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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 3⁄4 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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