Questions · Page 2 of 6

M.C.Q. [1 Marks Each]

MCQ 511 Mark
If $p$ is a prime number, find the number of factors of a number $p^3.$
  • A
    One
  • B
    Two
  • C
    Three
  • Four
Answer
Correct option: D.
Four
The answer must be true for any value of $p,$ so plug in an easy (prime) number for $p,$ such as $2.$
The factors of $2^3 = 8$ are $1, 2, 4,$ and $8,$ so answer $D$ is correct. In general, since $p$ is prime, the only numbers
that go into $p^3$ without a remainder are $1, p, p^21, $ and $p^3.$
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MCQ 521 Mark
The smallest fraction which should be subtracted from the sum of $1\frac{3}{4},2\frac{1}{2},5\frac{7}{12},3\frac{1}{3}$ and $2\frac{1}{4}$ to make the result a whole number, is _______.
  • $\frac{5}{12}$
  • B
    $\frac{7}{12}$
  • C
    $\frac{1}{2}$
  • D
    $7$
Answer
Correct option: A.
$\frac{5}{12}$
$\Rightarrow1\frac{3}{4},2\frac{1}{2},5\frac{7}{12},3\frac{1}{3}+2\frac{1}{4}$
$=\frac{7}{4}+\frac{5}{2}+\frac{67}{12}+\frac{10}{3}+\frac{9}{4}$
$=\frac{21+30+67+40+27}{12}$
$=\frac{185}{12}$
$=15\frac{5}{12}$
$\therefore$ The smallest fraction which should be subtracted from the sum of $1\frac{3}{4},2\frac{1}{2},5\frac{7}{12},3\frac{1}{3}$ and​ $2\frac{1}{4}$ to make the result a whole number is $\frac{5}{12}$.
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MCQ 531 Mark
A number $n$ is said to be perfect if the sum of all its divisors (excluding n itself) is equal to $n.$
An example of a perfect number is:
  • A
    $9$
  • B
    $15$
  • C
    $21$
  • $6$
Answer
Correct option: D.
$6$
$6$
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MCQ 541 Mark
Bhushan counted to $60$ using multiples of $6.$
Which statement is true about multiples of $6?$
  • A
    They are all odd numbers.
  • B
    They all have $66$ in the ones place.
  • They can all be divided evenly by $3.$
  • D
    They can all be divided evenly by $12.$
Answer
Correct option: C.
They can all be divided evenly by $3.$

Multiple of $6$ like $6, 12, 18,24, 30$ and they can all be divided evenly by $3.$
So option $C$ is correct.

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MCQ 551 Mark
The factors of $a^4 -$ $4a^2$ are
  • $a^2$ $(a - 2) (a + 2)$
  • B
    $a (a - 2) (a + 2)$
  • C
    $a (a + 2) (a + 2)$
  • D
    $a^2$ $(a - 2)^2$
Answer
Correct option: A.
$a^2$ $(a - 2) (a + 2)$
$a^4 - 4a^2 = a^2 (a^2 - 4)$
$= a^2$ $(a + 2) (a - 2)$ So correct answer will be option $A$
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MCQ 561 Mark
Mark the correct alternative in the following:
The number of factors of $1080$ is:
  • $32$
  • B
    $28$
  • C
    $24$
  • D
    $36$
Answer
Correct option: A.
$32$
$1080$ $= 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$ $= 2^3 \times 3^3 \times 5^1$
Thus, the total number of factors ig given by
$(3 + 1)(3 + 1)(1 + 1) = 32$
Hence, the correct answer is option $(a).$
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MCQ 571 Mark
$x$ is twice the difference between the $6th$ and $10th$ multiple of $7.$
Find the value of $x.$
  • A
    $38$
  • $56$
  • C
    $60$
  • D
    $28$
Answer
Correct option: B.
$56$

$6th$ multiple of $7 = 42$
$10th$ multiple of $7 = 70$
Now, $x = 2 × (70 − 42) = 2 × 28 = 56$

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MCQ 581 Mark
Choose the most appropriate option. The number of three digit numbers which are multiples of $9$ are$?$
  • A
    $98$
  • B
    $101$
  • $100$
  • D
    $99$
Answer
Correct option: C.
$100$
Smallest $2$ digit number which is divisible of $9$ is $108$ and highest $3$ digit number which is divisible of $9$ is $999.$
$108,117,126,..... 999.$
Now, it becomes $A. P.$ where we can easily find nn
$T_{n = a + (n - 1)d}$
Where, $T_{n = n^{th}}$ term of $A.P.$
$a =$ First term
$d =$ Common difference.
$\therefore999=108+(\text{n}-1)9$
$\Rightarrow\text{n}-1=\frac{999-108}{9}$
$\Rightarrow\text{n}-1=\frac{891}{9}$
$\Rightarrow\text{n}-1=99$
$\therefore\text{n}=100$
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MCQ 591 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following is a composite number?
  • A
    $23$
  • B
    $29$
  • $32$
  • D
    None of these.
Answer
Correct option: C.
$32$
$a.\ 23$ is not a composite number as it cannot be broken into factors.
$b.\ 29$ is not a composite number as it cannot be broken into factors.
$c.\ 32$ is a composite number as it can be broken into factors, which are $2 \times 2 \times 2 \times 2 \times 2.$
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MCQ 601 Mark
Mark the correct alternative in the following:
The $HCF$ of first $100$ natural numbers is:
  • A
    $2$
  • B
    $100$
  • $1$
  • D
    None of these.
Answer
Correct option: C.
$1$
The $HCF$ of first $100$ natural numbers is $1$ because there are some prime numbers like $2, 3, 5$ and so on which can't have common factor other than $1.$
Hence, the correct answer is option $(c).$
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MCQ 611 Mark
Mark the correct alternative in the following:
The least number exactly divisible by $36$ and $24$ is:
  • A
    $144$
  • $72$
  • C
    $64$
  • D
    $324$
Answer
Correct option: B.
$72$
$36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$
$24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1$
$LCM$ of $36$ and $24 = 2^3 \times 3^2 = 72$
Hence, the correct answer is option $(b).$
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MCQ 621 Mark
What is the reciprocal of $-3?$
  • A
    $-3$
  • $-\frac{1}{3}$
  • C
    $\frac{1}{3}$
  • D
    $3$
Answer
Correct option: B.
$-\frac{1}{3}$
 Reciprocal of $-3=\frac{1}{-3} $ or $\frac{-1}{3}$
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MCQ 631 Mark
In a unit fraction, the numerator is ________
  • A
    $0$
  • $1$
  • C
    $2$
  • D
    $3$
Answer
Correct option: B.
$1$

A unit fraction is a rational number written as a fraction in which numerator is $1$ and denominator is a positive integer.
Example- $\frac{1}{2},\frac{1}{3},\frac{1}{5}$​ etc.

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MCQ 641 Mark
Find a possible value of $v,$ if the least common multiple of $9,10,12$ nand $v$ is $540.$
  • A
    $18$
  • B
    $24$
  • $27$
  • D
    $36$
Answer
Correct option: C.
$27$

$LCM$ of $9,10,129 = 3 \times 310 = 2 \times 512 = 2 = 2 \times 2 \times 3$
$LCM (9,10,12) = 2 \times 2 \times 3 \times 3 \times 5 = 180$
$180$ is also multiply of $18,36,45.$
If $v$ is $18,36$ and $45$ than $LCM$ of all the number would be $180$ but its $540$

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MCQ 651 Mark
Which of the following is greatest?
  • A
    $4th$ multiple of $52$
  • B
    $8th$ multiple of $37$
  • $5th$ multiple of $25$
  • D
    $7th$ multiple of $50$
Answer
Correct option: C.
$5th$ multiple of $25$

$(a) 4th$ multiple of $52 = 52 \times 4 = 208$
$(b) 8th$ multiple of $37=37 \times 8 = 296$
$(c) 5th$ multiple of $25=25 \times 5 = 125$
$(d) 7th$ multiple of $50 = 50 \times 7 = 350$
So, $350$ is the greatest value amongst all.

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MCQ 661 Mark
$LCM$ of the numbers $12, 24$ and $36$ is:
  • A
    $36$
  • B
    $24$
  • $72$
  • D
    $108$
Answer
Correct option: C.
$72$

Taking out the factors of the given numbers, $12 = 2 \times 2 \times 3$
$24 = 2 \times 2 \times 2 \times 3$
$36 = 2 \times 2 \times 3 \times 3$
$\therefore LCM$ of $12, 24$ and $36 = 2 \times 2 \times 2 \times 3 \times 3 = 72$

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MCQ 671 Mark
If $p$ and $n$ are integers such that $p > n > 0$ and $p^2− n^2= 12,$ which of the following can be the value of pn? $i.\ 1$ $ii.\ 2$ $iii.\ 4$
  • A
    $I$ only
  • B
    $II$ only
  • C
    $I$ and $II$ only
  • None of these
Answer
Correct option: D.
None of these
Since $p$ are integers so $p + n$ and $p - n$ will also be integers.
$p^2 − n^2 = (p + n) (p - n) = 4 \times 3$ On comparing wee get $p + n = 4$ and $p - n = 3$
On solving these two equation we get $p = 3.5$ and $n = 0.5pn = 3.5 \times 0.5 = 1.75$
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MCQ 681 Mark
Mark the correct alternatiue in the following:
If the $HCF$ of two number is $16$ and their product is $3072$, then their $LCM$ is:
  • A
    $182$
  • $192$
  • C
    $12$
  • D
    None of these.
Answer
Correct option: B.
$192$

 We know:
$HCF \times LCM =$ Product of two numbers
$\because 16 \times LCM = 3,072$
$\therefore LCM = 3,07216=192$

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MCQ 691 Mark
Decimal expansion of a rational number cannot be ..........
  • Non-terminating and non-recrring
  • B
    Non-terminating and recurring
  • C
    Terminating
  • D
    None of these
Answer
Correct option: A.
Non-terminating and non-recrring

The decimal expansion of a rational number always either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over.
Moreover, any repeating or terminating decimal represents a rational number.
So,Decimal expansion of a rational number cannot be
Answer $(A)$ Non-terminating and non-recrring

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MCQ 701 Mark
Find a number which has a multiple of all the numbers from $1$ to $10?$
  • $5040$
  • B
    $1260$
  • C
    $720$
  • D
    $1440$
Answer
Correct option: A.
$5040$
Number which has a multiple of all the numbers from $1$ to $10$ will be multiple of their $LCM.$
$LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10) = 2520$
The only multiple of $2520$ from the options is $5040$ which is option $A$ so correct answer will be option $A$
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MCQ 711 Mark
$LCM$ of the numbers $4$ and $9$ is:
  • $32$
  • B
    $40$
  • C
    $45$
  • D
    $36$
Answer
Correct option: A.
$32$

Factors of the given numbers are, $4 = 2 \times 2$
$9 = 3 \times 3$
$\therefore LCM$ of $4$ and $9 = 2 \times 2 \times 3 \times 3 = 36$

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MCQ 721 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following are co$-$primes?
  • A
    $39, 91$
  • $161, 192$
  • C
    $385, 462$
  • D
    None of these.
Answer
Correct option: B.
$161, 192$
$a.\ 39$ and $91$ are not co$-$primes as $39$ and $91$ have a common factor, i.e. $13.$
$b.\ 161$ and $192$ are co$-$primes as $161$ and $192$ have no common factor other than $1.$
$c.\ 385$ and $462$ are not co$-$primes as $385$ and $462$ have common factors $7 $ and $11.$
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MCQ 731 Mark
Mark the correct alternative in the following:
From the numbers $2, 3, 4, 5, 6, 7, 8, 9$ how many pairs of co-primes can be formed$?$
  • $19$
  • B
    $18$
  • C
    $20$
  • D
    $21$
Answer
Correct option: A.
$19$
We can form $19$ pairs of co primes from the $2, 3, 4, 5, 6, 7, 8, 9$ which are given below,
$(2, 3), (2, 5), (2, 7), (2, 9), (3, 4),(3, 5), (3, 7), (3, 8), (4, 5), (4, 7), (4, 9), (5, 6), (5, 7), (5, 8), (5, 9), (6, 7), (7, 8), (7, 9)$ and $(8, 9)$
Hence, the correct answer is option $(a).$
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MCQ 741 Mark
Factorise $(3 − 4y − 7y^2)^2− (4y + 1)^2$
  • A
    $(7y^2− 4) (7y^2− 8y − 2)$
  • $(4 − 7y^2) (2 − 8y − 7y^2)$
  • C
    $(7y^2− 4) (2 − 8y − 7y^2)$
  • D
    $(4 − 7y^2) (7y^2+ 8y − 2)$
Answer
Correct option: B.
$(4 − 7y^2) (2 − 8y − 7y^2)$
We have, $(3 − 4y − 7y^2) 2 − (4y + 1)^2$
We know that $a^2− b^2= (a + b) (a − b)$
$\therefore \Rightarrow (3 − 4y − 7y^2+ 4y + 1) (3 − 4y − 7y^2− 4y − 1)$
$\Rightarrow (4 − 7y^2) (2 − 8y − 7y^2)$
Hence, this is the answer.
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MCQ 751 Mark
When $31513$ ad $34369$ are divided by a certain three digit number, the remainders are equal, then the remainder is ______.
  • A
    $86$
  • $97$
  • C
    $374$
  • D
    $113$
Answer
Correct option: B.
$97$

Let the divisor be $a$ and remainder be $r$
Let $31513 = am + r$ and $34369 = an + r$
Then, $a (n - m) = 2856 = 24 \times 119 = 12 \times 238 = 8 \times 357 = 6 \times 476 = 4 \times 714$
All three digit numbers give same remainder $= 97$

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MCQ 761 Mark
How many factors does the number $2424$ has $?$
  • A
    $2$ factors
  • B
    $4$ factors
  • C
    $6$ factors
  • $8$ factors
Answer
Correct option: D.
$8$ factors

Factors of $24$ are $1, 2, 3, 4, 6, 8, 12, 24$

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MCQ 771 Mark
Mark the correct alternatiue in the following:
The smallest number which when diminished by $3$ is divisible by $11,28,36$ and $45$ is:
  • A
    $1257$
  • B
    $1260$
  • C
    $1263$
  • None of these.
Answer
Correct option: D.
None of these.

Required smallest number $= LCM$ of $(11, 28, 36, 45) + 3 = 13,860 + 3 = 13,863$

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MCQ 781 Mark
The number of even prime factor(s) in $1955$ is/are
  • A
    $5$
  • $0$
  • C
    $3$
  • D
    $2$
Answer
Correct option: B.
$0$

Factor of $1955 = 5 × 17 × 23$
$1, 5, 17, 23$ are not even, so there are no even factors.

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MCQ 791 Mark
$5$ thousandths is:
  • A
    $0.05$
  • $0.005$
  • C
    $5.000$
  • D
    $0.056$
Answer
Correct option: B.
$0.005$

A decimal is a fractional number and is indicated by digits after a period which is called a decimal point.
Tenths have one digit after the decimal point. The decimal $0.8$ is pronounced as eight tenths.
Hundredths have two digits after the decimal point.
The decimal $0.06$ is pronounced as six hundredths.
Thousandths follow a similar pattern.
They have three digits after the decimal point.
The decimal $0.005$ is pronounced as five thousandths.
Hence, five thousandths is $0.005.$

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MCQ 801 Mark
Convert $75\%$ into regular fraction.
  • A
    $\frac{1}{4}$
  • B
    $\frac{2}{4}$
  • $\frac{3}{4}$
  • D
    $\frac{4}{5}$
Answer
Correct option: C.
$\frac{3}{4}$

$75\%$ can be written in regular fraction as $75\%=\frac{75}{100}$
On reducing the fraction, we get
$\frac{75}{100}=\frac{3}{4}$

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MCQ 811 Mark
Factors of $(a+b)^3 - (a-b)^3$ are:
  • A
    $2ab(3a^2+b^2)$
  • B
    $ab(3a^2+b^2)$
  • $2b(3a^2+b^2)$
  • D
    $3a^2+b^2$
Answer
Correct option: C.
$2b(3a^2+b^2)$
$(a + b)^3 - (a - b)^3$
$\Rightarrow a^3 + 3a^2b + 3ab^2 + b^3 - (a^3 - 3a^2b + 3ab^2 - b^3)$
$\Rightarrow a^3 + 3a^2b + 3ab^2 + b^3 - a^3 - 3a^2b + 3ab^2 - b^3$
$\Rightarrow 6a^2b + 2b^3$
$\Rightarrow 2b (3a^2 + b^2)$
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MCQ 821 Mark
Which of the following fraction has denominator $4?$
  • A
    $\frac{42}{7}$
  • B
    $\frac{7}{24}$
  • $\frac{9}{4}$
  • D
    $\frac{4}{9}$
Answer
Correct option: C.
$\frac{9}{4}$

Denominator is the number at the bottomOut of the given options, only $\frac{9}{4}$ has denominator as $4.$
Hence, the answer is $\frac{9}{4}.$

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MCQ 831 Mark
The prime number that comes just after $43$ is .............
  • A
    $49$
  • B
    $45$
  • $47$
  • D
    none of these
Answer
Correct option: C.
$47$

A prime number is a whole number greater than $1$ whose only factors are $1$ and itself.
A factor is a whole numbers that can be divided evenly into another number.
The first few prime numbers are $2, 3, 5, 7, 11, 13, 17, 19, 23$ and $29.$
The next prime number after $43$ is $47,$So option C is the correct answer.

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MCQ 841 Mark
The number of divisors of $2^6.3^5.5^3.7^4.11$ is equal to:
  • A
    $11^2 - 1$
  • B
    $21^2 - 1$
  • C
    $31^2 - 1$
  • $41^2 - 1$
Answer
Correct option: D.
$41^2 - 1$
Consider the factor $2^6$
$2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6.$
Any of the seven could be divisors. Arguing as above for the other factors we can say that the number of divisors are
$7 \times 6 \times 4 \times 5 \times 2 = 42 \times 40=1680=1600+80$
$=40^2 + 2\times40 +1- 1 = (40+1)^2 -1 = 41^2 -1$
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MCQ 851 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following numbers is divisible by $11?$
  • A
    $3333333$
  • B
    $1111111$
  • $22222222$
  • D
    None of these.
Answer
Correct option: C.
$22222222$
A number is divisible by $11,$ if the difference of the sum of its digits in odd places and the sum of the digits in even places $($starting from ones place$)$ is either $0$ or a multiple of $11.$
$a.\ 3333333$
Consider the number $3333333.$
Sum of its digits in odd places $(3 + 3 + 3 + 3) = 12$
Sum of its digits in even places $(3 + 3 + 3) = 9$
Difference of the two sums $= 12 - 9 = 3$
Since this number $(3)$ is not divisible by $11, 3333333$ is not divisible by $11.$
$b.\ 1111111$
Consider the number $1111111.$
Sum of its digits in odd places $(1 + 1 + 1 + 1) = 4$
Sum of its digits in even places $(1 + 1 + 1) = 3$
Difference of the two sums $= 4 - 3 = 1$
Since this number $(1) $ is not divisible by $11, 1111111$ is also not divisible by $11.$
$c.\ 22222222$
Consider the number $22222222.$
Sum of its digits in odd places $(2 + 2 + 2 + 2)= 8$
Sum of its digits in even places $(2 + 2 + 2 + 2) = 8$
Difference of the two sums $= 8 - 8 = 0$
Since this number $(0)$ is divisible by $11, 22222222$ is also divisible by $11.$
 
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MCQ 861 Mark
Mark $(\checkmark)$ against the correct answer in the following: The number which is neither prime nor composite is:
  • A
    $0$
  • $1$
  • C
    $2$
  • D
    $3$
Answer
Correct option: B.
$1$
$1$ is neither prime nor composite.
Is not correct because composite numbers are defined for positive numbers, but $0$ is neither a positive number nor a negative number.
​Is not correct because $2$ is a prime number.
Is not correct because $3$ is a prime number.
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MCQ 871 Mark
Convert the following into fraction. $56\%$
  • A
    $\frac{50}{100}$
  • $\frac{14}{25}$
  • C
    $56\times100$
  • D
    $\frac{23}{50}$
Answer
Correct option: B.
$\frac{14}{25}$
One percent is equal to one hundredth part:
$1\%\frac{1}{100}$
So in order to convert percent to fraction, divide the percent by $100\%$ and reduce the fraction.
For example $56\%$ is equal to $\frac{56}{100}$ with gcd $= 4$ is equal to $\frac{14}{25}:$
$56\%\frac{56}{100}=\frac{14}{25}$
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MCQ 881 Mark
The value of $\Bigg[\bigg(-2\frac{3}{4}\bigg)-\bigg(-1\frac{3}{4}\bigg)\Bigg]+\Bigg[\bigg(-2\frac{3}{4}\bigg)-\bigg(-1\frac{3}{4}\bigg)\Bigg]+ ....$ upto $30$ times is:
  • A
    $-1$
  • B
    $1$
  • C
    $30$
  • $-30$
Answer
Correct option: D.
$-30$

 Consider the given expression.
$\Bigg[\bigg(-2\frac{3}{4}\bigg)-\bigg(-1\frac{3}{4}\bigg)\Bigg]+\Bigg[\bigg(-2\frac{3}{4}\bigg)-\bigg(-1\frac{3}{4}\bigg)\Bigg]+ ....$ upto $30$ times
Sum $=\Bigg[\bigg(\frac{-11}{4}\bigg)-\bigg(\frac{-7}{4}\bigg)\Bigg]+\Bigg[\bigg(\frac{-11}{4}\bigg)-\bigg(\frac{-7}{4}\bigg)\Bigg]+ ....$ upto $30$ times
$=\big[-1\big]+\big[-1\big]+....$ upto $30$ times
$=-30$
Hence, the value of the expression is $-30.$

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MCQ 891 Mark
The number of prime factors in $1955$ are
  • A
    $5$
  • B
    $8$
  • $3$
  • D
    $2$
Answer
Correct option: C.
$3$

 Factors of $1955 = 5 \times 17 \times 23$
Number of factors are $3.$

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MCQ 901 Mark
Mark the correct alternative in the following:
If $1*548$ is divisible by $3, $which of the following digits can replace $*?$
  • $0$
  • B
    $2$
  • C
    $7$
  • D
    $9$
Answer
Correct option: A.
$0$

Sum of the given digits $= 1 + 5 + 4 + 8 = 18$
Since $18$ is a multiple of $3,$ the required digit is $0.$

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MCQ 911 Mark
Place value and face value are always equal for
  • $0$
  • B
    $10$
  • C
    any digit
  • D
    $100$
Answer
Correct option: A.
$0$

 The place value of every one-digit number is the same as and equal to its face value.
$(i)$ Place value and face value of $1, 2, 3, 4, 5, 6, 7, 8$ and $9$ are $1, 2, 3, 4, 5, 6, 7, 8$ and $9$ respectively.
$(ii)$ The place value of zero $(0)$ is always $0.$ As, in $105, 350, 42017, 90218$ the place value of $0$ in each number is $0.$
So option $A$ is the correct answer.

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MCQ 921 Mark
Example for an improper fraction is:
  • A
    $\frac{25}{26}$
  • B
    $\frac{12}{13}$
  • $\frac{15}{14}$
  • D
    $\frac{19}{20}$
Answer
Correct option: C.
$\frac{15}{14}$

Improper fraction is a fraction in which the numerator is greater than the denominator, such as $\frac{3}{2}$
Hence, $\frac{15}{14}$ is an improper fraction.

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MCQ 931 Mark
Evaluate the following: $0.8\times\frac{\frac{7}{12}}{\frac{5}{24}}$
  • $2\frac{6}{25}$
  • B
    $3\frac{6}{25}$
  • C
    $\frac{6}{25}$
  • D
    $\frac{26}{25}$
Answer
Correct option: A.
$2\frac{6}{25}$

 Given, $0.8\times\frac{\frac{7}{12}}{\frac{5}{24}}=0.8\times\frac{7}{12}\times\frac{24}{5}=0.8\times\frac{14}{5}$
$=\frac{8}{10}\times\frac{14}{5}$
$=\frac{56}{25}$
$=2\frac{6}{25}$

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MCQ 941 Mark
Which of the following integers has most number of divisors?
  • A
    $176$
  • B
    $182$
  • C
    $99$
  • $101$
Answer
Correct option: D.
$101$
 The answer must be true for any value of $p,$ so plug in an easy (prime) number for $p,$ such as $2.$
The factors of $2^3$ $= 8$ are $1, 2, 4,$ and $8,$ so answer $D$ is correct. In general, since $p$ is prime, the only numbers
that go into $p^3$ without a remainder are $1, p, p^21,$ and $p^3.$
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MCQ 951 Mark
How many two-digit prime numbers are there having the digit $3$ in their units place$?$
  • A
    $10$
  • B
    $8$
  • C
    $6$
  • $5$
Answer
Correct option: D.
$5$
$ 2$ digit prime nos. having $3$ in their units place are-
$13, 23, 43, 53, 73$
$\therefore $ There are $55$ such nos.
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MCQ 961 Mark
Mark $(\checkmark)$ against the correct answer in the following:
If $a$ and $b$ are co-primes, then their $LCM$ is:
  • A
    $1$
  • $\frac{\text{a}}{\text{b}}$
  • C
    $ab.$
  • D
    None of these.
Answer
Correct option: B.
$\frac{\text{a}}{\text{b}}$
 If $a$ and $b$ are co-primes then their $LCM$ will be $ab.$
For example, $4$ and $9$ are co-primes.
$LCM$ of $4$ and $9$ is $4 \times 9.$
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MCQ 971 Mark
Mark the correct alternative in the following:
If $1*548$ is divisible by $3,$ then $*$ can take the value:
  • $0$
  • B
    $2$
  • C
    $7$
  • D
    $8$
Answer
Correct option: A.
$0$

Sum of the given digits $= 1 + 5 + 4 + 8 = 18$
Since $18$ is a multiple of $3,$ the required digit is $0.$

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MCQ 981 Mark
Numerator in the fraction $\frac{5}{6}$​ is ___
  • $5$
  • B
    $6$
  • C
    $\frac{1}{5}$
  • D
    $\frac{1}{6}$
Answer
Correct option: A.
$5$

Numerator of a fraction is the upper number.
So, the numerator of the fraction $\frac{5}{6}$ is $5.$
Hence, the answer is $5.$

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MCQ 991 Mark
$LCM$ of the numbers $12, 24$ and $36$ is
  • A
    $36$
  • B
    $24$
  • $72$
  • D
    $108$
Answer
Correct option: C.
$72$

Taking out the factors of the given numbers,$12 = 2 \times 2 \times 3$
$24 = 2 \times 2 \times 2 \times 3$
$36 = 2 \times 2 \times 3 \times 3$
$\therefore LCM$ of 12, 24 and $36 = 2 \times 2 \times 2 \times 3 \times 3 = 72$

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MCQ 1001 Mark
Mark the correct alternatiue in the following:
The least number divisible by $15,20,24,32$ and $36$ is:
  • $1440$
  • B
    $1660$
  • C
    $2880$
  • D
    None of these.
Answer
Correct option: A.
$1440$

The least number divisible by $15, 20, 24, 32,$ and $36$ can be found by taking their $LCM$ as:

$\therefore LCM$ of $15, 20, 24, 32$ and $36 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 1,440$
Hence, $1,440$ is the least number that is divisible by $15, 20, 24, 32$ and $36.$

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