore, 11X = 77
  X = 77/11
  X = 7
  Thus, number of five-rupee coins = X = 7 and 
        number of two-rupee coins = 3X = 21 
    

Example-9 :- The sum of three consecutive multiples of 11 is 363. Find these multiples.

Solution :-
  Let three Consecutive numbers x, x + 11, x + 22 which are multiple of 11.
  Sum of multiples of 11 = 363
  x + x + 11 + x + 22 = 363
  3x + 33 = 363
  3x = 363 - 33
  3x = 330
   x = 330/3
   x = 110 
  Now, first number is x = 110, 
  second number is x + 11 = 110 + 11 = 121,
  third number is x + 22 = 110 + 22 = 132 
    

Example-10 :-  The difference between two whole numbers is 66. The ratio of the two numbers is 2 : 5. What are the two numbers?

Solution :-
  Since the ratio of the two numbers is 2 : 5, 
  we may take one number to be 2x and the other to be 5x. 
  The difference between the two numbers is (5x – 2x). 
  It is given that the difference is 66. 
  Therefore, 
  5x – 2x = 66  
  3x = 66 
   x = 22 
  Since the numbers are 2x and 5x, they are 2 × 22 = 44 and 5 × 22 = 110, respectively. 
  The difference between the two numbers is 110 – 44 = 66 as desired. 
    

Example-11 :-  Deveshi has a total of ₹ 590 as currency notes in the denominations of ₹ 50, ₹ 20 and ₹ 10. The ratio of the number of ₹ 50 notes and ₹ 20 notes is 3:5. If she has a total of 25 notes, how many notes of each denomination she has?

Solution :-
   Let the number of ₹ 50 notes and ₹ 20 notes be 3x and 5x, respectively. 
   But she has 25 notes in total. 
   Therefore, the number of ₹ 10 notes = 25 – (3x + 5x) = 25 – 8x. 
   The amount she has 
   from ₹ 50 notes : 3x × 50 = ₹ 150x 
   from ₹ 20 notes : 5x × 20 = ₹ 100x 
   from ₹ 10 notes : (25 – 8x) × 10 = ₹ (250 – 80x)
   Hence the total money  she has = 150x + 100x + (250 – 80x) = ₹ (170x + 250) 
   But she has ₹ 590. 
   Therefore, 
   170x + 250 = 590  
   170x = 590 – 250 
   170x = 340
      x = 340/170
      x = 2
   The number of ₹ 50 notes she has = 3x = 3 × 2 = 6 
   The number of ₹ 20 notes she has = 5x = 5 × 2 = 10 
   The number of ₹ 10 notes she has = 25 – 8x = 25 – (8 × 2) = 25 – 16 = 9
    

Example-12 :-  Solve 2x – 3 = x + 2.

Solution :-
  2x – 3 = x + 2
  2x - x = 2 + 3
       x = 5
  Verification :
  L.H.S = 2x – 3 = 2 x 5 - 3 = 10 - 3 = 7
  R.H.S = x + 2 = 5 + 2 = 7 
  L.H.S = R.H.S
    

Example-13 :-  Solve 5x + 7/2 = 3x/2 - 14.

Solution :-
  5x + 7/2 = 3x/2 - 14
  5x - 3x/2 = -14 - 7/2
  (10x - 3x)/2 = (-28 - 7)/2
  7x/2 = -35/2
  7x = -35
   x = -35/7
   x = -5
  Verification :
  L.H.S = 5x + 7/2 = 5 x (-5) + 7/2 = -25 + 7/2 = (-50 + 7)/2 = -43/2
  R.H.S = 3x/2 - 14 = [3 x (-5)]/2 - 14 = -15/2 -14 = (-15 - 28)/2 = -43/2 
  L.H.S = R.H.S
    

Example-14 :- The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can be the original number?

Solution :-
   Let us take the two digit number such that the digit in the units place is b. 
   The digit in the tens place differs from b by 3.  
   Let us take it as b + 3.  
   So the two-digit number is 10 (b + 3) + b = 10b + 30 + b = 11b + 30. 
   With interchange of digits, the resulting two-digit number will be 10b + (b + 3) = 11b + 3. 
   If we add these two two-digit numbers, their sum is (11b + 30) + (11b + 3) = 11b + 11b + 30 + 3 = 22b + 33. 
   It is given that the sum is 143. 
   Therefore,  
   22b + 33 = 143  
   22b = 143 – 33 
   22b = 110
     b = 110/22
     b = 5
   The units digit is 5 and therefore the tens digit is 5 + 3 which is 8. 
   The number is 85.
    

Example-15 :-  Arjun is twice as old as Shriya. Five years ago his age was three times Shriya’s age. Find their present ages.

Solution :-
   Let us take Shriya’s present age to be x years. 
   Then Arjun’s present age would be 2x years. 
   Shriya’s age five years ago was (x – 5) years. 
   Arjun’s age five years ago was (2x – 5) years. 
   It is given that Arjun’s age five years ago was three times Shriya’s age. 
   Thus, 2x – 5 = 3(x – 5)  
   2x – 5 = 3x – 15 
   15 – 5 = 3x – 2x 
   10 = x
    x = 10 
   So, Shriya’s present age = x = 10 years. 
   Therefore, Arjun’s present age = 2x = 2 × 10 = 20 years.
    

Example-16 :-  Solve (6x + 1)/3 + 1 = (x - 3)/6.

Solution :-
  (6x + 1)/3 + 1 = (x - 3)/6
  (6x + 1 + 3)/3 = (x - 3)/6
  (6x + 4)/3 = (x - 3)/6
  6 x (6x + 4) = 3 x (x - 3)
  36x + 24 = 3x - 9
  36x - 3x = -9 - 24
       33x = -33
         x = -1
  Verification : 
  L.H.S = (6x + 1)/3 + 1 
        = [6 x (-1) + 1]/3 + 1 
        = (-6 + 1)/3 + 1 
        = -5/3 + 1 
        = (-5 + 3)/3
        = -2/3
  R.H.S = (x - 3)/6
        = (-1 - 3)/6
        = -4/6
        = -2/3
  L.H.S = R.H.S
    

Example-17 :- Solve 5x – 2(2x – 7) = 2(3x – 1) + 7/2.

Solution :-
  5x – 2(2x – 7) = 2(3x – 1) + 7/2
  5x - 4x + 14 = 6x - 2 + 7/2
  x - 6x = -14 - 2 + 7/2
    -5x = -16 + 7/2
    -5x = (-32 + 7)/2
    -5x = -25/2
      x = 5/2
  Verification : 
  L.H.S = 5x – 2(2x – 7)
        = 5 x 5/2 - 2(2 x 5/2 - 7)
        = 25/2 - 2(5 - 7)
        = 25/2 + 4
        = (25 + 8)/4
        = 33/4
  R.H.S = 2(3x – 1) + 7/2
        = 2(3 x 5/2 - 1) + 7/2
        = 2(15/2 - 1) + 7/2
        = 2(15 - 2)/2 + 7/2
        = 13 + 7/2
        = (26 + 7)/2

    

Example-18 :-  Solve (x + 1)/(2x + 3) = 3/8.

Solution :-
   (x + 1)/(2x + 3) = 3/8
   8 (x + 1) = 3 (2x + 3)
   8x + 8 = 6x + 9
   8x - 6x = 9 - 8
        2x = 1
         x = 1/2
   Verification : 
   L.H.S = (x + 1)/(2x + 3)
         = (1/2 + 1)/(2 x 1/2 + 3)
         = (1 + 2)/2 ÷ (1 + 3)
         = 3/2 ÷ 4
         = 3/8 = R.H.S
    

Example-19 :-  Present ages of Anu and Raj are in the ratio 4:5. Eight years from now the ratio of their ages will be 5:6. Find their present ages.

Solution :-
   Let the present ages of Anu and Raj be 4x years and 5x years respectively. 
   After eight years, Anu’s age = (4x + 8) years; 
   After eight years, Raj’s age = (5x + 8) years.
   Therefore, the ratio of their ages after eight years = (4x + 8)/(5x + 8) 
   This is given to be 5 : 6
   Therefore,
   (4x + 8)/(5x + 8) = 5/6
   6 (4x + 8) = 5 (5x + 8)
   24x + 48 = 25x + 40
   24x - 25x = 40 - 48
          -x = -8
           x = 8
   Therefore, 
   Anu’s present age = 4x = 4 × 8 = 32 years 
   Raj’s present age = 5x = 5 × 8 = 40 years
    
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