Linear Equations in One Variable - 8th math

Question (7) The sum of three consecutive multiples of 8 is 888. Find the multiples.

Solution:

Let first multiple of 8 = 8a

Therefore, second consecutive multiple of 8 = 8 ( a + 1)

And, third consecutive multiple of 8 = 8 ( a + 2 )

As given in the question, sum of three consecutive multiples of 8 = 888

Therefore,

8a + [8 (a + 1)] + [8 (a + 2)] = 888

⇒ 8a + (8a + 8) + (8a + 16) = 888

⇒ 8a + 8a + 8 + 8a + 16 = 888

After rearranging the above expression

⇒ 8a + 8a + 8a + 8 + 16 = 888

⇒ 24a + 24 = 888

After transposing 24 to RHS, we get

24a = 888 – 24

24a = 864

After dividing both sides by 24, we get

Now, since first multiple of 8 = 8a

Thus, first multiple = 8 × 36= 288

[After substituting the value of a = 36]

And, second multiple of 8 = 8(a + 1)

Thus, second multiple = 8 (36 + 1)

[After substituting the value of a = 36]

= 8 × 37 = 296

And third multiple of 8 = 8 (a + 2)

[After substituting the value of a = 36]

Therefore, third multiple = 8 ( 36 + 3)

= 8 × 39 = 304

Thus, three required consecutive multiples of 8 = 288, 296 and 304 Answer

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 first integer = a

 

Therefore, second consecutive integer = a + 1

 

And, similarly, the third consecutive integer = a + 2

Now, According to question

(First integer × 2) + (second consecutive integer × 3) + (third consecutive integer × 4) = 74

⇒ (a × 02)+[(a+1) × 3]+[(a+2) × 4]=74

⇒ 2a + (3a + 3) + (4a + 8) = 74

⇒ 2 a + 3 a + 3 + 4 a + 8 = 74

After rearranging the above expression

⇒ 2 a + 3 a + 4 a + 3 + 8 = 74

⇒ 9 a + 11 = 74

After transposing 11 to RHS, we get

9 a = 74 – 11

⇒ 9 a = 63

After transposing 9 to RHS, we get

⇒ a = 7

Now, after substituting the value of a = 7 in the rest two consecutive integers we get

The second consecutive integer

= a + 1 = 7 + 1 = 8

And, the third consecutive integer

= a + 2 = 7 + 2 = 9

Thus, three required consecutive integers are 7, 8 and 9 Answer

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:

Given, The ages of Rahul and Harron are in the ratio of 5:7

Therefore, Let the age of Rahul = 5 x

And the age of Haroon = 7 x

Now, After 4 years

The age of Rahul = 5 x + 4

And, the age of Haroon = 7 x + 4

Now, according to question, the sum of ages of Rahul and Haroon after 4 years = 56 years

Thus, (5 x + 4) + (7 x + 4) = 56

⇒ 5 x + 7 x + 4 + 4 = 56

⇒ 12 x + 8 = 56

Now, after transposing 8 to RHS

⇒ 12 x = 56 – 8

⇒ 12 x = 48

Now, after transposing 12 to RHS we get

⇒ x = 4

Now, after Substituting the value of x we can find the ages of Rahul and Haroon

Thus, the age of Rahul

= 5 x = 5 × 4 = 20 years

And, the age of Haroon

= 7 x = 7 × 4 = 28 years

Thus, ages of Rahul and Haroon are 20 years and 28 years respectively Answer

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:

 

Given, the ratio of number of boys and girls in the class = 7:5

And, the number of boys = number of girls + 8

Thus, total class strength = ?

Since, according to question the number of boys : girls = 7:5

Thus, Let the number of boys = 7 x

And the number of girls = 5 x

According to question, number of boys is 8 more than the number of girls. This means the number of girls is 8 less than the number of boys.

Thus, since the number of boys = 7x

Therefore, the number of girls = 7x – 8

⇒ 5x = 7x – 8

After rearranging the above expression

⇒ 7x – 8 = 5x

After transposing –8 to RHS we get

⇒ 7 x = 5x + 8

After transposing 5x to LHS, we get

⇒ 7 x – 5 x = 8

⇒ 2 x = 8

After transposing 2 to RHS, we get

⇒

⇒ x = 4

Now, since the number of boys = 7 x

Thus, after substituting the value of x, we get

The number of boys = 7 × 4 = 28

And since the number of girls = number of boys – 8

Thus, number of girls = 28 – 8 = 20

And thus, the total strength of class = number of boys + number of girls

= 28 + 20 = 48

Thus, total strength of class = 48 Answer

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:

Given, Age of Baichung's father = Age of Baichung + 29

And, age of Baichung's grand father = Age of Baichung's father + 26

And the sum of the ages of all the three = 135 years

Then, age of each one of them = ?

Let the age of Baichung = b year

Therefore, age of Baichung's father = Age of Baichung + 29

⇒ age of Baichung's father = b + 29

And, age of Baichung's grand father = Age of Baichung's father + 26

Therefore, age of Baichung's grand father = (b + 29) + 26

Now, according to question, ages of all the three = 135 years

⇒ b + (b+29) + [(b + 29) + 26] = 135

⇒ b + b + 29 + b + 29 + 26 = 135

After rearranging the above expression

⇒ b + b + b + 29 + 29 + 26 = 135

⇒ 3 b + 84 = 135

After transposing 84 to RHS, we get

3 b = 135 – 84

⇒ 3 b = 51

After transposing 3 to RHS, we get

b = 51/3 = 17

Thus, age of Baichung = 17 years

And, age of Baichung's father = b + 29

After replacing the value of b in the above expression

Thus, age of Baichung's father = 17 + 29 = 46 years

And the age of Baichung's grand father = [(b + 29) + 26]

After replacing the value of b in the above expression

Thus, the age of Baichung's grand father = [(17 + 29) + 26]

⇒ The age of Baichung's grand father = 46 + 26

⇒ The age of Baichung's grand father = 72 years

Thus, ages of Baichung, his father and his grand father are 17 years, 46 years and 72 years respectively. Answer