Class VII Mathematics – Periodic Test-I Solutions

Solution:

Following the order of operations (BODMAS), we first solve the expression inside the brackets:

$$(-1 + 37) = 36$$

Now, substitute this back into the expression:

$$36 \div (-12) + (-3)$$

Next, perform the division:

$$36 \div (-12) = -3$$

Finally, perform the addition:

$$-3 + (-3) = -3 – 3 = -6$$

Correct Option: (b) (-6)

Solution:

First, evaluate the terms inside the parentheses and brackets:

Numerator/First term: $$(-10 – 10) = -20$$

Denominator/Second term: $$[10 + (-10 – 10)] = [10 + (-20)] = 10 – 20 = -10$$

Now, divide the first term by the second term:

$$\frac{-20}{-10} = 2$$

Correct Option: (b) 2

Solution:

According to the left-to-right rule for multiplication and division (BODMAS/PEMDAS), we perform the division first:

$$24 \div (-6) = -4$$

Now, multiply the result by 4:

$$-4 \times 4 = -16$$

Correct Option: (a) -16

Solution:

The additive inverse of any number or expression $x$ is $-x$.

Therefore, the additive inverse of $(-7 – 2b)$ is:

$$-(-7 – 2b)$$

Multiplying the negative sign inside the bracket yields:

$$7 + 2b$$

Correct Option: (c) $(7+2b)$

Solution:

Let the integer be $x$. If it is its own additive inverse, then:

$$x = -x$$

$$x + x = 0$$

$$2x = 0 \implies x = 0$$

Zero is the only integer that is its own additive inverse.

Correct Option: (b) Zero

Solution:

(a.) $0 \div (-12) = 0$ (Zero divided by any non-zero integer is zero)

(b.) $(-49) \div (49) = -1$

(c.) $[(-6) + 5] \div [(-2) + 1]$
Solving the brackets: $(-1) \div (-1) = 1$

(d.) $(-35) \times 0 \times (-448) = 0$ (Any integer multiplied by zero is zero)

Solution:

(a) $625 \times (-35) + (-625)$

We can rewrite $(-625)$ as $625 \times (-1)$. Substituting this into the expression:

$$= 625 \times (-35) + 625 \times (-1)$$

By using the distributive property of multiplication over addition $[a \times b + a \times c = a \times (b + c)]$, we can take $625$ as common:

$$= 625 \times [(-35) + (-1)]$$

$$= 625 \times (-36)$$

$$= -22500$$

(b) $15 \times (-25) \times (-4) \times (-10)$

Using the commutative and associative properties of multiplication, we can group the terms to make the calculation easier:

$$= 15 \times [(-25) \times (-4)] \times (-10)$$

$$= 15 \times [100] \times (-10)$$

$$= 1500 \times (-10)$$

$$= -15000$$

Solution:

Initial room temperature $= 40^\circ\text{C}$

Rate of change in temperature $= -5^\circ\text{C}$ per hour (negative because it is lowering)

Total change in temperature over 10 hours $= 10 \times (-5^\circ\text{C}) = -50^\circ\text{C}$

Final room temperature $= \text{Initial temperature} + \text{Total change}$

Final room temperature $= 40^\circ\text{C} + (-50^\circ\text{C}) = -10^\circ\text{C}$

Therefore, the room temperature will be $-10^\circ\text{C}$.

Solution:

(a.) Let the negative integer be $x$ and the positive integer be $y$. We need $x – y = -5$.

If we choose $x = -2$ and $y = 3$ (where $-2$ is negative and $3$ is positive):

Difference $= (-2) – (3) = -5$.

Hence, one such pair is $-2$ and $3$.

(b.) Let the two negative numbers be $a$ and $b$. We need $a – b = 28$.

This can be written as $a = b + 28$.

Let us choose $b = -30$ (a negative integer). Then:

$a = -30 + 28 = -2$ (which is also a negative integer).

Let’s verify the difference: $(-2) – (-30) = -2 + 30 = 28$.

Hence, one such pair is $-2$ and $-30$.

Solution:

Initial temperature at 12 noon $= +10^\circ\text{C}$

Rate of decrease $= 2^\circ\text{C}$ per hour

Part 1: Time when temperature is $-8^\circ\text{C}$ ($8^\circ\text{C}$ below zero)

Total difference in temperature required $= (\text{Final Temperature}) – (\text{Initial Temperature})$

Difference $= (-8^\circ\text{C}) – (10^\circ\text{C}) = -18^\circ\text{C}$

Since the temperature decreases by $2^\circ\text{C}$ every hour, the time required will be:

$$\text{Time taken} = \frac{-18}{-2} = 9 \text{ hours}$$

9 hours from 12 noon is 9:00 PM.

Part 2: Temperature at midnight

Midnight is exactly 12 hours after 12 noon.

Total decrease in temperature in 12 hours $= 12 \times (-2^\circ\text{C}) = -24^\circ\text{C}$

Temperature at midnight $= 10^\circ\text{C} + (-24^\circ\text{C}) = -14^\circ\text{C}$

Therefore, the temperature will be $-8^\circ\text{C}$ at 9:00 PM, and the temperature at midnight will be $-14^\circ\text{C}$.

Solution:

Let profit be represented by positive values and loss by negative values.

Profit on 1 pen $= \text{₹ } 1$

Loss on 1 pencil $= 40 \text{ paise} = \text{₹ } 0.40$ (represented as $-0.40$)

(a.) Net loss in the month $= \text{₹ } 5$ (represented as $-5$)

Number of pens sold $= 45$. Total profit from pens $= 45 \times 1 = \text{₹ } 45$

Let the number of pencils sold be $x$. Total loss from pencils $= -0.40 \times x$

$$\text{Total Profit/Loss} = \text{Profit from pens} + \text{Loss from pencils}$$

$$-5 = 45 + (-0.40x)$$

$$-0.40x = -5 – 45$$

$$-0.40x = -50$$

$$x = \frac{-50}{-0.40} = \frac{500}{4} = 125$$

He sold 125 pencils in that month.

(b.) In the next month, he earns neither profit nor loss. Therefore, Net Profit $= 0$.

Number of pens sold $= 70$. Total profit from pens $= 70 \times 1 = \text{₹ } 70$

Let the number of pencils sold be $y$. Total loss from pencils $= -0.40 \times y$

$$0 = 70 + (-0.40y)$$

$$0.40y = 70$$

$$y = \frac{70}{0.40} = \frac{700}{4} = 175$$

He sold 175 pencils in the next month.

Solution:

Marks awarded for 1 correct answer $= +5$

Marks awarded for 1 incorrect answer $= -2$

Marks awarded for an unattempted question $= 0$

(a.) Mohan’s Score:

Correct answers $= 4$. Marks from correct answers $= 4 \times 5 = 20$

Incorrect answers $= 6$. Marks from incorrect answers $= 6 \times (-2) = -12$

Total Score $= 20 + (-12) = 20 – 12 = 8$

Mohan’s total score is 8.

(c.) Heena’s Score:

Correct answers $= 2$. Marks from correct answers $= 2 \times 5 = 10$

Incorrect answers $= 5$. Marks from incorrect answers $= 5 \times (-2) = -10$

Unattempted questions $= 10 – (2 + 5) = 10 – 7 = 3$. Marks $= 3 \times 0 = 0$

Total Score $= 10 + (-10) + 0 = 0$

Heena’s total score is 0.