Model Question Paper 1

Using the order of operations (BODMAS), we carry out multiplication and division (from left to right) before addition.

(-10) × 2 + 0 ÷ (-2)

= -20 + 0 ÷ (-2)

= -20 + 0

= -20.

Hence, Option 1 is the correct option.

The multiplicative inverse of $-\frac{1}{2}$ is $-\frac{2}{1}$ i.e. -2.

$\Rightarrow -\frac{1}{2} + (-2)$

$\Rightarrow \frac{-1}{2} + \frac{-2}{1}$

LCM of 2 and 1 is 2

$$\begin{aligned}\Rightarrow \frac{-1 \times 1}{2 \times 1} + \frac{-2 \times 2}{1 \times 2}\\[1em] \Rightarrow \frac{-1}{2} + \frac{-4}{2}\\[1em] \Rightarrow \frac{-1 + (-4)}{2}\\[1em] \Rightarrow \frac{-5}{2}\\[1em] \Rightarrow -2\frac{1}{2}\end{aligned}$$

Hence, Option 3 is the correct option.

Solving,

(-36) ÷ ((-14) + 2)

= (-36) ÷ (-12)

= 3

Hence, (-36) ÷ ((-14) + 2) = 3.

Length of the rectangle = 8.26 cm

Breadth of the rectangle = 5.5 cm

Area of rectangle = length × breadth

= 8.26 × 5.5

= 45.43 sq. cm

Hence, the area of the rectangle is 45.43 sq. cm.

The denominator of $\frac{105}{-168}$ is negative. To make it positive, multiply its numerator and denominator by -1.

$\frac{105}{-168} = \frac{105 \times (-1)}{(-168) \times (-1)} = \frac{-105}{168}$

HCF of 105 and 168 is 21

So, divide its numerator and denominator by 21

$\frac{-105 \div 21}{168 \div 21} = \frac{-5}{8}$

Hence, the standard form of $\frac{105}{-168}$ is $\frac{-5}{8}$.

Number of questions attempted by Vishal = 17

Number of correct answers = 11

Number of incorrect answers = 17 - 11 = 6

Marks awarded for every correct answer = 5

Marks deducted for every incorrect answer = 2

Vishal's score = $(11 \times 5) + (6 \times (-2))$

= 55 + (-12)

= 55 - 12

= 43.

Hence, Vishal's score is 43.

Quantity of rice bought by Sara = $20\frac{3}{8}$ kg = $\frac{163}{8}$ kg

Rate of rice = ₹$52\frac{1}{2}$ per kg = ₹$\frac{105}{2}$ per kg

$$\begin{aligned}\text{Amount spent by Sara} = \frac{163}{8} \times \frac{105}{2}\\[1em] = \frac{163 \times 105}{8 \times 2}\\[1em] = \frac{17115}{16}\\[1em] = 1069\frac{11}{16}\end{aligned}$$

Hence, the amount spent by Sara is ₹$1069\frac{11}{16}$.

The given rational numbers are $-5\frac{5}{9}$ and $-5\frac{7}{12}$.

$-5\frac{5}{9} = -\frac{50}{9}$ and

$-5\frac{7}{12} = -\frac{67}{12}$

LCM of 9 and 12 is 2 × 2 × 3 × 3 = 36

$$\begin{aligned}-\frac{50}{9} = -\frac{50 \times 4}{9 \times 4} = -\frac{200}{36}\\[1em] -\frac{67}{12} = -\frac{67 \times 3}{12 \times 3} = -\frac{201}{36}\end{aligned}$$

Their numerators are -200 and -201

As -200 > -201, so $-\frac{200}{36} \gt -\frac{201}{36}$

\Rightarrow -5\dfrac{5}{9} &ampgt -5\dfrac{7}{12}

Hence, $-5\frac{5}{9}$ is the greater rational number.

Let us denote the distance walked towards north by positive sign, then the distance walked towards south will be denoted by minus sign.

$$\begin{aligned}\text{Simran's final distance from A} = 1\frac{5}{12} + \Big(-2\frac{7}{9}\Big)\\[1em] = \frac{17}{12} + \Big(-\frac{25}{9}\Big)\\[1em] = \frac{17}{12} - \frac{25}{9}\end{aligned}$$

LCM of 12 and 9 is 2 × 2 × 3 × 3 = 36

$$\begin{aligned}= \frac{17 \times 3}{12 \times 3} - \frac{25 \times 4}{9 \times 4}\\[1em] = \frac{51}{36} - \frac{100}{36}\\[1em] = \frac{51 - 100}{36}\\[1em] = \frac{-49}{36}\\[1em] = -1\frac{13}{36}\end{aligned}$$

As the final distance is negative, therefore Simran is at a distance of $1\frac{13}{36}$ km towards south from place A.

Hence, Simran is at a distance of $1\frac{13}{36}$ km towards south from place A.

Product of two decimal numbers = 17.55

One of the numbers = 2.7

$$\begin{aligned}\text{Other number} = 17.55 ÷ 2.7\\[1em] = \frac{17.55}{2.7}\\[1em] = \frac{175.5}{27}\\[1em] = 6.5\end{aligned}$$

Hence, the other number is 6.5.