Model Question Paper 2

Every composite number has atleast

1. 1 factor

2. 2 factors

3. 3 factors

4. 4 factors

Answer

A composite number is a natural number greater than 1 which has more factors than just 1 and itself.

Every composite number has at least 3 factors: 1, the number itself, and at least one more factor.

For example, 4 is the smallest composite number and its factors are 1, 2 and 4 (i.e., 3 factors).

Hence, option 3 is the correct option.

Which of the following collections form a set?

1. Collection of 5 odd prime numbers

2. Collection of 3 most intelligent students of your class

3. Collection of 4 vowels in English alphabet

4. Collection of months of a year having less than 31 days.

Answer

A set is a well-defined collection of objects. Let us examine each option:

1. Collection of 5 odd prime numbers — There are many odd prime numbers (3, 5, 7, 11, 13, 17, 19, …). It is not specified which 5 odd primes are to be taken. So, it is not well-defined.

2. Collection of 3 most intelligent students of your class — The word "intelligent" is subjective and varies from person to person. So, it is not well-defined.

3. Collection of 4 vowels in English alphabet — There are 5 vowels (a, e, i, o, u) in the English alphabet. It is not specified which 4 vowels are to be taken. So, it is not well-defined.

4. Collection of months of a year having less than 31 days — These months are clearly known (February, April, June, September and November). So, it is well-defined. Thus it forms a set.

Hence, option 4 is the correct option.

Find the prime factorisation of 1320.

Answer

To find the prime factorisation of 1320, we keep dividing it by the smallest prime number that divides it, until we get 1.

$\begin{array}{l|l} 2 & 1320 \\ \hline 2 & 660 \\ \hline 2 & 330 \\ \hline 3 & 165 \\ \hline 5 & 55 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

∴ 1320 = 2 × 2 × 2 × 3 × 5 × 11

Hence, the prime factorisation of 1320 is 2 × 2 × 2 × 3 × 5 × 11.

If A = {x : x is a positive multiple of 3 less than 20} and B = {x : x is an odd prime number less than 20}, then find n(A) + n(B).

Answer

Given:

A = {x : x is a positive multiple of 3 less than 20}

The positive multiples of 3 less than 20 are: 3, 6, 9, 12, 15, 18.

∴ A = {3, 6, 9, 12, 15, 18}

n(A) = 6

B = {x : x is an odd prime number less than 20}

The odd prime numbers less than 20 are: 3, 5, 7, 11, 13, 17, 19.

∴ B = {3, 5, 7, 11, 13, 17, 19}

n(B) = 7

∴ n(A) + n(B) = 6 + 7 = 13

Hence, n(A) + n(B) = 13.

Reduce the fraction $\dfrac{714}{1386}$ in its simplest form.

Answer

To reduce the given fraction in its simplest form, we find the prime factorisation of the numerator and denominator and cancel the common factors.

Prime factorisation of 714:

$\begin{array}{l|l} 2 & 714 \\ \hline 3 & 357 \\ \hline 7 & 119 \\ \hline 17 & 17 \\ \hline & 1 \end{array}$

∴ 714 = 2 × 3 × 7 × 17

Prime factorisation of 1386:

$\begin{array}{l|l} 2 & 1386 \\ \hline 3 & 693 \\ \hline 3 & 231 \\ \hline 7 & 77 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

∴ 1386 = 2 × 3 × 3 × 7 × 11

$\dfrac{714}{1386} = \dfrac{2 \times 3 \times 7 \times 17}{2 \times 3 \times 3 \times 7 \times 11} = \dfrac{17}{3 \times 11} = \dfrac{17}{33}$

Hence, the simplest form of $\dfrac{714}{1386}$ is $\dfrac{17}{33}$.

Simplify the following: $2\dfrac{3}{14} - 3\dfrac{5}{6} - \dfrac{2}{5} + 2\dfrac{1}{2}$

Answer

Convert the mixed fractions into improper fractions:

$2\dfrac{3}{14} - 3\dfrac{5}{6} - \dfrac{2}{5} + 2\dfrac{1}{2} = \dfrac{31}{14} - \dfrac{23}{6} - \dfrac{2}{5} + \dfrac{5}{2}$

Let us find L.C.M. of 14, 6, 5 and 2:

$\begin{array}{l|l} 2 & 14, 6, 5, 2 \\ \hline 3 & 7, 3, 5, 1 \\ \hline 5 & 7, 1, 5, 1 \\ \hline 7 & 7, 1, 1, 1 \\ \hline & 1, 1, 1, 1 \end{array}$

L.C.M. = 2 × 3 × 5 × 7 = 210

Convert the given fractions into equivalent like fractions with denominator 210:

$\Rightarrow \dfrac{31}{14} = \dfrac{31 \times 15}{14 \times 15} = \dfrac{465}{210} \\[1em] \Rightarrow \dfrac{23}{6} = \dfrac{23 \times 35}{6 \times 35} = \dfrac{805}{210}\\[1em] \Rightarrow \dfrac{2}{5} = \dfrac{2 \times 42}{5 \times 42} = \dfrac{84}{210}\\[1em] \Rightarrow \dfrac{5}{2} = \dfrac{5 \times 105}{2 \times 105} = \dfrac{525}{210}\\[1em] \therefore \dfrac{31}{14} - \dfrac{23}{6} - \dfrac{2}{5} + \dfrac{5}{2} \\[1em] = \dfrac{465}{210} - \dfrac{805}{210} - \dfrac{84}{210} + \dfrac{525}{210} \\[1em] = \dfrac{465 - 805 - 84 + 525}{210} \\[1em] = \dfrac{990 - 889}{210} \\[1em] = \dfrac{101}{210}.$

Hence, the answer is $\dfrac{101}{210}$.

A number is divisible by 5 and 8 both. By what other numbers will that number be always divisible?

Answer

If a number is divisible by both 5 and 8, then it must be divisible by the L.C.M. of 5 and 8.

Let us find L.C.M. of 5 and 8:

$\begin{array}{l|l} 2 & 5, 8 \\ \hline 2 & 5, 4 \\ \hline 2 & 5, 2 \\ \hline 5 & 5, 1 \\ \hline & 1, 1 \end{array}$

L.C.M. = 2 × 2 × 2 × 5 = 40

So, the number is divisible by 40. Therefore, the number will also be divisible by every factor of 40.

Factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40.

Excluding 5 and 8 (which are already given), the other numbers by which the number will always be divisible are: 1, 2, 4, 10, 20 and 40.

Hence, the number will always be divisible by 1, 2, 4, 10, 20 and 40.

Arrange the fractions $\dfrac{2}{3}, \dfrac{7}{9}, \dfrac{5}{8}, \dfrac{3}{5}$ in ascending order.

Answer

To compare the given fractions, we convert them into equivalent like fractions.

Let us find L.C.M. of 3, 9, 8 and 5:

$\begin{array}{l|l} 2 & 3, 9, 8, 5 \\ \hline 2 & 3, 9, 4, 5 \\ \hline 2 & 3, 9, 2, 5 \\ \hline 3 & 3, 9, 1, 5 \\ \hline 3 & 1, 3, 1, 5 \\ \hline 5 & 1, 1, 1, 5 \\ \hline & 1, 1, 1, 1 \end{array}$

L.C.M. = 2 × 2 × 2 × 3 × 3 × 5 = 360

Convert the given fractions into equivalent like fractions with denominator 360:

$\Rightarrow \dfrac{2}{3} = \dfrac{2 \times 120}{3 \times 120} = \dfrac{240}{360} \\[1em] \Rightarrow \dfrac{7}{9} = \dfrac{7 \times 40}{9 \times 40} = \dfrac{280}{360} \\[1em] \Rightarrow \dfrac{5}{8} = \dfrac{5 \times 45}{8 \times 45} = \dfrac{225}{360} \\[1em] \Rightarrow \dfrac{3}{5} = \dfrac{3 \times 72}{5 \times 72} = \dfrac{216}{360}$

As 216 < 225 < 240 < 280,

$\dfrac{216}{360} \lt \dfrac{225}{360} \lt \dfrac{240}{360} \lt \dfrac{280}{360}$

⇒ $\dfrac{3}{5} \lt \dfrac{5}{8} \lt \dfrac{2}{3} \lt \dfrac{7}{9}$

Hence, the given fractions in ascending order are $\dfrac{3}{5}, \dfrac{5}{8}, \dfrac{2}{3}, \dfrac{7}{9}$.

Find the smallest number of 5-digits which is divisible by 12, 15 and 18.

Answer

The smallest number which is divisible by 12, 15 and 18 is the L.C.M. of 12, 15 and 18.

Let us find L.C.M. of 12, 15 and 18:

$\begin{array}{l|l} 2 & 12, 15, 18 \\ \hline 2 & 6, 15, 9 \\ \hline 3 & 3, 15, 9 \\ \hline 3 & 1, 5, 3 \\ \hline 5 & 1, 5, 1 \\ \hline & 1, 1, 1 \end{array}$

L.C.M. = 2 × 2 × 3 × 3 × 5 = 180

So, every number divisible by 12, 15 and 18 is a multiple of 180.

The smallest 5-digit number = 10000.

Now divide 10000 by 180:

$\begin{array}{l} 180\overline{\smash{\big)}10000\smash{\big(}}\phantom{}55 \\ \phantom{x}\phantom{+)}\underline{-900} \\ \phantom{{x^2 } 1+}1000 \\ \phantom{12341}\underline{-900} \\ \phantom{123456)} 100 \\ \end{array}$

On dividing 10000 by 180, we get quotient = 55 and remainder = 100.

So, 10000 is not divisible by 180.

To get the smallest 5-digit number which is divisible by 180, we add (180 − 100) = 80 to 10000.

∴ Required smallest 5-digit number = 10000 + 80 = 10080.

Hence, the smallest 5-digit number divisible by 12, 15 and 18 is 10080.

Three bells are ringing continuously at intervals of 30, 40 and 45 minutes respectively. At what time will they ring together if they ring simultaneously at 5 A.M.?

Answer

The three bells will ring together again after a time equal to the L.C.M. of 30, 40 and 45 minutes.

Let us find L.C.M. of 30, 40 and 45:

$\begin{array}{l|l} 2 & 30, 40, 45 \\ \hline 2 & 15, 20, 45 \\ \hline 2 & 15, 10, 45 \\ \hline 3 & 15, 5, 45 \\ \hline 3 & 5, 5, 15 \\ \hline 5 & 5, 5, 5 \\ \hline & 1, 1, 1 \end{array}$

L.C.M. = 2 × 2 × 2 × 3 × 3 × 5 = 360

So, the three bells will ring together again after 360 minutes.

Convert 360 minutes into hours:

360 minutes = $\dfrac{360}{60}$ hours = 6 hours.

The bells ring together at 5 A.M., so they will ring together again after 6 hours.

∴ Required time = 5 A.M. + 6 hours = 11 A.M.

Hence, the three bells will ring together again at 11 A.M.