MCQ 11 Mark
If $n$ is a natural number, then $9^{2n}- 4^{2n}$ is always divisible by:
- A
$5$
- B
$3$
- ✓
both $5$ and $13$
- D
AnswerCorrect option: C. both $5$ and $13$
$n$ is natural number, and $9^{2 n}-4^{2 n}$ is the form of $a^{2 n}-b^{2 n}$ is or $\left(a^n\right)^2-\left(b^n\right)^2$ which is divisibel by $(a + b)$ and $(a - b)$ or $9 + 4$ and $9 - 4$ or $13$ and $5$ both.
View full question & answer→MCQ 21 Mark
The exponent of $2$ in the prime factorisation of $144$, is:
Answer$\begin{array}{c|c}2 &144\\\hline 2 & 72\\\hline 2 & 36\\\hline2 & 18\\\hline3 & 9\\\hline3 & 3 \\\hline&1 \end{array}$
$144 = 2^4× 3^2$
$\therefore$ Exponant of $2$ is $4$
View full question & answer→MCQ 31 Mark
The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after:
AnswerDecimal expansion of $\frac{14587}{1250}$ is terminate after 4 decimal place.
$\Bigg\{\because\frac{14587}{1250}=\frac{14587\times8}{1250\times8}=\frac{14587\times8}{10000}\Bigg\}$
View full question & answer→MCQ 41 Mark
For some integer $q$, every odd integer is of the form:
- A
$q$
- B
$q + 1$
- C
$2q$
- ✓
$2q + 1$
AnswerCorrect option: D. $2q + 1$
We know that, all numbers that are not the multiple of $2$ are odd numbers.
Odd integers are $..., -3, -1, 1, 3, 5,...$
So, odd numbers can be written as $2m + 1$, where m is an integer.
m can be $..., -2, -1, 0, 1, 2,...$
$\therefore$ $2m + 1$ can be $..., -3, -1, 1, 3,...$
Hence, the correct answer is option $D$.
View full question & answer→MCQ 51 Mark
The smallest rational number by which $\frac{1}{3}$ should be multiplied so that its decimal expansion terminates after one place of decimal, is:
- ✓
$\frac{3}{10}$
- B
$\frac{1}{10}$
- C
$3$
- D
$\frac{3}{100}$
AnswerCorrect option: A. $\frac{3}{10}$
The smallest rational number which should be multiplied by $\frac{1}{3}$ to get a terminating.
$\text{decimals }=\frac{3}{10}$
$\because\ \frac{1}{3}\times\frac{3}{10}=\frac{1}{10}=0.1$
View full question & answer→MCQ 61 Mark
If $ n=2^3 \times 3^4 \times 5^4 \times 7 $, then the number of consecutive zeroes in n, where n is a natural number, is:
AnswerSince, it is given that
$ n=2^3 \times 3^4 \times 5^4 \times 7 $
$ =2^3 \times 5^4 \times 3^4 \times 7$
$ =2^3 \times 5^3 \times 5 \times 3^4 \times 7 $
$ =(2 \times 5)^3 \times 5 \times 3^4 \times 7$
$ =5 \times 3^4 \times 7 \times(10)^3$
So, this means the given number $n$ will end with $3$ consecutive zeroes.
View full question & answer→MCQ 71 Mark
The number of decimal places after which the decimal expansion of the rational number $\frac{23}{2^2\times5}$ will terminate, is:
AnswerDecimal expansion of $\frac{23}{2^2\times5}=\frac{23}{20}$
$=\frac{23\times5}{20\times5}=\frac{115}{100}=1.15$
$\therefore$ Number of decimal places $= 2$
View full question & answer→MCQ 81 Mark
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 $m = …, -1, 0, 1, 2, 3, …$
$2m = …, -2, 0, 2, 4, 6, …$
View full question & answer→MCQ 91 Mark
The $HCF$ of $95$ and $152$, is:
Answer
$HCF$ of $95$ and $152 = 19$
View full question & answer→MCQ 101 Mark
If $p$ and $q$ are co-prime numbers, then $p^2$ and $q^2$ are:
AnswerWe know that the co-prime numbers have no factor in common, or, their $HCF$ is $1$.
Thus, $p^2$ and $q^2$ have the same factors with twice of the exponents of $p$ and $q$ respectively, which again will not have any common factor.
Thus we can conclude that $p^2$ and $q^2$ are co-prime numbers.
Hence, the correct choice is (a).
View full question & answer→MCQ 111 Mark
If the $LCM$ of a and $18$ is $36$ and the $HCF$ of a and $18$ is $2$, then $a =$
Answer$LCM (a, 18) = 36$
$HCF (a, 18) = 2$
We know that the product of numbers is equal to the product of their $HCF$ and $LCM$.
Therefore,
$18a = 2(36)$
$\text{a}=\frac{2(36)}{18}$
$a = 4$
Hence the correct choice is $(c)$.
View full question & answer→MCQ 121 Mark
If two positive integers $a$ and b are expressible in the form $a = pq^2$ and $b = p^2q; p, q$ being prime numbers, then $HCF (a, b)$ is:
- ✓
$pq$
- B
$p^3 q^3$
- C
$p^3 q^2$
- D
${p}^2{q}^2$
Answer$a = pq^2$ and $b = p^3q$ where $a$ and $b$ are positive integers and $p, q$ are prime numbers, then $HCF = pq$.
View full question & answer→MCQ 131 Mark
If $HCF (26, 169) = 13$, then $LCM (26, 169) =$
Answer$HCF (26, 169) = 13$
$LCM (26, 169)$ $=\frac{26\times169}{13}=338$
View full question & answer→MCQ 141 Mark
Euclid’s division lemma states that for two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a = bq + r$, where $r$ must satisfy:
- A
$1 < r < b$
- B
$0 < r ≤ b$
- ✓
$0 ≤ r < b$
- D
$0 < r < b$
AnswerCorrect option: C. $0 ≤ r < b$
According to Euclid’s Division lemma, for a positive pair of integers there exists unique integers $q$ and $r$, such that,
$a = bq + r$, where $0 ≤ r < b$
View full question & answer→MCQ 151 Mark
The remainder when the square of any prime number greater than $3$ is divided by $6$, is:
Answer$\because$ The given prime number is greater than $3$
Let the prime number be $=6\text{k}\pm1$
When k is a natural number
$\therefore\ (6\text{k}\pm1)^2=36\text{k}^2\pm12\text{k}+1$
$=6\text{k}(6\text{k}\pm2)+1$
$\therefore$ Remainder $= 1$
View full question & answer→MCQ 161 Mark
The largest number which divides $70$ and $125$, leaving remainders $5$ and $8$, respectively, is:
AnswerSince, $5$ and $8$ are the remainders of $70$ and $125$, respectively.
Thus, after subtracting these remainders from the numbers, we have the numbers $65 = (70 - 5), 117 = (125 - 8)$, which is divisible by the required number.
Now, required number $= HCF$ of $65, 117$
[For the largest number]
For this, $117 = 65 \times 1 + 52$ $[$Dividend = divisor $\times $ quotient $+$ remainder$]$
$\Rightarrow 65 = 52 \times 1 + 13$
$\Rightarrow 52 = 13 \times 4 + 0$
$HCF = 13$
Hence, $13$ is the largest number which divides $70$ and $125$, leaving remainders $5$ and $8$.
View full question & answer→MCQ 171 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:
Answer$3$ is the least prime factor of a $7$ is the least prime factor of b, then sum of a $a$ and $b$ will be divisible by $2, 2$ is the least prime factor of $a + b$.
View full question & answer→MCQ 181 Mark
If two positive integers $a$ and $b$ are written as $a = x^3y^2$ and $b = xy^3; x, y$ are prime numbers, then $HCF (a, b)$ is:
- A
$ x y$
- ✓
$ x y^2$
- C
$ x^3 y^3$
- D
$x^2 y^2$
AnswerCorrect option: B. $ x y^2$
It is given that,
$\text{a}=\text{x}^3\text{y}^2=\text{x}\times\text{x}\times\text{x}\times\text{y}\times\text{y}$
$\text{b}=\text{xy}^3=\text{x}\times\text{y}\times\text{y}\times\text{y}$
$\text{HCF(a, b)}=\text{HCF}(\text{x}^3\text{y}^2,\text{xy}^3)=\text{x}\times\text{y}\times\text{y}=\text{xy}^2$
Hence, the correct answer is option $B$.
View full question & answer→MCQ 191 Mark
Which of the following rational numbers have terminating decimal?
- $\frac{16}{225}$
- $\frac{5}{18}$
- $\frac{2}{21}$
- $\frac{7}{250}$
- A
$(i)$ and $(ii)$
- B
$(ii)$ and $(iii)$
- C
$(i)$ and $(iii)$
- ✓
$(i)$ and $(iv)$
AnswerCorrect option: D. $(i)$ and $(iv)$
We know that a rational number has terminating decimal if the prime factors of its denominator are in the form $2^m× 5^n$ $\frac{16}{225}$ and $\frac{7}{250}$ has terminating decimals.
View full question & answer→MCQ 201 Mark
If two positive integers $a$ and $b$ are expressible in the form $a=p q^2$ and $b=p^2 q ; p, q$ being prime numbers, then $LCM (a, b)$ is:
- A
$p q$
- B
$p^3 q^3$
- C
$p^3 q^2$
- ✓
$p^2 q^2$
AnswerCorrect option: D. $p^2 q^2$
$A$ and $b$ are two positive integers and $a=p q^2$ and $b=p^2 q$, where $p$ and $q$ are prime numbers, then $L C M=p^2 q^2$.
View full question & answer→MCQ 211 Mark
If two positive integers $tn$ and $n$ are expressible in the form $m=p q^3$ and $n=p^3 q^2$, where $p, q$ are prime numbers, then $HCF (m, n) =$
- A
$pq$
- ✓
$p q^2$
- C
$p^3 q^3$
- D
$p^2 q^3$
AnswerCorrect option: B. $p q^2$
$m$ and $n$ are two positive integers and $m=p q^3$ and $n=p q^2$, where $p$ and $q$ are prime numbers, then $H C F=p q^2$.
View full question & answer→MCQ 221 Mark
If $n$ is any natural number, then $6^n- 5^n$ always ends with:
Answer$n$ is any natural number and $6^n- 5^n$
We know that $6^n$ ends with $6$ and $5^n$ ends with $5$
$6^n- 5^n$ will end with $6 - 5 = 1$
View full question & answer→MCQ 231 Mark
The least number that is divisible by all the numbers from $1$ to $10$ (both inclusive) is:
AnswerCorrect option: D. $2520$
Factors of $1$ to $10$ numbers
$1 = 1$
$2 = 1 \times 2$
$3 = 1 \times 3$
$4 = 1 \times 2 \times 2$
$5 = 1 \times 5$
$6 = 1 \times 2 \times 3$
$7 = 1 \times 7$
$8 = 1 \times 2 \times 2 \times 2$
$9 = 1 \times 3 \times 3$
$10 = 1 \times 2 \times 5$
$LCM$ of number $1$ to $10 = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)$
$= 1 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 2520$
View full question & answer→MCQ 241 Mark
The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after:
AnswerRational number $=\frac{14587}{1250}=\frac{14587}{2^1\times5^4}$
$\begin{array}{c|c}2 &1250\\\hline 5 & 625\\\hline 5 & 125\\\hline5 & 25\\\hline5&5\\\hline&1 \end{array}$
$=\frac{14587}{10\times5^3}\times\frac{(2)^3}{(2)^3}$
$=\frac{14587\times8}{10\times1000}$
$=\frac{116696}{10000}=11.6696$
Hence, given rational number will terminate after four decimal places.
View full question & answer→MCQ 251 Mark
The sum of the exponents of the prime factors in the prime factorisation of $196$, is:
Answer$\begin{array}{c|c}2 &196\\\hline 2 & 98\\\hline 7 & 49\\\hline7 & 7\\\hline&1 \end{array}$
$= 2 \times 2 \times 7 \times 7$
$= 2^2\times 7^2$
Sum of exponents $= 2 + 2 = 4$
View full question & answer→MCQ 261 Mark
$\text { If } a=2^3 \times 3, b=2 \times 3 \times 5, c=3^n \times 5 \text { and } \operatorname{LCM}(a, b, c)=2^3 \times 3^2 \times 5 \text {, then } n=$
Answer$ a=2^3 \times 3, b=2 \times 3 \times 5, c=3^n \times 5 \text { and } \operatorname{LCm}(a, b, c)=2^3 \times 3^2 \times 5 $
$ \therefore 3^n=3^2 \Rightarrow n=2$
View full question & answer→MCQ 271 Mark
The $LCM$ of two numbers is $1200$. Which of the following cannot be their $HCF?$
Answer$LCM$ of two number $= 1200$
Their $HCF$ of these two numbers will be the factor of $1200$
$500$ cannot be its $HCF$.
View full question & answer→MCQ 281 Mark
$3.\overline{27}$ is:
Answer$3.\overline{27}$ is a rational number.
View full question & answer→MCQ 291 Mark
The decimal expansion of the rational number $\frac{33}{2^2\times5}$ will terminate after:
Answer$\frac{33}{2^2\times5}$
Multiply and divide the expansion by $5$
$\frac{33\times5}{2^2\times5^2}=\frac{165}{10^2}=1.65$
Hence, the decimal expansion of the rational number $\frac{33}{2^3\times5}$ will terminate after two decimal places.
View full question & answer→MCQ 301 Mark
If the $HCF$ of $65$ and $117$ is expressible in the form $65m - 117$, then the value of $m$ is:
AnswerUse Euclid's algorithm to find the $HCF$ of $65$ and $117$.
By Euclid's algorithm,
$b = aq + r, 0 \leq r < a$
$\Rightarrow 117 = 65 \times 1 + 32$
$\Rightarrow 65 = 52 \times 1 + 13$
$\Rightarrow 52 = 13 \times 4 + 0$
$\therefore$ $HCF (65, 117) = 13$
It is given that $HCF (65, 117) = 65m - 117$.
$\Rightarrow 65m - 117 = 13$
$\Rightarrow 65m = 130$
$\Rightarrow m = 2$
Hence, the correct option is option $B$.
View full question & answer→MCQ 311 Mark
The $LCM$ and $HCF$ of two rational numbers are equal, then the numbers must be:
Answer$LCM$ and $HCF$ of two rational numbers are equal. Then those must be equal.
View full question & answer→MCQ 321 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:
- A
$203400$
- ✓
$194400$
- C
$198400$
- D
$205400$
AnswerCorrect option: B. $194400$
Given that sum of $LCM$ and $HCF = 1260$
$LCM + HCF = 1260 .....(1)$
Let two numbers be $a$ and $b$ and $HCF (a, b) = x$
According to question:
Put value of $HCF$ and $LCM$ in equation $(1)$
$\Rightarrow 900 + x + x = 1260$
$\Rightarrow 2x = 1260 - 900$
$\Rightarrow 2x = 360$
$\Rightarrow\ \text{x}=\frac{360}{2}$
$\Rightarrow x = 180 ......(2)$
Now, $LCM \times HCF =$ Product of two numbers
Product of two number $= (x + 900)(x)$
$= (180 + 900)(180)$
$= 1080 \times 180$
$= 194400$
View full question & answer→MCQ 331 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 341 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.
$\Rightarrow a=(2 k)^2-1$
$ \Rightarrow a=4 k^2-1$
At $\mathrm{k}=-1,4(-1)^2-1=4-1=3$, which is not divisible by $8$ .
At $\mathrm{k}=0, \mathrm{a}=4(0)^2-1=0-1=-1$, 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=2 k+1 $
$ \Rightarrow a=(2 k+1)^2-1 $
$ \Rightarrow a=4 k^2+4 k+1-1 $
$ \Rightarrow a=4 k^2+4 k$
$ \Rightarrow a=4 k(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 351 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→