Model Question Paper 3

The number of integers between -16 and 5 is

1. 19

2. 20

3. 21

4. 22

Answer

The integers between -16 and 5 are -15, -14, -13, ..., 0, 1, 2, 3, 4.

Number of integers = 4 - (-15) + 1 = 20.

Hence, Option 2 is the correct option.

50 m 5 cm is same as

1. 50.5 m

2. 50.05 m

3. 50.005 m

4. 5.05 m

Answer

As 100 cm = 1 m, 5 cm = $\dfrac{5}{100}$ m = 0.05 m.

∴ 50 m 5 cm = 50 m + 0.05 m = 50.05 m.

Hence, Option 2 is the correct option.

$\left(-3\dfrac{1}{3}\right) ÷ 1\dfrac{4}{5}$ is equal to

1. $-1\dfrac{23}{27}$

2. $-3\dfrac{5}{12}$

3. $1\dfrac{23}{27}$

4. -6

Answer

Solving,

$\Rightarrow \left(-3\dfrac{1}{3}\right) ÷ 1\dfrac{4}{5}\\[1em] = -\dfrac{10}{3} ÷ \dfrac{9}{5}\\[1em] = -\dfrac{10}{3} \times \dfrac{5}{9}\\[1em] = -\dfrac{50}{27}\\[1em] = -1\dfrac{23}{27}$

Hence, Option 1 is the correct option.

The number 5,540,000,000,000 in the scientific notation can be written as:

1. 554 × 10¹⁰

2. 55.4 × 10¹¹

3. 5.54 × 10¹²

4. 5.54 × 10¹¹

Answer

In scientific notation, the number is written as a number between 1 and 10 multiplied by a power of 10.

5,540,000,000,000 = 5.54 × 10¹².

Hence, Option 3 is the correct option.

The number of unlike terms in the expression
5x²y - 2xy² - 2yx² + 3y(xy + y²) + 7 is

1. 3

2. 4

3. 5

4. 6

Answer

Simplifying the expression,

5x²y - 2xy² - 2yx² + 3y(xy + y²) + 7

= 5x²y - 2xy² - 2x²y + 3xy² + 3y³ + 7

= (5x²y - 2x²y) + (-2xy² + 3xy²) + 3y³ + 7

= 3x²y + xy² + 3y³ + 7

The unlike terms are 3x²y, xy², 3y³ and 7, which are 4 in number.

Hence, Option 2 is the correct option.

x = -2 is a solution of the equation

1. 2x + 5 = 9

2. 3x - 1 = 5

3. 4x + 3 = 1

4. 5x + 12 = 2

Answer

Given Equation : 5x + 12,

Substituting x = -2 in L.H.S., we get :

5(-2) + 12 = -10 + 12 = 2.

R.H.S = 2

So, x = -2 satisfies the equation 5x + 12 = 2.

Hence, Option 4 is the correct option.

The ratio of the number of girls to the number of boys in a class is 5 : 4. If there are 16 boys in the class, then the number of students in the class is

1. 20

2. 32

3. 36

4. 45

Answer

Let the number of girls be x.

$\therefore \dfrac{x}{16} = \dfrac{5}{4}\\[1em] \Rightarrow x = \dfrac{5 \times 16}{4}\\[1em] \Rightarrow x = 20$

∴ The number of students = 20 + 16 = 36.

Hence, Option 3 is the correct option.

If 12% of a number is 9, then the number is

1. 36

2. 48

3. 60

4. 75

Answer

Let the number be x.

$\dfrac{12}{100} \times x = 9\\[1em] \Rightarrow x = 9 \times \dfrac{100}{12}\\[1em] \Rightarrow x = \dfrac{900}{12}\\[1em] \Rightarrow x = 75$

Hence, Option 4 is the correct option.

Using suitable properties, evaluate: 238 × (-44) + (-238) × 56

Answer

Using the distributive property,

238 × (-44) + (-238) × 56

= -238 × 44 - 238 × 56

= -238 × (44 + 56)

= -238 × 100

= -23800

Hence, 238 × (-44) + (-238) × 56 = -23800.

State whether each of the following statement is true or false for the sets P and Q where

P = {letters of TITLE} and Q = {letters of LITTLE}

(i) P ↔ Q

(ii) P = Q

Answer

Writing the sets by listing distinct letters,

P = {letters of TITLE} = {T, I, L, E}

Q = {letters of LITTLE} = {L, I, T, E}

(i) Both sets P and Q have 4 elements each, so they are equivalent. ∴ P ↔ Q is True.

(ii) Both sets P and Q have exactly the same elements {T, I, L, E}, so they are equal. ∴ P = Q is True.

Evaluate: $-3\dfrac{3}{8} - \left(-2\dfrac{1}{6}\right)$

Answer

Solving,

$\Rightarrow -3\dfrac{3}{8} - \left(-2\dfrac{1}{6}\right)\\[1em] = -\dfrac{27}{8} + \dfrac{13}{6}$

LCM of 8 and 6:

$\begin{array}{l|r} 2 & 8, 6 \\ \hline 2 & 4, 3 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$

LCM of 8 and 6 = 2 x 2 x 2 x 3 = 24

$= -\dfrac{27 \times 3}{24} + \dfrac{13 \times 4}{24}\\[1em] = -\dfrac{81}{24} + \dfrac{52}{24}\\[1em] = \dfrac{-81 + 52}{24}\\[1em] = -\dfrac{29}{24}\\[1em] = -1\dfrac{5}{24}$

Hence,$-3\dfrac{3}{8} - \left(-2\dfrac{1}{6}\right) = -1\dfrac{5}{24}$.

Simplify and express in the exponential form: (4³ × 3⁶) ÷ (16 × 9²)

Answer

Solving,

(4³ × 3⁶) ÷ (16 × 9²)

= $\dfrac{4^3 × 3^6}{4^2 × (3^2)^2}$

= $\dfrac{4^3 × 3^6}{4^2 × 3^4}$

= 4^(3-2) × 3^(6-4)

= 4¹ × 3²

= 4 × 9

= 36

= 6².

Hence, (4³ × 3⁶) ÷ (16 × 9²) = 6².

If I earn ₹75,000 per month and spend ₹40,000 per year for helping poor students, then find the ratio of the money spent for helping poor students and the annual income.

Answer

Monthly income = ₹75,000.

∴ Annual income = ₹(75,000 × 12) = ₹9,00,000.

Money spent per year for helping poor students = ₹40,000.

∴ Required ratio = $\dfrac{40,000}{9,00,000} = \dfrac{4}{90} = \dfrac{2}{45}$ = 2 : 45.

Hence, the required ratio is 2 : 45.

If ₹4,000 amounts to ₹5,000 in 2 years, find the rate of simple interest per annum.

Answer

Here, Principal (P) = ₹4,000, Amount (A) = ₹5,000 and Time (T) = 2 years.

Simple Interest (SI) = Amount - Principal = ₹5,000 - ₹4,000 = ₹1,000.

$\text{Rate} = \dfrac{SI × 100}{P × T}\\[1em] = \dfrac{1,000 × 100}{4,000 × 2}\\[1em] = \dfrac{1,00,000}{8,000}\\[1em] = 12.5$

Hence, the rate of simple interest is 12.5% per annum.

Simplify: (-4) × 7 - [13 - {49 + 40 ÷ (11 - $\overline{17 - 4}$)}]

Answer

Solving step by step, removing the bar (vinculum) first,

$(-4) × 7 - [13 - {49 + 40 ÷ (11 - \overline{17 - 4})}]$

= (-4) × 7 - [13 - {49 + 40 ÷ (11 - 13)}]

= (-4) × 7 - [13 - {49 + 40 ÷ (-2)}]

= (-4) × 7 - [13 - {49 + (-20)}]

= (-4) × 7 - [13 - 29]

= (-4) × 7 - (-16)

= -28 + 16

= -12

Hence, the value is -12.

Vikram's monthly salary is ₹12,750. He spends $\dfrac{1}{5}$ of his salary on food and out of the remaining, he spends $\dfrac{1}{4}$ on rent and $\dfrac{1}{6}$ on the education of children. Find

(i) how much he spends on each item?

(ii) how much money is still left with him?

Answer

(i) Vikram's monthly salary = ₹12,750.

Amount spent on food = $\dfrac{1}{5} × 12,750$ = ₹2,550.

Remaining amount = ₹12,750 - ₹2,550 = ₹10,200.

Amount spent on rent = $\dfrac{1}{4} × 10,200$ = ₹2,550.

Amount spent on education = $\dfrac{1}{6} × 10,200$ = ₹1,700.

Hence, he spends ₹2,550 on food, ₹2,550 on rent and ₹1,700 on education.

(ii) Money still left with him = ₹10,200 - ₹2,550 - ₹1,700 = ₹5,950.

Hence, ₹5,950 is still left with him.

Arrange the rational numbers $\dfrac{7}{-10}, \dfrac{-3}{5}, \dfrac{5}{-6} \text{and} \dfrac{-11}{-12}$ in descending order.

Answer

Writing each rational number with a positive denominator,

$\dfrac{7}{-10} = \dfrac{-7}{10}, \quad \dfrac{-3}{5} = \dfrac{-3}{5}, \quad \dfrac{5}{-6} = \dfrac{-5}{6}, \quad \dfrac{-11}{-12} = \dfrac{11}{12}$

LCM of 10, 5, 6 and 12:

$\begin{array}{l|r} 2 & 10, 5, 6, 12 \\ \hline 2 & 5, 5, 3, 6 \\ \hline 3 & 5, 5, 3, 3 \\ \hline 5 & 5, 5, 1, 1 \\ \hline & 1, 1, 1, 1 \end{array}$

The LCM of the denominators 10, 5, 6 and 12 = 2 x 2 x 3 x 5 = 60.

Expressing each with denominator 60,

$\dfrac{-7}{10} = \dfrac{-42}{60}, \quad \dfrac{-3}{5} = \dfrac{-36}{60}, \quad \dfrac{-5}{6} = \dfrac{-50}{60}, \quad \dfrac{11}{12} = \dfrac{55}{60}$

Arranging the numerators in descending order: 55 > -36 > -42 > -50.

∴ Descending order is $\dfrac{55}{60}, \dfrac{-36}{60}, \dfrac{-42}{60}, \dfrac{-50}{60}$.

Hence, the descending order is $\dfrac{-11}{-12}, \dfrac{-3}{5}, \dfrac{7}{-10}, \dfrac{5}{-6}$.

Afzal can walk $5\dfrac{3}{4}$ km in one hour. How much distance will he cover in 2 hours 40 minutes?

Answer

Distance covered in one hour = $5\dfrac{3}{4}$ km = $\dfrac{23}{4}$ km.

Time = 2 hours 40 minutes = $2\dfrac{40}{60}$ hours = $2\dfrac{2}{3}$ hours = $\dfrac{8}{3}$ hours.

∴ Distance covered = Speed × Time

$\text{Distance covered }= \dfrac{23}{4} × \dfrac{8}{3}\\[1em] = \dfrac{23 × 2}{3}\\[1em] = \dfrac{46}{3}\\[1em] = 15\dfrac{1}{3}$

Hence, Afzal will cover $15\dfrac{1}{3}$ km.

If a vehicle covers a distance of 57.72 km in 3.7 litres of petrol. How much distance will it cover in one litre of petrol?

Answer

Distance covered in 3.7 litres = 57.72 km.

∴ Distance covered in one litre = $\dfrac{57.72}{3.7}$ km.

$\dfrac{57.72}{3.7} = \dfrac{577.2}{37} = 15.6$

Hence, the vehicle will cover 15.6 km in one litre of petrol.

The perimeter of a triangle is 5 - 3x + 7x² and two of its sides are 2x² + 3x - 2 and 3x² - x + 3. Find the third side of the triangle.

Answer

Sum of the two given sides = (2x² + 3x - 2) + (3x² - x + 3)

= 2x² + 3x - 2 + 3x² - x + 3

= (2x² + 3x²) + (3x - x) + (-2 + 3)

= 5x² + 2x + 1

Third side = Perimeter - (sum of the two sides)

= (7x² - 3x + 5) - (5x² + 2x + 1)

= 7x² - 3x + 5 - 5x² - 2x - 1

= (7x² - 5x²) + (-3x - 2x) + (5 - 1)

= 2x² - 5x + 4

Hence, the third side of the triangle is 2x² - 5x + 4.

If $\left(-\dfrac{2}{3}\right)^x = -\dfrac{32}{243}$, then find the value of x.

Answer

Solving,

$\left(-\dfrac{2}{3}\right)^x = -\dfrac{32}{243}\\[1em] \Rightarrow \left(-\dfrac{2}{3}\right)^x = \dfrac{(-2) × (-2) × (-2) × (-2) × (-2)}{3 × 3 × 3 × 3 × 3}\\[1em] \Rightarrow \left(-\dfrac{2}{3}\right)^x = \left(-\dfrac{2}{3}\right)^5\\[1em] \Rightarrow x = 5$

Hence, x = 5.

Solve the equation: 3(2x - 1) - 2(2 - 5x) = 1

Answer

Solving,

⇒ 3(2x - 1) - 2(2 - 5x) = 1

⇒ 6x - 3 - 4 + 10x = 1

⇒ 16x - 7 = 1

⇒ 16x = 1 + 7

⇒ 16x = 8

⇒ x = $\dfrac{8}{16}$

⇒ x = $\dfrac{1}{2}$

Hence, x = $\dfrac{1}{2}$.

If 74% of the population of a village is illiterate and the number of literate people is 2158, then find the population of the village.

Answer

As 74% of the population is illiterate, the percentage of literate people = (100 - 74)% = 26%.

Let the population of the village be x.

According to the problem,

$26$

Hence, the population of the village is 8300.

Simplify: $\left(\dfrac{4}{5} - \dfrac{1}{3}\right) ÷ 4\dfrac{1}{5} + \dfrac{2}{3} \text { of } \left(5\dfrac{1}{6} - 4\dfrac{3}{8}\right)$

Answer

Solving the brackets first,

$\dfrac{4}{5} - \dfrac{1}{3} = \dfrac{12 - 5}{15} = \dfrac{7}{15}\\[1em] 5\dfrac{1}{6} - 4\dfrac{3}{8} = \dfrac{31}{6} - \dfrac{35}{8} = \dfrac{124 - 105}{24} = \dfrac{19}{24}$

Now, substituting these values,

$\dfrac{7}{15} ÷ 4\dfrac{1}{5} + \dfrac{2}{3} \text{ of } \dfrac{19}{24}\\[1em] = \dfrac{7}{15} ÷ \dfrac{21}{5} + \dfrac{2}{3} × \dfrac{19}{24}\\[1em] = \dfrac{7}{15} × \dfrac{5}{21} + \dfrac{19}{36}\\[1em] = \dfrac{1}{9} + \dfrac{19}{36} \text{ (L.C.M. of 9 and 36 = 36)} \\[1em] = \dfrac{4}{36} + \dfrac{19}{36}\\[1em] = \dfrac{23}{36}$

Hence, the value is $\dfrac{23}{36}$.

If we represent the distance above the ground by a positive rational number and that below the ground by a negative rational number, then answer the following question: An elevator descends into a mine shaft at the rate of $4\dfrac{3}{4}$ metre per minute. If it begins to descend from $7\dfrac{1}{2}$ metre above the ground, what will be its position after 18 minutes from the ground?

Answer

Rate of descent = $4\dfrac{3}{4}$ metre per minute = $\dfrac{19}{4}$ metre per minute.

Distance descended in 18 minutes = $\dfrac{19}{4} × 18$ metres.

$\dfrac{19}{4} × 18 = \dfrac{19 × 18}{4} = \dfrac{342}{4} = 85.5 \text{ m}$

As the elevator descends (moves down), this is represented by a negative number, i.e. -85.5 m.

The starting position = $7\dfrac{1}{2}$ m above the ground = +7.5 m.

∴ Final position = 7.5 + (-85.5) = -78 m.

The negative sign shows the position is below the ground.

Hence, the elevator will be 78 m below the ground after 18 minutes.

In a competition, question paper consists of 25 questions. 4 marks are awarded for every correct answer, 2 marks are deducted for every incorrect answer and no marks for not attempting a question. If Vaishali scored 58 marks and got 17 correct answers, how many questions she attempted incorrectly? How many questions she did not attempt?

Answer

Marks awarded for 17 correct answers = 17 × 4 = 68 marks.

Let the number of questions attempted incorrectly be x.

Marks deducted for x incorrect answers = 2x.

According to the problem,

⇒ 68 - 2x = 58

⇒ -2x = 58 - 68

⇒ -2x = -10

⇒ x = 5

∴ The number of questions not attempted = Total - correct - incorrect = 25 - 17 - 5 = 3.

Hence, Vaishali attempted 5 questions incorrectly and did not attempt 3 questions.

Divide ₹2,16,000 into two parts such that one-fourth of one part is equal to one-fifth of the other part. Find the two parts.

Answer

Let one part be ₹x, then the other part = ₹(2,16,000 - x).

According to the problem,

$\dfrac{1}{4}x = \dfrac{1}{5}(216000 - x)\\[1em] \Rightarrow 5x = 4(216000 - x)\\[1em] \Rightarrow 5x = 864000 - 4x\\[1em] \Rightarrow 5x + 4x = 864000\\[1em] \Rightarrow 9x = 864000\\[1em] \Rightarrow x = \dfrac{864000}{9}\\[1em] \Rightarrow x = 96000$

∴ One part = ₹96,000 and the other part = ₹(2,16,000 - 96,000) = ₹1,20,000.

Hence, the two parts are ₹96,000 and ₹1,20,000.

If a table is sold for ₹437 at a loss of 8%, find its cost price. At what price must it be sold to gain 10%?

Answer

Let the cost price of the table be ₹x.

As there is a loss of 8%, the selling price = (100 - 8)% of x = 92% of x.

According to the problem,

$\dfrac{92}{100} × x = 437\\[1em] \Rightarrow x = 437 × \dfrac{100}{92}\\[1em] \Rightarrow x = \dfrac{43700}{92}\\[1em] \Rightarrow x = 475$

∴ The cost price of the table = ₹475.

To gain 10%, the selling price = (100 + 10)% of cost price = 110% of ₹475.

$\dfrac{110}{100} × 475 = \dfrac{52250}{100} = 522.50$

Hence, the cost price is ₹475 and it must be sold for ₹522.50 to gain 10%.

Solve the inequality: 3 - 2x ≥ x - 10, x ∈ N

Also represent its solution set on the number line.

Answer

Solving 3 - 2x ≥ x - 10, x ∈ N,

⇒ 3 - 2x ≥ x - 10

⇒ 3 + 10 ≥ x + 2x

⇒ 13 ≥ 3x

⇒ $x \le \dfrac{13}{3}$

As x ∈ N, the natural numbers less than or equal to $\dfrac{13}{3}$ (≈ 4.33) are 1, 2, 3 and 4.

∴ The solution set is {1, 2, 3, 4}.

The solution set is shown by thick dots on the number line.