MCQ 511 Mark
If the $\text{LCM}$ of two numbers is $45$ times their $\text{HCF}$ and the sum of $\text{LCM}$ and $\text{HCF}$ is $1150,$ then $\text{HCF} =$
AnswerGiven : $\text{LCM} = 45 \times \text{HCF} ...(i)$
And $\text{LCM} + \text{HCF} = 1150 ...(ii)$
Putting the value of $\text{LCM}$ from eq. $(i)$ in eq. $(ii),$ we get
$45 \times \text{HCF} + \text{HCF} = 1150$
$\text{HCF} (45 + 1) = 1150$
$\Rightarrow 46 \times \text{HCF} = 1150$
$\Rightarrow \text{HCF} = 25$
View full question & answer→MCQ 521 Mark
Two tankers contain $850$ litres and $680$ litres of petrol. The maximum capacity of a container which can measure the petrol of each tanker in exact number of times is :
- A
$200$ litres
- B
$180$ litres
- ✓
$170$ litres
- D
$190$ litres
AnswerCorrect option: C. $170$ litres
Here, the maximum capacity of measuring container $= \text{HCF} \ (680, 850)$
$\therefore$ Applying Euclid’s division algorithm to $850$ and $680$
$850 = 680 \times 1 + 170$
$\Rightarrow 680 = 170 \times 4 + 0\ [$Zero remainder$]$
$\therefore \text{HCF} = (680, 850) = 170$
View full question & answer→MCQ 531 Mark
The number $3.24636363...$ is :
Answer$3.24636363 ... $
Which is repeating decimal number, and hence is a rational number.
View full question & answer→MCQ 541 Mark
Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non $-$ terminating decimal expansion $\frac{7}{210}$ :
AnswerCorrect option: B. Non $-$ terminating decimal expansion
Simplify it by dividing nominator and denominator both by $7$ we get $\frac{1}{30}$
Factorize the denominator we get $30 = 2 \times 3 \times 5$
Denominator has $3$ also in denominator
So denominator is not in form of $2^n \times 5^n$
View full question & answer→MCQ 551 Mark
If $p$ and $q$ are co $-$ prime numbers, then $p^2$ and $q^2$ are :
- ✓
Co $-$ prime.
- B
Not co $-$ prime.
- C
- D
AnswerCorrect option: A. Co $-$ prime.
We know that the co $-$ prime numbers have no factor in common, or, their $\text{HCF}$ 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 561 Mark
A card is drawn at random from a pack of $52$ cards. The probability that the card is drawn is neither an ace nor a king is :
- A
$\frac{1}{26}$
- B
$\frac{4}{13}$
- ✓
$\frac{11}{13}$
- D
$\frac{11}{26}$
AnswerCorrect option: C. $\frac{11}{13}$
total number of outcomes $=52$
favourable outcomes in this case $= 52 -{4+4}=44\ [52-{4}$ aces $+ \ 4$ kings$]$
$\therefore P\ ($neither an ace nor a king $)=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{44}{11}=\frac{11}{13}$
View full question & answer→MCQ 571 Mark
The decimal expansion of the rational number $\frac{37}{2^2\times5}$ will terminate after :
AnswerThe prime factorisation of the denominator is $2^2 \times 5$
Since $2 > 1,$
The decimal expansion will terminate after $2$ decimal places.
View full question & answer→MCQ 581 Mark
If the $\text{HCF}$ of $65$ and $117$ is expressible in the form $65m – 117,$ then the value of $'m’$ is
AnswerFirst, find the $\text{HCF}$ of $65$ and $117$
$117 = 65 \times 1 + 52$
$65 = 52 \times 1 + 13$
$52 = 13 \times 4 + 0\ ($zero remainder$)$
Therefore, $\text{HCF}\ (117 , 65)$ is $13$
Now,
$\therefore 65m – 117 = 13$
$\Rightarrow 65m = 13 + 117$
$\Rightarrow 65m = 130$
$\Rightarrow m = 2$
View full question & answer→MCQ 591 Mark
Choose the correct answer from the given four options in the following questions : If two positive integers $p$ and $q$ can be expressed as $p=a b^2$ and $q=a^3 b ; a, b$ being prime numbers, then $\text{LCM}\ (p, q)$ is :
- A
$ab.$
- B
$a^2 b^2$
- ✓
$a^3 b^2$
- D
$a^3 b^3$
AnswerCorrect option: C. $a^3 b^2$
Given that, $p=a b^2=a \times b \times b$
and $q=a^3 b=a \times a \times a \times b$
$\therefore \text{LCM}$ of pandq $= \text{LCM} \left(a b^2, a^3 b\right)$
$=a \times b \times b \times a \times a=a^3 b^2$
$($since, $\text{LCM}$ is the product of the greatest power of each prime factor Invotved in the numbers$)$
View full question & answer→MCQ 601 Mark
$\sqrt2$ is :
- A
- ✓
- C
- D
A non-terminating repeating decimal.
AnswerAn irrational number is a number that is non $-$ terminating and non $-$ repeating.
$\sqrt2=1.4142135\dots$ which is neither terminating nor repeating, and hence is an irrational number.
View full question & answer→MCQ 611 Mark
If the $\text{LCM}$ of $a$ and $18$ is $36$ and the $\text{HCF}$ of $a$ and $18$ is $2,$ then $a =$
Answer$\text{LCM}\ (a, 18) = 36$
$\text{HCF}\ (a, 18) = 2$
We know that the product of numbers is equal to the product of their $\text{HCF}$ and $\text{LCM}$.
Therefore,
$18a = 2(36)$
$\text{a}=\frac{2(36)}{18}$
$a = 4$
Hence the correct choice is $(c)$.
View full question & answer→MCQ 621 Mark
If $112 = q \times 6 + r,$ then the possible values of $r,$ are :
- A
$\{0, 1, 2, 3\}$
- B
$\{2, 3, 5\}$
- C
$\{1, 2, 3, 4\}$
- ✓
$\{0, 1, 2, 3, 4, 5\}$
AnswerCorrect option: D. $\{0, 1, 2, 3, 4, 5\}$
For the relation $\text{x}=\text{qy}+\text{r},0\leq\text{r}<\text{y}$
So, here $r$ lies between $0\leq\text{r}<6$
Hence, $r = \{0, 1, 2, 3, 4, 5\}$
View full question & answer→MCQ 631 Mark
A die is thrown once. Find the probability of getting a number between $3$ and $6$.
- A
$\frac{2}{3}$
- B
$\frac{1}{2}$
- C
$\frac{1}{4}$
- ✓
$\frac{1}{3}$
AnswerCorrect option: D. $\frac{1}{3}$
Total number of outcomes $=\{1, 2, 3, 4, 5, 6, \} = 6$
Favourable outcomes in this case $= \{4, 5\} = 2$
$\therefore p \ (a$ number between $3$ and $6)$
$=\frac{\text{favourable outcomes}}{\text{total outcomes}}=\frac{2}{6}=\frac{1}{3}$
View full question & answer→MCQ 641 Mark
The smallest number of $4$ digits exactly divisible by $12, 15, 18$ and $27$ is :
- A
$1002$
- B
$1001$
- C
$1000$
- ✓
$1080$
AnswerCorrect option: D. $1080$
$\text{LCM}\ (12, 15, 18, 27) = 540$
Now, smallest four $-$ digit number $= 1000$
$\therefore 1000 \div 540 = 1 \times 540 + 460\ ($Remainder $= 460)$
Therefore, the smallest number of $4$ digits exactly divisible by $12, 15, 18$ and $27$ is $1000 + (540 - 460) $
$= 1000 + 80 = 1080$
View full question & answer→MCQ 651 Mark
The decimal expansion of $\frac{246}{2^7\times5^{-3}}$ terminates after :
- ✓
$6$ places of decimals
- B
$7$ places of decimals
- C
$10$ places of decimals
- D
AnswerCorrect option: A. $6$ places of decimals
$\frac{246}{2^7\times{5^{-3}}}=\frac{123}{2^6\times5^{-3}}$
Therefore,$\frac{246}{2^7\times5^{-3}}$
will terminate after $6$ places of decimals.
View full question & answer→MCQ 661 Mark
Which of the following is not a rational number ?
- A
$\sqrt{25}$
- B
$\sqrt{9}$
- ✓
$\sqrt{8}$
- D
$\sqrt{16}$
AnswerCorrect option: C. $\sqrt{8}$
$\sqrt{16}=\sqrt{4}\times\sqrt{4}=4 $ is a rational number
$\sqrt{9}=\sqrt{3} \times \sqrt{3}=3 $ is rational number
$\sqrt{25}=\sqrt{5}\times\sqrt{5}=5 $ is also a rational number
but $ \sqrt{8}$ is not a rational number
because $\sqrt{8}=2\sqrt{2}$ and $\sqrt{2} $ is an irrational
therefore. $\sqrt{8}$ is also an irrational number.
View full question & answer→MCQ 671 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 $\text{HCF}\ (a, b)$ is :
- ✓
$pq$
- B
$p^3 q^3$
- C
$p^3 q^2$
- D
$p^2 q^2$
Answer$\mathrm{a}=\mathrm{pq}^2$ and $\mathrm{b}=\mathrm{p}^3 \mathrm{q}$ where $a$ and $b$ are positive integers and $p, q$ are prime numbers,
then $\text{HCF = pq}$.
View full question & answer→MCQ 681 Mark
If $\text{HCF} \ (26, 169) = 13,$ then $\text{LCM}\ (26, 169) =$
- A
$26.$
- B
$52.$
- ✓
$338.$
- D
$13.$
AnswerCorrect option: C. $338.$
$\text{HCF}\ (26, 169) = 13$
$\text{LCM}\ (26, 169) =\frac{26\times169}{13}=338$
View full question & answer→MCQ 691 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 701 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 711 Mark
$0.\overline{68}+0.\overline{73}=?$
- A
$1.\overline{41}$
- ✓
$1.\overline{42}$
- C
$0.\overline{141}$
- D
AnswerCorrect option: B. $1.\overline{42}$
Consider, $\text{x}=0.\overline{68}$
$\Rightarrow\text{x}=0.6868\dots \dots(\text{i})$
Multiply by $100$
$\Rightarrow\text{100x}=68.68\dots \dots(\text{ii})$
Subtracting $(i)$ from $(ii),$ we get
$\text{99x}=68$
$\Rightarrow\text{x}=\frac{68}{99}\dots(\text{A})$
Consider, $\text{x}=0.\overline{73}$
$\Rightarrow x = 0.7373 ...... (iii)$
Multiply by $100$
$\Rightarrow 100x = 73.73 ...... (iv)$
Subtracting $(iii)$ from $(iv),$ we get
$\text{99x}=73$
$\Rightarrow\text{x}=\frac{73}{99}\dots(\text{B})$
Adding $(A)$ and $(B),$ gives us
$\frac{68}{99}+\frac{73}{99}=\frac{141}{99}=1.42424\dots$
$\Rightarrow0.\overline{68}+0.\overline{73}=1.42424\dots$
$=1.\overline{42}$
View full question & answer→MCQ 721 Mark
The $\text{HCF}$ of $256,442$ and $940$ is :
AnswerPrime factors of numbers are
$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
$442 = 2 \times 13 \times 17$
$940 = 2 \times 2 \times 5 \times 47$
$\text{HCF}$ is the highest common factor among the numbers.
Thus among the given numbers $2$ is the common factor.
Hence $\text{HCF}$ of given numbers is $2$.
View full question & answer→MCQ 731 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 $= \text{HCF}$ of $65, 117$
$[$For the largest number$]$
For this, $117 = 65 \times 1 + 52\ [$Dividend $=$ divisor $\times$ quotient $+$ remainder$]$
$\Rightarrow 65 = 52 \times 1 + 13$
$\Rightarrow 52 = 13 \times 4 + 0$
$\text{HCF} = 13$
Hence, $13$ is the largest number which divides $70$ and $125,$ leaving remainders $5$ and $8$.
View full question & answer→MCQ 741 Mark
Which of the following numbers have the non $-$ terminating repeating decimal expansion ?
AnswerCorrect option: A. $\frac{77}{210}$
$\frac{77}{210}=\frac{11}{30}=\frac{11}{2\times3\times5}$
Because non $-$ terminating repeating decimal expansion should have the denominator other than $2$ or $5$.
View full question & answer→MCQ 751 Mark
Classify the following numbers as rational or irrational : $2-\sqrt{5}$
Answer$2$ is rational
$\sqrt{5}=2.035 ........$ which is non terminating and non repeating hence irrational number.
We know that, rational $−$ irrational $=$ irrational number.
Hence $2 - 5 = $ irrational number
View full question & answer→MCQ 761 Mark
Which of the following numbers has terminating decimal expansion ?
- A
$\frac{3}{11}$
- ✓
$\frac{3}{5}$
- C
$\frac{5}{3}$
- D
$\frac{3}{7}$
AnswerCorrect option: B. $\frac{3}{5}$
$\frac{3}{5}$ has terminal decimal expansion because terminal decimal expansion should have the denominator $2$ or $5$ only.
View full question & answer→MCQ 771 Mark
The decimal representation of $\frac{71}{150}$ is :
- A
- ✓
A non-terminating, repeating decimal.
- C
A non-terminating and non-repeating decimal.
- D
AnswerCorrect option: B. A non-terminating, repeating decimal.
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.
The prime factorisation of the denominator is $2 \times 3 \times 50^2$
So, the denominator will be non $-$ terminating.
Since $\frac{71}{150}$ is a rational number, it will surely be repeating.
View full question & answer→MCQ 781 Mark
For any two positive integers $a$ and $b,$ such that $a > b$. There exist $($unique$)$ whole numbers $q$ and $r$ such that:
- A
$\text{a}=\text{qbr}$
- B
$\text{b}=\text{aq}+\text{r},0\leq\text{r}<\text{b}$
- ✓
$\text{a}=\text{bq}+\text{r},0\leq\text{r}<\text{b}$
- D
$\text{q}=\text{ar}+\text{b},0\leq\text{r}<\text{b}$
AnswerCorrect option: C. $\text{a}=\text{bq}+\text{r},0\leq\text{r}<\text{b}$
Euclid’s Division Lemma states that for given positive integer $a$ and $b,$
There exist unique integers $q$ and $r$ satisfying a $=\text{bq}+\text{r},0\leq\text{r}<\text{b}$
View full question & answer→MCQ 791 Mark
Choose the correct answer from the given four options in the following questions : The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after :
AnswerRational number $=\frac{14587}{1250}=\frac{14587}{2^{1}\times5^{4}}$
$=\frac{14587}{10\times5^{3}}=\frac{(2)^{3}}{(2)^{3}}$
$=\frac{14587\times8}{10\times1000}$
$=\frac{1166969}{10000}=11.6696$
Hence, given rational number will terminate after four decimal places.
$\begin{array}{c|c} 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$
View full question & answer→MCQ 801 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 811 Mark
Every prime number has exactly $.........$ factors.
- A
more than $4$
- B
$3$
- C
$4$
- ✓
$2$
AnswerPrime numbers are the numbers which have only two factors,
i.e., $1$ and number itself.
View full question & answer→MCQ 821 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 $\text{HCF} \ (a, b) $ is :
- A
$xy$
- ✓
$x y^2$
- 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 831 Mark
Which of the following numbers has non $-$ terminating repeating decimal expansion ?
- A
$\frac{35}{50}$
- B
$\frac{15}{1600}$
- ✓
$\frac{17}{6}$
- D
$\frac{23}{2}$
AnswerCorrect option: C. $\frac{17}{6}$
$\frac{17}{6}$ has a non $-$ terminal repeating decimal expansion.
$\frac{17}{6}=2.6333 ...$
View full question & answer→MCQ 841 Mark
What is the largest number that divides each one of $1152$ and $1664$ exactly ?
AnswerThe largest number that divides each one of $1152$ and $1664$ exactly will be the $\text{HCF}$ of the numbers.
Using Euclid's Division Algorithm,
$1664 = 1152 \times 1 + 512$
$1152 = 512 \times 2 + 128$
$512 = 128 \times 4 + 0$
So $, \text{HCF}\ (1152, 1664) = 128$
Hence, the largest number is $128$
View full question & answer→MCQ 851 Mark
The $\text{LCM}$ of two numbers is $14$ times their $\text{HCF}$. The sum of $\text{LCM}$ and $\text{HCF}$ is $600,$ then $\text{LCM} =$
AnswerAccording to question,
${\text{MCU=14 }} \times {\text{HCF}}$
$ \Rightarrow\frac{\text{LCM}}{14}$
and $\text{LCM} +\text{ HCF} = 600$
$\Rightarrow {\text{LCM}}+\frac{\text{LCM}}{14}=600\ \big[$from eq $.({\text{i}})\big]$
$\Rightarrow15 \times{\text{LCM}}=600 \times 14$
$\Rightarrow {\text{LCM}}=\frac{600\times14}{15}=560$
View full question & answer→MCQ 861 Mark
The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after
- A
$1$ decimal place
- ✓
$4$ decimal places
- C
$2$ decimal places
- D
$3$ decimal places
AnswerCorrect option: B. $4$ decimal places
$\frac{14587}{1250}=\frac{14587}{2\times5^4}$
Here, in the denominator of the given fraction the highest power of prime factor $5$ is $4,$
therefore, the decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after $4$ decimal places.
View full question & answer→MCQ 871 Mark
The $\text{HCF}$ and $\text{LCM}$ of two numbers are $9$ and $90$ respectively. If one numberis $18,$ then the other number is :
AnswerUsing the result,
$\text{HCF} \times \text{LCM } =$ product of two natural numbers,
$\Rightarrow $ The other number $=\frac{9\times90}{18}=45$
View full question & answer→MCQ 881 Mark
The sum of two irrational numbers is always
- A
- ✓
a rational number or an irrational number
- C
- D
AnswerCorrect option: B. a rational number or an irrational number
The sum of two irrational numbers can be either a rational number or an irrational number.
e.g $5\sqrt{3}+3\sqrt{2}=5\sqrt{3}+3\sqrt{2}$ sum is irrational
$(2+6\sqrt{7})+(-6\sqrt{7})=2$ sum is rational Hence sum can be either rational or irrational
View full question & answer→MCQ 891 Mark
Which of the following rational numbers have terminating decimal?
$(i) \ \frac{16}{225}$
$(ii) \ \frac{5}{18}$
$(iii) \ \frac{2}{21}$
$(iv) \ \frac{7}{250}$
- 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
$2^m \times 5^n \frac{16}{225}$ and $\frac{7}{250}$ has terminating decimals.
View full question & answer→MCQ 901 Mark
By Euclid’s division lemma $x = qy + r, x > y$ the value of $q$ and $r$ for $x = 27$ and $y = 5$ are :
- A
$q = 5, r = 3$
- B
$q = 6, r = 3$
- C
- ✓
$q = 5, r = 2$
AnswerCorrect option: D. $q = 5, r = 2$
$x = qy + r$
$\Rightarrow 27 = 5 \times 5 + 2$
$\Rightarrow q = 5, r = 2$
View full question & answer→MCQ 911 Mark
$\pi$ is :
AnswerAn irrational number is a number that is non $-$ terminating and non $-$ repeating.
$\pi=3.1415926\dots$
Which is neither terminating nor repeating, and hence is an irrational number.
View full question & answer→MCQ 921 Mark
If $m^2-1$ is divisible by $8,$ then $m$ is :
AnswerLet $a=m^2-1$
Here $m$ can be ever or odd.
Case $I$ :
$m=$ Even i.e., $m=2 k$, where $k$ is an integer,
$\Rightarrow a=(2 k)^2-1$
$\Rightarrow a=4 k^2-1$
At $k=-1,=4(-1)^2-1=4-1=3$, 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 $II$ :
$\mathrm{m}=$ Odd i.e., $\mathrm{m}=2 \mathrm{k}+1$, 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 the above two cases, if $m$ is odd, then $m^2-1$ is divisible by $8$ .
View full question & answer→MCQ 931 Mark
The decimal expansion of the number $\frac{441}{2^2\times5^3\times7^2}$ has :
AnswerTo check if the number is terminating :
we will find the lowest form of the number.
$\frac{441}{2^2\times5^7\times7^2}$
Here $441=49\times9=7^2\times3^2$
$\frac{7^2\times3^2}{2^2\times5^2\times7^2}=\frac{3^2}{2^2\times5^7}$
Here denominator $= 2^2 \times 5^7$
Here the denominator is of the form $2^{\mathrm{m}}, 5^n$
$m = 2, n = 7$
Hence, the number has a terminal decimal representation.
View full question & answer→MCQ 941 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 $\text{LCM}\ (a, b)$ is :
- A
$pq$
- 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 $\text{LCM} =p^2 q^2$.
View full question & answer→MCQ 951 Mark
If two positive integers $m$ and $n$ can be expressed as $m=x^2 y^5$ and $n=x^3 y^2,$ where $x$ and $y$ are prime numbers, then $\operatorname{HCF}(m, n)=$
- A
$x^2 y^3$
- ✓
$x^2 y^2$
- C
$x^3 y^2$
- D
$x^3 y^3$
AnswerCorrect option: B. $x^2 y^2$
$x^2 y^5=y^3\left(x^2 y^2\right)$
$x^3 y^3=x\left(x^2 y^2\right)$
Therefore $\text{HCF}\ (m, n)$ is $x^2 y^2$
View full question & answer→MCQ 961 Mark
The $\text{HCF}$ of two consecutive numbers is :
AnswerThe $\text{HCF}$ of two consecutive numbers is always $1.$
$ ($e.g. $\text{HCF}$ of $24, 25$ is $1).$
View full question & answer→MCQ 971 Mark
Choose the correct answer from the given four options in the following questions : The least number that is divisible by all the numbers from $1$ to $10\ ($both inclusive$)$ is :
- A
$10.$
- B
$100.$
- C
$504.$
- ✓
$2520.$
AnswerCorrect option: D. $2520.$
Factors of $1$ to $10$ numbers,
$1 = 1$
$2 = 1 \times 2$
$3 = 1 \times 3$
$4 = 1 \times 2 \times 2$
$5 = 1 \times 5$
$6 = 1 \times 2 \times 3$
$7 = 1 \times 7$
$8 = 1 \times 2 \times 2 \times 2$
$9 = 1 \times 3 \times 3$
$10 = 1 \times 2 \times 5$
$\therefore \text{LCM}$ 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 981 Mark
If two positive integers $'a\ ’$ and $'b\ ’$ are written as $a=pq^2$ and $b=p^3q^2,$ where $'p\ ’$ and $'q\ ’ $ are prime numbers, then $\text{LCM} \ (a, b) =$
- ✓
$p^3 q^2$
- B
$pq$
- C
$p^2 q^3$
- D
$p^2 q^2$
AnswerCorrect option: A. $p^3 q^2$
We know that $\text{LCM} =$ product of the highest powers of all the prime factors of the numbers $pq^2, p^3 q^2$
$\operatorname{LCM}=p^3 q^2$
View full question & answer→MCQ 991 Mark
What is the least number that divisible by all the natural numbers from $1$ to $10 \ ($both inclusive$)$ ?
- A
$100$
- B
$1260$
- ✓
$2520$
- D
$5040$
AnswerCorrect option: C. $2520$
To find the least number divisible by all the natural numbers is the $\text{LCM}$ of the numbers from $1$ to $10$
Find the prime factorization of each of the numbers to find the $\text{LCM}$.
$1,2,3,5,7,4=2^2, 6=2 \times 3,8=2^3, 9=3^2, 10=2 \times 5$
$\text{LCM} =2^3 \times 3^2 \times 5 \times 7=2520$
View full question & answer→MCQ 1001 Mark
If $9^x+2=240+9^x$, then the value of $x$ is:
Answer$9^{\text{x}+2} = 240 + 9\text{x}$
$\Rightarrow 9^\text{x}. 9^2 = 240 + 9\text{x}$
$\Rightarrow 9^\text{x} (81 - 1) = 240$
$\Rightarrow 9^\text{x} = 3$
$\Rightarrow 9^\text{x} = 9\frac{1}{2}$
$\Rightarrow \text{ x} =\frac{1}{2}= 0.5$
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