MCQ 11 Mark
Mark $(\checkmark)$ against the correct answer in the following:Which of the following are co$-$primes?
- ✓
$91$ and $72$
- B
$34$ and $51$
- C
$21$ and $36$
- D
$15$ and $20$
AnswerCorrect option: A. $91$ and $72$
The $HCF$ of $72$ and $91$ is $1.$
So, they are co-primes.
$a.$ Is not correct because $34$ and $51$ have $17$ as their $\text{HCF}.$
$b.$ Is not correct because $21$ and $56$ have $3$ as their $\text{HCF}.$
$c.$ Is not correct because $15$ and $20$ have $5$ as their $\text{HCF}.$
View full question & answer→MCQ 21 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following is a prime number?
Answer$a.\ 323$ is not a prime number because $323$ can be written as $17 \times 19.$
$b.\ 361$ is not a prime number because $361$ can be written as $19 \times 19.$
$c.\ 263$ is a prime number.
View full question & answer→MCQ 31 Mark
Mark $(\checkmark)$ against the correct answer in the following: The greatest number which divides $134$ and $167$ leaving $2$ as remainder in each case is:
AnswerSince we need $2$ as the remainder, we will subtract $2$ from each of the numbers.
$167 - 2 = 165$
$134 - 2 = 132$
Now, any of the common factors of $165$ and $132$ will be the required divisor.
On factorising:
$165 = 3 \times 5 \times 11$
$132 = 2 \times 2 \times 3 \times 11$
Their common factors are $11$ and $3.$
So, $3 \times 11 = 33$ is the required divisor.
View full question & answer→MCQ 41 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following is a composite number?
Answer$a.\ 23$ is not a composite number as it cannot be broken into factors.
$b.\ 29$ is not a composite number as it cannot be broken into factors.
$c.\ 32$ is a composite number as it can be broken into factors, which are $2 \times 2 \times 2 \times 2 \times 2.$
View full question & answer→MCQ 51 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following are co-primes?
- A
$39, 91$
- ✓
$161, 192$
- C
$385, 462$
- D
AnswerCorrect option: B. $161, 192$
a. $39$ and $91$ are not co-primes as $39$ and $91$ have a common factor, i.e. $13.$
b. $161$ and $192$ are co-primes as $161$ and $192$ have no common factor other than $1$.
c. $385$ and $462$ are not co-primes as $385$ and $462$ have common factors $7$ and $11.$
View full question & answer→MCQ 61 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following numbers is divisible by $11?$
- A
$3333333$
- B
$1111111$
- ✓
$22222222$
- D
AnswerCorrect option: C. $22222222$
A number is divisible by $11$, if the difference of the sum of its digits in odd places and the sum of the digits in even places (starting from ones place) is either 0 or a multiple of $11.$
$a.\ 3333333$
Consider the number $3333333.$
Sum of its digits in odd places $(3 + 3 + 3 + 3) = 12$
Sum of its digits in even places $(3 + 3 + 3) = 9$
Difference of the two sums $= 12 - 9 = 3$
Since this number $(3)$ is not divisible by $11, 3333333$ is not divisible by $11.$
$b.\ 1111111$
Consider the number $1111111.$
Sum of its digits in odd places $(1 + 1 + 1 + 1) = 4$
Sum of its digits in even places $(1 + 1 + 1) = 3$
Difference of the two sums $= 4 - 3 = 1$
Since this number (1) is not divisible by $11, 1111111 $is also not divisible by $11.$
$c.\ 22222222$
Consider the number $22222222.$
Sum of its digits in odd places $(2 + 2 + 2 + 2)= 8$
Sum of its digits in even places $(2 + 2 + 2 + 2) = 8$
Difference of the two sums $= 8 - 8 = 0$
Since this number $(0)$ is divisible by $11, 22222222$ is also divisible by $11.$
View full question & answer→MCQ 71 Mark
Mark $(\checkmark)$ against the correct answer in the following:The number which is neither prime nor composite is:
Answer$1$ is neither prime nor composite.
$a.\ $Is not correct because composite numbers are defined for positive numbers, but $0$ is neither a positive number nor a negative number.
$b.\ $Is not correct because $2$ is a prime number.
$c.\ $Is not correct because $3$ is a prime number.
View full question & answer→MCQ 81 Mark
Mark $(\checkmark)$ against the correct answer in the following:Which of the following is a prime number?
AnswerTo find a prime number between $100$ and $200$, we have to check whether the given number is divisible by any prime number less than $15$. If yes, it is not prime, otherwise it is.
By examining, we find that $131$ is a prime number.
View full question & answer→MCQ 91 Mark
Mark $(\checkmark)$ against the correct answer in the following:Which of the following numbers is divisible by $9?$
- ✓
$8576901$
- B
$96345210$
- C
$67594310$
- D
AnswerCorrect option: A. $8576901$
A number is divisible by $9$ if the sum of its digits is divisible by $9.$
$a.\ $Consider the number $8576901.$
Sum of its digits $= 8 + 5 + 7 + 6 + 9 + 0 + 1 = 36,$ which is divisible by $9.$
So, $8576901$ is divisible by $ 9.$
$b.\ $Consider the number $96345210$.
Sum of its digits $= 9 + 6 + 3 + 4 + 5 + 2 + 1 + 0 = 30$, which is not divisible by $ 9.$
So, $96345210$ is not divisible by $9.$
$c.\ $Consider the number $67594310.$
Sum of its digits $= 6 + 7 + 5 + 9 + 4 + 3 + 1 + 0 = 35$, which is not divisible by $ 9.$
So, $67594310$ is not divisible by $ 9.$
View full question & answer→MCQ 101 Mark
Mark $(\checkmark)$ against the correct answer in the following: If a and b are co-primes, then their $LCM$ is:
AnswerCorrect option: B. $\frac{\text{a}}{\text{b}}$
If a and b are co-primes then their $LCM$ will be ab.
For example, $4$ and $9$ are co-primes.
$LCM$ of $4$ and $9$ is $4 × 9.$
View full question & answer→MCQ 111 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following are co$-$primes?
- A
$8, 12$
- ✓
$9, 10$
- C
$6, 8$
- D
$15, 18$
AnswerCorrect option: B. $9, 10$
$a.\ 8, 12$ are not co$-$primes as they have a common factor $4.$
$b.\ 9, 10$ are co$-$primes as they do not have a common factor.
$c.\ 6, 8$ are not co$-$primes as they have a common factor $2.$
$d.\ 15,18$ are not co$-$primes as they have a common factor $3.$
View full question & answer→MCQ 121 Mark
Mark $(\checkmark)$ against the correct answer in the following:The $LCM$ of two co-prime numbers is their:
AnswerThe $LCM$ of two co-prime numbers is their product.
View full question & answer→MCQ 131 Mark
Mark $(\checkmark)$ against the correct answer in the following: The $\text{HCF}$ of $144, 180$ and $192$ is:
AnswerWe will first factorise the two numbers:
$\begin{array}{c|c}2&144\\\hline2&72\\\hline2&36\\\hline2&18\\\hline3&9\\\hline3&3\\\hline&1\end{array}$
$\begin{array}{c|c}2&8188\\\hline2&90\\\hline3&45\\\hline3&15\\\hline5&5\\\hline&1\end{array}$
$\begin{array}{c|c}2&192\\\hline2&96\\\hline2&48\\\hline2&24\\\hline2&12\\\hline2&6\\\hline3&3\\\hline&1\end{array}$
$144=2\times2\times2\times2\times3\times3=2^4\times3^2$
$180=2\times2\times3\times3\times5=2^2\times3^2\times5$
$192=2\times2\times2\times2\times2\times3=2^6\times3$
Here, $12($i.e.$2^2\times 3 = 12)$ is the highest common factor of the three numbers.
View full question & answer→MCQ 141 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following is a prime number?
Answer$a.\ 81$ is not a prime number because $81$ can be written as $9 × 9.$
$b.\ 87$ is not a prime number because $87$ can be written as $29 × 3.$
$c.\ 91$ is not a prime number because $91$ can be written as $13 × 7.$
$d.\ 97$ is a prime number.
View full question & answer→MCQ 151 Mark
Mark $(\checkmark)$ against the correct answer in the following: The product of two numbers is $2160$ and their $HCF$ is $12$. The $LCM$ of these numbers is:
AnswerHere, $HCF = 12$
Product of two number $= 2160$
We know:
$LCM × HCF =$ Product of the two numbers
$LCM$ $=\frac{2160}{\text{HCF}}$
$=\frac{2160}{12}$
$= 180$
$LCM = 180$
View full question & answer→MCQ 161 Mark
Mark $(\checkmark)$ against the correct answer in the following:What least number should be replaced for * so that the number $67301*2$ is exactly divisible by $9?$
Answer$6 + 7 + 3 + 0 + 1 + * + 2 = 19 + *$
$8$ is the least number that should be added to 19 such that number will be divisible by $9.$
Sum of the digits:
$6 + 7 + 3 + 0 + 1 + 8 + 2 = 27$
$27$ is divisible by $9.$
View full question & answer→MCQ 171 Mark
Mark $(\checkmark)$ against the correct answer in the following:Which of the following numbers is divisible by $6?$
- A
$67821$
- B
$78134$
- ✓
$87432$
- D
AnswerCorrect option: C. $87432$
A number is divisible by $6$ if it is divisible by both $2$ and $3.$
Since the ones digit of $87432 $ is $2$, it is divisible by $2.$
Now, $8 + 7 + 4 + 3 + 2 = 24$
$24$ is divisible by $3.$
Hence, $87432$ is divisible by $6$ because it is divisible by both $2$ and $3.$
$a.$ Is not correct because $67821$ is not divisible by $2.$
$b.$ Is not correct because $78134$ is not divisible by $3.$
$7 + 8 + 1 + 3 + 4 = 23$
$23$ is not divisible by $3.$
View full question & answer→MCQ 181 Mark
Find which of the following numbers are prime:
Answer$89 $ is a prime number.
View full question & answer→MCQ 191 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following numbers is divisible by $3?$
- A
$24357806$
- B
$35769812$
- ✓
$83479560$
- D
$3336433$
AnswerCorrect option: C. $83479560$
A number is divisible by $3$ if the sum of its digits is divisible by $3$
$a.\ $Consider the number $24357806$
Sum of its digits $= 2 + 4 + 3 + 5 + 7 + 8 + 0 + 6 = 35$, which is not divisible by $3.$
So, $2357806$ is not divisible by $3.$
$b.\ $Consider the number $35769812$
Sumofitsdigits $3 + 5 + 7 + 6 + 9 + 8 + 1 + 2 = 41$, which is not divisible by $3.$
So, $35769812$ is not divisible by $3$
$c.\ $Consider the number $83479560$
Sum of its digits $= 8 + 3 + 4 + 7 + 9 + 5 + 6 + 0 = 42$, which is divisible by $3.$
So, $2357806$ is divisible by $3$
$d.\ $Consider the number 333643$3.$
Sum of its digits $= 3 + 3 + 3 + 6 + 4 + 3 + 3 = 25$, which is not divisible by $3.$
So, $3336433$ is not divisible by $3.$ View full question & answer→MCQ 201 Mark
Mark $(\checkmark)$ against the correct answer in the following:$\frac{289}{391}$ when reduced to lowest term is:
- A
$\frac{13}{17}$
- B
$\frac{17}{19}$
- ✓
$\frac{17}{23}$
- D
$\frac{17}{21}$
AnswerCorrect option: C. $\frac{17}{23}$
$\begin{array}{c|c}17&289\\\hline17&17\\\hline&1\end{array}$
$\begin{array}{c|c}17&391\\\hline23&23\\\hline&1\end{array}$
$289 = 17 × 17$
$391 = 17 × 23$
The $HCF$ of $289$ and $391$ is $17.$
Dividing both the numerator and the denominator by $17:$
$\frac{289\div17}{391\div17}=\frac{17}{23}$
View full question & answer→MCQ 211 Mark
Mark $(\checkmark)$ against the correct answer in the following: The least number divisible by each of the numbers $15, 20, 24, 32$ and $36$ is:
- A
$1660$
- B
$2880$
- ✓
$1440$
- D
AnswerCorrect option: C. $1440$
The least number divisible by each of the numbers $15, 20, 24, 32$ and $36$ is their $LCM.$
$\begin{array}{c|c}2&15,20,24,32,36\\\hline2&15,10,12,16,18\\\hline2&15,5,6,8,9\\\hline2&15,5,3,4,9\\\hline2&15,5,3,2,9\\\hline3&15,5,3,1,9\\\hline3&5,5,1,1,3\\\hline5&5,5,1,1,1\\\hline&1,1,1,1,1\end{array}$
$LCM = 2^5× 3^2× 5$$= 1440$
View full question & answer→MCQ 221 Mark
Mark $(\checkmark)$ against the correct answer in the following: $\frac{289}{391}$ When reduced to the lowest terms is:
- A
$\frac{11}{23}$
- B
$\frac{13}{31}$
- C
$\frac{17}{31}$
- ✓
$\frac{17}{23}$
AnswerCorrect option: D. $\frac{17}{23}$
$HCF = 17$
Dividing both the numerator and the denominator by the $HCF$ of $289$ and $391:$
$\begin{array}{c|c}17&289\\\hline17&17\\\hline&1\end{array}$
$\begin{array}{c|c}17&391\\\hline23&23\\\hline&1\end{array}$
$\frac{289\div17}{391\div17}=\frac{17}{23}$
View full question & answer→MCQ 231 Mark
Mark $(\checkmark)$ against the correct answer in the following:Every counting number has an infinite number of:
AnswerEvery counting number has an infinite number of multiples.
If $p$ is a counting number, its multiples are $1p, 2p, 3p....$
View full question & answer→MCQ 241 Mark
Mark $(\checkmark)$ against the correct answer in the following: The $LCM$ of $12, 15, 20, 27$ is:
- A
$270$
- B
$360$
- C
$480$
- ✓
$540$
Answer$\begin{array}{c|c}2&12,15,20,27\\\hline2&6,15,10,27\\\hline3&3,15,5,27\\\hline3&1,5,5,9\\\hline3&1,5,5,3\\\hline5&1,5,5,1\\\hline&1,1,1,1\end{array}$
$LCM$ $2^2× 3^3× 5 = 540$
View full question & answer→MCQ 251 Mark
Mark $(\checkmark)$ against the correct answer in the following: The smallest number which when diminished by $3$ is divisible by $14, 28, 36$ and $45$, is:
AnswerThe smallest number that is exactly divisible by $14, 28, 36$ and $45$ will be their $\text{LCM.}$
So, the required number will be the $\text{LCM}$ plus $3.$
$\begin{array}{c|c}2&11,28,36,45\\\hline2&11,14,18,45\\\hline3&11,7,9,45\\\hline3&11,7,3,15\\\hline5&11,7,1,5\\\hline7&11,7,1,1\\\hline11&11,1,1,1\\\hline&1,1,1,1\end{array}$
$\text{LCM}$ of the three numbers $ = 2^2\times 3^2\times 5 \times 7 \times 11$
$= 13860$
$\therefore $ Required number $= 13860 + 3 = 13863$
View full question & answer→MCQ 261 Mark
Mark $(\checkmark)$ against the correct answer in the following: The $LCM$ of $24, 36, 40$ is:
Answer$\begin{array}{c|c}2&24,36,40\\\hline2&12,18,20\\\hline2&6,9,10\\\hline3&3,9,5\\\hline3&1,3,5\\\hline5&1,1,5\\\hline&1,1,1\end{array}$
$LCM$ $= 2^3× 3^2× 5$$= 360$
View full question & answer→MCQ 271 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following numbers is divisible by $4?$
- A
$78653234$
- B
$98765042$
- C
$24689602$
- ✓
$87941032$
AnswerCorrect option: D. $87941032$
A number is divisible by $4$ if the number formed by its digits in the tens and ones places is divisible by $4.$
$a.\ 78653234$
Consider the number $78653234.$
Here, the number formed by the tens and the ones digit is $34$, which is not divisible by 1$4.$
Therefore, $78653234$ is not divisible by $4.$
$b.\ 98765042$
Consider the number $98765042$.
Here, the number formed by the tens and the ones digit is $42$, which is not divisible by $4.$
Therefore, $98765042$ is not divisible by $4.$
$c.\ 24689602$
Consider the number $24689602$.
Here, the number formed by the tens and the ones digit is $02$, which is not divisible by $4.$
Therefore, $24689602$ is not divisible by $4$
$d.\ 87941032$
Consider the number $87941032.$
Here, the number formed by the tens and ones digit is $32$, which is divisible by $4.$
Therefore, $87941032$ is divisible by $4.$
View full question & answer→MCQ 281 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following is a prime number?
Answer$a.117$ is not a prime number because $117$ can be written as $3 \times 39.$
$b.171$ is not a prime number because $171$ can be written as $19 \times 9.$
$c.179$ is prime number.
View full question & answer→MCQ 291 Mark
Mark $(\checkmark)$ against the correct answer in the following: The $HCF$ of $144$ and $198$ is:
AnswerWe first factorise the two numkbers:
$\begin{array}{c|c}2&144\\\hline2&72\\\hline2&36\\\hline2&18\\\hline3&9\\\hline3&3\\\hline&1\end{array}$
$\begin{array}{c|c}2&198\\\hline3&99\\\hline3&33\\\hline11&11\\\hline&1\end{array}$
$144=2\times2\times2\times2\times3\times3=2^4\times3^2$
$198=2\times3\times3\times11=2\times3^2\times11$
Here, $18(2 × 3^2= 18)$ is the highest common factor of the two numbers.
View full question & answer→MCQ 301 Mark
Mark $(\checkmark)$ against the correct answer in the following:
The $HCF$ of two co-primes is:
Answer$HCF$ of two co-primes is$ 1.$
This is because two co-prime numbers do not have any common factor.
For example, $15$ and $16$ are co-primes.Their.
$HCF$ is $1.$
View full question & answer→MCQ 311 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following numbers is divisible by $8?$
- A
$96354142$
- ✓
$37450176$
- C
$57064214$
- D
AnswerCorrect option: B. $37450176$
A number is divisible by $8$ if the number formed by its digits in hundreds, tens and ones places is divisible by $8.$
$a.\ 96354142$
Consider the number $96354142.$
Here, the number formed by the digits in hundreds, tens and ones places is $142,$ which is clearly not divisible by $8.$
Therefore, $96354142$ is not divisible by $8.$
$b.\ 37450176$
Consider the number $37450176.$
The number formed by the digits in hundreds, tens and ones places is $176$, which is clearly divisible by $8.$
Therefore, $37450176$ is divisible by $8.$
$c.\ 57064214$
Consider the number $57064214.$
Here, the number formed by the digits in hundreds, tens and ones places is $214$, which is clearly not divisible by $8.$
Therefore, $57064214$ is not divisible by $8.$
View full question & answer→MCQ 321 Mark
Mark $(\checkmark)$ against the correct answer in the following: Which of the following numbers is divisible by $6?$
- ✓
$8790432$
- B
$98671402$
- C
$85492014$
- D
AnswerCorrect option: A. $8790432$
A number is divisible by $6$, if it is divisible by both $2$ and $3.$
$a.\ 8790432$
Consider the number $8790432.$
The number in the ones digit is $2.$
Therefore, 8790432 is divisible by $2.$
Now, the sum of its digits $(8 + 7 + 9 + 0 + 2 + 3 + 2)$ is $33.$
Since $33$ is divisible by $3$, we can say that 8790432 is also divisible by $3.$
Since $8790432$ is divisible by both $2$ and $3$, it is also divisible by $6.$
$b.\ 98671402$
Consider the number $98671402.$
The number in the ones digit is $2$.
Therefore, $98671402$ is divisible by $2$.
Now, the sum of its digits $(9 + 8 + 6 + 7 + 1 + 4 + 0 + 2)$ is $37$.
Since $37$ is not divisible by $3,$ we can say that 98671402 is also not divisible by $3.$
Since $98671402$ is not divisible by both $2$ and $3$, it is not divisible by $6.$
$c.\ 85492014$
Consider the number $85492014.$
The number in the ones digit is $4.$
Therefore, $85492014$ is divisible by $2$.
Now, the sum of its digits $(8 + 5 + 4 + 9 + 2 + 0 + 1 + 4)$ is 3$3.$
Since $33$ is divisible by $3$, we can say that 85492014 is also divisible by $3.$
Since $85492014$ is divisible by both $2$ and $3$, it is also divisible by $6.$
View full question & answer→MCQ 331 Mark
Mark $(\checkmark)$ against the correct answer in the following: Three bells toll together at intervals of $9, 12, 15$ minutes. If they start tolling together, after what time will they next toll together?
- A
$1\text{ hour}$
- B
$1\frac12\text{ hours}$
- C
$2\frac12\text{hours}$
- ✓
$3\text{ hours}$
AnswerCorrect option: D. $3\text{ hours}$
The $L.C.M$. of $9, 12$ and $15$ will give us the minutes after which the bells will next toll together.
$\begin{array}{c|c}2&9,12,15\\\hline2&9,6,15\\\hline3&9,6,15\\\hline3&3,1,5\\\hline5&1,1,5\\\hline&1,1,1\end{array}$
LCM $= 2^2× 3^2× 5$
$= 180$
So,the bells will toll together after $180$ min.
On converting into hours:
$180/60 = 3$ hours
View full question & answer→MCQ 341 Mark
Mark $(\checkmark)$ against the correct answer in the following: The $HCF $of two numbers is $145$ and their $LCM$ is $2175$. If one of the numbers is $725$, the other number is:
AnswerOne of the numbers is $725.$
$HCF = 145$
$LCM = 2175$
We know:
$LCM \times HCF =$ Product of the two numbers
$\therefore$ Product of the two numbers $= 145 \times 2175$
$= 315375$
$\therefore$ Other number $=\frac{315375}{725}$
$=435$
View full question & answer→