CBSE Class 7 Mathematics Practice Worksheet Set 11

Question 1. ______________ is the most common representative value of a group of data.
Answer: Arithmetic mean (or Mean / Average)
In simple words: The mean is the most common number used to show the central value of a data set.

Exam Tip: Remember that arithmetic mean is found by dividing the sum of all observations by the total count.

Question 2. The difference between the highest and lowest observation is called ______________.
Answer: Range
In simple words: Range tells you how spread out the numbers are from the smallest to the largest.

Exam Tip: Always subtract the minimum value from the maximum value to find the range.

Question 3. The observation that occurs most often is called ______________.
Answer: Mode
In simple words: Mode is the number that shows up the most times in a list.

Exam Tip: A data set can have more than one mode if two or more values share the highest frequency.

Question 4. ______________ refers the value which lies in the middle of the data.
Answer: Median
In simple words: Median is the exact middle score after putting all numbers in order.

Exam Tip: Never calculate the median before arranging the numbers in ascending or descending order first.

Question 5. Find the mean of : -
a. First 5 mean even numbers
b. Odd Prime numbers below 10
c. First 5 prime numbers
Answer:
a. The first 5 positive even numbers are 2, 4, 6, 8, and 10.
\[ \text{Mean} = \frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6 \]

b. The odd prime numbers less than 10 are 3, 5, and 7.
\[ \text{Mean} = \frac{3 + 5 + 7}{3} = \frac{15}{3} = 5 \]

c. The first 5 prime numbers are 2, 3, 5, 7, and 11.
\[ \text{Mean} = \frac{2 + 3 + 5 + 7 + 11}{5} = \frac{28}{5} = 5.6 \]
In simple words: Add all the numbers in each group and divide by how many numbers there are.

Exam Tip: Remember that 2 is an even number, so part (b) excludes 2 because it asks only for odd prime numbers.

Question 6. Find the mean, median and mode of the following data.
35, 32, 35, 42, 38, 32, 31
Answer:
Given data: 35, 32, 35, 42, 38, 32, 31

1. Mean:
Total number of observations = 7
\[ \text{Sum} = 35 + 32 + 35 + 42 + 38 + 32 + 31 = 245 \]
\[ \text{Mean} = \frac{245}{7} = 35 \]

2. Median:
Arrange the data in ascending order:
31, 32, 32, 35, 35, 38, 42
Since \( n = 7 \) (odd), the median is the \( \left(\frac{7 + 1}{2}\right)^{\text{th}} = 4^{\text{th}} \) observation.
\[ \text{Median} = 35 \]

3. Mode:
Both 32 and 35 repeat twice, which is the highest frequency.
\[ \text{Mode} = 32 \text{ and } 35 \]
In simple words: The mean is 35, the middle number is 35, and both 32 and 35 appear most often.

Exam Tip: When two values appear with the same maximum frequency, state both as modes.

Question 7. Following are the number of members in 20 families in a village.
6, 8, 6, 3, 2, 5, 7, 8, 6, 5, 5, 7, 7, 8, 6, 6, 7, 7, 6, 5
Organize the data in a tabular form.
i) What is the smallest family size?
ii) What is the most common family size ?
Answer:
Frequency distribution table for the given data:

i) The smallest family size is 2.
ii) The most common family size is 6 (it has the highest frequency of 6 families).
In simple words: The table groups the family sizes. The smallest family has 2 members, and a size of 6 members occurs the most.

Exam Tip: Verify that the total of your frequency column matches the total number of families given in the question (20).

Question 8. When a die is thrown, what is the probability of getting
a. an even number
b. an odd number
c. zero
d. 5
Answer:
A standard die has 6 possible outcomes: \( \{1, 2, 3, 4, 5, 6\} \).
Total number of outcomes = 6.

a. Favourable outcomes for an even number are \( \{2, 4, 6\} \) (3 outcomes):
\[ P(\text{even number}) = \frac{3}{6} = \frac{1}{2} \]

b. Favourable outcomes for an odd number are \( \{1, 3, 5\} \) (3 outcomes):
\[ P(\text{odd number}) = \frac{3}{6} = \frac{1}{2} \]

c. A standard die has no face with 0 (0 outcomes):
\[ P(\text{zero}) = \frac{0}{6} = 0 \]

d. Favourable outcome for getting 5 is \( \{5\} \) (1 outcome):
\[ P(5) = \frac{1}{6} \]
In simple words: Divide the number of chances for each event by 6, which is the total count of faces on a die.

Exam Tip: An impossible event has a probability of 0, so getting a zero on a standard die is simply 0.

Question 9. A man with a monthly salary of Rs 6400 plans his budget for a month as given below.

Represent the above data by a bar graph.
Answer:
To draw the bar graph, take items on the horizontal axis (X-axis) and amount in Rs. on the vertical axis (Y-axis). Choose a scale of 1 unit length = Rs. 300 (or 1 cm = Rs. 300). In simple words: The bar graph shows five bars. The tallest bar is for food, and the shortest bar is for clothing.

Exam Tip: Always keep equal gaps between the bars and maintain a uniform width for each bar.

Question 10. Given is the data of school going students (boy and girls)

Draw a double bar graph to represent the above data.
Answer:
To draw the double bar graph, place the modes of transport on the X-axis and student count on the Y-axis. Choose a scale of 1 unit length = 30 students. In simple words: The chart places two bars side by side for each travel type, one for boys and one for girls.

Exam Tip: Always include a legend indicating which bar represents boys and which represents girls.

Ch. 4: Simple Equations

Question 1. Solve the following equations
a. \frac{x}{3} - 5 = 2
b. \frac{y}{2} - 3 = 8
c. \frac{4a}{3} - 1 = -9
d. 3y -2 = 1
e. 5 + 4 (P-1) = 29
f. 44 – 5 (P-1) = 4
g. 7 = 5 (P– 2)
h. 12 = 4 + 3 (t + 3)
Answer:
a. \( \frac{x}{3} - 5 = 2 \)
Transposing \( -5 \) to RHS:
\( \frac{x}{3} = 2 + 5 = 7 \)
Multiplying both sides by 3:
\( x = 7 \times 3 = 21 \)

b. \( \frac{y}{2} - 3 = 8 \)
Transposing \( -3 \) to RHS:
\( \frac{y}{2} = 8 + 3 = 11 \)
Multiplying both sides by 2:
\( y = 11 \times 2 = 22 \)

c. \( \frac{4a}{3} - 1 = -9 \)
Transposing \( -1 \) to RHS:
\( \frac{4a}{3} = -9 + 1 = -8 \)
Multiplying by 3:
\( 4a = -8 \times 3 = -24 \)
Dividing by 4:
\( a = \frac{-24}{4} = -6 \)

d. \( 3y - 2 = 1 \)
Transposing \( -2 \) to RHS:
\( 3y = 1 + 2 = 3 \)
Dividing by 3:
\( y = \frac{3}{3} = 1 \)

e. \( 5 + 4(P - 1) = 29 \)
Transposing 5 to RHS:
\( 4(P - 1) = 29 - 5 = 24 \)
Dividing by 4:
\( P - 1 = \frac{24}{4} = 6 \)
\( P = 6 + 1 = 7 \)

f. \( 44 - 5(P - 1) = 4 \)
Transposing 44 to RHS:
\( -5(P - 1) = 4 - 44 = -40 \)
Dividing by \( -5 \):
\( P - 1 = \frac{-40}{-5} = 8 \)
\( P = 8 + 1 = 9 \)

g. \( 7 = 5(P - 2) \)
Expanding brackets:
\( 7 = 5P - 10 \)
Transposing \( -10 \) to LHS:
\( 5P = 7 + 10 = 17 \)
\( P = \frac{17}{5} = 3.4 \)

h. \( 12 = 4 + 3(t + 3) \)
Transposing 4 to LHS:
\( 12 - 4 = 3(t + 3) \)
\( 8 = 3t + 9 \)
Transposing 9 to LHS:
\( 3t = 8 - 9 = -1 \)
\( t = -\frac{1}{3} \)
In simple words: Move the numbers without letters to the other side first. Then divide to leave the variable completely on its own.

Exam Tip: Change a plus to a minus and a minus to a plus whenever moving a term across the equals sign.

Question 2. Construct 4 equations starting with
a. x = 3
b. x = -3
c. x = 1
d. x = -1
Answer:
a. Starting with \( x = 3 \):
1. Multiply both sides by 2:
\( 2x = 6 \)
2. Add 5 to both sides:
\( x + 5 = 8 \)
3. Multiply by 3 and subtract 2:
\( 3x - 2 = 7 \)
4. Multiply by 4 and add 1:
\( 4x + 1 = 13 \)

b. Starting with \( x = -3 \):
1. Multiply both sides by 2:
\( 2x = -6 \)
2. Add 3 to both sides:
\( x + 3 = 0 \)
3. Multiply by 5 and add 4:
\( 5x + 4 = -11 \)
4. Subtract 2 from both sides:
\( x - 2 = -5 \)

c. Starting with \( x = 1 \):
1. Multiply both sides by 3:
\( 3x = 3 \)
2. Add 7 to both sides:
\( x + 7 = 8 \)
3. Multiply by 2 and subtract 1:
\( 2x - 1 = 1 \)
4. Multiply by 5 and add 3:
\( 5x + 3 = 8 \)

d. Starting with \( x = -1 \):
1. Multiply both sides by 4:
\( 4x = -4 \)
2. Add 1 to both sides:
\( x + 1 = 0 \)
3. Multiply by 3 and add 5:
\( 3x + 5 = 2 \)
4. Subtract 6 from both sides:
\( x - 6 = -7 \)
In simple words: Do the same calculation to both sides of the equal sign to make brand new equations.

Exam Tip: As long as you perform identical operations on both the left and right sides, your constructed equation is valid.

Question 3. If 5 is subtracted from a number and the result is multiplied by 10 the answer is 20. Find the number.
Answer:
Let the unknown number be \( x \).
Subtracting 5 from the number gives \( (x - 5) \).
Multiplying this result by 10 gives \( 10(x - 5) \).

According to the question:
\[ 10(x - 5) = 20 \]
Divide both sides by 10:
\[ x - 5 = \frac{20}{10} = 2 \]
Add 5 to both sides:
\[ x = 2 + 5 = 7 \]
Hence, the number is 7.
In simple words: Set up the equation ten times x minus five equals twenty. Solve it to find the number is 7.

Exam Tip: Put \( (x - 5) \) inside brackets so the whole difference is multiplied by 10.

Question 4. Raju has some money. If he gives on fifth of it to Raheem he is still left with ₹ 20. How much had he at first?
Answer:
Let the total money Raju had at first be Rs. \( m \).
Share given to Raheem = \( \frac{1}{5}m \)
Money remaining with Raju:
\[ m - \frac{1}{5}m = \frac{4}{5}m \]

According to the given condition:
\[ \frac{4}{5}m = 20 \]
Multiply by 5:
\[ 4m = 20 \times 5 = 100 \]
Divide by 4:
\[ m = \frac{100}{4} = 25 \]
Therefore, Raju had Rs. 25 at first.
In simple words: Raju keeps four parts out of five, which equals 20 rupees. This means he started with 25 rupees.

Exam Tip: If \( \frac{1}{5} \) is given away, the remaining fraction is always \( 1 - \frac{1}{5} = \frac{4}{5} \).

Question 5. Maya says that she has 9 marbles more than 5 times the marble Roopa has. Maya has 49 marbles. How many marbles does Roopa has?
Answer:
Let the count of marbles Roopa has be \( r \).
Five times Roopa's marbles = \( 5r \)
Maya has 9 marbles more than this, which gives \( 5r + 9 \).

According to the question:
\[ 5r + 9 = 49 \]
Transposing 9 to RHS:
\[ 5r = 49 - 9 = 40 \]
Dividing by 5:
\[ r = \frac{40}{5} = 8 \]
Hence, Roopa has 8 marbles.
In simple words: Take 9 away from Maya's 49 marbles to get 40. Then divide 40 by 5 to find Roopa has 8 marbles.

Exam Tip: Define the unknown quantity clearly with a variable before building your linear equation.

Question 6. The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 97. What is the lowest score?
Answer:
Let the lowest score be \( l \).
Twice the lowest marks plus 7 gives \( 2l + 7 \).

According to the question:
\[ 2l + 7 = 97 \]
Transposing 7 to RHS:
\[ 2l = 97 - 7 = 90 \]
Dividing by 2:
\[ l = \frac{90}{2} = 45 \]
Therefore, the lowest score is 45.
In simple words: Subtract 7 from 97 to get 90. Divide 90 by 2 to find the lowest mark, which is 45.

Exam Tip: Check your final result: \( 2 \times 45 + 7 = 90 + 7 = 97 \), which confirms the answer is correct.

Question 7. Length of a rectangle is twice its breadth. If perimeter is 54 cm. Find the length and breadth.
Answer:
Let the breadth of the rectangle be \( b\text{ cm} \).
Then the length is \( l = 2b\text{ cm} \).

The formula for the perimeter of a rectangle is:
\[ \text{Perimeter} = 2(l + b) \]
Given perimeter = \( 54\text{ cm} \):
\[ 2(2b + b) = 54 \]
\[ 2(3b) = 54 \]
\[ 6b = 54 \]
Dividing by 6:
\[ b = \frac{54}{6} = 9\text{ cm} \]

Now, find the length:
\[ l = 2 \times 9 = 18\text{ cm} \]
Therefore, the length is \( 18\text{ cm} \) and the breadth is \( 9\text{ cm} \).
In simple words: Six times the breadth equals 54, so the breadth is 9 cm. Double 9 gives the length of 18 cm.

Exam Tip: Include the correct unit (cm) with both the length and breadth values in your final conclusion.