Class 7 Maths Worksheet on Integers - Word Problems and Calculations.

Let's solve each problem step by step.

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Problem 1:

Shyak has overdrawn his checking account by Rs.38. The bank debited him Rs.20 for an overdraft fee. Later, he deposited Rs.150. What is his current balance?

Solution:
1. Initial overdrawn amount: \(-38\) (negative because it's a debt).
2. Overdraft fee: \(-20\) (additional debit).
3. Deposit: \(+150\) (credit).

Current balance = Initial overdrawn amount + Overdraft fee + Deposit
\[
-38 + (-20) + 150 = -38 - 20 + 150 = -58 + 150 = 92
\]

Answer: \(\boxed{92}\)

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Problem 2:

In a school exam, Rohan’s score was up to 500 points. Then he scored \(-100\) points in the words category. What was his score then?

Solution:
1. Initial score: \(500\).
2. Score in the words category: \(-100\) (deduction).

Final score = Initial score + Score in the words category
\[
500 + (-100) = 500 - 100 = 400
\]

Answer: \(\boxed{400}\)

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Problem 3:

The melting point of mercury is \(-39^\circ \text{C}\). The freezing point of alcohol is \(-98^\circ \text{C}\). How much more is the melting point of mercury as compared to the freezing point of alcohol?

Solution:
1. Melting point of mercury: \(-39^\circ \text{C}\).
2. Freezing point of alcohol: \(-98^\circ \text{C}\).

Difference = Melting point of mercury - Freezing point of alcohol
\[
-39 - (-98) = -39 + 98 = 59
\]

Answer: \(\boxed{59}\)

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Problem 4:

An elevator descends into a mine shaft at the rate of \(6 \, \text{m/min}\). If the descent starts from \(10 \, \text{m}\) above the ground level, how long will it take to reach \(-350 \, \text{m}\)?

Solution:
1. Starting position: \(10 \, \text{m}\) above ground.
2. Final position: \(-350 \, \text{m}\) below ground.
3. Total distance to descend = Final position - Starting position
\[
-350 - 10 = -360 \, \text{m}
\]
(The negative sign indicates the direction of descent.)

4. Rate of descent: \(6 \, \text{m/min}\).

Time taken = Total distance to descend ÷ Rate of descent
\[
\frac{-360}{-6} = 60 \, \text{minutes}
\]

Convert minutes to hours:
\[
60 \, \text{minutes} = 1 \, \text{hour}
\]

Answer: \(\boxed{1}\)

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Problem 5:

Find the product with suitable properties for the following:
\[ 16 \times (-34) + (-34) \times (-18) \]

Solution:
Use the distributive property:
\[
a \times b + a \times c = a \times (b + c)
\]
Here, \(a = -34\), \(b = 16\), and \(c = -18\).

\[
16 \times (-34) + (-34) \times (-18) = (-34) \times (16 + (-18))
\]

Simplify inside the parentheses:
\[
16 + (-18) = 16 - 18 = -2
\]

Now multiply:
\[
(-34) \times (-2) = 68
\]

Answer: \(\boxed{68}\)

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Problem 6:

\[ 23 \times -36 \times 10 \]

Solution:
Multiply the numbers step by step:
\[
23 \times -36 = -828
\]
\[
-828 \times 10 = -8280
\]

Answer: \(\boxed{-8280}\)

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Problem 7:

\[ 296 \div \_\_ = -148 \]

Solution:
Let the missing number be \(x\). Then:
\[
296 \div x = -148
\]

Rewrite as:
\[
\frac{296}{x} = -148
\]

Solve for \(x\):
\[
x = \frac{296}{-148} = -2
\]

Answer: \(\boxed{-2}\)

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Problem 8:

\[ -88 \div \_\_ = 11 \]

Solution:
Let the missing number be \(x\). Then:
\[
-88 \div x = 11
\]

Rewrite as:
\[
\frac{-88}{x} = 11
\]

Solve for \(x\):
\[
x = \frac{-88}{11} = -8
\]

Answer: \(\boxed{-8}\)

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Problem 9:

\[ \_\_ \div 156 = -2 \]

Solution:
Let the missing number be \(x\). Then:
\[
x \div 156 = -2
\]

Rewrite as:
\[
\frac{x}{156} = -2
\]

Solve for \(x\):
\[
x = -2 \times 156 = -312
\]

Answer: \(\boxed{-312}\)

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Problem 10:

\[ \_\_ \div 567 = -1 \]

Solution:
Let the missing number be \(x\). Then:
\[
x \div 567 = -1
\]

Rewrite as:
\[
\frac{x}{567} = -1
\]

Solve for \(x\):
\[
x = -1 \times 567 = -567
\]

Answer: \(\boxed{-567}\)

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Final Answers:

1. \(\boxed{92}\)
2. \(\boxed{400}\)
3. \(\boxed{59}\)
4. \(\boxed{1}\)
5. \(\boxed{68}\)
6. \(\boxed{-8280}\)
7. \(\boxed{-2}\)
8. \(\boxed{-8}\)
9. \(\boxed{-312}\)
10. \(\boxed{-567}\)