2 Marks Questions

Question 12 Marks

$\ \ \ \underline{\ \ \ \ \ \text{A}\ \ \ \text{A}{}\\\times \ \ \ \ \ \ \ \text{A}\ }\\\ \ \underline{\text{C}\ \ \text{ A}\ \ \ \text{ B}\ \ }$ and $B - A =1$.

Answer

Here, $AA \times A$ is three digit number, whose unit’s digit is $B$,
Therefore $A$ can take values form $4$ to $9$ as $A = 0, 1, 2, 3$ give a single digit or a two-digit number.
Since, ten’s digit of the product is $A$, itself.
So, A cannot take values $4, 5, 6, 7$ and $8$ Hence, $A = 9$ and then $B = 1, C = 8$

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Question 22 Marks

$\underline{\ \ \ \ \ \text{B}\ \ \ \ \ 6\\\text{+}\ \ 8\ \ \ \ \ \text{A}\ }\\\underline{\ \ \ \ \text{CA}\ \ \ 2\ }$

Answer

Here, $6 + A = 2$, therefore the possible value of $A$ is $6$.
Now, $CA = B + 8 + 1$ $C 6 = B + 9$ i.e. $B + 9$ is a number whose one’s digit is $6$.
Therefore, $B = 7$ and $C = 1$.

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Question 32 Marks

A three-digit number $2$ a $3$ is added to the number $326$ to give a three-digit number $5b9$ which is divisible by $9$. Find the value of $b – a$

Answer

In one’s column, $3 + 6 = 9$ (true)
In third column, $2 + 3 = 5$ (true)
$a + 2$ is a single digit number b as there is no carry over in the addition Thus, $a + 2 = b$
$b - a = 2$

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Question 42 Marks

If $\ \ \ \ \ 1\ \ \ \ \ \ \text{P}\\\ \ \ \ \underline{{\times}\ \ \ \\ \ \ \ \text{P}}\\\ \ \ \ \text{Q}\ \ \ \ \ 6\ $ where $Q - P = 3$, then find the values of $P$ and $Q$.

Answer

Here, $P \times P$ is $6$, so the value of ? is either $4$ or $6$.
But if $P = 4,Q = 5$ which does not satisfy the relation $Q - 0 P = 3$
Hence, $P = 6$ and then $Q = 9$

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Question 52 Marks

$\underline{\ \ \ \ \ \text{A}\ \ 0 \ \ 1\ \ \text{B}\\\text{+}\ \ 1\ \ 0 \ \ \text{A}\ \ \ \text{B}}\\\underline{\ \ \ \ \text{B}\ \ \ 1\ \ \ 0\ \ \ 8}$

Answer

In first column, $B + B = 8 B = 9$ In second column, we have $A + 1 + 1 = 0$
So, A should be $8$ Third column is true for these values. Also, the fourth column is satisfied.
Hence, $A = 8$ and $B = 9$

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Question 62 Marks

If $123123A4$ is divisible by $11$, find the value of $A$.

Answer

Given, $12312344$ is divisible by $11$, then we have $(1 + 3 + 2 + 4) - (2 + 1 + 3 + 4)$ is a multiple of $11$.
i.e. $(6 + 4) - 10 = 0, 11, 22,..... = A - 4 = 0, 11, 22,..... = A - 4 = 0$
$[A$ is a digit of the given number$] = A = 4$

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Question 72 Marks

Find the value of the letters in the following: $\underline{\ \ \ \ \ \text{PQ}\ \\\ {\ \ }\times\ 6}\\\underline{\ \ \text{QQQ}\ \ }$

Answer

Here, in first column, we see that $6$ $\times Qp = ?.$
Therefore, the possible values of $Q$ are $2, 4, 6$ & $8$.
For $Q = 2, 6 \times p + 1$ can not be equal to $22$ for any value of $P$.
So, $Q = 2$ is not possible. $Q = 4 6 \times P + 2 = 44$ $6P = 42\ P = 7$ Hence, $P = 7$ and $Q = 4$

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Question 82 Marks

$\underline{\ \ \ \ \ \ 8\ \ \ \ \ 5\\\text{+}\ \ 4\ \ \ \ \ \text{A}\ }\\\underline{\ \ \ \ \text{B}\text{C}\ \ \ 3\ }$

Answer

Here, $5 + A = 3\ 5 + A$ can be a single digit number.
So, $5 + A$ is a two-digit number whose one’s digit is $3$.
$ A = 8 B = 1, C = 3 [BC = 8 + 4 + 1 = 10B + C = 13 = 10 × 1 + 3 = B = 1, C = 3]$

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Question 92 Marks

If $2A7 ÷ A = 33$, then find the value of $A$.

Answer

We observe that, $4 \times 3$ can never be a single digit number $2$, so $4 \times 3$ must be a twodigit number, whose ten’s digit is $2$ and unit’s digit is the number less than or equal to $4$.
Therefore, the value of $4$ can be $9$, as the values of $4$ from $1$ to $8$ do not $t$.

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Question 102 Marks

$\underline{\ \ \ \ \ \text{B}\ \ \text{A} \ \ \text{A}\\\text{+}\ \ \text{B}\ \ \text{A} \ \ \text{A}\ }\\\underline{\ \ \ \ 3\ \ \ \text{A}\ \ \ 8\ }$

Answer

Here,
$A + A = 8$
$A = 9$
Now, in third column,
we have,
$B + B + 1 = 3$
$2B = 2$
$B = 1$
Hence, $a = 9$ and $B = 1$

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Question 112 Marks

Find the value of the letters in the following questions
$\underline{\ \ \ \ \ \ \text{A}\ \ \ \ \ \text{A}\\\text{+}\ \ \text{A}\ \ \ \ \ \text{A}\ }\\\underline{\ \ \ \text{XA}\ \ \ \ \ \text{Z}\ }$

Answer

Here,$ A + A = Z$, therefore $A$ can take any value between $0$ to $9$.
Since, the sum in second column is a two digit number, possible values of $A$ are $5$ to $9$.
The values $A = $5 to $8$ are not fitted in second column.
Hence, $A = 9\ Z = 8$ and $X = 1$

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