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MATHS 3 Marks Question

Playing with Numbers · Rajasthan Board (RBSE) STD 8 (Rajasthan - English Medium). 10 questions.

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Question 13 Marks
If is a digit of the number $\overline{66784\text{x}}$ such that it is divisible by $9$, find the possible values of $x.$
Answer
$\because$ The number $\overline{66784\text{x}}$ is divisible by $9$
$\therefore$ The sum of its digits will also be divisible by $9$
$\Rightarrow 6 + 6 + 7 + 8 + 4 + x$ is divisible by $9$
$\Rightarrow 31 + x$ is divisible by $9$
Sum greater than $31$, are $36, 45, 54.....$ which are divisible by $9$
$\therefore$ Values are divisible by $9$
$\therefore x = 5$
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Question 23 Marks
Given that the number $\overline{67\text{y}19}$ is divisible by $9$, where y is a digit, what are the possible valuse of y?
Answer
It is given that $\overline{67\text{y}19}$ is a multiple of $9$
$\therefore (6 + 7 + y + 1 + 9) $is a multiple of $9.$
$\therefore (23 + y)$ is a multiple of $9.$
$23 + y = 0, 9, 18, 27, 36...$ But $x$ is a digit,
So, $x$ can take values $0, 1, 2, 3, 4...9. 23 + y = 27$
$\Rightarrow y = 4$
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Question 33 Marks
If x denotes the digit at hundreds place of the number $\overline{67\text{x}19}$ such that the number is divisible by $11.$ Find all possible values of $x.$
Answer
$\because$ The number $\overline{67 \times 19}$ is divisible by $11$
$\therefore$ The difference of the sums its alternate digits will be $0$ or divisible by $11$
$\therefore$ Difference of $(9+x+6)$ and $(1+7)$ is zero divisible by $11$
$\Rightarrow 15+x-8=0$, or multiple of $11,7+x=0$
$\Rightarrow=-7$, which in not possible
$\therefore 7+x=11,7+x=22$ etc,
$\Rightarrow x=11-7=4, x 2=22-7$
$\Rightarrow x=15$ which is not a digit $\therefore x=4$
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Question 43 Marks
Find the quotient when difference of $985$ and $958$ is divided by $9.$
Answer
The numbers of three digits are $985$ and $958$ in which tens and ones digits are reversed,
then. $\overline{\text{abc}}-\overline{\text{acb}}=9(\text{b}-\text{c})$
$985 - 958 = 9(8 - 5) = 9 \times 3$
i. e., it is divisible by $9$,
then quotient $= b - c = 8 - 5 = 3$
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Question 53 Marks
Solve Cryptarithms. $\ \ \ \ \ \ \ \ \ \ \ 3\ \ \ \ 7\\\ \ \ \ \ \underline{+\text{A}\ \ \ \ \text{B}\ }\\\ \ \ \ \ {\ \ \ \ \ \ 9\ \ \ \ \text{A}\ }$
Answer
Two possible values of $A$ are:
$i.$ If $7+\text{B}\leq9,3+\text{A}=9$
$\therefore\text{A}=6$
But if $A = 6, 7 + B$ must be larger than $9.$
Hence, it is impossible.
$ii.$ If $7+\text{B}\geq9$
$\therefore1+3+\text{A}=9$
$\Rightarrow\text{A}=5$
If $A = 5$ and $7 + B = 5,$
$B$ must be $8$
$\therefore A = 5, B = 8$
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Question 63 Marks
If $\overline{3\text{x}2}$ is a multiple of $11$, where v is a digit, what is the value of $x?$
Answer
$\because$ The number $\overline{3\text{x}2}$ is multiple of $11$
$\therefore$ It is divisible by $11$
$\therefore$ Difference of the sum of its alternate digits is zero or multiple of $11$
$\therefore$ Difference of $(2 + 3)$ and $x$ is zero or multiple of $11$
$\Rightarrow If x - (2 + 3) = 0$
$\Rightarrow x - 5 = 0$
Then $x = 5$
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Question 73 Marks
If sum of the number $985$ and two other numbers by arranging the digits of $985$ in cyclic order is divided by $111, 22$ and $37$ respectively. Find the quotient in each case.
Answer
The given number is $985$ The other two numbers by arranging its digits in cyclic order,
will be $859, 598$ of the form, $\overline{\text{abc}},\overline{\text{bca}},\overline{\text{cba}}$
Therefore, If $985 + 598 + 598$ is divided by $111,$
then quotient will be $a + 6 + c = 9 + 8 + 5 = 22$ If this sum is divided by $22,$
then quotient $= 111$ And if it is divided by $37,$
then quotient $= 3(a + 2 + c) = 3(22) = 66$
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Question 83 Marks
Solve Cryptarithms. $\ \ \ \ \ \ \ \ \ \ \ \text{A}\ \ \ \ \text{B}\\\ \ \ \ \ \underline{+\ \ \ 3\ \ \ \ \ 7\ }\\\ \ \ \ \ {\ \ \ \ \ \ 9\ \ \ \ \text{A}\ }$
Answer
Two possibilities of $A$ are:
$i.$ If $\text{B}+7<9,$
$\text{A}=6$
But clearly, if $A = 6,$
$\text{B}+7\geq9;$
It is impossible
$ii.$ If $\text{B}+7\geq9,$
$A = 5$ and $B + 7 = 5$
Clearly, $B = 8$
$\therefore A = 5, B = 8$
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Question 93 Marks
If $x$ is a digit such that the number $\overline{18\text{x}71}$ is divisible by $3$, find possible values of $x$.
Answer
$\because$ The number $\overline{18 \times 71}$ is divisible by $3$
$\therefore$ The sum of its digits will also be divisible by $3$
$\Rightarrow I+8+x+7+1$ is divisible by $3$
$ \Rightarrow 17+ x$ is divisible by $3$ The sum greater than $17$ , can be $18,21,24,27 \therefore x$ can be $1,4,7$ which are divisible by $3.$
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Question 103 Marks
Given that the number $\overline{35\text{a}64}$ is divisible by $3$, where a is a digit, what are the possible volues of a?
Answer
The number $\overline{35\text{a}64}$ is divisible by $3$
$\because$ The sum of its digits will also be divisible by $3$
$\therefore 3 + 5 + a + b + 4$ is divisible by $3$
$\Rightarrow 18 + a$ is divisible by $3$
$\Rightarrow a$ is divisible by $3$
$(\because$ is divisible by $3 )$
$\therefore$ Values of a can be $0, 3, 6, 9.$
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