3 Marks Question - Maths STD 8 Questions - Vidyadip

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3 Marks Question

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15 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

Find the value of the letters in the following: $\\\underline{\ \ \ \ \ 2 \ \ \ \text{L}\ \ \ \text{M} \\\ {+}\ \text{L}\ \ \text{M}\ \ \ \ \text{I}}\ \ \ \ \ \ \ \ \\\\\underline{\ \ \ \ \text{M}\ \ \ \text{I}\ \ \ \ 8\ }$

Answer

In first column, $M + 1 = 8$ Clearly,
$M = 7$ In second column,
$L + M = 1 L + 7 = 1$ The value of $L$ can be $4$ In third column,
$2 + L + 1 = M 2 + 4 + 1 = 7 7 = 7$
so the third column is satisfied for $L = 4, M = 7$
Hence, $L = 4$ and $M = 7$

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Question 23 Marks

A five-digit number $AABAA$ is divisible by $33$. Write all the numbers of this form.

Answer

Given, a number of the form $AABAA$ is divisible by $33.$
Then, it is also divisible by $3$ and $11$, as if a number ?
is divisible by ?, then it is also divisible by each factor of $b.$
Since, $AABAA$ is divisible by $3$,
sum its digits is also divisible by $3.$
i.e. $4 + 4 + 8 + A + 4 = 0, 3, 6, 9$ or $\frac{4}{4}+8 = 0, 3, 6 9, (i)$ From Eq. $(i),$
we have Further, the given number is also divisible by $11,$
therefore $\Big(\frac{\text{z}}{4}+8\Big)-2\text{A} = 0, 11, 22 B = Q 11, 22, 8 = 0 [V8$ is a digit of the given number] $\frac{4}{4}\ = 12$ or $24$ or $36 A = 3, 6 9$
Hence, the required numbers are $33033, 66066$ and $99099.$

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Question 33 Marks

$1y3y6$ is divisible by $11$. Find the value of $y.$

Answer

It is given that, $1y3y6$ is divisible by $11$. Then, we have $(1 + 3 + 6) – (y + y) = 0, 11, 22 = 10 - 2y = 0, 11, 22 = 10 - 2y = 0$ [other values give a negative number] $= 2y = 10 = y = 5$

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Question 43 Marks

Let $E = 3, B = 7$ and $A = 4$. Find the other digits in the sum
$\underline{\ \ \ \ \ \text{B}\ \ \text{A} \ \ \text{S}\ \ \ \text{E}\\\text{+}\ \ \text{B}\ \ \text{A} \ \ \text{L}\ \ \ \text{L} }\\\underline{\ \text{G} \ \text{A}\ \text{M} \ \ \text{E}\ \ \ \text{L}\ }$

Answer

$E = 3, B = 7$ and $A = 4$
Now, In one’s column, we have $3 + L = S$ or $S - L = 3$
In ten’s column, we have $S + L = 3$
On solving eq. $(i) \& (ii)$ we get,
$S = 3$ and $L = 0$
In hundred’s column, we have
$4 + 4 = M$
$M = 8$
& in thousand’s column, we have
$7 + 7 = G4$
$14 = G4$
$G = 1$
Hence, $L = 0, S + 3, M = 8$ and $G = 1$

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Question 53 Marks

$\underline{\ \ \ \ \ \text{C}\ \ \text{B} \ \ \text{A}\\\text{+}\ \ \text{C}\ \ \text{B} \ \ \text{A}\ }\\\underline{\ 1 \ \ \text{A}\ \ \ 3\ \ \ 0\ }$

Answer

In 1st column $A + A = 0A = 0$ or $5$
For $A = 0$, the second & third column, sums are not satisfied
So, $A = 5$
Now, in second column
$B + B + 1 = 3$
For $B = 1$, third column sum is not satisfied.
So, $B = 6$
Again, in third column, $C + C + 1 = 1A$
$C + C = 15 − 1$
$2C = 14$
$C = 7$
Hence, $A = 5, B = 6$ and $C = 7$

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Question 63 Marks

$212 \times 5$ is a multiple of $3$ and $11$. Find the value of $x.$

Answer

Since, $212 \times 5$ is a multiple of $3, 2 + 1 + 2 + x + 5 = 0, 3, 6, 9, 12, 15, 18, $
$= 10 + ? = 0, 3, 6, … … … .. = x = 2, 5, 8 …(i)$
Again, 212 × 5 is a multiple of $11, (2 + 2 + 5) − (1 + x) $
$= 0, 11, 22, 33 = 8 – x = 0, 11, 22, … = x = 8 …(ii)$
From Eqs. $(i)$ and $(ii),$
we have $x = 8$

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Question 73 Marks

Answer

$AB \times AB = 6AB..... (i)$
Here,
$B \times B$ is a number whose unit’s digit is $B.$
Therefore, $B = 1$ or $5$
Again,
$AB \times AB = 6AB$
The square of a two digit number is a three digit number.
So, A can take values $1, 2 \& 3.$
For $A = 1, 2, 3$ and $B = 1$ Eq. $(i)$ is not satisfied.
Now, for $? = 1, B = 5$ Eq. $(i)$ is not satisfied.
$A = 2, B = 5$ satisfies the Eq. $(i)$
Hence, $A = 2, B = 5$

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Question 83 Marks

$\underline{\ \ \ \ \ \text{A}\ \ \text{B} \ \ \\\\ \ \ \ \times \ \ \ 6\ }\\\underline{\\ \ \text{C}\ \ \ 6\ \ \ 8\ }$

Answer

Here, $6 \times B$ is a number, whose unit’s digit is $8.$
Therefore, the possible values of $B$ are $3 \& 8$ If $B = 3,$
then $A \times 6 + 1 = C6$ which is not possible for any value of A between $0$ to $9.$
$B = 8$ and then $A = 7$
The values of ? and ?
also satisfies the given condition i.e. $8 - 7 = 1$ If $A = 7$
then $7 \times 6 + 4 = C6 46 = C6 C = 4$
Hence, $A = 7, B = 8$ and $C = 4$

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Question 93 Marks

$\underline{\ \ \ \ \ 1\ \ \text{B} \ \ \text{A}\\\text{+}\ \ \text{A}\ \text{B} \ \ \text{A}\ }\\\underline{\ \ \ \ 8\ \ \ \text{A}\ \ \ 2\ }$

Answer

Here, $A + A$ is a number whose one’s digit is $2,$
therefore $A = 6$
Now,$ B + B + 1 = B$
So, the possible value of $B$ is $9$. Again,
in third column $A + 1 + 1 = 8 A = 6$, which is true.
Hence, $A = 6$ and $B = 9$

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Question 103 Marks

If $56x32y$ is divisible by $18$, find the least value of $y.$

Answer

It is given that, the number $56 \times 32?$ is divisible by $18.$
Then, it is also divisible by each factor of $18.$
Thus, it is divisible by $2$ as well as $3.$
Now, the number is divisible by $-2,$
its unit’s digit must be an even number that is $0, 2, 4, 6,$
Therefore, the least value of y is $0.$
Again, the number is divisible by $3$ also,
sum of its digits is a multiple of $3.$
i.e. $5 + 6 + x + 3 + 2 + y$ is a multiple of $3 $
$=> 16 + x + y = 0, 3, 6, 9,..... = 16 + x = 18 = x = 2,$
which is the least value of $x.$

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Question 113 Marks

$x$ is a multiple of $11$, find the value of $x.$

Answer

We are given that, $756x$ is a multiple of $11.$
Then, we have to find the value of $x.$
Since, 756x is divisible by $11$, then $(7 + 6) – (5 + x)$ is a multiple of $11$,
i.e. $8 - x = 0, 11, 22$
$= 8 - x = 0$
$= x = 8$

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Question 123 Marks

Let $D = 3, L = 7$ and $A = 8$. Find the other digits in the sum $\underline{\ \ \ \text{M}\ \ \ \text{A} \ \ \ \text{ D} \\\ \ \ \text{+}\ \ \ \ \text{A}\ \ \ \ \text{S} \ \ \ \ \ \ \ \\ \ \ \text{ +}\ \ \ \ \ \ \ \ \ \text{ A}}\\\underline{\ \text{B}\ \text{U}\ \ \ \text{L} \ \ \ \ \text{L}\ \ \ }$

Answer

In the first column, we have $3 + A + 8,$
which is definitely a two digits number whose unit’s digit is $7.$
S must be $6.$
Now, in second column,
$2A + 1 = 16 + 1 = 7 [1$ is carry forward] n third column,
$M + 1$ is $a 2$ digit number, therefore $M$ must be $9.$
Then, $M + 1 = 9 + 1 = 10 6 = 1,U = 0$
Hence, $S = 6, M = 9 6 = 1$ and $U = 0$

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Question 133 Marks

Find the value of ? where $31k 2$ is divisible by $6.$

Answer

Given, $31k2$ is divisible by $6.$
Then, it is also divisible by $2$ and $3$ both.
Now, $31k2$ is divisible by $3,$
sum of its digits is a multiple of $3.$
i.e. $3 + 1 + k + 2 = 0, 3, 6, 9, 12, … $
$= k + 6 = 0, 3, 6, 9, 12 = k = 0 or 3, 6, 9$

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Question 143 Marks

If $148101B095$ is divisible by $33$, find the value of $B.$

Answer

Given that the number $148101B095$ is divisible by $33$, therefore it is also divisible by $3$ and $11$ both.
Now, the number is divisible by $3$, its sum of digits is a multiple of $3.$
i.e. $1 + 4 + 8 + 1 + 0 + 1 + B + 0 + 9 + 5$ is a multiple of $3.$
$29 + B = 0, 3, 6, 9, = B = 1, 4, 7 .....(i)$
Also, given number is divisible by $11,$
therefore $(1 + 8 + 0 + B + 9) − (4 + 1 + 1 + 0 + 5) = 0, 11, 22, = (18 + B) - 11 = 0, 11, 22 B + 7 = 0, 11, 22 = B + 7 = 11 = B = 4 …..(ii)$
From Eqs. $(i)$ and $(ii)$, we have $B = 4$

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Question 153 Marks

If $1AB + CCA = 697$ and there is no carry–over in addition, find the value of $A + B + C.$

Answer

Since, there is no carry over in addition,
$1 + C = 6$
$C = 5$
$A + C = 9$
$A + 5 = 9$
$A = 4$
$& B + A = 7$
$B = 3$
Hence, $A + B + C = 4 + 3 + 5 = 12$

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