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Maths M.C.Q. [1 Marks Each]

Playing with Numbers · Jharkhand Board (JAC) STD 8 (Jharkhand - English Medium). 17 questions.

Questions

M.C.Q. [1 Marks Each]

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17 questions · auto-graded multiple-choice test.

MCQ 11 Mark
If $x + y + z = 6$ and $z$ is an odd digit, then the three-digit number $xyz$ is:
  • ✓
    an odd multiple of $3$
  • B
    odd multiple of $6$
  • C
    even multiple of $3$
  • D
    even multiple of $9$
Answer
Correct option: A.
an odd multiple of $3$
We have, $x + y + z = 6$ and $?$ is an odd digit. Since, sum of the digits is divisible by $3,$ it will also be divisible by $2$ and $3$ but unit digit is odd, so it is divisible by $3$ only.Hence, the number is an odd multiple of $3.$
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MCQ 21 Mark
A four-digit number aabb is divisible by $55.$ Then possible value $(s)$ of $b$ is/are
  • A
    $0$ and $2$
  • B
    $2$ and $5$
  • ✓
    $0$ and $5$
  • D
    $7$
Answer
Correct option: C.
$0$ and $5$
It is given that, aabb is divisible by $55.$
Then, it is also divisible by $5.$ Now, if a number is divisible by $5,$ then its unit digit is either $0$ or $5$
Hence, the possible values of $b$ are $0$ and $5.$
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MCQ 31 Mark
The sum of all the numbers formed by the digits $x, y$ and $z$ of the number $xyz$ is divisible by.
  • A
    $11$
  • B
    $33$
  • ✓
    $37$
  • D
    $74$
Answer
Correct option: C.
$37$
We have, $xyz + yzx + zxy$
$= (100x + 10y + z) + (100y + 10 z + x) + (100z + 10x + y).....(i)$
$= 100x + 10x + x + 10y + 100y + y + z + 100z + 10z$
$= 111x + 111y + 111z = 111 (x + y + z)$
$= 3 × 37 × (x + y + z)$
Hence, Eq. $(i)$ is divisible by $37,$ but not divisible by $11, 33$ and $74.$
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MCQ 41 Mark
If abc is a three digit number, then the number $abc – a – b – c$ is divisible by
  • A
    $9$
  • B
    $90$
  • C
    $10$
  • ✓
    $11$
Answer
Correct option: D.
$11$
We have, $abc = 100a + 10b + c$
$\therefore abc - a - b - c = (100a + 10b + c) - a - b - c = 100a - a + 10b - b = 99a + 9b = 9(11a + b)$
Hence, the given number $abc - a - b - c$ is divisible by $9.$
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MCQ 51 Mark
The usual form of $1000a + 10b + c$ is
  • A
    $abc.$
  • B
    $abco.$
  • ✓
    $aobc.$
  • D
    $aboc.$
Answer
Correct option: C.
$aobc.$
Given expanded (or generalised) form of a number is $1000? + 10? + ?$. Then, we have to find its usual form. We can write it as $1000 × ? + 100 × 0 + 10 × ? + ?$ i.e. $?0??,$ which is the usual form.
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MCQ 61 Mark
A four-digit number $4ab5$ is divisible by $55.$ Then the value of $b – a$ is
  • A
    $0$
  • ✓
    $1$
  • C
    $4$
  • D
    $5$
Answer
Correct option: B.
$1$
Given, a four-digit number $4??5$ is divisible by $55.$
Then, it is also divisible by $11.$
The difference of sum of its digits in odd places and sum of its digits in even places is either $0$ or multiple of $11.$
i.e. $(4 + b)– (a + 5)$ is 0 or a multiple of $11,$ if $4 + b - a - 5 = 0$
$b - a = 1$
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MCQ 71 Mark
If $6 A × B = A 8 B,$ then the value of $A – B$ is:
  • A
    $-2$
  • B
    $2$
  • ✓
    $-3$
  • D
    $3$
Answer
Correct option: C.
$-3$
Given, $6A × B = A86$
Let us assume, $A = 1$ and $S = 3$ Then,
$LHS = 61 × 3 = 183$ and $RHS = 183$
Thus, our assumption is true. $A - 6 = 1 - 3 = -2$
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MCQ 81 Mark
A six-digit number is formed by repeating a three-digit number. For example $256256, 678678,$ etc. Any number of this form is divisible by
  • A
    $7$ only
  • B
    $11$ only
  • C
    $13$ only
  • ✓
    $1001$
Answer
Correct option: D.
$1001$
Let the six-digit number be $abcabc,$ then
$= 100000 × a + 10000b + 1000c + 100a + 10b + c$
$= a(100000 + 100) + b(10000 + 10) + c(1000 + 1)$
$= a(100100) + b(10010) + c(1001) = 1001(a × 100 + b × 10 + c)$
Hence, it is divisible by $1001$
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MCQ 91 Mark
Let $abc$ be a three digit number. Then $abc + bca + cab$ is not divisible by
  • A
    $a + b + c$
  • B
    $3$
  • C
    $37$
  • ✓
    $9$
Answer
Correct option: D.
$9$
We know that, the sum of three-digit numbers taken in cyclic order can be written as $111 (a + b + c)$
i.e. $abc + pea + cab = 3 × 37 × (a + b + c)$
Hence, the sum is divisible by $3, 37$ and $(a + b + c)$ but not divisible by $9.$
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MCQ 101 Mark
Generalised form of a four-digit number $abdc$ is:
  • A
    $1000 a + 100 b + 10 c + d$
  • B
    $1000 a + 100 c + 10 b + d$
  • ✓
    $1000 a + 100 b + 10 d + c$
  • D
    $a × b × c × d$
Answer
Correct option: C.
$1000 a + 100 b + 10 d + c$
In generalised form, we express a number as the sum of the products of its digits with their respective place values. $????$ is written in generalised form as $1000? + 100? + 10? + ?.$ i.e. $???? = 1000? + 100? + 10? + c$
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MCQ 111 Mark
If $A 3 + 8 B = 150,$ then the value of $A + B$ is:
  • A
    $13$
  • ✓
    $12$
  • C
    $17$
  • D
    $15$
Answer
Correct option: B.
$12$
Here, $3 + B = 0,$ so $3 + B$ is a two-digit number whose unit’s digit is zero.
$\therefore 3 + B = 10$
$B = 7$
Now, considering ten’s column, $A + 8 + 1 = 15$
$= A + 9 = 15$
$= A = 6$
Hence, $A + B = 6 + 7 = 13$
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MCQ 121 Mark
If $5 A + B 3 = 65,$ then the value of $A$ and $B$ is:
  • A
    $A = 2, B = 3$
  • B
    $A = 3, B = 2$
  • ✓
    $A = 2, B = 1$
  • D
    $A = 1, B = 2$
Answer
Correct option: C.
$A = 2, B = 1$
We have, $\frac{5\text{A}+\text{B}3}{65}$
Evidently, $A + 3$ is a number taking values from $3$ to $12.$ So,
Either $A + 3$ is $5$ or it is a two-digit number whose unit digit is $5.$ But, $A+3$ is less than or equal to $12.$
$A + 3 = 5$
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MCQ 131 Mark
If $5 A × A = 399,$ then the value of $A$ is:
  • A
    $3$
  • B
    $6$
  • ✓
    $7$
  • D
    $9$
Answer
Correct option: C.
$7$
We have, $5? × ? = 399$
Here, $A × A = 9$ i.e. $A × A$ is the number $9$ or a number whose unit’s digit is $9.$ Thus, the number whose product with itself produces a two-digit number having its unit’s digit as $9$ is $7.$ i.e. $A × A = 49$
$= A = 7$
Now, $5 × A + 4 = 39$
$= 5 × 7 + 4 = 39$
So, A satisfies the product. Hence, the value of $A$ is $7.$
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MCQ 141 Mark
Which of the following numbers is divisible by $99:$
  • A
    $913462$
  • ✓
    $114345$
  • C
    $135792$
  • D
    $3572406$
Answer
Correct option: B.
$114345$
Given a number is divisible by $99.$
Now, going through the options, we observe that the number $(b)$ is divisible by $9$ and $11$ both as the sum of digits of the number is divisible by $9$ and sum of digits at odd places $=$ sum of digits at even places.
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MCQ 151 Mark
Generalised form of a two-digit number $xy$ is
  • A
    $x + y$
  • ✓
    $10x + y$
  • C
    $10x – y$
  • D
    $10y + x$
Answer
Correct option: B.
$10x + y$
In generalised form, $??$ can be written as the sum of the products of its digits with their respective place values, i.e.$?? = 10? + y$
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MCQ 161 Mark
If the sum of digits of a number is divisible by three, then the number is always divisible by:
  • A
    $2$
  • B
    $3$
  • ✓
    $6$
  • D
    $9$
Answer
Correct option: C.
$6$
We know that, if sum of digits of a number is divisible by three, then the number must be divisible by $3,$ i.e. the remainder obtained by dividing the number by $3$ is same as the remainder obtained by dividing the sum of its digits by $3.$
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MCQ 171 Mark
Let $abc$ be a three-digit number. Then $abc – cba$ is not divisible by
  • A
    $9$
  • B
    $11$
  • ✓
    $18$
  • D
    $33$
Answer
Correct option: C.
$18$
Given, abc is a three-digit number. Then,
$abc = 100a + 10b + c and$
$cba = 100c + 10b + a$
$abc − eba = (100a + 10b + c) − (100c + 10b + a)$
$= 100a - a + 10b - 10b + c - 100$
$= 99a - 99c = 99 (a - c)$
$= abc - cba$ is divisible by $99$
$abc - cba$ is divisible by $9, 11, 33,$ but it is not divisible by $18.$
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