MCQ 11 Mark
The common difference of an $AP,$ whose $\mathrm{n}^{\text {th }}$ term is $a_n=(3 n+7)$, is :
Answer$a_n=3 n+7$
$a_1=3 \times 1+7=10$
$a_2=3 \times 2+7=13$
$d=a_2-a_1$
$d=13-10$
$d=3$
common difference $=3$
View full question & answer→MCQ 21 Mark
The $\text{HCF}$ of $135$ and $225$ is :
Answer$135=5 \times 3^3$
$225=5^2 \times 3^2$
$\text{H.C.F}(135,225)=5 \times 3^3$
$=45$
View full question & answer→MCQ 31 Mark
The sum of exponents of prime factors in the prime-factorisation of $196$ is :
Answer$196=2 \times 2 \times 7 \times 7$
$=2^2 \times 7^7$
$=2+2 \text { (exponent is the power of a number) }$
$=4$
View full question & answer→MCQ 41 Mark
Euclid’s division Lemma states that for two positive integers $a$ and $b,$ there exists unique integer $q$ and $r$ satisfying $a = bq + r,$ and.
- A
$0 < r < b$
- B
$0 < r ≤ b$
- ✓
$0 ≤ r < b$
- D
$0 ≤ r ≤ b$
AnswerCorrect option: C. $0 ≤ r < b$
Euclid's division Lemma states that for any two positive integers $'a\ '$ and $'b\ '$ there exist two unique whole numbers $'q\ '$ and $'r\ '$ such that, $a = bq + r,$ where $0 \leq r < b$.
Here, $a =$ Dividend, $b =$ Divisor, $q=$ quotient and $r =$ Remainder.
Hence, the values $'r'$ can take $0 \leq r < b$.
View full question & answer→MCQ 51 Mark
The $\text{HCF}$ and the $\text{LCM}$ of $12, 21, 15$ respectively are :
- A
$3, 140$
- B
$12, 420$
- ✓
$3, 420$
- D
$420, 3$
AnswerCorrect option: C. $3, 420$
$12 = 2 \times 2 \times 3$
$21 = 3 \times 7$
$15 = 5 \times 3$
$\text{HCF} = 3$
$\text{L.C.M} = 2 \times 2 \times 3 \times 5 \times 7 = 420$
View full question & answer→MCQ 61 Mark
The total number of factors of a prime number is :
AnswerTotal number of factors of a prime number is $2.$
View full question & answer→MCQ 71 Mark
If $n$ is a natural number, then $9^{2 n}-4^{2 n}$ 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 81 Mark
The decimal expansion of $\pi$ :
AnswerCorrect option: D. is non $-$ terminating and non $-$ recurring
The value of $\pi=3.141592653589...$
Therefore the value of $\pi$ is not $-$ repeating decimal, non $-$ terminating and non $-$ recurring numbers.
View full question & answer→MCQ 91 Mark
The number $(\sqrt{3}+\sqrt{5})^2$ is :
Answer$\big(\sqrt{3}+\sqrt{5}\big)^2$
$=\big(\sqrt{3}\big)^2+\big(\sqrt{5} \big)^2+2\times\sqrt{3}\times\sqrt{5}$
$= 3 + 5 + 2\sqrt{15}$
$=8+2\sqrt{15}$
Here, $\sqrt{15}=\sqrt{3}\times\sqrt{5}$
Since $\sqrt{3}$ and $\sqrt{5}$ both are an irrational number.
Therefore, $\big(\sqrt{3}+\sqrt{5}\big)^2$ is an irrational number.
View full question & answer→MCQ 101 Mark
The exponent of $2$ in the prime factorisation of $144,$ is :
Answer$\begin{array}{c|c}2 &144\\\hline 2 & 72\\\hline 2 & 36\\\hline2 & 18\\\hline3 & 9\\\hline3 & 3 \\\hline&1 \end{array}$
$144=2^4 \times 3^2$
$\therefore$ Exponant of $2$ is $4$
View full question & answer→MCQ 111 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 121 Mark
$ (+\sqrt{2})+(-1\sqrt{2})$ is
Answer$(1+\sqrt{2})+(1-\sqrt{2)}$
$=1+\sqrt{2}+1-\sqrt{2}$
$=1+1=2$ and $2$ is a rational number.
Therefore the given number is rational number.
View full question & answer→MCQ 131 Mark
The least number $n$ so that $5n$ is divisible by $3,$ where n is :
AnswerSince $5 $ is a prime number so it is not divisible by $3$.
Therefore there is no natural number $n$ such that $5n$ is divisible by $3$.
View full question & answer→MCQ 141 Mark
Two tankers contain $135$ and $225$ litres of water respectively. The maximum capacity of a container which can measure the water of each tanker in the exact number of times is
- A
$35$ litres
- B
$25$ litres
- C
$55$ litres
- ✓
$45$ litres
AnswerCorrect option: D. $45$ litres
Here, maximum capacity of measuring container $= \text{HCF}\ (135, 225)$
$\therefore$ Applying Euclid’s division algorithm to $225$ and $135$
$225 = 135 \times 1 + 90$
$\Rightarrow 135 = 90 \times 1 + 45$
$90 = 45 \times 2 + 0\ [$Zero remainder$]$
$\therefore \text{HCF} \ (135, 225) = 45$
Therefore, the maximum capacity of a container which can measure the water of each tanker in an exact number of times is $45$ litres
View full question & answer→MCQ 151 Mark
If $\text{HCF}\ (a, b) = 12$ and $a \times b = 1800, $ then $\text{LCM}\ (a, b$) is :
AnswerUsing the result,
$\text{HCF} \times \text{LCM} =$ product of two natural numbers
$\Rightarrow \text{LCM}\ (a.b)$
$=\frac{1800}{12}=150$
View full question & answer→MCQ 161 Mark
The $\text{LCM}$ of two numbers $a^2 b^3 c^9 d^6 e^{11}$ and $g^5 f^{21} a^3 b^1 c^{10}$ where $a, b, c, d, e, f, g$ are prime numbers is :
AnswerCorrect option: D. $a^3 b^3 c^{10} d^6 e^{11} f^{21} g^5$
The $\text{LCM}$ of two numbers is their prime factors with the greatest power.
$\therefore \text{LCM}$ of given numbers is $a^3 b^3 c^{10} d^6 e^{11} f^{21} g^5$
View full question & answer→MCQ 171 Mark
A number when divided by $61$ gives $27$ as quotient and $32$ as the remainder, then the number is :
- A
$1796$
- B
$1967$
- ✓
$1679$
- D
$1569$
AnswerCorrect option: C. $1679$
The dividend is equal to Divisor $\times$ Quotient $+$ Remainder
Number $($dividend$) = D \times Q + R$
$\therefore$ the number $($Dividend$) = 61 \times 27 + 32$
$= 1647 + 32$
$= 1679$
View full question & answer→MCQ 181 Mark
Choose the correct answer from the given four options in the following questions : 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 $\text{HCF}\ (a, b)$ is :
- A
$xy.$
- B
$x y^2$
- C
$x^3 y^3$
- ✓
$x^2 y^2$
AnswerCorrect option: D. $x^2 y^2$
Given that, $a=x^3 y^2=x \times x \times x \times y \times y$
and $b=x y^3=x \times y \times y \times y$
$\therefore \text{HCF}$ of $a$ and $b=\operatorname{HCF}\left(\mathrm{x}^3 \mathrm{y}^2 x \mathrm{y}^2\right)=\mathrm{x} \times \mathrm{y} \times \mathrm{y}=x \mathrm{y}^2$
$[$since, $\text{HCF}$ is the product of the smallest power of each common prime facter involved in the numbers$]$
View full question & answer→MCQ 191 Mark
A number when divided by $143$ leaves $31$ as remainder. What will be the remainder when the same number is divided by $13$ ?
AnswerLet the number be $n$.
When the number is divided by $143,$ leaves $31$ as remainder.
$\Rightarrow$ The given number is of the form, $143x + 31$
$\Rightarrow n = 143x + 31,$ where $x$ is the quotient
$\Rightarrow n = 13(11x) + 13(2) + 5$
$\Rightarrow n = 13(11x + 2) + 5$
So, here the remainder will be $5$ when divided by $13$
View full question & answer→MCQ 201 Mark
The largest number which divides $33$ and $75,$ leaving remainders $1$ and $3$ respectively is :
AnswerLet us subtract $1$ from $33$ and $3$ from $75$ in order to find their $\text{HCF}$
$33 - 1 = 32$
$75 - 3 = 72$
To find $\text{HCF}$ of $32, 72$
$72 = 32 \times 2 + 8$
$32 = 8 \times 4 + 0$
$8$ is the largest number which divides $33$ and $75$, leaving remainders $1$ and $3$ respectively.
View full question & answer→MCQ 211 Mark
The difference of a rational and an irrational number is always :
AnswerRational Numbers say $\frac{4}9,{\frac{\text{p}}{\text{q}}},\sqrt{4,}$ fraction, whole numbers, terminating decimal, repeating decimal, perfect square, can be expressed as a ratio of two integers provided the denominator is not equal to zero Irrational Numbers, $\sqrt{2},\sqrt{5},\sqrt{7},\pi$ not a fraction, decimal does not repeat, decimal does not end, non $-$ perfect square, we cannot express as a ratio but both can be expressed as decimal numbers.
The difference between a rational and an irrational number is always an irrational number.
e.g. rational $-$ irrational $=$ irrational say $2-\sqrt{2} =$ irrational
View full question & answer→MCQ 221 Mark
Choose the correct answer from the given four options in the following questions : $n^2-1$ is divisible by $8,$ if $n$ is :
AnswerLet $a=n^2-1$
Here $n$ can be ever or odd.
Case $I$ :
$\mathrm{n}=$ Even i.e., $\mathrm{n}=2 k$. 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 $k=0, a=4(0)^2-1=0-1=-1$, which is not divisible by $8,$ which is not
Case $II$ :
$\mathrm{n}=$ Odd i.e., $\mathrm{n}=2 \mathrm{k}+1$, 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 231 Mark
$\text{HCF}$ 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$\left(2^3 \times 3^2 \times 5\right),\left(2^2 \times 3^3 \times 5^2\right)$ and $\left(2^4 \times 3 \times 5^3 \times 7\right)$
$\mathrm{HCF}=2^2 \times 3 \times 5=60$
View full question & answer→MCQ 241 Mark
The $\text{HCF}$ of two consecutive even numbers is.
AnswerThe $\text{HCF}$ of two consecutive even numbers is $2.($e.g the $\text{HCF}$ of $24, 26$ is $2).$
View full question & answer→MCQ 251 Mark
The number $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$
Answer$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$
$=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times$$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}$
$=\frac{\big(\sqrt{5}+\sqrt{2}\big)^2}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}$
$=\frac{\big(\sqrt{5}\big)^2+\big(\sqrt{2}\big)^2+2\times\sqrt{5}\times\sqrt{2}}{5-2}$
$=\frac{5+2+2\sqrt{10}}{3}$
$=\frac{7+2\sqrt{10}}{3}$
Here$\sqrt{10}=\sqrt{2}\times\sqrt{5}$
since $\sqrt{2}$ and $\sqrt{5}$ both are an irrational number
Therefore, $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5-\sqrt{2}}}$ is an irrational number.
View full question & answer→MCQ 261 Mark
$2.13113111311113 ... $ is :
AnswerAn irrational number is a number that is non $-$ terminating and non $-$ repeating.
$2.13113111311113 ... $ is neither terminating nor repeating, and hence is an irrational number.
View full question & answer→MCQ 271 Mark
The decimal expansion of $\frac{21}{24}$ will terminate after :
- A
$1$ decimal place
- B
- ✓
$3$ decimal places
- D
$2$ decimal places
AnswerCorrect option: C. $3$ decimal places
$\frac{21}{24}=\frac{7}{8}=\frac{7}{2^3},$
Here, in the denominator of the given fraction, the highest power of prime factor $2$ is $3,$
therefore, the decimal expansion of the rational number $\frac{7}{2^3}$ will terminate after $3$ decimal places.
View full question & answer→MCQ 281 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 291 Mark
The least positive integer divisible by $20$ and $24$ is :
AnswerLeast positive integer divisible by $20$ and $24$ is
$\text{LCM}$ of $(20,24)$.
$20=2^2 \times 5$
$24=2^3 \times 3$
$\therefore \operatorname{LCM}\ (20,24)=2^3 \times 3 \times 5=120$
Thus $120$ is divisible by $20$ and $24$.
View full question & answer→MCQ 301 Mark
$a$ and $b$ are two positive integers such that the least prime factor of $a$ is $3$ and the least prime factor of $b$ is $5$. Then, the least prime factor of $(a + b)$ is :
AnswerSince $3$ is the least prime factor of $a,$ and $5$ is the least prime factor of $b,$ so, $2$ cannot be a factor of either.
$\therefore a$ and $b$ are both odd.
We know that, sum of two odd numbers is alwayas even.
So, $a + b$ is even.
$\Rightarrow$ The least prime factor of $(a + b)$ is $2$
View full question & answer→MCQ 311 Mark
The product of a rational number and an irrational number is :
- A
an irrational number only
- B
- ✓
both rational and irrational number
- D
AnswerCorrect option: C. both rational and irrational number
The product of a rational number and an irrational number can be either a rational number or an irrational number.
$\text{e.g} \sqrt{5}\times\sqrt{2}=\sqrt{10}$ which irrational
but $\sqrt{8}\times\sqrt{2}=\sqrt{16} = 4$ which is a rational number
Thus, the product of two irrational numbers can be either rational or irrational
similarly, the product of rational and irrational numbers can be either rational or irrational
$5\times\sqrt{2}=5\sqrt{2}$ which is a rational number
but $4\times\sqrt{4}=4\times2=8$ which is rational.
View full question & answer→MCQ 321 Mark
What is the largest number that divides $70$ and $125,$ leaving remainders $5$ and $8$ respectively?
Answer$70$ and $125$ are divided by the largest number leaving remainders $5$ and $8$ respectively.
$70 - 5 = 65$
$125 - 8 = 117$
So, $65$ and $117$ are exactly divisible by the required number.
Thus, the required number is the $\text{HCF}$ of $65$ and $117$
$\text{HCF}\ (65, 117) = 13$
View full question & answer→MCQ 331 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 :
- A
$\frac{3}{10}$
- ✓
$\frac{1}{10}$
- C
$3$
- D
$\frac{3}{100}$
AnswerCorrect option: B. $\frac{1}{10}$
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 341 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 351 Mark
Which of the following has terminating decimal expansion ?
- A
$\frac{32}{91}$
- ✓
$\frac{19}{80}$
- C
$\frac{23}{45}$
- D
$\frac{25}{42}$
AnswerCorrect option: B. $\frac{19}{80}$
A number is a terminating decimal, if the denominator is of the form $2^m \times 5^n,$
where $m$ and $n$ are non $-$ negative integers.
$\frac{32}{91}=\frac{32}{7\times13}$
$\frac{19}{80}=\frac{19}{2^4\times5}$
$\frac{23}{45}=\frac{23}{3^2\times5}$
$\frac{25}{42}=\frac{25}{2\times3\times7}$
Clearly, option $(b)$ is a terminating decimal, since its denominator is of the form $2^m \times 5^n$
View full question & answer→MCQ 361 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 371 Mark
The $\text{HCF}$ of $867$ and $255$ is :
Answer$867 = 255 \times 3 + 102$
$255 = 102 \times 2 + 51$
$102 = 51 \times 2 + 0$
Hence, we got remainder as $0,$
therefore $\text{HCF}$ of $(867, 255)$ is $51$
View full question & answer→MCQ 381 Mark
$\frac{1}{\sqrt{2}}$ is :
AnswerAn irrational number is a number that is non $-$ terminating and non $-$ repeating.
$\frac{1}{\sqrt{2}}=\frac{1\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}} ... ($Rationalising the denominator$)$
$=\frac{\sqrt{2}}{2}$
$=\frac{1}2{}\times\sqrt{2}$
Now, $\frac{1}2{}$ is rational but $\sqrt2$ is irrational.
Product of a rational number and an irrational number is irrational.
Hence, $\frac{1}{\sqrt2}$ is an irrational number.
View full question & answer→MCQ 391 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 401 Mark
If $a$ and $b$ are both positive rational numbers, then $\big(\sqrt{\text{a}}+\sqrt{\text{b}}\big)\big(\sqrt{\text{a}}-\sqrt{\text{b}}\big)$ is :
- A
neither rational nor rational number
- B
- C
- ✓
Answer$\bigg(\sqrt{\text{a}}+\sqrt{\text{b}}\bigg)\bigg(\sqrt{\text{a}}-\sqrt{\text{b}}\bigg)$
$=\Big\{(\sqrt{\text{a}})^2-\big(\sqrt{\text{b}})^2\Big\}$
$= (a - b)$
Since $a$ and $b$ both are positive rational numbers,
Therefore, the difference of two positive rational numbers is also rational.
View full question & answer→MCQ 411 Mark
The largest number which divides $245$ and $1029$ leaving remainder $5$ in each case is :
AnswerLet us subtract $5 \ ($the remainder$)$ from each number in order to find their $\text{HCF}$.
$245 - 5 = 240$
$1029 - 5 = 1024$
Now, Let us find $\text{HCF}$ of $240, 1024$
$1024 = 240 × 4 + 64$
$240 = 64 × 3 + 48$
$64 = 48 × 1 + 16$
$48 = 16 × 3 + 0$
The largest number which divides $245$ and $1029$ leaving remainder $5$ in each case is $16$.
View full question & answer→MCQ 421 Mark
$0.515115111511115 ...$ is :
Answer$0.515115111511115 ...$ Because it is a non $-$ repeating and non $-$ terminating decimal expression,
Hence it is an irrational number.
View full question & answer→MCQ 431 Mark
Write whether every positive integer can be of the form $4q + 2, $ where $q$ is an integer :
AnswerNo, all positive integers cannot be written in the form of $4q + 2$
Because, $4q + 2 = 2(2q + 1)$
Therefore, $4q + 2$ is an even number.
So, we can't write odd numbers in the form $4q + 2 \ ($where $q$ is an integer$)$
Also $2q + 1$ is an odd number, hence.
the maximum power of $2$ that divides $4q + 1$ is $1$.
Therefore, we can't represent the numbers divisible by $4$ this form.
View full question & answer→MCQ 441 Mark
Which of the following rational numbers is expressible as a terminating decimal?
- A
$\frac{124}{165}$
- B
$\frac{131}{30}$
- ✓
$\frac{2027}{625}$
- D
$\frac{1625}{462}$
AnswerCorrect option: C. $\frac{2027}{625}$
A number is a terminating decimal, if the denominator is of the form $2^m \times 5^n,$
where $m$ and $n$ are non $-$ negative integers.
$\frac{124}{165}=\frac{124}{3\times5\times11}$
$\frac{131}{30}=\frac{131}{2\times3\times5}$
$\frac{2027}{625}=\frac{2027}{5^4}=\frac{2027}{2^0\times5^4}$
$\frac{1625}{462}=\frac{1625}{2\times3\times7\times11}$
Clearly, option $(c)$ is a terminating decimal, since its denominator is of the form $2^m \times 5^n$.
View full question & answer→MCQ 451 Mark
If $d$ is the $\text{HCF}$ of $56$ and $72,$ the values of $x, y$ satisfying $d = 56x + 72y$ :
- A
$x = -3, y = 4$
- ✓
$x = 4, y = -3$
- C
$x = 3, y = -4$
- D
$x = -4, y = 3$
AnswerCorrect option: B. $x = 4, y = -3$
Since, $\text{HCF}$ of $56$ and $72,$ by Euclid’s division lemma,
$72 = 56 \times 1 + 16 ...(i)$
$56 = 16 \times 3 + 8 ...(ii)$
$16 = 8 \times 2 + 0 ...(iii)$
$\therefore \text{HCF}$ of $56$ and $72$ is $8$.
$\therefore 8 = 56 - 16\times 3\ ($from eq. $(ii))$
$8 = 56 - (72 - 56 \times 1) \times 3\ [$From eq. $(i) : 16 = 72 - 56 \times 1]$
$8 = 56 - 3 \times 72 + 56 \times 3$
$8 = 56 \times 4 + (-3) \times 72$
$\therefore x = 4, y = -3$
View full question & answer→MCQ 461 Mark
The decimal form of $\frac{5}{8}$ is :
- A
$0.750$
- B
$0.375$
- C
$0.600$
- ✓
$0.625$
AnswerCorrect option: D. $0.625$
Use long division :
Thus $\frac {5}{8} = 0.625$ View full question & answer→MCQ 471 Mark
The $\text{LCM}$ of two co $-$ prime numbers is
AnswerThe $\text{LCM}$ of two co $-$ prime numbers is their product always.
For example, the $\text{LCM}$ of $3, 5$ is $= 3 \times 5 = 15$
View full question & answer→MCQ 481 Mark
The $\text{HCF}$ of $95$ and $152$, is :
Answer$\text{HCF}$ of $95$ and $152 = 19$
View full question & answer→MCQ 491 Mark
The exponent of $3$ in the prime factorization of $864$ is :
AnswerPrime factorization of $864=32 \times 27=2^5 \times 3^3$
Therefore the exponent of $3$ in the prime factorization of $864$ is $3$
View full question & answer→MCQ 501 Mark
$2.\overline{35}$ is :
Answer$2.\overline{35}=2.35353535\dots$
Which is repeating decimal number, and hence is a rational number.
View full question & answer→