Model Question Paper 3

Product of rational number $\dfrac{-4}{7}$ and its additive inverse is

1. $\dfrac{16}{49}$

2. $\dfrac{-16}{49}$

3. 1

4. 0

Answer

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

$\Rightarrow \text{Product} = \dfrac{-4}{7} \times \dfrac{4}{7}\\[1em] = \dfrac{-4 \times 4}{7 \times 7}\\[1em] = \dfrac{-16}{49}$

Hence, Option 2 is the correct option.

What should be subtracted from $\dfrac{-3}{5}$ to get $\dfrac{-2}{3}$?

1. $\dfrac{1}{15}$

2. $-\dfrac{1}{15}$

3. $\dfrac{19}{25}$

4. $\dfrac{-19}{15}$

Answer

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

$\Rightarrow \dfrac{-3}{5} - x = \dfrac{-2}{3}\\[1em] \Rightarrow x = \dfrac{-3}{5} + \dfrac{2}{3}\\[1em] \Rightarrow x = \dfrac{-9 + 10}{15}\\[1em] \Rightarrow x = \dfrac{1}{15}$

Hence, Option 1 is the correct option.

The value of $\left(\dfrac{2}{3}\right)^0 + \left(\dfrac{2}{3}\right)^{-1}$ is

1. $\dfrac{3}{2}$

2. $-\dfrac{1}{3}$

3. $\dfrac{5}{2}$

4. none of these

Answer

We know that,

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

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

Hence, Option 3 is the correct option.

The time in which ₹2,000 amount to ₹2,662 at 20% per annum compounded half-yearly is

1. 1 year

2. $1\dfrac{1}{2}$ years

3. 2 years

4. 3 years

Answer

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 = $\dfrac{20}{2}$% = 10%

Let the number of conversion periods be $n$.

We know that,

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

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

Hence, Option 2 is the correct option.

The volume of a cube is 216 m³. Length of its side is

1. 3 m

2. 6 m

3. 9 m

4. 12 m

Answer

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

We know that,

Volume of a cube = (side)³

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

Resolving 216 into prime factors :

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

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

Hence, Option 2 is the correct option.

When the difference of a 2-digit number $ab$ ($a$ > $b$) and the number obtained by reversing the digits is divided by 9, then quotient is

1. $a + b$

2. $b - a$

3. $a - b$

4. none of these

Answer

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 = $10b + a$

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

On dividing the difference by 9, we get :

$\Rightarrow \text{Quotient} = \dfrac{9(a - b)}{9}\\[1em] = a - b$

Hence, Option 3 is the correct option.

Amar buys a toy for ₹120 and sells it for ₹144, his profit % is

1. 5%

2. 10%

3. 20%

4. 24%

Answer

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

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

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

= 20%

Hence, Option 3 is the correct option.

If $567x8$ is divisible by 9, then value of $x$ is

1. 1

2. 2

3. 3

4. 4

Answer

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.

$\Rightarrow 26 + x = 27\\[1em] \Rightarrow x = 27 - 26\\[1em] \Rightarrow x = 1$

Hence, Option 1 is the correct option.

Virat bought a cricket bat for ₹1,320 including 10% GST. Find the price of the bat before GST was added.

Answer

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

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

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

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

Find A in the addition

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

Answer

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.

$\Rightarrow A + 5 = 12\\[1em] \Rightarrow A = 12 - 5\\[1em] \Rightarrow A = 7$

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.

Find the value of x for which $\left(-\dfrac{4}{5}\right)^2 \times \left(-\dfrac{5}{4}\right)^{2x} = \dfrac{625}{256}$.

Answer

We know that,

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

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

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

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

$\Rightarrow 2 - 2x = -4\\[1em] \Rightarrow -2x = -4 - 2\\[1em] \Rightarrow -2x = -6\\[1em] \Rightarrow x = \dfrac{-6}{-2}\\[1em] \Rightarrow x = 3$

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

Using appropriate properties of operations of rational numbers, evaluate:

$\dfrac{5}{12} \times \dfrac{2}{3} + \dfrac{5}{4} - \dfrac{2}{3} \times \dfrac{7}{6}$

Answer

Using the distributive property of multiplication over subtraction,

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

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

What number should be subtracted from the sum of $\dfrac{-5}{12}$ and $\dfrac{2}{3}$ to get $\dfrac{5}{4}$?

Answer

Calculating sum first,

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

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

$\Rightarrow \dfrac{1}{4} - x = \dfrac{5}{4}\\[1em] \Rightarrow x = \dfrac{1}{4} - \dfrac{5}{4}\\[1em] \Rightarrow x = \dfrac{1 - 5}{4}\\[1em] \Rightarrow x = \dfrac{-4}{4}\\[1em] \Rightarrow x = -1$

Hence, -1 should be subtracted from the sum.

If n(ξ) = 40, n(A) = 23, n(B′) = 28 and n((A ∪ B)′) = 9, find n(A ∩ B)

Answer

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.

In an auditorium the number of rows is equal to the number of chairs in each row. If there are 5184 chairs in the auditorium, find the number of rows.

Answer

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 = $x$

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

Resolving 5184 into prime factors :

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

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

Hence, there are 72 rows in the auditorium.

Insert 5 rational numbers between $\dfrac{1}{3}$ and $\dfrac{1}{2}$

Answer

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

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

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

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

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

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

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

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

Reducing these to their lowest terms,

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

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

Find the smallest number by which 10584 should be multiplied so that product is a perfect cube. Also find the cube root of the product.

Answer

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

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

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

In a 3-digit number unit's digit is 2 less than hundred's digit and hundred's digit is 3 less than the ten's digit. If the sum of the original number and number obtained by reversing the digits is 968, find the number.

Answer

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.

$\Rightarrow (100a + 10b + c) + (100c + 10b + a) = 968\\[1em] \Rightarrow 101a + 20b + 101c = 968$

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

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

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.

Rohit purchased some bananas at 3 for ₹5 and sold them at 5 for ₹12. Thus he gained ₹143 in all. How many bananas did he purchase?

Answer

C.P. of 3 bananas = ₹ 5

∴ C.P. of 1 banana = ₹ $\dfrac{5}{3}$

S.P. of 5 bananas = ₹ 12

∴ S.P. of 1 banana = ₹ $\dfrac{12}{5}$

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

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

= ₹ $\dfrac{36 - 25}{15}$

= ₹ $\dfrac{11}{15}$

Let the number of bananas purchased be $x$.

$\Rightarrow x \times \dfrac{11}{15} = 143\\[1em] \Rightarrow x = \dfrac{143 \times 15}{11}\\[1em] \Rightarrow x = 13 \times 15\\[1em] \Rightarrow x = 195$

Hence, Rohit purchased 195 bananas.

How much percent above the cost price should a shopkeeper mark his goods so that after allowing a discount of 20% he still earns a profit of 20%?

Answer

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

$\Rightarrow 120 = x - \dfrac{20}{100} \times x\\[1em] \Rightarrow 120 = x - \dfrac{x}{5}\\[1em] \Rightarrow 120 = \dfrac{5x - x}{5}\\[1em] \Rightarrow 120 = \dfrac{4x}{5}\\[1em] \Rightarrow x = \dfrac{120 \times 5}{4}\\[1em] \Rightarrow x = 150$

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

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

= 50%

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

Find the difference between S.I. and C.I. on ₹2,000 for $1\dfrac{1}{2}$ years at 6% p.a., compound interest being reckoned half yearly.

Answer

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

We know that,

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

As the compound interest is reckoned half yearly,

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

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

We know that,

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

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).

If 5 men or 7 women can earn ₹2,709 in a day, how much will 10 men and 5 women earn in a day?

Answer

Given, earning of 5 men = earning of 7 women

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

⇒ Earning of 10 men = earning of $\left(\dfrac{7}{5} \times 10\right)$ 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.

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

∴ It is a case of direct variation.

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

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

A hostel had rations for 150 students for 60 days. After 12 days, 30 more students join the hostel. How long will the remaining rations last?

Answer

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.

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

∴ It is a case of inverse variation.

$\Rightarrow 150 \times 48 = 180 \times x\\[1em] \Rightarrow 7200 = 180x\\[1em] \Rightarrow x = \dfrac{7200}{180}\\[1em] \Rightarrow x = 40$

Hence, the remaining rations will last for 40 days.

Simplify the following expression and write in the exponential form:

3^-5 × 3² ÷ 3^-6 + (2² × 3)² + $\left(\dfrac{2}{3}\right)^{-1}$ + 2^-1 + $\left(\dfrac{1}{19}\right)^{-1}$

Answer

We know that,

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

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

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.

A student scored 30% marks in an exam and failed by 20 marks, while another student scored 40% marks and got 40 marks more than the minimum marks required to pass the exam. Find the maximum marks and pass percentage.

Answer

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.

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

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

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

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

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

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

$\Rightarrow y = \dfrac{30}{100} \times 600 + 20\\[1em] \Rightarrow y = 180 + 20\\[1em] \Rightarrow y = 200$

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

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

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

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

If ξ = {x | x ∈ W, x < 20}, A = {factors of 18} and B = {multiples of 3}, find

(i) A ∪ B

(ii) A ∩ B

(iii) (A ∪ B)′.

Also verify that n(A ∪ B) = n(A) + n(B) - n(A ∩ B).

Answer

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

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

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

A and B together can do a piece of work in 15 days; B and C together can do it in 12 days; C and A can do it in 20 days. How long will they take to finish the work, working together? Also find the number of days taken by each to do the same work, working alone.

Answer

As,

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

Adding these three, we get :

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

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

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

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

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

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

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

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.

Is 31104 a perfect square number? If not then find the smallest number by which 31104 should be divided so that quotient is a perfect square. Also find the square root of the quotient.

Answer

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 = $\dfrac{31104}{6}$ = 5184

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

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.

Three numbers are in the ratio 1 : 3 : 5 and sum of their cubes is 52479. Find the numbers.

Answer

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

Given, the sum of their cubes is 52479.

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

∴ The three numbers are :

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

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