Page 3 of the Learners' Planet algebra worksheet featuring linear equation word problems.

Let's solve each problem step by step.

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Problem 11

Question: A number \( x \) when added to 43 gives 352. What is \( x \)?

#### Solution:
The equation can be written as:
\[
x + 43 = 352
\]

To solve for \( x \), subtract 43 from both sides:
\[
x = 352 - 43
\]
\[
x = 309
\]

Thus, the correct answer is:
\[
\boxed{A}
\]

---

Problem 12

Question: 73 exceeds a number \( x \) by 23. What is \( x \)?

#### Solution:
The phrase "73 exceeds \( x \) by 23" means:
\[
73 = x + 23
\]

To solve for \( x \), subtract 23 from both sides:
\[
x = 73 - 23
\]
\[
x = 50
\]

Thus, the correct answer is:
\[
\boxed{D}
\]

---

Problem 13

Question: Hari’s father gave him 70 rupees. Now he has 130 rupees. How much money did Hari have in the beginning? Let the amount of money in the beginning be \( x \).

#### Solution:
If Hari had \( x \) rupees initially and his father gave him 70 rupees, then:
\[
x + 70 = 130
\]

To solve for \( x \), subtract 70 from both sides:
\[
x = 130 - 70
\]
\[
x = 60
\]

Thus, the correct answer is:
\[
\boxed{A}
\]

---

Problem 14

Question: A number \( x \) is twice another number \( y \). If their sum is 96, what are the numbers \( y \) and \( x \) respectively?

#### Solution:
We are given two conditions:
1. \( x = 2y \)
2. \( x + y = 96 \)

Substitute \( x = 2y \) into the second equation:
\[
2y + y = 96
\]
\[
3y = 96
\]

Solve for \( y \):
\[
y = \frac{96}{3}
\]
\[
y = 32
\]

Now, find \( x \):
\[
x = 2y = 2 \times 32 = 64
\]

Thus, the numbers are \( y = 32 \) and \( x = 64 \). The correct answer is:
\[
\boxed{C}
\]

---

Problem 15

Question: The difference between two numbers is 18. If their sum is 86, what are the numbers?

#### Solution:
Let the two numbers be \( a \) and \( b \). We are given:
1. \( a - b = 18 \)
2. \( a + b = 86 \)

Add the two equations:
\[
(a - b) + (a + b) = 18 + 86
\]
\[
2a = 104
\]

Solve for \( a \):
\[
a = \frac{104}{2}
\]
\[
a = 52
\]

Substitute \( a = 52 \) into the second equation:
\[
52 + b = 86
\]
\[
b = 86 - 52
\]
\[
b = 34
\]

Thus, the numbers are \( a = 52 \) and \( b = 34 \). The correct answer is:
\[
\boxed{B}
\]

---

Problem 16

Question: Divide 72 into two parts so that the larger part exceeds the smaller part by 12. Find both the parts.

#### Solution:
Let the smaller part be \( x \). Then the larger part is \( x + 12 \). The sum of the two parts is 72:
\[
x + (x + 12) = 72
\]
\[
2x + 12 = 72
\]

Subtract 12 from both sides:
\[
2x = 60
\]

Solve for \( x \):
\[
x = \frac{60}{2}
\]
\[
x = 30
\]

The smaller part is \( x = 30 \), and the larger part is:
\[
x + 12 = 30 + 12 = 42
\]

Thus, the parts are 30 and 42. The correct answer is:
\[
\boxed{C}
\]

---

Problem 17

Question: When a number is multiplied by 4 and then diminished by 7, the result is 65. What is the number?

#### Solution:
Let the number be \( x \). According to the problem:
\[
4x - 7 = 65
\]

Add 7 to both sides:
\[
4x = 72
\]

Solve for \( x \):
\[
x = \frac{72}{4}
\]
\[
x = 18
\]

Thus, the number is:
\[
\boxed{C}
\]

---

Problem 18

Question: In 4 years' time, a baby will be 5 times as old as she is now. Find the age of the baby.

#### Solution:
Let the current age of the baby be \( x \). In 4 years, her age will be \( x + 4 \). According to the problem:
\[
x + 4 = 5x
\]

Subtract \( x \) from both sides:
\[
4 = 4x
\]

Solve for \( x \):
\[
x = \frac{4}{4}
\]
\[
x = 1
\]

Thus, the current age of the baby is:
\[
\boxed{B}
\]

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Final Answers:

11. \(\boxed{A}\)
12. \(\boxed{D}\)
13. \(\boxed{A}\)
14. \(\boxed{C}\)
15. \(\boxed{B}\)
16. \(\boxed{C}\)
17. \(\boxed{C}\)
18. \(\boxed{B}\)