Linear Equations in One Variable - 8th math

Question (1) Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?

Solution:

As given in the question

Thus, number = ?

Now, Let the required number = n

⇒ (2n – 5) × 4 = 3n

⇒ (2n × 4) – (5 × 4) = 3n

⇒ 8n – 20 = 3n

After transposing 20 to RHS, we get

8n = 3n + 20

Now, after transposing 3n to LHS, we get

8n – 3n = 20

⇒ 5n = 20

After transposing 5 to RHS, we get

n = 20/5 = 4

Thus, the required number = 4 Answer

Question (2) A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Solution:

Given, A positive number = 5 × another number

Again given, that when 21 is added to both the numbers,

The the one of the new number = 2 × other new number

Thus, numbers = ?

Let the given positive number = n

Therefore, another number which is 5 times of the given number = 5n

Now, again as given in the question,

When 21 is added to both of the number

That is, first given positive number + 21 and second number + 21

Then, (first given positive number + 21) × 2 = second positive number + 21

⇒ (n + 21) × 2 = 5n + 21

⇒ 2n + 21 × 2 = 5n + 21

⇒ 2n + 42 = 5n + 21

After transposing 21 to LHS, we get

2n + 42 – 21 = 5n

⇒ 2n + 21 = 5n

Again after transposing 2n to RHS, we get

21 = 5n – 2n

⇒ 21 = 3n

After rearranging the above expression

⇒ 3n = 21

After transposing 3 to RHS, we get

⇒ n =7

Thus, first number = 7

Now, since second number = 5n

Thus, after substituting the value of n

The second number = 5 × 7 = 35

Thus, required numbers are 7 and 35 Answer

Question (3) Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Solution:

Given, the sum of the digits of a two digit number = 9

And, when the digits of the number is interchanged,

The new number – 27 = Original number

Thus, number = ?

Let the number at ones place of two digit number = n

Now, as per question, the sum of digits of number = 9

Thus, digit at tens place = 9 – n

Thus, the given two digit number = 10(9 – n) + n

Now, after interchanging to the digits of number

The new number = 10n + (9 – n)

Now, according to question,

10(9 – n) + n + 27 = 10n + (9 – n)

⇒ 90 – 10n + n + 27 = 10n + 9 – n

⇒ 90 + 27 – 10n + n = 10n – n + 9

⇒ 117 – 9n = 9n + 9

After transposing –9 to RHS, we get

117 = 9n + 9 + 9n

⇒ 117 = 18n + 9

After transposing 9 to LHS, we get

117 – 9 = 18n

⇒ 108 = 18n

After rearranging the above expression, we get

18 n = 108

Now, after transposing 18 to RHS, we get

⇒ n = 108/18 = 6

Thus, digit at ones place (n) of the given number = 6

Now, since the digit at tens place = 9 – n

= 9 – 6 =3

Thus, digit at ones place = 6 and digit at tens place = 3

Thus, required number = 36 Answer

Question (4) One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Solution:

Given, One of the two digits in a two digit number = 3 × other digit

When digits are interchanged and added to the original digit, the sum is 88

Then, the original digit =?

Let one of the digits of two digit number, i.e. digit at ones place = n

Thus, the other digit, which is at tens place = 3n

Thus, digit = (10 × 3n) + n

And, after interchanging the digit = 10n + 3n

Now, according to question,

The sum of the both the digit, i.e. original digit + new one = 88

Thus, [(10 × 3n) + n] + [10n + 3n] = 88

⇒ [30n + n] + 13n = 88

⇒ 31n + 13n = 88

⇒ 44n = 88

After transposing n to RHS, we get

n = 88/44 = 2

Thus, digit at ones place (n) = 2

Now, since the digit at tens place = 3n

= 3 × 2 = 6

Thus, digit at ones place = 2 and digit at tens place = 6

Thus, original digit = 62

Check of Result

As per question, original digit + After interchange of digits = 88

⇒ 26 + 62 = 88 Proved

Thus, required digit = 62

Question (5) Shobo's mother's present age is six times Shobo's present age. Shobo's age five years from now will be one third of his mother's present age. What are their present ages?

Solution:

Given, The present age of Shobo's mother = Shobo's present age × 6

After five years from now, the age of Shobo's mother = Shobo's present age × 3

Thus, their present ages = ?

Let the present age of Shobo's mother = s × 6 year

And, after five years from now, the present age of Shobo's mother = (s + 5) × 3

Therefore, 6s = (s + 5) × 3

⇒ 6s = 3s + 15

After transposing 3s to LHS, we get

6s – 3s = 15

3s = 15

After transposing 3 to RHS, we get

s = 15/3 = 5

Thus, present age of Shobo = 5 years

Since, The present age of Shobo's mother = Age of Shobo × 6

= 5 × 6 = 30

Thus, the present age of Shobo = 5 and the present age of Shobo's mother = 30 years Answer