MCQ 511 Mark
If $p^2=\frac{32}{50}$, then $p$ is a/an
Answer(C) rational number
We have, $p^2=\frac{32}{50} \Rightarrow p^2=\frac{16}{25} \Rightarrow p= \pm \frac{4}{5}$, which is a rational number.
View full question & answer→MCQ 521 Mark
The ratio of LCM and HCF of the least composite number and the least prime number is
- A
$1: 2$
- ✓
$2: 1$
- C
$1: 1$
- D
$1: 3$
AnswerCorrect option: B. $2: 1$
View full question & answer→MCQ 531 Mark
The HCF and the LCM of $12,21,15$ respectively are
View full question & answer→MCQ 541 Mark
The total number of factors of a prime number is
MCQ 551 Mark
If $\operatorname{HCF}(x, 8)=4, \operatorname{LCM}(x, 8)=24$, then $x$ is
Answer(C) 12
We know that: $\operatorname{LCM}(x, 8) \times \operatorname{HCF}(x, 8)=x \times 8$
$
\Rightarrow \quad 24 \times 4=x \times 8 \Rightarrow x=12
$
View full question & answer→MCQ 561 Mark
The smallest number which when divided by 17, 23 and 29 leaves a remainder 11 in each case is
Answer(B) 11350
Let the required natural number be $n$. Then,
$n-11$ is the LCM of 17,23 and 629.
i.e. $\quad n-11=\operatorname{LCM}(17,23,29)$
$\Rightarrow \quad n-11=17 \times 23 \times 29 \qquad[\because 17,23,29$ are distinct primes $]$
$\Rightarrow \quad n=11+11339=11350$
View full question & answer→MCQ 571 Mark
If two positive integers $a$ and $b$ are written as $a=p^3 q^4$ and $b=p^2 q^3$, where $p$ and $q$ are prime numbers, such that $\operatorname{HCF}(a, b)=p^m q^n$ and $\operatorname{LCM}(a, b)=p^r q^5$, then $(m+n)(r+s)$ equal to
Answer(C) 35
We find that: $\operatorname{HCF}(a, b)=\operatorname{HCF}\left(p^3 q^4, p^2 q^3\right)=p^2 q^3$
and,
$
\operatorname{LCM}(a, b)=\operatorname{LCM}\left(p^3 q^4, p^2 q^3\right)=p^3 q^4
$
$
\therefore \quad m=2, n=3, r=3 \text { and } s=4
$
Hence,
$
(m+n)(r+s)=5 \times 7=35
$
View full question & answer→MCQ 581 Mark
If $a=2^2 \times 3^x, b=2^2 \times 3 \times 5, c=2^2 \times 3 \times 7$, and $L C M(a, b, c)=3780$, then $x=$
Answer(D) 3
We have, $\operatorname{LCM}(a, b, c)=3780$
$
\Rightarrow \quad\left(2^2 \times 3^x, 2^2 \times 3 \times 5,2^2 \times 3 \times 7\right)=3780 \Rightarrow 2^2 \times 3^x \times 5 \times 7=2^2 \times 3^3 \times 5 \times 7 \Rightarrow x=3
$
View full question & answer→MCQ 591 Mark
If the LCM of two numbers is 3600 , then which of the following numbers cannot be their HCF?
Answer(B) 500
The HCF of two natural numbers is always a factor of their LCM. Clealy, 600, 400 and 150 are factors of 3600 , but 500 is not its factor. Hence, 500 cannot be the HCF of numbers whose LCM is 3600 .
View full question & answer→MCQ 601 Mark
The sum of the HCF and LCM of $12,21,15$ is
Answer(A) 423
We find that: $12=2^2 \times 3,21=3 \times 7$ and $15=3 \times 5$
$
\begin{array}{l}
\therefore \quad \operatorname{HCF}(12,21,15)=3 \text { and } \operatorname{LCM}(12,21,15)=2^2 \times 3 \times 5 \times 7=420 \\
\text { Hence, } \operatorname{HCF}(12,21,15)+\operatorname{LCM}(12,21,15)=3+420=423
\end{array}
$
View full question & answer→MCQ 611 Mark
If the HCF of 408 and 1032 is expressible in the form $1032 m-408 \times 5$, then the value of $m$ is
Answer(D) 3
We find that: $408=2^3 \times 3 \times 17$ and $1032=2^3 \times 3 \times 43$.
$
\therefore \quad \operatorname{HCF}(408,1032)=2^3 \times 3=24
$
It is given that: $\operatorname{HCF}(408,1032)=1032 m-408 \times 5$
$
\therefore \quad 1032 m-408 \times 5=24 \Rightarrow 1032 m=2040+24 \Rightarrow 1032 m=2064 \Rightarrow m=2
$
View full question & answer→MCQ 621 Mark
If $p_1$ and $p_2$ are odd prime numbers such that $p_1>p_2$, then $p_1^2-p_2^2$ is
Answer(A) an even number
$p_1^2-p_2^2=\left(p_1+p_2\right)\left(p_1-p_2\right)$.
It is given that $p_1$ and $p_2$ are odd prime numbers. Therefore $p_1+p_2$ and $p_1-p_2$ are even numbers. Hence, $\left(p_1+p_2\right)\left(p_1-p_2\right)=p_1^2-p_2^2$ is an even number. In fact, $p_1^2-p_2^2$ is a multiple of 4 .
View full question & answer→MCQ 631 Mark
If the HCF of 85 and 153 is expressible in the form $85 n-153$, then the value of $n$ is
Answer(B) 2
We have, $85=5 \times 17$ and $153=3^2 \times 17$. Therefore, $\operatorname{HCF}(85,153)=17$.
It is given that $\operatorname{HCF}(85,153)=85 n-153$.
$
\therefore \quad 85 n-153=17 \Rightarrow 85 n=170 \Rightarrow n=2
$
View full question & answer→MCQ 641 Mark
If 3 is the least prime factor of $m$ and 5 is the least prime factor of $n$, then the least prime factor of $(m+n)$ is
Answer(B) 2
It is given that least prime factor of $m$ is 3 , so other prime factors of $m$ are greater than 3 . Consequently, 2 is not a prime factor of $m$ and hence $m$ is an odd integer. Similarly, $n$ is an odd integer. Therefore, $m+n$ is an even integer and hence its least prime factor is 2 .
View full question & answer→MCQ 651 Mark
$m=d n+r$ where $m, n$ are positive integers and $d$ and $r$ integers, then $n$ is the HCF of $m$ and $n$, if
- A
$r=1$
- B
$0$
- ✓
$r=0$
- D
$r$ is a real number
Answer(C) $r=0$
In $n$ is the HCF of $m$ and $n$, then $n$ is a factor of $m$ and hence $n$ is a factor of $d n+r$. This is possible only when $r=0$.
View full question & answer→MCQ 661 Mark
If two positive integers $p$ and $q$ can be expressed as $p=18 a^2 b^4$ and $q=20 a^3 b^2$ where $a$ and $b$ are prime numbers, then $\operatorname{LCM}(p, q)$ is
- A
$2 a^2 b^2$
- B
$180 a^2 b^2$
- C
$12 a^2 b^2$
- ✓
$180 a^3 b^4$
AnswerCorrect option: D. $180 a^3 b^4$
(D) $180 a^3 b^4$
We have,
$
\begin{aligned}
& p=2 \times 3^2 \times a^2 \times b^4 \text { and } q=2^2 \times 5 \times a^3 \times b^2 \\
\therefore \quad & \operatorname{LCM}(p, q)=2^2 \times 3^2 \times 5 \times a^3 \times b^4=180 a^3 b^4
\end{aligned}
$
View full question & answer→MCQ 671 Mark
Let $n$ be a natural number. Then, the $\operatorname{LCM}(n, n+1)$ is
AnswerCorrect option: C. $n(n+1)$
(C) $n(n+1)$
For any natural number $n$, natural numbers $n$ and $n+1$ are relatively prime.
$
\therefore \quad \operatorname{LCM}(n, n+1)=n(n+1)
$
View full question & answer→MCQ 681 Mark
The HCF of two consecutive positive integers is
Answer(B) 1
In any two consecutive positive integers one is even and other is odd. So, they have 1 as the only common factor. Hence, their HCF is 1 .
View full question & answer→MCQ 691 Mark
The smallest number divisible by all natural numbers between 1 and 10 (both inclusive) is
Answer(B) 2520
Required number is the LCM of $1,2,3,4,5,6,7,8,9,10$.
Now,
$
\begin{aligned}
\operatorname{LCM}(1,2,3,4,5,6,7,8,9,10) & =\operatorname{LCM}\left(2^0, 2^1, 3^1 2^2, 5^1, 2^1 \times 3^1, 7^1, 2^3, 3^2, 2^1 \times 5^1\right) \\
& =2^3 \times 3^2 \times 5^1 \times 7^1=2520
\end{aligned}
$
View full question & answer→MCQ 701 Mark
The HCF of smallest prime number and the smallest composite number is
Answer(A) 2
The smallest prime and the smallest composite numbers are 2 and 4 respectively.
$\operatorname{HCF}(2,4)=2$.
View full question & answer→MCQ 711 Mark
The LCM of the smallest two digit composite number and the smallest composite number is
Answer(B) 20
The smallest two digit composite number is 10 and the smallest composite number is 4 . Prime factorizations of these two are: $10=2^1 \times 5^1$ and $4=2^2$.
$
\therefore \quad \operatorname{LCM}(10,4)=2^2 \times 5^1=20
$
View full question & answer→MCQ 721 Mark
Let $p$ be a prime number. The quadratic equation having its factors as zeros is
- A
$x^2-p x+p=0$
- ✓
$x^2-(p+1) x+p=0$
- C
$x^2+(p+1) x+p=0$
- D
$x^2-p x+(p+1)=0$
AnswerCorrect option: B. $x^2-(p+1) x+p=0$
(B) $x^2-(p+1) x+p=0$
Factors of $p$ are 1 and $p$ only. So, the quadratic equation having 1 and $p$ as its zero is
$x^2-x(1+p)+1 \times p=0$ or, $x^2-(p+1) x+p=0$
View full question & answer→MCQ 731 Mark
Let $p$ be a prime number. The sum of its factors is
Answer(C) $p+1$
The factors of $p$ are 1 and $p$. So, their sum is $p+1$.
View full question & answer→MCQ 741 Mark
If $p$ and $q$ are two distinct prime numbers, then $\operatorname{LCM}(p, q)$ is
Answer(D) $p q$
Factors of $p$ are: 1 and $p$. Factors of $q$ are: 1 and $q$.
$
\begin{array}{ll}
\therefore & \operatorname{LCM}(p, q)=1 \times p \times q=p q \\
\operatorname{ALITER} & \operatorname{LCM}(p, q) \times \operatorname{HCF}(p, q)=p q \Rightarrow \operatorname{LCM}(p, q) \times 1=p q \Rightarrow \operatorname{LCM}(p, q)=p q
\end{array}
$
View full question & answer→MCQ 751 Mark
If $p$ and $q$ are two distinct prime numbers, then their HCF is
Answer(D) 1
A prime number has no factor other than 1 and the number itself. Therefore, Factors of $p$ are: 1 and $p$ only. Factors of $q$ are: 1 and $q$ only.
Therefore, 1 is the only common factor of $p$ and $q$. Hence, HCF $(p, q)=1$.
View full question & answer→MCQ 761 Mark
HCF of two positive integers is always
MCQ 771 Mark
For any natural numbers, $25^{2 n}-9^{2 n}$ is always divisible by
Answer(C) both 16 and 34
[Hint: $a^{2 n}-b^{2 n}$ is divisible by both $(a+b)$ and $(a-b)$ ]
View full question & answer→MCQ 781 Mark
If $(a \times 5)^n$ ends with the digit zero for every natural number $n$, then a is
View full question & answer→MCQ 791 Mark
View full question & answer→MCQ 801 Mark
The remainder when the square of any prime number greater than 3 is divided by 6 , is
Answer(A) 1
[Hint: Any prime number greater than 3 is of the form $6 k \pm 1$, where $k$ is a natural number and $\left.(6 k \pm 1)^2=36 k^2 \pm 12 k+1=6 k(6 k \pm 2)+1\right]$
View full question & answer→MCQ 811 Mark
If $n$ is any natural number, then $6^n-5^n$ always ends with
Answer(A) 1
[Hint: For any $n \in N, 6^n$ and $5^n$ end with 6 and 5 respectively. Therefore, $6^n-5^n$ always ends with $6-5=1$.]
View full question & answer→MCQ 821 Mark
If $n$ is a natural number, then $9^{2 n}-4^{2 n}$ is always divisible by
Answer(C) both 5 and 13
[Hint: $9^{2 n}-4^{2 n}$ is of the form $a^{2 n}-b^{2 n}$ which is divisible by both $a-b$ and $a+b$. So, $9^{2 n}-4^{2 n}$ is divisible by both $9-4=5$ and $9+4=13$.]
View full question & answer→MCQ 831 Mark
Three bells ring at intervals of 4,7 and 14 minutes. All the three rang at 6 AM . When will they ring together again?
- A
$6: 07 AM$
- B
$6: 14 AM$
- ✓
$6: 28 AM$
- D
$6: 25 AM$
AnswerCorrect option: C. $6: 28 AM$
View full question & answer→MCQ 841 Mark
The smallest number by which $\sqrt{27}$ should be multiplied so as to get a rational number is
- A
$\sqrt{27}$
- B
$3 \sqrt{3}$
- ✓
$\sqrt{3}$
- D
AnswerCorrect option: C. $\sqrt{3}$
View full question & answer→MCQ 851 Mark
If 3 is the least prime factor of number $a$ and 7 is the least prime factor of number $b$, then the least prime factor of $a+b$, is
View full question & answer→MCQ 861 Mark
If $p$ and $q$ are co-prime numbers, then $p^2$ and $q^2$ are
View full question & answer→MCQ 871 Mark
If the LCM of $a$ and 18 is 36 and the HCF of $a$ and 18 is 2 , then $a=$
View full question & answer→MCQ 881 Mark
Any one of the numbers $a, a+2$ and $a+4$ is a multiple of
View full question & answer→MCQ 891 Mark
How many prime numbers are of the form $10 n+1$, where $n$ is a natural number such that $1 \leq n<10$ ?
View full question & answer→MCQ 901 Mark
If the LCM of two prime number $p$ and $q(p>q)$ is 221 then the value of $3 p-q$ is
View full question & answer→MCQ 911 Mark
If the sum of two numbers is 1215 and their HCF is 81 , then the possible number of pairs of such number is
MCQ 921 Mark
If $\operatorname{LCM}(x, 18)=36$ and $\operatorname{HCF}(x, 18)=2$, then $x$ is
View full question & answer→MCQ 931 Mark
If $a^2=\frac{23}{25}$, then $a$ is
View full question & answer→MCQ 941 Mark
Prime factors of the denominator of a rational number with decimal expansion 44.123 are
MCQ 951 Mark
If the sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is
MCQ 961 Mark
The LCM and HCF of two rational numbers are equal, then the numbers must be
MCQ 971 Mark
$3 . \overline{27}$ is
View full question & answer→MCQ 981 Mark
$119^2-11^2$ is a
View full question & answer→MCQ 991 Mark
If $a=2^3 \times 3, b=2 \times 3 \times 5, c=3^n \times 5$ and $\operatorname{LCM}(a, b, c)=2^3 \times 3^2 \times 5$, then $n=$
View full question & answer→MCQ 1001 Mark
The largest number which divides 70 and 125 , leaving remainders 5 and 8 , respectively, is