Model Question Paper 3

Additive inverse of $\frac{-4}{7}$ is $\frac{4}{7}$.

$$\begin{aligned}\Rightarrow \text{Product} = \frac{-4}{7} \times \frac{4}{7}\\[1em] = \frac{-4 \times 4}{7 \times 7}\\[1em] = \frac{-16}{49}\end{aligned}$$

Hence, Option 2 is the correct option.

Let $x$ be subtracted from $\frac{-3}{5}$ to get $\frac{-2}{3}$.

$$\begin{aligned}\Rightarrow \frac{-3}{5} - x = \frac{-2}{3}\\[1em] \Rightarrow x = \frac{-3}{5} + \frac{2}{3}\\[1em] \Rightarrow x = \frac{-9 + 10}{15}\\[1em] \Rightarrow x = \frac{1}{15}\end{aligned}$$

Hence, Option 1 is the correct option.

We know that,

$a^0 = 1$ and $a^{-1} = \frac{1}{a}$

$$\begin{aligned}\Rightarrow \left(\frac{2}{3}\right)^0 + \left(\frac{2}{3}\right)^{-1} = 1 + \frac{3}{2}\\[1em] \Rightarrow \left(\frac{2}{3}\right)^0 + \left(\frac{2}{3}\right)^{-1} = \frac{2 + 3}{2}\\[1em] \Rightarrow \left(\frac{2}{3}\right)^0 + \left(\frac{2}{3}\right)^{-1} = \frac{5}{2}\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

P = ₹ 2,000, A = ₹ 2,662 and R = 20% per annum compounded half-yearly.

As the interest is compounded half-yearly,

Rate for each conversion period = $\frac{20}{2}$% = 10%

Let the number of conversion periods be $n$.

We know that,

$$\begin{aligned}A = P\left(1 + \frac{R}{100}\right)^n\\[1em] \Rightarrow 2{,}662 = 2{,}000\left(1 + \frac{10}{100}\right)^n\\[1em] \Rightarrow \frac{2{,}662}{2{,}000} = \left(\frac{11}{10}\right)^n\\[1em] \Rightarrow \frac{1331}{1000} = \left(\frac{11}{10}\right)^n\\[1em] \Rightarrow \left(\frac{11}{10}\right)^3 = \left(\frac{11}{10}\right)^n\\[1em] \Rightarrow n = 3\end{aligned}$$

∴ Time = 3 half years = $\frac{3}{2}$ years = $1\frac{1}{2}$ years

Hence, Option 2 is the correct option.

Let the length of the side of the cube be $x$ m.

We know that,

Volume of a cube = (side)³

$$\begin{aligned}\Rightarrow x^3 = 216\\[1em] \Rightarrow x = \sqrt[3]{216}\end{aligned}$$

Resolving 216 into prime factors :

$$\begin{array}{l|r} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

$$\begin{aligned}\Rightarrow x = \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow x = 2 \times 3\\[1em] \Rightarrow x = 6\end{aligned}$$

Hence, Option 2 is the correct option.

Let the 2-digit number $ab$ have $a$ as its tens digit and $b$ as its units digit.

∴ The number $ab = 10a + b$

The number obtained by reversing the digits = $10b + a$

$$\begin{aligned}\Rightarrow \text{Difference} = (10a + b) - (10b + a)\\[1em] = 10a + b - 10b - a\\[1em] = 9a - 9b\\[1em] = 9(a - b)\end{aligned}$$

On dividing the difference by 9, we get :

$$\begin{aligned}\Rightarrow \text{Quotient} = \frac{9(a - b)}{9}\\[1em] = a - b\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

C.P. = ₹ 120 and S.P. = ₹ 144.

As S.P. is more than C.P., there is a profit.

Profit = S.P. - C.P. = ₹ (144 - 120) = ₹ 24

We know that,

Profit Percentage = $\frac{\text{Profit}}{\text{C.P.}} \times 100$%

= $\frac{24}{120} \times 100$%

= $\frac{2{,}400}{120}$%

= 20%

Hence, Option 3 is the correct option.

A number is divisible by 9 if the sum of its digits is divisible by 9.

Sum of the digits of $567x8 = 5 + 6 + 7 + x + 8 = 26 + x$

As $x$ is a digit, its value lies from 0 to 9, so the value of $26 + x$ lies from 26 to 35.

The only multiple of 9 in this range is 27.

$$\begin{aligned}\Rightarrow 26 + x = 27\\[1em] \Rightarrow x = 27 - 26\\[1em] \Rightarrow x = 1\end{aligned}$$

Hence, Option 1 is the correct option.

Let the price of the bat before GST be ₹ $x$.

GST = 10% of ₹ $x$ = ₹ $\frac{10x}{100}$

$$\begin{aligned}\Rightarrow x + \frac{10x}{100} = 1{,}320\\[1em] \Rightarrow \frac{100x + 10x}{100} = 1{,}320\\[1em] \Rightarrow \frac{110x}{100} = 1{,}320\\[1em] \Rightarrow x = \frac{1{,}320 \times 100}{110}\\[1em] \Rightarrow x = \frac{1{,}32{,}000}{110}\\[1em] \Rightarrow x = 1{,}200\end{aligned}$$

Hence, the price of the bat before GST was added is ₹ 1,200.

Considering the units column,

A + 5 gives 2 in the units place, so A + 5 must be a 2-digit number ending in 2, i.e. A + 5 = 12.

$$\begin{aligned}\Rightarrow A + 5 = 12\\[1em] \Rightarrow A = 12 - 5\\[1em] \Rightarrow A = 7\end{aligned}$$

So, 1 is carried over to the tens column.

Verifying the addition by putting A = 7 :

$$\begin{array}{rrrr} &3 &4 &7 \\ + &3 &7 &5 \\ \hline &7 &2 &2 \end{array}$$

Tens column : 4 + 7 + 1 (carry) = 12, so 2 is written and 1 is carried over.

Hundreds column : 3 + 3 + 1 (carry) = 7 = A

Hence, A = 7.

We know that,

$\left(-\frac{5}{4}\right) = \left(-\frac{4}{5}\right)^{-1} \text{ and } \frac{625}{256} = \left(\frac{5}{4}\right)^4 = \left(-\frac{4}{5}\right)^{-4}$

Substituting these values in $\left(-\frac{4}{5}\right)^2 \times \left(-\frac{5}{4}\right)^{2x} = \frac{625}{256}$, we get :

$$\begin{aligned}\Rightarrow \left(-\frac{4}{5}\right)^2 \times \Big[\left(-\frac{4}{5}\right)^{-1}\Big]^{2x} = \left(-\frac{4}{5}\right)^{-4}\\[1em] \Rightarrow \left(-\frac{4}{5}\right)^2 \times \left(-\frac{4}{5}\right)^{-2x} = \left(-\frac{4}{5}\right)^{-4}\\[1em] \Rightarrow \left(-\frac{4}{5}\right)^{2 - 2x} = \left(-\frac{4}{5}\right)^{-4}\end{aligned}$$

On both the sides the bases are equal, so the powers must also be equal.

$$\begin{aligned}\Rightarrow 2 - 2x = -4\\[1em] \Rightarrow -2x = -4 - 2\\[1em] \Rightarrow -2x = -6\\[1em] \Rightarrow x = \frac{-6}{-2}\\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, $\bm{x = 3}$.

Using the distributive property of multiplication over subtraction,

$$\begin{aligned}\Rightarrow \frac{5}{12} \times \frac{2}{3} + \frac{5}{4} - \frac{2}{3} \times \frac{7}{6}\\[1em] \Rightarrow \frac{2}{3} \times \left(\frac{5}{12} - \frac{7}{6}\right) + \frac{5}{4}\\[1em] \Rightarrow \frac{2}{3} \times \left(\frac{5 - 14}{12}\right) + \frac{5}{4}\\[1em] \Rightarrow \frac{2}{3} \times \left(\frac{-9}{12}\right) + \frac{5}{4}\\[1em] \Rightarrow \frac{2}{3} \times \left(\frac{-3}{4}\right) + \frac{5}{4}\\[1em] \Rightarrow \frac{-1}{2} + \frac{5}{4}\\[1em] \Rightarrow \frac{-2 + 5}{4}\\[1em] \Rightarrow \frac{3}{4}\end{aligned}$$

Hence, the value of the given expression is $\mathbf{\frac{3}{4}}$.

Calculating sum first,

$$\begin{aligned}\Rightarrow \text{Sum} = \frac{-5}{12} + \frac{2}{3}\\[1em] = \frac{-5 + 8}{12}\\[1em] = \frac{3}{12}\\[1em] = \frac{1}{4}\end{aligned}$$

Let $x$ be subtracted from $\frac{1}{4}$ to get $\frac{5}{4}$.

$$\begin{aligned}\Rightarrow \frac{1}{4} - x = \frac{5}{4}\\[1em] \Rightarrow x = \frac{1}{4} - \frac{5}{4}\\[1em] \Rightarrow x = \frac{1 - 5}{4}\\[1em] \Rightarrow x = \frac{-4}{4}\\[1em] \Rightarrow x = -1\end{aligned}$$

Hence, -1 should be subtracted from the sum.

We know that,

n(B) = n(ξ) - n(B′)

⇒ n(B) = 40 - 28

⇒ n(B) = 12

Also,

n(A ∪ B) = n(ξ) - n((A ∪ B)′)

⇒ n(A ∪ B) = 40 - 9

⇒ n(A ∪ B) = 31

We know that,

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

⇒ 31 = 23 + 12 - n(A ∩ B)

⇒ 31 = 35 - n(A ∩ B)

⇒ n(A ∩ B) = 35 - 31

⇒ n(A ∩ B) = 4

Hence, n(A ∩ B) = 4.

Let the number of rows in the auditorium be $x$.

As the number of rows is equal to the number of chairs in each row,

∴ Number of chairs in each row = $x$

$$\begin{aligned}\Rightarrow x \times x = 5184\\[1em] \Rightarrow x^2 = 5184\\[1em] \Rightarrow x = \sqrt{5184}\end{aligned}$$

Resolving 5184 into prime factors :

$$\begin{array}{l|r} 2 & 5184 \\ \hline 2 & 2592 \\ \hline 2 & 1296 \\ \hline 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

$$\begin{aligned}\Rightarrow x = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)}\\[1em] \Rightarrow x = 2 \times 2 \times 2 \times 3 \times 3\\[1em] \Rightarrow x = 72\end{aligned}$$

Hence, there are 72 rows in the auditorium.

We have to insert 5 rational numbers between $\frac{1}{3}$ and $\frac{1}{2}$.

First, express the given rational numbers with a common denominator.

L.C.M. of 3 and 2 = 6

$\frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6}$ and $\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}$

As there is no integer between 2 and 3, we multiply the numerator and denominator of both by 8.

$\frac{2}{6} = \frac{2 \times 8}{6 \times 8} = \frac{16}{48}$ and $\frac{3}{6} = \frac{3 \times 8}{6 \times 8} = \frac{24}{48}$

Now, the rational numbers lying between $\frac{16}{48}$ and $\frac{24}{48}$ are :

$\frac{18}{48}, \frac{19}{48}, \frac{20}{48}, \frac{21}{48}$ and $\frac{22}{48}$

Reducing these to their lowest terms,

$\frac{18}{48} = \frac{3}{8}, \frac{20}{48} = \frac{5}{12}, \frac{21}{48} = \frac{7}{16}$ and $\frac{22}{48} = \frac{11}{24}$

Hence, the five rational numbers between $\mathbf{\frac{1}{3}}$ and $\mathbf{\frac{1}{2}}$ are $\mathbf{\frac{3}{8}, \frac{19}{48}, \frac{5}{12}, \frac{7}{16}}$ and $\mathbf{\frac{11}{24}}$.

Resolving 10584 into prime factors :

$$\begin{array}{l|r} 2 & 10584 \\ \hline 2 & 5292 \\ \hline 2 & 2646 \\ \hline 3 & 1323 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

∴ 10584 = (2 × 2 × 2) × (3 × 3 × 3) × (7 × 7)

Grouping the factors in triplets, we see that the factor 7 occurs only twice and does not form a triplet.

So, to make the product a perfect cube, 10584 should be multiplied by 7.

Product = 10584 × 7 = 74088

$$\begin{aligned}\Rightarrow \sqrt[3]{74088} = \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow \sqrt[3]{74088} = 2 \times 3 \times 7\\[1em] \Rightarrow \sqrt[3]{74088} = 42\end{aligned}$$

Hence, the smallest number is 7 and the cube root of the product is 42.

Let the hundred's digit of the 3-digit number be $a$, the ten's digit be $b$ and the unit's digit be $c$.

Given, the unit's digit is 2 less than the hundred's digit.

∴ c = a - 2 ....(1)

Given, the hundred's digit is 3 less than the ten's digit.

∴ a = b - 3

⇒ b = a + 3 ....(2)

Original number = 100a + 10b + c

Number obtained by reversing the digits = 100c + 10b + a

Given, the sum of the two numbers is 968.

$$\begin{aligned}\Rightarrow (100a + 10b + c) + (100c + 10b + a) = 968\\[1em] \Rightarrow 101a + 20b + 101c = 968\end{aligned}$$

Substituting the values of $b$ and $c$ from (2) and (1), we get :

$$\begin{aligned}\Rightarrow 101a + 20(a + 3) + 101(a - 2) = 968\\[1em] \Rightarrow 101a + 20a + 60 + 101a - 202 = 968\\[1em] \Rightarrow 222a - 142 = 968\\[1em] \Rightarrow 222a = 968 + 142\\[1em] \Rightarrow 222a = 1110\\[1em] \Rightarrow a = \frac{1110}{222}\\[1em] \Rightarrow a = 5\end{aligned}$$

Substituting a = 5 in (1) and (2), we get :

c = 5 - 2 = 3 and b = 5 + 3 = 8

∴ Original number = 100 × 5 + 10 × 8 + 3 = 500 + 80 + 3 = 583

Hence, the number is 583.

C.P. of 3 bananas = ₹ 5

∴ C.P. of 1 banana = ₹ $\frac{5}{3}$

S.P. of 5 bananas = ₹ 12

∴ S.P. of 1 banana = ₹ $\frac{12}{5}$

As S.P. is more than C.P., there is a gain.

Gain on 1 banana = ₹ $\left(\frac{12}{5} - \frac{5}{3}\right)$

= ₹ $\frac{36 - 25}{15}$

= ₹ $\frac{11}{15}$

Let the number of bananas purchased be $x$.

$$\begin{aligned}\Rightarrow x \times \frac{11}{15} = 143\\[1em] \Rightarrow x = \frac{143 \times 15}{11}\\[1em] \Rightarrow x = 13 \times 15\\[1em] \Rightarrow x = 195\end{aligned}$$

Hence, Rohit purchased 195 bananas.

Let the C.P. of the goods be ₹ 100.

Given, the shopkeeper earns a profit of 20%.

∴ S.P. = C.P. + Profit = ₹ (100 + 20) = ₹ 120

Let the marked price of the goods be ₹ $x$.

Given, a discount of 20% is allowed on the marked price.

∴ S.P. = M.P. - Discount

$$\begin{aligned}\Rightarrow 120 = x - \frac{20}{100} \times x\\[1em] \Rightarrow 120 = x - \frac{x}{5}\\[1em] \Rightarrow 120 = \frac{5x - x}{5}\\[1em] \Rightarrow 120 = \frac{4x}{5}\\[1em] \Rightarrow x = \frac{120 \times 5}{4}\\[1em] \Rightarrow x = 150\end{aligned}$$

∴ Amount by which the marked price is above the C.P. = ₹ (150 - 100) = ₹ 50

Required percent = $\frac{50}{100} \times 100$%

= 50%

Hence, the shopkeeper should mark his goods 50% above the cost price.

Given, P = ₹ 2,000, R = 6% p.a. and T = $1\frac{1}{2}$ years = $\frac{3}{2}$ years.

We know that,

$$\begin{aligned}\Rightarrow \text{S.I.} = \frac{P \times R \times T}{100}\\[1em] \Rightarrow \text{S.I.} = \frac{2{,}000 \times 6 \times \frac{3}{2}}{100}\\[1em] \Rightarrow \text{S.I.} = \frac{18{,}000}{100}\\[1em] \Rightarrow \text{S.I.} = 180\end{aligned}$$

As the compound interest is reckoned half yearly,

Rate for each conversion period = $\frac{6}{2}$% = 3%

Number of conversion periods = $\frac{3}{2} \times 2$ = 3

We know that,

$$\begin{aligned}A = P\left(1 + \frac{R}{100}\right)^n\\[1em] \Rightarrow A = 2{,}000\left(1 + \frac{3}{100}\right)^3\\[1em] \Rightarrow A = 2{,}000 \times \frac{103}{100} \times \frac{103}{100} \times \frac{103}{100}\\[1em] \Rightarrow A = \frac{10{,}92{,}727}{500}\\[1em] \Rightarrow A = 2{,}185.454\end{aligned}$$

C.I. = A - P = ₹ (2,185.454 - 2,000) = ₹ 185.454

∴ Difference between C.I. and S.I. = ₹ (185.454 - 180) = ₹ 5.454 ≈ ₹ 5.45

Hence, the difference between the S.I. and the C.I. is ₹ 5.45 (approx).

Given, earning of 5 men = earning of 7 women

⇒ Earning of 1 man = earning of $\frac{7}{5}$ women

⇒ Earning of 10 men = earning of $\left(\frac{7}{5} \times 10\right)$ women = earning of 14 women

⇒ Earning of 10 men and 5 women = earning of (14 + 5) women = earning of 19 women

Let 19 women earn ₹ $x$ in a day.

Number of women	7	19
Earning per day (in ₹)	2,709	x

More the number of women, more will be the earning per day.

∴ It is a case of direct variation.

$$\begin{aligned}\Rightarrow \frac{7}{2{,}709} = \frac{19}{x}\\[1em] \Rightarrow x = \frac{2{,}709 \times 19}{7}\\[1em] \Rightarrow x = \frac{51{,}471}{7}\\[1em] \Rightarrow x = 7{,}353\end{aligned}$$

Hence, 10 men and 5 women will earn ₹ 7,353 in a day.

After 12 days, the remaining rations are sufficient for 150 students for (60 - 12) days, i.e. 48 days.

After 12 days, 30 more students join the hostel.

∴ Number of students in the hostel = 150 + 30 = 180

Let the remaining rations last for $x$ days when there are 180 students.

Number of students	150	180
Number of days	48	x

More the number of students, less will be the number of days for which the rations last.

∴ It is a case of inverse variation.

$$\begin{aligned}\Rightarrow 150 \times 48 = 180 \times x\\[1em] \Rightarrow 7200 = 180x\\[1em] \Rightarrow x = \frac{7200}{180}\\[1em] \Rightarrow x = 40\end{aligned}$$

Hence, the remaining rations will last for 40 days.

We know that,

$a^m \times a^n = a^{m + n}, a^m \div a^n = a^{m - n}$ and $a^{-1} = \frac{1}{a}$

$$\begin{aligned}3^{-5} \times 3^2 \div 3^{-6} + (2^2 \times 3)^2 + \left(\frac{2}{3}\right)^{-1} + 2^{-1} + \left(\frac{1}{19}\right)^{-1}\\[1em] \Rightarrow 3^{-5 + 2 - (-6)} + (2^2)^2 \times 3^2 + \frac{3}{2} + \frac{1}{2} + 19\\[1em] \Rightarrow 3^3 + 2^4 \times 3^2 + \frac{3 + 1}{2} + 19\\[1em] \Rightarrow 27 + 144 + 2 + 19\\[1em] \Rightarrow 192\end{aligned}$$

Resolving 192 into prime factors :

$$\begin{array}{l|r} 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

∴ 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁶ × 3¹

Hence, the value of the given expression in exponential form is 2⁶ × 3¹.

Let the maximum marks of the exam be $x$ and the minimum marks required to pass the exam be $y$.

The first student scored 30% marks and failed by 20 marks.

∴ $\frac{30}{100} \times x + 20 = y$ ....(1)

The second student scored 40% marks and got 40 marks more than the minimum pass marks.

∴ $\frac{40}{100} \times x - 40 = y$ ....(2)

From (1) and (2), we get :

$$\begin{aligned}\Rightarrow \frac{30x}{100} + 20 = \frac{40x}{100} - 40\\[1em] \Rightarrow 20 + 40 = \frac{40x}{100} - \frac{30x}{100}\\[1em] \Rightarrow 60 = \frac{10x}{100}\\[1em] \Rightarrow 60 = \frac{x}{10}\\[1em] \Rightarrow x = 600\end{aligned}$$

Substituting $x = 600$ in (1), we get :

$$\begin{aligned}\Rightarrow y = \frac{30}{100} \times 600 + 20\\[1em] \Rightarrow y = 180 + 20\\[1em] \Rightarrow y = 200\end{aligned}$$

Pass percentage = $\frac{200}{600} \times 100$%

= $\frac{100}{3}$%

= $33\frac{1}{3}$%

Hence, the maximum marks are 600 and the pass percentage is $\mathbf{33\frac{1}{3}}$%.

Given,

ξ = {x | x ∈ W, x < 20}

∴ ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}

A = {factors of 18} = {1, 2, 3, 6, 9, 18}

B = {multiples of 3} = {0, 3, 6, 9, 12, 15, 18}

(i) A ∪ B is the set of all the elements which are either in A or in B or in both.

Hence, A ∪ B = {0, 1, 2, 3, 6, 9, 12, 15, 18}.

(ii) A ∩ B is the set of all the elements which are common to both A and B.

Hence, A ∩ B = {3, 6, 9, 18}.

(iii) (A ∪ B)′ is the set of all the elements of ξ which are not in A ∪ B.

Hence, (A ∪ B)′ = {4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19}.

Verification :

n(A ∪ B) = 9, n(A) = 6, n(B) = 7 and n(A ∩ B) = 4

$$\begin{aligned}\Rightarrow n(A) + n(B) - n(A \cap B) = 6 + 7 - 4\\[1em] \Rightarrow n(A) + n(B) - n(A \cap B) = 9\\[1em] \Rightarrow n(A) + n(B) - n(A \cap B) = n(A \cup B)\end{aligned}$$

Hence, n(A ∪ B) = n(A) + n(B) - n(A ∩ B) is verified.

As,

$$\begin{aligned}\text{(A + B)'s one day's work} = \frac{1}{15}\\[1em] \text{(B + C)'s one day's work} = \frac{1}{12}\\[1em] \text{(C + A)'s one day's work} = \frac{1}{20}\end{aligned}$$

Adding these three, we get :

$$\begin{aligned}\Rightarrow 2(\text{A + B + C})\text{'s one day's work} = \frac{1}{15} + \frac{1}{12} + \frac{1}{20}\\[1em] = \frac{4 + 5 + 3}{60}\\[1em] = \frac{12}{60}\\[1em] = \frac{1}{5}\\[1em] \Rightarrow (\text{A + B + C})\text{'s one day's work} = \frac{1}{5 \times 2}\\[1em] = \frac{1}{10}\\[1em] \Rightarrow \text{Number of days taken by A, B and C working together} = \frac{1}{\frac{1}{10}}\\[1em] = 10\end{aligned}$$

A's one day's work = (A + B + C)'s one day's work - (B + C)'s one day's work

$$\begin{aligned}= \frac{1}{10} - \frac{1}{12}\\[1em] = \frac{6 - 5}{60}\\[1em] = \frac{1}{60}\\[1em] \Rightarrow \text{Number of days taken by A alone} = \frac{1}{\frac{1}{60}}\\[1em] = 60\end{aligned}$$

B's one day's work = (A + B + C)'s one day's work - (C + A)'s one day's work

$$\begin{aligned}= \frac{1}{10} - \frac{1}{20}\\[1em] = \frac{2 - 1}{20}\\[1em] = \frac{1}{20}\\[1em] \Rightarrow \text{Number of days taken by B alone} = \frac{1}{\frac{1}{20}}\\[1em] = 20\end{aligned}$$

C's one day's work = (A + B + C)'s one day's work - (A + B)'s one day's work

$$\begin{aligned}= \frac{1}{10} - \frac{1}{15}\\[1em] = \frac{3 - 2}{30}\\[1em] = \frac{1}{30}\\[1em] \Rightarrow \text{Number of days taken by C alone} = \frac{1}{\frac{1}{30}}\\[1em] = 30\end{aligned}$$

Hence, working together they will finish the work in 10 days, and working alone A takes 60 days, B takes 20 days and C takes 30 days.

Resolving 31104 into prime factors :

$$\begin{array}{l|r} 2 & 31104 \\ \hline 2 & 15552 \\ \hline 2 & 7776 \\ \hline 2 & 3888 \\ \hline 2 & 1944 \\ \hline 2 & 972 \\ \hline 2 & 486 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

∴ 31104 = (2 × 2) × (2 × 2) × (2 × 2) × 2 × (3 × 3) × (3 × 3) × 3

Grouping the factors in pairs, we see that the factors 2 and 3 are left unpaired.

∴ 31104 is not a perfect square number.

To make the quotient a perfect square, 31104 should be divided by 2 × 3, i.e. by 6.

Quotient = $\frac{31104}{6}$ = 5184

$$\begin{aligned}\Rightarrow \sqrt{5184} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)}\\[1em] \Rightarrow \sqrt{5184} = 2 \times 2 \times 2 \times 3 \times 3\\[1em] \Rightarrow \sqrt{5184} = 72\end{aligned}$$

Hence, 31104 is not a perfect square number, the smallest number by which it should be divided is 6 and the square root of the quotient is 72.

Let the three numbers be $x, 3x$ and $5x$.

Given, the sum of their cubes is 52479.

$$\begin{aligned}\Rightarrow x^3 + (3x)^3 + (5x)^3 = 52479\\[1em] \Rightarrow x^3 + 27x^3 + 125x^3 = 52479\\[1em] \Rightarrow 153x^3 = 52479\\[1em] \Rightarrow x^3 = \frac{52479}{153}\\[1em] \Rightarrow x^3 = 343\\[1em] \Rightarrow x = \sqrt[3]{343}\\[1em] \Rightarrow x = \sqrt[3]{7 \times 7 \times 7}\\[1em] \Rightarrow x = 7\end{aligned}$$

∴ The three numbers are :

$x$ = 7, $3x$ = 3 × 7 = 21 and $5x$ = 5 × 7 = 35

Hence, the three numbers are 7, 21 and 35.