MCQ 11 Mark
Out of the following .............. is rational number.
- A
$\sqrt{3}$
- B
$\pi$
- ✓
$\frac{4}{0}$
- D
$\frac{0}{4}$
AnswerCorrect option: C. $\frac{4}{0}$
$\frac{4}{0}$
View full question & answer→MCQ 21 Mark
The least perfect square number which is divisible $3,4,5,6$ and $8$ is
- A
$900$
- B
$1200$
- C
$2500$
- ✓
$3600$
AnswerCorrect option: D. $3600$
$3600$
View full question & answer→MCQ 31 Mark
$\operatorname{LCM}(a, 18)=36 \operatorname{HCF}(a, 18)=2$ then $a=$ ..............
View full question & answer→MCQ 41 Mark
If $\operatorname{HCF}(a, b)=18$ then $\operatorname{LCM}(a, b)=\ldots \ldots \ldots$. is not possible.
View full question & answer→MCQ 51 Mark
Out of the following ................ is a square of any natural number.
- A
$136162$
- B
$126738$
- C
$410883$
- ✓
$385641$
AnswerCorrect option: D. $385641$
$385641$
View full question & answer→MCQ 61 Mark
If HCF $(x, y)=1$ then HCF $(x-y, x+y)=$ ................
- A
$4$
- ✓
$1$ or $2$
- C
$x$ or $y$
- D
$x+y$ or $x-y$
AnswerCorrect option: B. $1$ or $2$
$1$ or $2$
View full question & answer→MCQ 71 Mark
The L.C.M is _______ of the smallest prime number and the smallest composite number.
MCQ 81 Mark
If $p$ and $q$ are co-prime numbers, then $p^2$ and $q^2$ are
Answer(a) : Since $p$ and $q$ are co-prime. So, their squares i.e., $p^2$ and $q^2$ will also have no common factor i.e., they are also co-prime.
View full question & answer→MCQ 91 Mark
If $p$ and $q$ are primes, then $\operatorname{HCF}(p, q)$ will be
View full question & answer→MCQ 101 Mark
If LCM of $a$ and 18 is 36 and HCF of $a$ and 18 is 2 , then $a=$
Answer(c) : Product of two numbers $= HCF \times LCM$ of two numbers
$\therefore \quad a \times 18=2 \times 36 \Rightarrow a=\frac{72}{18}=4 .$
View full question & answer→MCQ 111 Mark
The $\text{HCF}$ of two numbers is $23$ and their $\text{LCM}$ is $1449.$ If one of the numbers is $161,$ then the other number is
AnswerLet the other number be $x$.
Product of two numbers $= \text{HCF} \times \text{LCM}$ of two numbers
$\therefore x \times 161=23 \times 1449$
$\Rightarrow x=\frac{23 \times 1449}{161}$
$\Rightarrow x=207$
View full question & answer→MCQ 121 Mark
If $p$ and $q$ are positive integers such that $p=a b^2$ and $q=a^3 b$, where $a, b$ are prime numbers, then $\operatorname{HCF}(p, q)=$
- ✓
$a b$
- B
$a^2 b^2$
- C
$a^3 b^2$
- D
$a^3 b^3$
Answer(a) : $\operatorname{HCF}(p, q)=\operatorname{HCF}\left(a b^2, a^3 b\right)=a b$.
View full question & answer→MCQ 131 Mark
The least number which when divided by $18,24,30$ and $42$ will leave in each case the same remainder $1$ , would be
AnswerCorrect option: C. $2521$
We have, $18=2 \times 3^2,$
$24=2^3 \times 3,$
$30=2 \times 3 \times 5,$
$42=2 \times 3 \times 7$
$\text{LCM}=2^3 \times 3^2 \times 5 \times 7=2520$
So, required least number is $2520+1=2521$.
View full question & answer→MCQ 141 Mark
HCF of $\left(2^3 \times 3^2 \times 5\right),\left(2^2 \times 3^3 \times 5^2\right)$ and $\left(2^4 \times 3 \times 5^3 \times 7\right)$ is
Answer(c) : $HCF = 2^2 \times 3 \times 5=4 \times 3 \times 5=60$.
View full question & answer→MCQ 151 Mark
If the LCM of 12 and 42 is $10 m+4$, then the value of $m$ is
Answer(b) : We have, $12=2 \times 2 \times 3=2^2 \times 3,42=2 \times 3 \times 7$
$\therefore \quad \operatorname{LCM}(12,42)=2^2 \times 3 \times 7=84$
$\Rightarrow 10 m+4=84 \Rightarrow 10 m=84-4=80 \Rightarrow m=\frac{80}{10}=8$
View full question & answer→MCQ 161 Mark
Find $\operatorname{HCF}(8,9,25) \times \operatorname{LCM}(8,9,25)$.
Answer(b) : We have, $8=2 \times 2 \times 2$,
$9=3 \times 3,25=5 \times 5$
$\operatorname{HCF}(8,9,25)=1$ and $\operatorname{LCM}(8,9,25)=2^3 \times 3^2 \times 5^2=1800$
$\therefore \operatorname{HCF}(8,9,25) \times \operatorname{LCM}(8,9,25)=1 \times 1800=1800$
View full question & answer→MCQ 171 Mark
If $\operatorname{HCF}(26,169)=13$, then $\operatorname{LCM}(26,169)$ equal to
Answer(c) : Given, $\operatorname{HCF}(26,169)=13$
We know that, Product of two numbers $= HCF \times LCM$ of two numbers
$\Rightarrow 26 \times 169=13 \times \text { LCM } \Rightarrow \text { LCM }=338$
View full question & answer→MCQ 181 Mark
The LCM and HCF of two non-zero positive numbers are equal, then the numbers must be
Answer(d) : Let the two numbers be $a$ and $b$.
Given, $\operatorname{HCF}(a, b)=\operatorname{LCM}(a, b)=k$ (say)
Since, $\operatorname{HCF}(a, b)=k \Rightarrow a=k m$ and $b=k n$, for some natural numbers $m, n$.
We know, $HCF \times LCM =$ Product of two numbers
$\therefore k \times k=k m \times k n$
$\Rightarrow 1=m \cdot n$
$\Rightarrow m=n=1$, since, $m, n$ are natural numbers.
Therefore, $a=k m=k$ and $b=k n=k$
$\Rightarrow a=b=k$ i.e., the numbers must be equal.
View full question & answer→MCQ 191 Mark
What is the HCF of smallest prime number and smallest composite number?
Answer(b) : Smallest prime number $=2$
Smallest composite number $=4$
$\therefore \quad \operatorname{HCF}(2,4)=2$
View full question & answer→MCQ 201 Mark
The HCF and the LCM of 12,21 and 15 respectively, are
Answer(c) : We have, $12=2 \times 2 \times 3=2^2 \times 3$
$\begin{aligned}
21 & =3 \times 7 \\
15 & =3 \times 5 \\
\therefore \quad & \operatorname{HCF}(12,21,15)=3 \text { and } \\
\operatorname{LCM}(12,21,15) & =2^2 \times 3 \times 5 \times 7=420
\end{aligned}$
View full question & answer→MCQ 211 Mark
If the $\text{HCF}$ of $65$ and $117$ is expressible in the form $65 m-117$, then the value of $m$ is
AnswerHere, $65=13 \times 5$
$117=13 \times 3^2$
$\therefore \operatorname{HCF}(65,117)=13\ldots(i)$
Also, given that, $\operatorname{HCF}(65,117)=65 m-117\ldots(ii)$
From $(i)$ and $(ii),$
$65 m-117=13$
$\Rightarrow 65 m=130$
$\Rightarrow m=2$
View full question & answer→MCQ 221 Mark
The largest number which divides 70 and 125 , leaving remainders 5 and 8 , respectively, is
Answer(a): Since 5 and 8 are the remainders of 70 and 125 , respectively. So, after subtracting these remainders from the numbers, we have the numbers $(70-5)=65$, $(125-8)=117$ which are divisible by the required number.
Now, required number $=$ HCF of 65 and $117=13$
View full question & answer→MCQ 231 Mark
If two positive integers $a$ and $b$ are written as $a=x^3 y^2$ and $b=x y^3 ; x, y$ are prime numbers, then $\operatorname{HCF}(a, b)$ is
- A
$x y$
- ✓
$x y^2$
- C
$x^3 y^3$
- D
$x^2 y^2$
AnswerCorrect option: B. $x y^2$
(b) : Given that, $a=x^3 y^2=x \times x \times x \times y \times y$
and $b=x y^3=x \times y \times y \times y$
$\therefore \quad$ HCF of $a$ and $b=\operatorname{HCF}\left(x^3 y^2, x y^3\right)=x \times y \times y$ $=x y^2$
View full question & answer→MCQ 241 Mark
Find the smallest number which when increased by 17 is exactly divisible by 520 and 468.
Answer(b) : Required number $=\operatorname{LCM}$ of $(520,468)-17$
$=4680-17=4663$
View full question & answer→MCQ 251 Mark
The HCF of two numbers is 27 and their LCM is 162 . If one of the numbers is 54 , find the other.
Answer(c) : For two numbers $a$ and $b$, we know that
$
(a \times b)=\operatorname{HCF}(a, b) \times \operatorname{LCM}(a, b)
$
Here $a=54, HCF =27$ and LCM $=162$
$
\therefore \quad 54 \times b=27 \times 162 \Rightarrow b=\frac{27 \times 162}{54}=81
$
Hence, the other number is 81 .
View full question & answer→MCQ 261 Mark
Find the least positive integer divisible by 20 and 24.
Answer(d): We have, $20=2^2 \times 5 ; 24=2^3 \times 3$
$\therefore \quad$ Required number $=\operatorname{LCM}(20,24)=2^3 \times 3 \times 5=120$
View full question & answer→MCQ 271 Mark
If the HCF of 150 and 100 is 50 , find the LCM of 150 and 100 .
Answer(c) : Given, $\operatorname{HCF}(150,100)=50$
$\therefore \quad \operatorname{LCM}(150,100)=\frac{150 \times 100}{\operatorname{HCF}(150,100)}=\frac{150 \times 100}{50}=300$
View full question & answer→MCQ 281 Mark
The exponent of 2 in the prime factorisation of 144 , is
Answer(d) : Prime factorisation of $144=2^4 \times 3^2$
Hence, exponent of 2 is 4 .
View full question & answer→MCQ 291 Mark
The HCF of $2^2 \times 3^2 \times 5^3 \times 7,2^3 \times 3^3 \times 5^2 \times 7^2$ and $3 \times 5 \times 7 \times 11$ is
Answer(d) : HCF = Product of lowest powers of each common prime factors $=3 \times 5 \times 7=105$
View full question & answer→MCQ 301 Mark
A rectangular courtyard 3.78 metres long and 5.25 metres wide is to be paved exactly with square tiles, all of the same size. Then the largest size of the tile which could be used for the purpose is equal to
Answer(b) : Largest size of the tile $= HCF$ of $378 cm$ and $525 cm =21 cm$
View full question & answer→MCQ 311 Mark
If the HCF of 45 and 105 is 15 , then their LCM is
Answer(c) : We have, $\operatorname{HCF}(45,105)=15$
As, $\operatorname{HCF}(a, b) \times \operatorname{LCM}(a, b)=a \times b$
$\therefore \quad \operatorname{LCM}(45,105)=\frac{45 \times 105}{\operatorname{HCF}(45,105)}=\frac{45 \times 105}{15}=315$
View full question & answer→MCQ 321 Mark
The prime factorisation of $3^{13}-3^{10}$ is
- A
$3^{13} \times 2 \times 13$
- B
$3^{11} \times 2 \times 13$
- C
$3^{10} \times 2 \times 13$
- D
Answer$\begin{aligned}
(c) : 3^{13}-3^{10} & =3^{10}\left(3^3-1\right)=3^{10}(26) \\
& =3^{10} \times 2 \times 13
\end{aligned}$
View full question & answer→MCQ 331 Mark
If $\operatorname{HCF}(a, b)=12$ and $a \times b=1800$, then $\operatorname{LCM}(a, b)=$
Answer(c) : $\operatorname{LCM}(a, b)=\frac{a \times b}{\operatorname{HCF}(a, b)}=\frac{1800}{12}=150$
View full question & answer→MCQ 341 Mark
If HCF $(306,1314)=18$, then LCM $(306$, 1314) is
Answer(a) : $LCM =\frac{306 \times 1314}{18}=22338$
View full question & answer→MCQ 351 Mark
If $p$ is prime, then HCF and LCM of $p$ and $p+1$ would be
- A
$HCF =p, LCM =p+1$
- B
$HCF =p(p+1), LCM =1$
- ✓
$HCF =1, LCM =p(p+1)$
- D
AnswerCorrect option: C. $HCF =1, LCM =p(p+1)$
(c) : Since, $p$ is prime
$\therefore \quad p$ and $p+1$ has no common factor other than 1 .
$\therefore \quad$ HCF of $p$ and $p+1=1$
& LCM of $p$ and $p+1=p \times(p+1)=p(p+1)$
View full question & answer→MCQ 361 Mark
There is a circular path around a sports field. Priya takes 21 minutes to drive one round of the field, while Ravish takes 28 minutes for the same. Suppose they both started at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Answer(c) : Required number of minutes is the LCM of 21 and 28 .
$\begin{aligned}
& 21=7 \times 3 \text { and } 28=2^2 \times 7 \\
\therefore & \operatorname{LCM}(21,28)=2^2 \times 3 \times 7=84
\end{aligned}$
View full question & answer→MCQ 371 Mark
Which of the following is a pair of co-primes?
- A
$(14,35)$
- ✓
$(18,25)$
- C
$(31,93)$
- D
$(32,62)$
AnswerCorrect option: B. $(18,25)$
(b) : $\operatorname{HCF}(18,25)=1$. So, $(18,25)$ is a pair of coprimes.
View full question & answer→MCQ 381 Mark
Find the least number which when divided by 12 , leaves a remainder of 7 , when divided by 15 , leaves a remainder of 10 and when divided by 16 , leaves a remainder of 11 .
Answer(b) : $12-7=5,15-10=5$ and $16-11=5$
Hence, the desired number is 5 short for divisibility by 12,15 and 16 . LCM of $12,15,16$ is 240 .
Hence, the least number $=240-5=235$
View full question & answer→MCQ 391 Mark
Two numbers are in the ratio of $15: 11$. If their HCF is 13 , then numbers will be
Answer(a) : Let the required numbers be $15 x$ and $11 x$.
Then, their HCF is $x$. So, $x=13$
$\therefore \quad$ The numbers are $15 \times 13$ and $11 \times 13$ i.e., 195 and 143.
View full question & answer→MCQ 401 Mark
Three farmers have $490 \ kg , 588 \ kg$ and $882 \ kg$ weights of wheat respectively. Find the maximum capacity of a bag so that the wheat can be packed in exact number of bags.
- A
$96 \ kg$
- B
$95 \ kg$
- C
$94 \ kg$
- ✓
$98 \ kg$
AnswerCorrect option: D. $98 \ kg$
Maximum capacity of a bag so that the wheat can be packed in exact number of bags is $\text{HCF} (490,588,882)$.
$490=2 \times 5 \times 7^2 ;$
$588=2^2 \times 3 \times 7^2 ;$
$882=2 \times 3^2 \times 7^2$
$\therefore \text { HCF }(490,588,882)$
$=2 \times 7^2$
$=98$
View full question & answer→MCQ 411 Mark
The product of two numbers is 4107 . If the HCF of these numbers is 37 , then find the greater number.
Answer(a) : Let the numbers be $37 a$ and $37 b$. Then $37 a \times 37 b=4107 \Rightarrow a b=3$
Now, co-primes with product 3 are $(1,3)$
So, the required numbers are $(37 \times 1,37 \times 3)$
i.e., $(37,111) \quad \therefore$ Greater number $=111$
View full question & answer→MCQ 421 Mark
Three numbers are in the ratio $1: 2: 3$ and their HCF is 12. Then the positive square root of largest number is
Answer(c) : Let the required numbers be $x, 2 x$ and $3 x$. Then their $HCF =x$. So, $x=12$
$\therefore \quad$ The numbers are 12,24 and 36 .
$\therefore \quad$ Required number $=\sqrt{36}=6$
View full question & answer→MCQ 431 Mark
If least prime factor of $p$ is 5 and least prime factor of $q$ is 7 , then the least prime factor of $(p+q)$ is
Answer(a) : Clearly, 2 is neither a factor of $p$ nor that of $q$.
$\therefore \quad p$ and $q$ are both odd.
So, $(p+q)$ must be an even number, which is divisible by 2 . Hence, the least prime factor of $(p+q)$ is 2 .
View full question & answer→MCQ 441 Mark
Pens are sold in pack of 8 and notepads are sold in pack of 12. Find the least number of pack of each type that one should buy so that there are equal number of pens and notepads.
Answer(a) : LCM of 8 and 12 is 24 .
$\therefore \quad$ The least number of pack of pens $=24 / 8=3$
$\therefore \quad$ The least number of pack of notepads $=24 / 12=2$
View full question & answer→MCQ 451 Mark
In a classroom activity on real numbers, the students have to pick a number card from a pile and frame question on it for the rest of the class. Shreya picked up 9 and her question was : Which of the following is not irrational?
- A
$9-4 \sqrt{5}$
- B
$\sqrt{7}-9$
- ✓
$2+2 \sqrt{9}$
- D
$4 \sqrt{11}-9$
AnswerCorrect option: C. $2+2 \sqrt{9}$
(c) : Here, $\sqrt{9}=3$
So, $2+2 \sqrt{9}=2+6=8$, which is not irrational.
View full question & answer→