MCQ 1511 Mark
If $p_1$ and $p_2$ are two odd prime numbers such that $p_1>p_2$, then $\text{p}^2_1-\text{p}^2_2$ is :
AnswerLet the two odd prime numbers $p_1$ and $p_2$ be $5$ and $3$.
Then, $\text{p}^2_1=5^2$
$=25$
And
$\text{p}^2_2=3^2$
$=9$
Thus $, \text{p}^2_1-\text{p}^2_2=25-9$
$=16$
$16$ is even number.
Take another example, with $p_1$ and $p_2$ be $11$ and $7$.
Then $, \text{p}^2_1=11^2$
$=121$
And
$\text{p}^2_2=7^2$
$=49$
Thus $, \text{p}^2_1-\text{p}^2_2=121-49$
$=72$
$72$ is even number.
Thus, we can say that $\text{p}^2_1-\text{p}^2_2$ is even number
In general the square of odd prime number is odd.
Hence the difference of square of two prime numbers is odd
Hence the correct choice is $(a)$.
View full question & answer→MCQ 1521 Mark
$n^2- 1$ is divisible by $8,$ if $n$ is :
AnswerLet $a = n^2- 1$
Here $n$ can be even or odd.
Case $I : n =$ Even
i.e., $n = 2k,$ where $k$ is an integer.
$\Rightarrow a = (2k)^2- 1$
$\Rightarrow a = 4k^2- 1$
At $k = -1, 4(-1)^2-1 = 4 - 1 = 3$, which is not divisible by $8$.
At $k = 0, a = 4(0)^2- 1 = 0 - 1 = -1$, which is not divisible by $8,$ which is not.
Case $ II : n =$ Odd i.e., $n = 2k + 1,$ where $k$ is an odd integer.
$\Rightarrow a = 2k + 1$
$\Rightarrow a = (2k + 1)^2- 1$
$\Rightarrow a = 4k^2+ 4k + 1 - 1$
$\Rightarrow a = 4k^2+ 4k$
$\Rightarrow a = 4k(k + 1)$
At $k = -1, a = 4(-1)(-1 + 1) = 0$ which is divisible by $8$.
At $k = 0, a = 4(0)(0 + 1) = 4$ which is divisible by $8$.
At $k = 1, a = 4(1)(1 + 1) = 8$ which is divisible by $ 8$.
Hence, we can conclude from above two cases, if $n$ is odd, then $n^2- 1$ is divisible by $8$.
View full question & answer→MCQ 1531 Mark
The product of three consecutive positive integers is divisible by :
AnswerLet $n$ be a positive integer,
then three consecutive positive integers are $(n + 1)(n + 2)(n + 3) $
$= n(n + 1)(n + 2) +3 (n + 1)(n + 2)$
Here, the first term is divisible by $6$ and the second term is also divisible by $6$
Because it contains a factor $3$ and one of the two consecutive integers $(n + 1)$ or $(n + 2$) is even and thus is divisible by $2$.
$\therefore,$ the sum of multiple of $6$ is also a multiple of $6$.
View full question & answer→MCQ 1541 Mark
Choose the correct answer from the given four options in the following questions : For some integer $m,$ every even integer is of the form :
- A
$m.$
- B
$m + 1.$
- ✓
$2m.$
- D
$2m + 1$.
AnswerWe know that, even integers are $2, 4, 6, ...$
So, it can be written in the form of $2m.$
where, $m =$ Integer $= Z\ $[since, integer is represented by $Z]$
or $m = ···, -1, 0, 1, 2, 3, ...$
$\therefore 2m = ···, -2, 0, 2, 4, 6, ...$
Alternate Answer
Let $'a\ '$ be a positive integer.
On dividing $'a\ '$ by $2,$ let m be the quotient and be the remainder.
Then, by Euclid's divlslon algonthm, we have
$a = 2m + r,$ where
$\text{a}\leq\text{r}<2\text{ i.e.,}$
$r = 0$ and $r = 1$.
$\Rightarrow a = 2m$ or $a = 2m + 1$
when, $a = 2m$ for some integer $m,$ then clearly a is even.
View full question & answer→MCQ 1551 Mark
The multiplicative inverse of zero
AnswerAll numbers except zero have a multiplicative inverse because we cannot multiply any number by it to get $1$.
View full question & answer→MCQ 1561 Mark
A rational number can be expressed as a terminating decimal if the denominator has the factors :
- A
$2$ only
- ✓
$2$ or $5$ only
- C
$2$ or $3$ only
- D
$2, 3$ or $5$ only
AnswerCorrect option: B. $2$ or $5$ only
A rational number can be expressed as a terminating decimal if the denominator has the factors $2$ or $5$ or both.
Any other factors in the denominator yield a non $-$ terminating decimal expansion.
View full question & answer→MCQ 1571 Mark
$...........$ is neither prime nor composite.
Answer$1$ is neither prime nor composite.
A prime is a natural number greater than $1$ that has no positive divisors other than $1$ and itself
e.g. $5$ is prime because $1$ and $5$ are its only positive integers factors but $6$ is composite because it has divisors $2$ and $3$ in addition to $1$ and $6$.
View full question & answer→MCQ 1581 Mark
The $\text{LCM}$ of $x$ and $18 $ is $36$. The $\text{HCF}$ of $x$ and $18$ is $2$. What is the number $x$ ?
AnswerWe know that $\text{LCM}\ \times \text{HCF} =$ First number $\times$ Second number
$\text{HCF}\ (x, 18) \times \text{LCM }\ (x, 18) = x \times 18$
$2 \times 36 = x \times 18$
View full question & answer→MCQ 1591 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
$3$
AnswerCorrect option: C. $\sqrt{3}$
$\sqrt{27}=\sqrt{3\times3\times3}$
$=3\sqrt{3}$
Out of the given choices $\sqrt{3}$ is the only smallest number by which if we multiply $\sqrt{27}$ we get a rational number.
Hence, the correct choice is $(c)$.
View full question & answer→MCQ 1601 Mark
The prime factors of $196$ are
- A
$2 \times 7$
- B
$2^2 \times 7$
- C
$2 \times 7^2$
- ✓
$2^2 \times 7^2$
AnswerCorrect option: D. $2^2 \times 7^2$
$196=2 \times 2 \times 7 \times 7=2^2 \times 7^2$
View full question & answer→MCQ 1611 Mark
For any positive integer $a$ and $3,$ there exist unique integers $q$ and $r$ such that $a = 3q + r$ where $r$ must satisfy
- A
$0<\text{r}<3$
- B
$0<\text{r}\leq3$
- ✓
$0\leq\text{r}<3$
- D
$1<\text{r}<3$
AnswerCorrect option: C. $0\leq\text{r}<3$
Since a is a positive integer,
therefore $, r = 0, 1, 2$ only.
So, that $a =3\text{q},3\text{q}+,3\text{q}+2.$
View full question & answer→MCQ 1621 Mark
The $\text{LCM}$ of two consecutive numbers is :
AnswerThe $\text{LCM}$ of two consecutive numbers is their product always.
For example the $\text{LCM}$ of $24, 25$ is equal to $24 \times 25 = 600$
View full question & answer→MCQ 1631 Mark
Which of the following is a rational number ?
- A
$\sqrt{10}$
- ✓
$\sqrt{9}$
- C
$\sqrt{15}$
- D
$\sqrt{12}$
AnswerCorrect option: B. $\sqrt{9}$
$\sqrt{9}$ is an irrational number but because $\sqrt{9}=\sqrt{3}^2=3$ and $3$ is a rational
View full question & answer→MCQ 1641 Mark
$20$ is written as the product of primes as :
AnswerCorrect option: C. $2 \times 2 \times 5$
To write a number as product of its primes, we divide it by various prime numbers $\{2, 3, 5, 7 \}$ etc one by one and check by which prime numbers it is divisible with and how many times.
View full question & answer→MCQ 1651 Mark
$\text{H.C.F}.$ of $26$ and $91$ is :
AnswerPrime factors of $26 = 2 \times 13$
Prime factors of $91 = 7 \times 13$
As $13$ is the only common factor.
Therefore, $= \text{HCF}\ (26, 91) = 13z$
View full question & answer→MCQ 1661 Mark
The decimal expansion of $\frac{987}{10500}$ will terminate after :
- A
$1$ decimal place
- ✓
$3$ decimal places
- C
- D
$2$ decimal places
AnswerCorrect option: B. $3$ decimal places
$\frac{987}{10500}=\frac{47}{500}=\frac{47}{2^2\times5^3}$
Here, in the denominator of the given fraction the highest power of prime factor $5$ is $3$,
therefore, the decimal expansion of the rational number $\frac{47}{2^2\times5^3}$ will terminate after $3$ decimal places.
View full question & answer→MCQ 1671 Mark
$3+2\sqrt{5}$ is a/ an :
AnswerHere, $3$ is rational and $2\sqrt{5}$ is irrational.
We know that the sum of a rational and an irrational is an irrational number,
therefore, $3+2\sqrt{5}$ is irrational.
View full question & answer→MCQ 1681 Mark
An army contingent of $616$ members is to march behind an army band of $32$ members in a parade on the occasion of Republic Day. The two groups are to march in the same number of the column. The maximum number of column in which they can march is :
AnswerWe know that maximum number of columns $= \text{HCF}$ of $(616, 32)$
Applying Euclid’s division algorithm to find $\text{HCF}$ of two numbers.
View full question & answer→MCQ 1691 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→MCQ 1701 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 1711 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 1721 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 1731 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 1741 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 1751 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 1761 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 1771 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 1781 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 1791 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 1801 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 1811 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 1821 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 1831 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 1841 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 1851 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 1861 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 1871 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 1881 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 1891 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 1901 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 1911 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 1921 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 1931 Mark
If the $\text{HCF}$ of $65$ and $117$ is expressible in the form $65 m-117,$ then the value of $m$ is
Answer Here, $65=13 \times 5$
$117=13 \times 3^2$
$\therefore \operatorname{HCF}(65,117)=13\ldots(i)$
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 1941 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 1951 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 1961 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 1971 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 1981 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 1991 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 2001 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→