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TOPICS
Exercise - 2.2

Question-1 :-  If you subtract 1/2, from a number and multiply the result by 1/2, you get 1/8. What is the number?

Solution :-
```  Let the number is x.
According to question :
1/2 (x - 1/2) = 1/8
x/2 - 1/4 = 1/8
x/2 = 1/8 + 1/4
x/2 = (1 + 2)/8
x/2 = 3/8
x = (3 x 2)/8
x = 6/8
x = 3/4
The number is 3/4.
```

Question-2 :-  The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?.

Solution :-
```  Breadth = x m
Length = 2x + 2 m
The perimeter of a rectangular pool = 154 m
Accordding to question :
2(Length + Breadth) = 154 m
2(2x + 2 + x) = 154
3x + 2 = 154/2
3x = 77 - 2
3x = 75
x = 75/3
x = 25
Breadth of rectangular swimming pool = x = 25 m
Length of rectangular swimming pool = 2x + 2 = 2 x 25 + 2 = 50 + 2 = 52 m
```

Question-3 :-  The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 4 215 cm . What is the length of either of the remaining equal sides?

Solution :-
```  Let each of equal sides of an isosceles triangle = x cm.
The base of an isosceles triangle = 4/3 cm
Perimeter of a triangle = 4 2⁄15 cm
Sum of three sides = 4 2⁄15 cm
Accordding to question :
x + 4/3 + x = (60 + 2)/15
2x = 62/15 - 4/3
2x = (62 - 20)/15
2x = 42/15
x = 42/30
x = 7/5
Each of equal sides of an isosceles triangle = x = 7/5 cm.
```

Question-4 :-  Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Solution :-
```  Let the first number = x and
the second number = x + 15
Sum of two numbers = 95
Accordding to question :
x + x + 15 = 95
2x + 15 = 95
2x = 95 - 15
2x = 80
x = 80/2
x = 40
First number = x = 40
Second number = x + 15 = 40 + 15 = 55
```

Question-5 :-  Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Solution :-
```  Since the ratio of the two numbers = 5 : 3,
Let first number = 5x and
second number = 3x.
The difference between the two numbers = (5x – 3x).
It is given that the difference = 18.
Accordding to question :
5x – 3x = 618
2x = 18
x = 18/2
x = 9
First number = 5x = 5 × 9 = 45 and
Second number = 3x = 3 × 9 = 27.
```

Question-6 :- Three consecutive integers add up to 51. What are these integers?

Solution :-
```  Three consecutive integer numbers are x, x + 1 and x + 2.
Sum of three consecutive integer numbers = 51
Accordding to question :
x + x + 1 + x + 2 = 51
3x + 3 = 51
3x = 51 - 3
3x = 48
x = 48/3
x = 16
First integer number = x = 16
Second integer number = x + 1 = 16 + 1 = 17
Third integer number = x + 2 = 16 + 2 = 18
```

Question-7 :-  The sum of three consecutive multiples of 8 is 888. Find the multiples.

Solution :-
```  Let Three consecutive numbers x, x + 8 and x + 16 are multiples of 8.
sum of three consecutive multiples of 8 = 888
Accordding to question :
x + x + 8 + x + 16 = 888
3x + 24 = 888
3x = 888 - 24
3x = 864
x = 864/3
x = 288
First multiple number of 8 = x = 288
Second multiple number of 8 = x + 8 = 288 + 8 = 296
Third multiple number of 8 = x + 16 = 288 + 16 = 304
```

Question-8 :-  Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

Solution :-
```  Let three consecutive numbers are x, x + 1 and x + 2.
multiplied y 2 in first number  = 2x
multiplied y 3 in second number  = 3(x + 1)
multiplied y 4 in first number  = 4(x + 2)
Sum of numbers = 74
Accordding to question :
2x + 3(x + 1) + 4(x + 2) = 74
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 74
9x = 74 - 11
9x = 63
x = 63/9
x = 7
First number = x =  7
Second number = x + 1 = 7 + 1 = 8
Third number = x + 2 = 7 + 2 = 9
```

Question-9 :- The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

Solution :-
```  Let the present ages of Rahul = 5x years and
the present ages of Haroon = 7x years
Now, 4 years later age of Rahul = 5x + 4
Now, 4 years later age of Haroon = 7x + 4
Sum of their ages = 56 years
Accordding to question :
5x + 4 + 7x + 4 = 56
12x + 8 = 56
12x = 56 - 8
12x = 48
x = 48/12
x = 4
The present ages of Rahul = 5x = 5 x 4 = 20 years
The present ages of Haroon = 7x = 7 x 4 = 28 years
```

Question-10 :-  The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Solution :-
```  Let, The number of girls = x
The number of boys = x + 8
Accordding to question :
(x + 8)/x = 7/5
5(x + 8) = 7x
5x + 40 = 7x
40 = 7x - 5x
40 = 2x
x = 40/2
x = 20
The number of girls = x = 20
The number of boys = 20 + 8 = 28
```

Question-11 :-  Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Solution :-
```  Let the age of Baichung's  = x years
the age of Baichung's father = x + 29 years
the age of Baichung's grandfather = x + 29 + 26 = x + 55 years
The sum of the ages of all the three = 135 years
Accordding to question :
x + x + 29 + x + 55 = 135
3x + 84 = 135
3x = 135 - 84
3x = 51
x = 51/3
x = 17
The age of Baichung's  = x = 17 years
The age of Baichung's father = x + 29 = 17 + 29 = 46 years
The age of Baichung's grandfather = x + 55 = 17 + 55 = 72 years
```

Question-12 :- Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Solution :-
```  Let the present age of Ravi = x years
Now, after 15 years later age of Ravi = 4x
Fifteen years from now, Ravi’s age  = x + 15 years
Accordding to question :
4x = x + 15
4x - x = 15
3x = 15
x = 15/3
x = 5
The present age of Ravi = x = 5 years
```

Question-13 :-  A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number ?

Solution :-
```  Let, The rational number = x
Accordding to question :
5x/2 + 2/3 = -7/12
5x/2 = -7/12 - 2/3
5x/2 = (-7 - 8)/12
5x/2 = -15/12
5x/2 = -5/4
x = (-5 x 2)/(4 x 5)
x = -10/20
x = -1/2
The rational number = x = -1/2
```

Question-14 :-  Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have?

Solution :-
```  Let the number of notes of ₹ 100 = 2x
the number of notes of ₹ 50 = 3x
the number of notes of ₹ 10 = 5x
The total cash with Lakshmi = ₹ 4,00,000
According to question :
100 x 2x +  50 x 3x + 10 x 5x = 400000
200x + 150x + 50x = 400000
400x = 400000
x = 400000/400
x = 1000
The number of notes of ₹ 100 = 2x = 2 x 1000 = 2000
The number of notes of ₹ 50 = 3x = 3 x 1000 = 3000
The number of notes of ₹ 10 = 5x = 5 x 1000 = 5000
```

Question-15 :-  I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Solution :-
```  Let the number of ₹ 5 coins = x
the number of ₹ 2 coins = 3x
the number of ₹ 1 coins = 160 - (x + 3x) = 160 - 4x
total money = ₹ 300
According to question :
5x + 2 x 3x + 160 - 4x = 300
7x = 300 - 160
7x = 140
x = 140/7
x = 20
The number of ₹ 5 coins = x = 20
The number of ₹ 2 coins = 3x = 3 x 20 = 60
The number of ₹ 1 coins = 160 - 4x = 160 - 4 x 20 = 160 - 80 = 80
```

Question-16 :-  The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.

Solution :-
```  Let The number of winners of ₹ 100 = x
The number of not winners of ₹ 25 = 63 - x
Total money = ₹ 3000
According to question :
100x + 25 x (63 - x) = 3000
100x + 1575 - 25x = 3000
75x = 3000 - 1575
75x = 1425
x = 1425/75
x = 19
The number of winners of ₹ 100 = x = 19
```
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