MCQ 1011 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. $(18, 25)$
Two numbers are said to be co $-$ prime
If the $\text{HCF}$ between them is $1$
$14 = 2 \times 7$
$35 = 5 \times 7$
$\text{HCF}\ (14, 35) = 7$
$18 = 2 \times 3 \times 3$
$25 = 5 \times 5$
$\text{HCF}\ (18, 25) = 1$
$31 = 1 \times 31$
$93 = 3 \times 31$
$\text{HCF}\ (31, 93) = 31$
$32 = 2 \times 2 \times 2 \times 2 \times 2$
$62 = 2 \times 31$
$\text{HCF}\ (32, 62) = 2$
Since the $\text{HCF}\ (18, 25)$ is $1, 18$ and $25$ is the pair of co $-$ primes.
View full question & answer→MCQ 1021 Mark
The largest number of $4$ digits exactly divisible by $12, 15, 18$ and $27$ is :
- ✓
$9720$
- B
$9999$
- C
$9270$
- D
$1000$
AnswerCorrect option: A. $9720$
$\text{LCM} \ (12, 15, 18, 27) = 540$
Now, largest four digit number $= 9999$
$\therefore 9999 \div 540 = 18 \times 540 + 279 \ ($Remainder $= 279)$
Therefore, the largest number of $4$ digits exactly divisible by $12, 15, 18$ and $27$ is $9999 – 279 = 9720$
View full question & answer→MCQ 1031 Mark
For every positive integer $n, n^2-n$ is divisible by
Answer$n^2-n = n(n - 1)$.
Since $n$ and $(n - 1)$ are consecutive integers.
Therefore, one of them must be divisible by $2.$
View full question & answer→MCQ 1041 Mark
If two positive integers tn and $n$ arc expressible in the form $m = pq^3$ and $n = p^3q^2$, where $p, q$ are prime numbers, then $\text{HCF} (m, n) =$
- A
$pq$
- ✓
${pq}^2$
- C
$p^3 q^3$
- D
$p^2 q^3$
AnswerCorrect option: B. ${pq}^2$
$m$ and $n$ are two positive integers and $m = {pq}^3$ and $ n = {pq}^2,$
where $p$ and $q$ are prime numbers, then $\text{HCF} = {pq}^2$
View full question & answer→MCQ 1051 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 1061 Mark
Euclid's division lemma sates that for any positive integers $a$ and $b,$ there exist unique integers $q$ and $r$ such that $a = bq + r,$ where r must satisfy :
- A
$1<\text{r}<\text{b}$
- B
$0<\text{r}\le\text{b}$
- ✓
$0\le\text{r}<\text{b}$
- D
$0<\text{r}<\text{b}$
AnswerCorrect option: C. $0\le\text{r}<\text{b}$
Euclid's division lemma states that,
For any positive integers $a$ and $b,$ there exist unique integers $q$ and $r$ such that
$\text{a}=\text{bq}+\text{r},$ where $0\le\text{r}<\text{b}$
View full question & answer→MCQ 1071 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
$1 = 1$
$2 = 1 \times 2$
$3 = 1 \times 3$
$4 = 1 \times 2 \times 2$
$5 = 1 \times 5$
$6 = 1 \times 2 \times 3$
$7 = 1 \times 7$
$8 = 1 \times 2 \times 2 \times 2$
$9 = 1 \times 3 \times 3$
$10 = 1 \times 2 \times 5$
$\text{LCM}$ of number $1 $ to $10 = \text{LCM}\ (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)$
$= 1 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 2520$
View full question & answer→MCQ 1081 Mark
View full question & answer→MCQ 1091 Mark
The relationship between $\text{HCF}$ and $\text{LCM}$ of two natural numbers is :
- A
$\text{HCF} \times \text{LCM} = a - b$
- ✓
$\text{HCF} \times \text{LCM} = a \times b$
- C
$\text{HCF} \times \text{LCM} = a + b$
- D
AnswerCorrect option: B. $\text{HCF} \times \text{LCM} = a \times b$
The product of $\text{HCF}$ and $\text{LCM}$ of two natural numbers is equal to the product of these numbers
$\text{HCF}\ (a, b) \times \text{LCM}\ (a, b) = $ Product of two natural numbers $(a \times b)$
View full question & answer→MCQ 1101 Mark
Which of the following statement is false?
- A
$\text{H.C.F} \ (p, q, r) \times \text{LCM}\ (p, q, r) = p \times q \times r$
- ✓
$\text{LCM}\ (p, q, r) = p \times q \times r$; if $p, q, r$ are prime numbers
- C
$\text{HCF}\ (p, q, r) = 1$; if $p, q, r$ are prime numbers
- D
$\text{HCF}\ (a, b) \times LCM (a, b) = a \times b$
AnswerCorrect option: B. $\text{LCM}\ (p, q, r) = p \times q \times r$; if $p, q, r$ are prime numbers
$\text{H.C.F} \ (p, q, r) \times \text{LCM }\ (p, q, r) \neq p \times q \times r$
This condition is applied on $\text{HCF}$ and $\text{LCM}$ of two numbers $($either $(p, q )$ or $(q, r )$ or $(r, p))$
For three numbers $p,q$ and $r$ the formula is different.
View full question & answer→MCQ 1111 Mark
The $\text{HCF}$ of two numbers is $27$ and their $\text{LCM}$ is $162$. If one of the numbers is $54,$ what is the other number ?
AnswerLet the two $n$ umbers be $a$ and $b$.
$\text{HCF} \times \text{LCM} = ab$
$\Rightarrow 27 \times 162 = 54 \times b$
$\Rightarrow b = 81$
View full question & answer→MCQ 1121 Mark
Choose the correct answer from the given four options in the following questions : The product of a non $-$ zero rational and an irrational number is :
AnswerProduct of a non $-$ zero rational and an irrational number is always irrational
i.e., $\frac{3}{4}\times\sqrt{2}=\frac{3\sqrt{2}}{4} \ ($irrational$).$
View full question & answer→MCQ 1131 Mark
What is the largest number that divides $245$ and $1029, $ leaving remainder $5$ in each case?
Answer$245 $ and $1029$ are divided by the largest number leaving remainders $5$ in each case.
$245 - 5 = 240$
$1029 - 5 = 1024$
So, $240$ and $1024$ are exactly divisible by the required number.
Thus, the required number is the $\text{HCF}$ of $240$ and $102$4
$\text{HCF}\ (240, 1024) = 16$
View full question & answer→MCQ 1141 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 1151 Mark
The $\text{LCM}$ of two numbers is $14$ times their $\text{HCF}$. The sum of $\text{LCM}$ and $\text{HCF} $ is $600$. If one number is $280,$ then the other number is :
AnswerGiven: $\text{LCM} = 14 \times \text{HCF}$
and $\text{LCM} + \text{HCF} = 600 ...(i)$
Putting $\text{LCM} = 14 \times \text{HCF}$ in eq. $(i),$
$14 \times \text{HCF} + \text{HCF} = 600$
$\Rightarrow 15 \times \text{HCF} = 600$
$\Rightarrow \text{HCF} = 40$
And $\text{LCM} = 14 \times 40 = 560$
$\therefore$ Using the result,
$\text{HCF} \times \text{LCM} =$ Product of two natural numbers Other number
$ =\frac{40\times560}{280}=80$
View full question & answer→MCQ 1161 Mark
Every positive even integer is of the form $........$ for some integer $'q\ '$.
AnswerLet a be any positive integer and $b = 2$
Then by applying Euclid’s Division Lemma, we have,
$a = 2q + r$ where $0\leq\text{r}<2\text{ r}=0\text{ r}1$
Therefore, $a = 2q$ or $2q + 1$
Thus, it is clear that $a = 2q$
i.e. $a$ is an even integer in the form of $2q$
View full question & answer→MCQ 1171 Mark
The LCM of $ 2^3 \times 3^2 $ and $ 2^2 \times 3^3 $
- A
$ 2^2 \times 3 $
- B
$ 2 \times 3^2 $
- C
$ 2^2 \times 3^2 $
- ✓
$ 2^3 \times 3^3 $
AnswerCorrect option: D. $ 2^3 \times 3^3 $
Solution:
L.C.M. of $ 2^3 \times 3^2 $ and $ 2^2 \times 3^3 $ is the product of all prime numbers with the greatest
power of every given number, hence it will be $ 2^3 \times 3^3 $
View full question & answer→MCQ 1181 Mark
If $\text{LCM}\ (26, 91) = 182,$ then $\text{HCF}\ (26, 91) =$
AnswerUsing the result, $\text{HCF} \times \text{LCM} =$ product of two natural numbers
$\Rightarrow\text{HCF}\ (26,91)=\frac{26\times91}{182}=13$
View full question & answer→MCQ 1191 Mark
The number of possible pairs of number, whose product is $540$0 and the $\text{HCF}$ is $30$ is :
AnswerGiven that product of the number is $5400 = 30 \times 3 \times 2 \times 30.$
$\therefore$ Possible pairs as per the requirment are
$(1) 30 \times (3 \times 2 \times 30) = 30 \times 180$
$(2) (30 \times 3) \times (2 \times 30) = 90 \times 60$
$\therefore$ Total number of pairs $= 2$
View full question & answer→MCQ 1201 Mark
Choose the correct answer from the given four options in the following questions : 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, odd integers are $1, 3, 5, ...$
So, it can be written in the form of $2q + 1.$
where, $q =$ integer $= Z$
or $q = ···, -1, 0, 1, 2, 3, ...$
$\therefore 2q + 1 = ... -3, -1, 1, 3, 5,$
Alternate Answer
Let $'a\ '$ be given positive integer.
On dividing $'a\ '$ by $2,$ let $q$ be the quotient and $r$ be the remainder.
Then, by Euclid's division algorithm, we have
$a = 2q + r, $ where
$0\leq\text{r}<2$
$\Rightarrow a = 2q + r,$ where $r = 0$ or $r = 1$
$\Rightarrow a = 2q$ or $2q + 1$
when $a= 2q + 1$ for some integer $q,$ then clearly a is odd.
View full question & answer→MCQ 1211 Mark
On dividing a positive integer $n$ by $9, $ we get $7$ as remainder. What will be the remainder if $(3n - 1)$ is divided by $9$?
AnswerOn dividing $n$ by $9$ the remainder is $7$
$\Rightarrow n = 9q + 7,$ where $q$ is the quotient
$\Rightarrow 3n = 3(9q + 7)$
$\Rightarrow 3n = 27q + 21$
$\Rightarrow 3n - 1 = 27q + 21 - 1$
$\Rightarrow 3n - 1 = 27q + 20$
$\Rightarrow 3n - 1 = 27q + 18 + 2$
$\Rightarrow 3n - 1 = 9(3q + 2) + 2$
So, the remainder will be $2$
View full question & answer→MCQ 1221 Mark
Every positive odd integer is of the form $.........$ where $'q\ ’$ is some integer.
- ✓
$2q+1$
- B
$5q+1$
- C
$2q+2$
- D
$3q+1$
AnswerCorrect option: A. $2q+1$
Let a be any positive integer and $b = 2$
Then by applying Euclid’s Division Lemma,
we have $, a = 2q + r,$
where $0\leq\text{r}<2$
$\Rightarrow r = 0$ or $1$
$\therefore\text{a }2\text{q}$ or $2\text{q}+1.$
Therefore, it is clear that $a = 2q$
i.e., a is an even integer.
Also, $2q$ and $2q + 1$ are consecutive integers,
therefore $2q + 1$ is an odd integer.
View full question & answer→MCQ 1231 Mark
If one of the zeroes of the cubic polynomial $x^3+ ax^2+ bx + c$ is $-1,$ then the product of other two zeroes is :
- ✓
$b - a + 1$
- B
$b - a – 1$
- C
$a - b + 1$
- D
$a - b - 1$
AnswerCorrect option: A. $b - a + 1$
Let $p(x) = x^3+ ax^2+ bx + c$
Now, $-1$ is a zero of the polynomial
So $, p(0) = 0$
$\Rightarrow (-1)^3+ a(-1)^2+ b(-1) + c = 0$
$\Rightarrow -1 + a - b + c = 0$
$\Rightarrow a - b + c = 1$
$\Rightarrow c = 1 - a + b$
Now, if $\alpha,\beta,\gamma$ are the zeroes of the cubic polynomial $ax^3+ bx^2+ cx + d,$
then product of zeroes is given by $\alpha\beta\gamma=-\frac{\text{d}}{\text{a}}$
So, for the given polynomial, $p(x) = x^3+ ax^2+ bx + c$
$\alpha\beta(-1)=\frac{-\text{c}}{1}=\frac{-(1-\text{a}+\text{b})}{1}$
$\Rightarrow\alpha\beta=1-\text{a + b}$
Hence, the correct answer is option $(a)$
View full question & answer→MCQ 1241 Mark
Choose the correct answer from the given four options in the following questions : The largest number which divides $70$ and $125,$ leaving remainders $5$ and $8,$ respectively, is :
- ✓
$13.$
- B
$65.$
- C
$875.$
- D
$1750.$
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 $= \text{HCF}$ of $65, 117\ [$for the largest number$]$
$117 = 65 \times 1 + 52 \ [\because$ dividend $=$ divisor $\times$ quotient $+$ remainder$]$
$65 = 52 \times 1 + 13$
$52 = 13 \times 4 + 0$
$\therefore \text{HCF} = 13$
Hence, $13$ is the largest number which divides $70$ and $125$, leaving remainders $5$ and $8$.
View full question & answer→MCQ 1251 Mark
Which of the following is an irrational number ?
- A
$\frac{22}{7}$
- B
$3.1416$
- C
$3.\overline{1416}$
- ✓
$3.141141114...$
AnswerCorrect option: D. $3.141141114...$
An irrational number is a number that is non $-$ terminating and non $-$ repeating.
Option $(a$) is a rational number, while
Option $(c)$ is a repeating decimal number, and so are rational numbers.
Option $(d)$ is an irrational number.
View full question & answer→MCQ 1261 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 $n =$
Answer$a=2^3 \times 3, b=2 \times 3 \times 5, c=3^n \times 5$ and
$\operatorname{LCM}(a, b, c)=2^3 \times 3^2 \times 5$
$\therefore 3^n=3^2$
$ \Rightarrow n=2$
View full question & answer→MCQ 1271 Mark
The simplest form of $\frac{1095}{1168}$ is :
- A
$\frac{17}{26}$
- B
$\frac{25}{26}$
- C
$\frac{13}{16}$
- ✓
$\frac{15}{16}$
AnswerCorrect option: D. $\frac{15}{16}$
$\frac{1095}{1168}=\frac{5\times3\times73}{2\times2\times2\times2\times73}$
$=\frac{5\times3}{2\times2\times2\times2}$
$=\frac{15}{16}$
View full question & answer→MCQ 1281 Mark
The $\text{HCF}$ and $\text{LCM}$ of two numbers are $9$ and $459$ respectively. If one of the
number is $27,$ then the other number is
AnswerUsing the result,
$\text{HCF} \times \text{LCM} =$ product of two natural numbers
$\Rightarrow $ the other number $=\frac{9\times459}{27}=153$
View full question & answer→MCQ 1291 Mark
All non $-$ terminating and non $-$ recurring decimal numbers are
AnswerAll non $-$ terminating and non $-$ recurring decimal numbers are irrational numbers.
A number is rational if and only if its decimal representation is repeating or terminating.
View full question & answer→MCQ 1301 Mark
If a is rational and $\sqrt{\text{b}}$ is irrational ,than $a +\sqrt{\text{b}}$ is :
Answerlet a be rational and $\sqrt{\text{b}}$ is irrational.
If possible let $\text{a}+\sqrt{\text{b}}$ be rational .
then $\text{a}+\sqrt{\text{b}}$ is rational and a is rational.
$\Rightarrow\Big[\Big(\text{a}+\sqrt{\text{b}}\Big)-\text{a}\Big]$ is rational $[$Difference of two rationals is rationals$]$
$\Rightarrow\sqrt{\text{b}}$ is rational.
This contradicts the fact that $\sqrt{\text{b}}$ is irrational.
The contradication arises by assuming that $\text{a}+\sqrt{\text{b}}$ is rational.
Therefore, $\text{a}+\sqrt{\text{b}}$ is irrational .
View full question & answer→MCQ 1311 Mark
The product of two numbers is $1600$ and their $\text{HCF}$ is $5$. The $\text{LCM}$ of the numbers is :
- A
$8000$
- B
$1600$
- ✓
$320$
- D
$1605$
AnswerLet the two $n$ umbers be $a$ and $b$.
$\text{HCF} \times \text{LCM} = ab$
$\Rightarrow 5 \times \text{LCM} = 1600$
$\Rightarrow \text{LCM} = 320$
View full question & answer→MCQ 1321 Mark
If $p$ is a prime number, then $\sqrt{\text{p}}$ is
Answer$\sqrt{\text{p}}$ is an irrational number because the square root of every prime number is an irrational number.
$($for example $\sqrt{3}$ is an irrational number$)$
View full question & answer→MCQ 1331 Mark
The $\text{LCM}$ of two numbers is $1200$. Which of the following cannot be their $\text{HCF}$?
Answer$\text{LCM}$ of two number $= 1200$
Their $\text{HCF}$ of these two numbers will be the factor of $1200$
$500$ cannot be its $\text{HCF}$.
View full question & answer→MCQ 1341 Mark
If the $\text{HCF}$ of $65$ and $117$ is expressible in the form $65m – 117$, then the value of $m$ is :
AnswerBy Euclid's division algorithm,
$\text{b}=\text{aq}+\text{r, }0\leq\text{r}<\text{a}\ [\because$ dividend $=$ divisor $\times$ quotient $+$ remainder$]$
$\Rightarrow 117 = 65 \times 1 + 52$
$\Rightarrow 65 = 52 \times 1 + 13$
$\therefore \text{ HCF}\ (65, 117) = 13 .....(i)$
Also, given that, $\text{HCF}\ (65, 117) = 65m - 117 .....(ii)$
From Eqs. $(i)$ and $(ii),$
$65m - 117=13$
$65m = 130$
$m = 2.$
View full question & answer→MCQ 1351 Mark
$\text{LCM}$ of $\left(2^3 \times 3 \times 5\right)$ and $\left(2^4 \times 5 \times 7\right)$ is :
- A
$40$
- B
$560$
- C
$1120$
- ✓
$1680$
AnswerCorrect option: D. $1680$
$\left(2^3 \times 3 \times 5\right)$ and $\left(2^4 \times 5 \times 7\right)$
$\text{LCM } =2^4 \times 3 \times 5 \times 7=1680$
View full question & answer→MCQ 1361 Mark
The decimal expansion of the number $\frac{14753}{1250}$ will terminate after :
AnswerThe prime factorisation of the denominator is $2 \times 5^2$
Since $4 > 1,$
The decimal expansion will terminate after $4$ decimal places.
View full question & answer→MCQ 1371 Mark
The difference between two distinct irrational numbers is always
- ✓
both rational and irrational number
- B
- C
- D
AnswerCorrect option: A. both rational and irrational number
The difference between two distinct irrational numbers can be either a rational number or an irrational number.
e.g difference between $\pi$ and $(\pi - 3)$ is equal to $3$ which is rational
$\sqrt{2}$ and $\sqrt{2}+1$ both are irrational but their difference is $1$ which is rational
similarly $,\sqrt{2}$ and $\sqrt{3}$ are irrational and their differnce $ (\sqrt{3}-\sqrt{2})$ is also irrational
View full question & answer→MCQ 1381 Mark
$\big(2+\sqrt{2}\big)$ is :
AnswerAn irrational number is a number that is non $-$ terminating and non $-$ repeating.
Now, $2$ is a rational number and $\sqrt2$ is an irrational number.
Sum of a rational number and an irrational number is irrational.
Hence, $\big(2+\sqrt{2}\big)$ is an irrational number.
View full question & answer→MCQ 1391 Mark
If $\frac{241}{4000}=\frac{241}{2^\text{m}\times5^\text{n}},$ Then :
- A
$m = 4$ and $n = 5$
- B
$m = 3$ and $n = 2$
- ✓
$m = 5$ and $n = 3$
- D
$m = 2$ and $n = 5$
AnswerCorrect option: C. $m = 5$ and $n = 3$
$\frac{241}{4000}=\frac{241}{2^\text{m}\times5^\text{m}}$
$\Rightarrow\frac{241}{2^5\times5^3}=\frac{241}{2^\text{m}\times5^\text{n}}$
Comparing the denominators of both fractions, we have $m = 5$ and $n = 3$
View full question & answer→MCQ 1401 Mark
If two numbers do not have common factor $($other than $1),$ then they are called :
AnswerCorrect option: D. co $-$ prime numbers
If two numbers do not have a common factor $($other than $1),$ then they are called co $-$ primer numbers.
We know that two numbers are co $-$ prime if their common factor $($greatest common divisor) is $1$.
e.g. co $-$ prime of $12$ are $11, 13$.
View full question & answer→MCQ 1411 Mark
$3.\overline{27}$ is :
Answer$3.\overline{27}$ is a rational number.
View full question & answer→MCQ 1421 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 1431 Mark
If the $\text{HCF}$ of $65$ and $117$ is expressible in the form $65m - 117,$ then the value of $m$ is :
AnswerUse Euclid's algorithm to find the $\text{HCF}$ of $65$ and $117$.
By Euclid's algorithm,
$b = aq + r, 0 \leq r < a$
$\Rightarrow 117 = 65 \times 1 + 32$
$\Rightarrow 65 = 52 \times 1 + 13$
$\Rightarrow 52 = 13 \times 4 + 0$
$\therefore \text{HCF}\ (65, 117) = 13$
It is given that $\text{HCF}\ (65, 117) = 65m - 117$.
$\Rightarrow 65m - 117 = 13$
$\Rightarrow 65m = 130$
$\Rightarrow m = 2$
Hence, the correct option is option $B$.
View full question & answer→MCQ 1441 Mark
If $a=\left(2^2 \times 3^3 \times 5^4\right)$ and $b=\left(2^3 \times 3^2 \times 5\right)$, then $\text{HCF} \ (a, b) =$ ?
Answer$a=2^2 \times 3^3 \times 5^4$
$b=2^3 \times 3^2 \times 5$
$\operatorname{HCF}\ (a, b)=2^2 \times 3^2 \times 5=180$
View full question & answer→MCQ 1451 Mark
If $\text{HCF}\ (72, 120) = 24,$ then $\text{LCM }\ (72, 120)$ is
- A
$240$
- ✓
$360$
- C
$1728$
- D
$2880$
AnswerUsing the result, $\text{HCF} \times \text{LCM} =$ product of two natural numbers
$\Rightarrow\text{LCM}\ (72,120)=\frac{72\times120}{24}=360$
View full question & answer→MCQ 1461 Mark
Let $\text{x}=\frac{\text{p}}{\text{q}}$ be a rational number, such that the prime factorization of $q$ is of the form $2^{\mathrm{n}} 5^{\mathrm{m}}$, where $n,m$ are non $-$ negative integers. Then $x$ has a decimal expansion which terminates :
AnswerThe form of $q$ is $2^n \times 5^m$
$q$ can be $\{1, 2, 5, 10, 20, 40\} ....$
Any integer divided by these numbers will always give a terminating decimal number.
View full question & answer→MCQ 1471 Mark
$\pi$ is
AnswerThe value of $\pi = 3.141592653589 ……….$
$\therefore$ Value of $\pi$ is not $-$ repeating decimal number
Therefore $, \pi$ is an irrational number.
View full question & answer→MCQ 1481 Mark
The $\text{LCM}$ and $\text{HCF}$ of two rational numbers are equal, then the numbers must be :
Answer$\text{LCM}$ and $\text{HCF}$ of two rational numbers are equal.
Then those must be equal.
View full question & answer→MCQ 1491 Mark
If the sum of $\text{LCM}$ and $\text{HCF}$ of two numbers is $1260$ and their $\text{LCM}$ is $900$ more than their $\text{HCF},$ then the product of two numbers is :
- A
$203400$
- ✓
$194400$
- C
$198400$
- D
$205400$
AnswerCorrect option: B. $194400$
Given that sum of $\text{LCM}$ and $\text{HCF} = 1260$
$\text{LCM} + \text{HCF} = 1260 .....(1)$
Let two numbers be $a$ and $b$ and $\text{HCF}\ (a, b) = x$
According to question :
Put value of $\text{HCF}$ and $\text{LCM}$ 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, $\text{LCM} \times \text{HCF} =$ Product of two numbers
Product of two number $= (x + 900)(x)$
$= (180 + 900)(180)$
$= 1080 \times 180$
$= 194400$
View full question & answer→MCQ 1501 Mark
If the $\text{HCF}$ of $210$ and $55$ is expressible in the form $210 \times 5 + 55y,$ then $y =$
AnswerFirst, find the $\text{HCF}$ of $210$ and $55$ by Euclid's Division Algorithm
$210 = 55 \times 3 + 45$
$55 = 45 \times 1 + 10$
$45 = 10 \times 4 + 5$
$10 = 5 \times 2 + 0\ ($zero remainder$)$
therefore $ \text{HCF}\ (210, 55) = 5$
Now,
$\therefore 5 = 210 \times 5 + 55y$
$\Rightarrow 5 - 1050 = 55y$
$\Rightarrow -1045 = 55y$
$\Rightarrow y = -19$
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