CBSE Class 7 Mathematics Practice Worksheet Set 08

Maths

Note:
(1) Students should do the assignment questions themselves.
(2) Students will do the assignment on ruled sheets and submit in file.
(3) Draw diagrams wherever required using pencil and scale.

Question 1. I am equivalent to 2/5. The product of my numerator and denominator is 40. What fraction am I.
Answer: Let the required fraction be \( \frac{2k}{5k} \), where \( k \) is a common positive multiplier.
Numerator = \( 2k \)
Denominator = \( 5k \)
According to the question, the product of the numerator and denominator is 40:
\( 2k \times 5k = 40 \)
\( 10k^2 = 40 \)

\( \implies k^2 = \frac{40}{10} = 4 \)

\( \implies k = 2 \)
Now, find the numerator and denominator by putting \( k = 2 \):
Numerator = \( 2 \times 2 = 4 \)
Denominator = \( 5 \times 2 = 10 \)
Therefore, the required fraction is \( \frac{4}{10} \).
In simple words: The fraction is 4/10. It simplifies to 2/5, and multiplying 4 by 10 gives 40.

Exam Tip: Set the unknown fraction as \( \frac{2k}{5k} \) to keep the ratio equal to \( \frac{2}{5} \) while solving for the multiplier \( k \).

Question 2. Take two integers (both should not be positive) show their addition, subtraction and multiplication on number line.
Answer: Let us select two integers: \( -3 \) (negative) and \( 2 \) (positive). Since both are not positive, this satisfies the given condition.
1. Addition: \( (-3) + 2 = -1 \)
Start at 0, move 3 steps to the left to reach \( -3 \). Then, move 2 steps to the right to land on \( -1 \).2. Subtraction: \( (-3) - 2 = -5 \)
Start at \( -3 \) and move 2 steps to the left to reach \( -5 \).3. Multiplication: \( 2 \times (-3) = -6 \)
Multiplication means repeated addition. Take 2 jumps of 3 units to the left starting from 0 to reach \( -6 \).In simple words: Moving right adds numbers, while moving left subtracts or multiplies negative numbers on the number line.

Exam Tip: Remember to draw arrows on the jumps to show whether you are moving left or right.

Question 3. Select four alphabets from English alphabets whose order of rotational symmetry is two. Explain the rotational symmetry of the alphabets with the help of diagram.
Answer: Four English capital letters that have an order of rotational symmetry equal to 2 are H, I, N, and S.
A letter has rotational symmetry of order 2 if it looks exactly the same in two positions during one complete turn of \( 360^\circ \) (at \( 180^\circ \) and at \( 360^\circ \)).
1. Letter H: Rotating H by \( 180^\circ \) gives back the letter H. After another \( 180^\circ \) (total \( 360^\circ \)), it returns to its original look.
2. Letter I: Turning I by half a turn (\( 180^\circ \)) keeps it identical. A full turn (\( 360^\circ \)) also matches the start.
3. Letter N: After rotating N by \( 180^\circ \), the diagonal connects the parallel bars the same way. A \( 360^\circ \) turn returns to the original shape.
4. Letter S: An upside-down turn of \( 180^\circ \) produces the letter S again, and another \( 180^\circ \) completes the full circle.In simple words: Turning these letters upside down by 180 degrees leaves them looking identical to how they started.

Exam Tip: State the angle of rotation (\( 180^\circ \)) and the number of matching positions (2) clearly in your answer.

Question 4. Place positive and negative numbers in the boxes, so that the total of each row and column is O.
Answer: We can place positive and negative numbers into the 8 boxes so that the sum of every row and every column equals 0.
One correct arrangement is:
- Top row: \( 1 \) and \( -1 \) (Sum = \( 1 + (-1) = 0 \))
- Middle row: \( -2 \), \( -3 \), \( 3 \), and \( 2 \) (Sum = \( -2 + (-3) + 3 + 2 = 0 \))
- Bottom row: \( 2 \) and \( -2 \) (Sum = \( 2 + (-2) = 0 \))
Checking the vertical columns:
- Left central column: \( 1 + (-3) + 2 = 0 \)
- Right central column: \( -1 + 3 + (-2) = 0 \)
- Left outer box and right outer box balance each other across the middle row: \( -2 + 2 = 0 \).In simple words: Pair positive numbers with matching negative numbers so that adding up any line gives zero.

Exam Tip: Use opposite pairs like 1 and -1, or 2 and -2, so that row and column sums cancel out to 0 easily.

Question 5. Solve the puzzle given below by solving the given equation. Fill one digit in 1 box.
Puzzle

Down
1. \( 2x = 22 \)
3. \( 2x = 16 + x \)
6. \( 4(x - 3) = 48 \)
7. \( \frac{5x - 20}{2} = 50 \)
Left to right
1. \( 2x + 20 = 40 \)
2. \( 3x + 25 = 100 \)
5. \( 5y - 9 = 36 \)
6. \( \frac{8x}{2} = 48 \)
7. \( \frac{48}{y} = 2 \)
Answer:
Step 1: Solve the equations across (Left to right):
1. \( 2x + 20 = 40 \implies 2x = 20 \implies x = 10 \)
2. \( 3x + 25 = 100 \implies 3x = 75 \implies x = 25 \)
5. \( 5y - 9 = 36 \implies 5y = 45 \implies y = 9 \)
6. \( \frac{8x}{2} = 48 \implies 4x = 48 \implies x = 12 \)
7. \( \frac{48}{y} = 2 \implies 2y = 48 \implies y = 24 \)
Step 2: Solve the equations down:
1. \( 2x = 22 \implies x = 11 \)
3. \( 2x = 16 + x \implies 2x - x = 16 \implies x = 16 \)
6. \( 4(x - 3) = 48 \implies x - 3 = 12 \implies x = 15 \)
7. \( \frac{5x - 20}{2} = 50 \implies 5x - 20 = 100 \implies 5x = 120 \implies x = 24 \)
Step 3: Fill each digit into its corresponding box in the grid:
- Row 1: Across 1 is \( 10 \) (Row 1, Columns 1 and 2); Across 2 is \( 25 \) (Row 1, Columns 4 and 5).
- Column 1: Down 1 gives \( 11 \) (Row 1 and Row 2, Column 1); Down 3 starts at Row 2 giving \( 16 \) (Row 2 and Row 3, Column 1); Down 6 gives \( 15 \) (Row 4 and Row 5, Column 1).
- Row 4: Across 6 gives \( 12 \) (Row 4, Columns 1 and 2); Across 7 gives \( 24 \) (Row 4, Columns 4 and 5).
- Column 4: Down 7 gives \( 24 \) (Row 4 and Row 5, Column 4).
- Cell 5: Across 5 gives \( 9 \) (Row 3, Column 5).

In simple words: Solve each simple linear equation first. Then write one digit of each answer in its numbered box.

Exam Tip: Verify that shared cells between across and down answers contain the same digit.

Question 6. Ritu plans to make a display by stacking cans. The top 3 rows are shown here. The display will be a total of 7 rows. How many cans in all will Rohan need to make the display?
Answer: Looking at the pattern of cans in the display:
- Row 1 (top row) has 1 can.
- Row 2 has 2 cans.
- Row 3 has 3 cans.
This forms an arithmetic series where row \( n \) contains \( n \) cans.
For a total of 7 rows, the total number of cans needed is the sum of the first 7 natural numbers:
\( \text{Total cans} = 1 + 2 + 3 + 4 + 5 + 6 + 7 \)
Using the formula for the sum of first \( n \) numbers, \( S = \frac{n(n + 1)}{2} \):
\( S = \frac{7 \times (7 + 1)}{2} = \frac{7 \times 8}{2} = \frac{56}{2} = 28 \text{ cans} \).
Therefore, 28 cans in all are needed to make the complete display.In simple words: Each row adds one more can than the row above. Adding numbers from 1 to 7 gives a total of 28 cans.

Exam Tip: Use the formula \( \frac{n(n + 1)}{2} \) to quickly sum consecutive counting numbers without adding them one by one.

Question 7. A mistake was made in simplifying the expression given below :
Simplify : 5 ÷ 2(6 + 4) − 8
Step 1 = 5 + 2(10) − 8
Step 2 = 7(10) − 8
Step 3 = 70 − 8
Step 4 = 62
Find the step in which mistake is made.
Answer: The mistake was made in Step 1.
Reason:
In the original expression \( 5 \div 2(6 + 4) - 8 \), the operation after 5 is division (\( \div \)). However, in Step 1, the division symbol was incorrectly replaced with an addition sign (\( + \)).
(Additionally, in Step 2, another rule of order of operations was broken: addition was performed before multiplication, adding \( 5 + 2 = 7 \) instead of multiplying \( 2 \times 10 \) first).
Correct simplification following BODMAS rules:
Original expression: \( 5 \div 2(6 + 4) - 8 \)
1. Solve brackets first: \( 6 + 4 = 10 \), giving \( 5 \div 2(10) - 8 \).
2. Perform division and multiplication from left to right: \( \frac{5}{2} \times 10 - 8 = 2.5 \times 10 - 8 = 25 - 8 \).
3. Perform subtraction: \( 25 - 8 = 17 \).
In simple words: The first error is in Step 1, where the division sign was mistakenly copied as a plus sign.

Exam Tip: Check each copied step against the original problem to catch copied sign errors immediately.