Question

1.
Solve the following equations:
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agad
(*3 **
2.
3.
The digit in the tens place of a two-digit number is three times that in the ones
place. If the digits are reversed, the new number will be 36 less than the original
number. Find the number.
The numerator of a rational number is less than its denominator by 3. If the
numerator becomes three times and the denominator is increased by 20, the
new number becomes . Find the original number. IP​

Answer 1

Answer:

Required number is 62.

Step-by-step explanation:

Question 1

Let,

Required two-digit number be ab, where a is at tens place and b is at ones place.

Given,

Digit in the tens place of a two-digit number is three times that in the ones place.

So,

= > Digit at tens place = 3 times that digit at ones place

= > a = 3 x b

= > a = 3b ... ( 1 )

Number : ab which can be written as 10a + b, since both are same, when we talk about numbers.

So,

= > Required number is : 10a + b

= > Required number : 10( 3b ) + b { from ( 1 ) }

= > Required number : 30b + b

= > Required number : 31b

According to the question : If digits are reversed, the new number will be 36 less than the original number.

It means : Original number - 36 = Original number with reversed digits

= > 10a + b - 36 = 10b + a { new number is ba, that's 10b + a }

= > 10a - a + b - 10b = 36

= > 9a - 9b = 36

= > 9( a - b ) = 36

= > 9( 3b - b ) = 36 { from ( 1 ) }

= > 2b = 36 / 9

= > 2 b = 4

= > b = 2

Hence,

= > Required number is : 31b

= > Required number : 31( 2 )

= > Required number : 62

Hence the required number is 62.

Question 2

Let,

Denominator of the fraction be a.

Since, denominator is a, numerator should be a - 3 , as given that the numerator of the fraction is 3 less than the denominator.

So, required fraction should be ( a - 3 ) / a.

According to the question : For new fraction, new numerator is 3 times the old one and denominator is increased by 15.

So,

= > New fraction : { 3( a - 3 ) } / ( a + 15 )

On the basis of the new fraction, we can find the value of a and then that fraction.

Here, new fraction is not given, so solution can't be explained further.

Answer 2

• Let ten's digit be M and one's digit be N.

Original number = 10 × M + N

$\implies$ 10M + N

» The digit in the tens (M) place of a two-digit number is three times that in the ones (N) place.

• A.T.Q.

$\implies$ M = 3N ____ (eq 1)

We have original number = 10M + N

$\implies$ 10(3N) + N

[From (eq 1)]

$\implies$ 31N

______________________________

» If the digits are reversed, the new number will be 36 less than the original

number.

Reversed number = 10N + M

• A.T.Q.

$\implies$ 10M + N - 36 = 10N + M

$\implies$ 10M - M + N - 10N = - 36

$\implies$ 9M - 9N = + 36

$\implies$ M - N = 4

$\implies$ 3N - N = 4 [From (eq 1)]

$\implies$ 2N = 4

$\implies$ N = 2

______________________________

We have to find the number..

Original number = 10M + N

From above calculations we have M = 6 and N = 2

$\implies$ 10 × 6 + 2

$\implies$ 60 + 2 = 62

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$\textbf{Number is 62}$

____________ $\bold{[ANSWER]}$

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