4 Mark Question

Question 14 Marks

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number.

Answer

Let the tens place digit be a and the units place digit be b.
Then, number is (10a + b).
According to the question:
4(a + b) + 3 = (10 a + b)
4a + 4b + 3 = 10a + b
6a - 3b = 3
3(2a - b) = 3
2a - b = 1... (1)
Given:
If 18 is added to the number, its digits are reversed.
The reverse of the number is (10b + a).
$\therefore$ (10a + b) + 18 = 10b + a
10a - a + b -10b = -18
9a - 9b = -18
9(a - b) = -18
a - b = -2... (2)
Subtracting equation (2) from equation (1):
Using a = 3 in equation (1):
$\therefore$ (2 × 3) - b = 1
⇒ 6 - b = 1$$
$$⇒ b = 6 - 1
⇒ b = 5
Therefore, the number ={(10 × 3)+5} = 35.

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Question 24 Marks

Replace A, B, C by suitable numerals.
$ \ \ \ \ \ \ \ \ \ \ \ \text{A B}\\\underline{ \ \ \ \ \ \ \times\text{B A}}\\\underline{\text{(B+1)}\text{C B}}$

Answer

A × B = B
⇒ A = 1
$\ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \text{B}\\\underline{\times\ \ \ \ \ \text{B}\ \ \ \ \ \ \ 1\ \ \ \ }\\\underline{ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \text{B}\\ \ \text{B }\ \ \ \text{B}^{2}\ \ \ \ \ \ \times\ }\\\text{B}\ \ (1+\text{B}^{2}) \ \text{B}$
In the equation:
First digit = B + 1
Thus 1 will be carried from $1+B^2$ and becomes $\left.(B+1)\left(B^2-9\right) B\right]$ $\$$ therefore\$ C = $B ^2-1$
Now all $b , B +1$ and $B ^2-9$ are one digit number.
This condition is satisfied for $B =3$ or $B =4$.
For, $B <3, B-9$ will be negative.
For, $B >3, B^2-9$ will become digit number.
For $B =3, C =3^2-9=9-9=0$
For, $B=4, C=4^2-9=16-9=7$
$\$$ therefore $\$ A=1, B=4, C =7$

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Question 34 Marks

The sum of the digits of a two-digit number is 15. The number obtained by interchanging its digits exceeds the given number by 9. Find the original number.

Answer

a + b = 15 ...(1)
Original number = 10a + b
Number obtained by interchanging its digits = 10b + a.
(10a + b) + 9 = (10b + a)
⇒ 10a + b + 9 = 10b + a
⇒ 10a + b - 10b - a = -9
⇒ 9a - 9b = -9
⇒ 9(a - b) = -9
a - b = -1 ...(2)
Adding the equation (1) and (2)
a + b = 15
$\text{a}+\text{b}=15\\\underline{\text{a}-\text{b}=-1}\\2\text{a} \ \ \ \ \ =14$
$\Rightarrow\text{a}=7$
Using $\text{a} = 7$ in equation (2)
$\text{a} + \text{b} = 15$
$⇒ 7 + \text{b} = 15$
$\Rightarrow\text{b}=15-7$
$\Rightarrow\text{b}=18$
Therefore, the number $=\{(10\times7)+8\}=78$

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