CBSE Class 7 Mathematics Practice Assignment Set 04

Q1. Arrange the following numbers in descending order: -25, -18, -34, 16, 35, 5
Answer: Descending order means arranging numbers from the largest to the smallest.
Positive numbers are always greater than negative numbers. Among the positive numbers (\(16, 35, 5\)), the order from largest to smallest is \(35, 16, 5\).
Among the negative numbers (\(-25, -18, -34\)), the number with the smaller absolute value is greater: \(-18 > -25 > -34\).
Combining both, the descending order is: \(35, 16, 5, -18, -25, -34\)
In simple words: Write the biggest positive number first, followed by smaller positive numbers, then zero if any, and finally negative numbers starting from the one closest to zero.

Exam Tip: Always pay close attention to negative signs. A smaller negative number is greater in value than a larger negative number.

Q2. Add:
a) (-125) + (-200)
b) 632 + (-232) + 750
Answer:
a) \((-125) + (-200)\)
\(= -125 - 200\)
\(= -325\)
b) \(632 + (-232) + 750\)
\(= 632 - 232 + 750\)
\(= 400 + 750\)
\(= 1150\)
In simple words: When adding two negative numbers, add their values and keep the negative sign. When adding positive and negative numbers, subtract the smaller number from the larger one and give it the sign of the larger number.

Exam Tip: Simplify signs first (\(+(-)\) becomes \(-\)) before performing addition or subtraction.

Q3. Is 30 \(\div\) 2 + 4 = 30 \(\div\) (2+4)
Answer:
Let us evaluate the Left Hand Side (LHS) using the BODMAS rule:
LHS = \(30 \div 2 + 4\)
\(= 15 + 4\)
\(= 19\)
Now let us evaluate the Right Hand Side (RHS) by solving inside the bracket first:
RHS = \(30 \div (2 + 4)\)
\(= 30 \div 6\)
\(= 5\)
Since LHS (\(19\)) \(\neq\) RHS (\(5\)), the given statement is false. No, they are not equal.
In simple words: Always perform operations inside brackets first. Without brackets, division is done before addition.

Exam Tip: Never ignore brackets; they completely change the order of operations and the final result.

Q4. Subtract 3x – y – 4z from 5x + 2y – 3z
Answer:
To subtract the first expression from the second, write the second expression first and change the signs of all terms in the first expression:
\(= (5x + 2y - 3z) - (3x - y - 4z)\)
\(= 5x + 2y - 3z - 3x + y + 4z\)
Group like terms together:
\(= (5x - 3x) + (2y + y) + (-3z + 4z)\)
\(= 2x + 3y + z\)
In simple words: Write the 'from' expression first, then distribute the negative sign to every term of the second expression and combine the matching variables.

Exam Tip: Be very careful with signs when opening brackets preceded by a minus sign.

Q5. The sum of two integers is -12 . If one of them is 8 , find the other.
Answer:
Let the unknown integer be \(x\).
According to the question:
\(8 + x = -12\)
\(x = -12 - 8\)
\(x = -20\)
The other integer is \(-20\).
In simple words: Subtract the known number from the total sum to find the missing number.

Exam Tip: Transposing a positive term from the left side to the right side changes its sign to negative.

Q6. Arrange the following in ascending order: 5/6, 11/12, 7/18, 1/2
Answer:
First, find the Least Common Multiple (LCM) of the denominators (\(6, 12, 18, 2\)).
LCM = \(36\)
Now, convert each fraction to an equivalent fraction with denominator \(36\):
\(\frac{5}{6} = \frac{5 \times 6}{6 \times 6} = \frac{30}{36}\)
\(\frac{11}{12} = \frac{11 \times 3}{12 \times 3} = \frac{33}{36}\)
\(\frac{7}{18} = \frac{7 \times 2}{18 \times 2} = \frac{14}{36}\)
\(\frac{1}{2} = \frac{1 \times 18}{2 \times 18} = \frac{18}{36}\)
Comparing the numerators of fractions with the same denominator (\(\frac{30}{36}, \frac{33}{36}, \frac{14}{36}, \frac{18}{36}\)) in ascending order:
\(14 < 18 < 30 < 33\)
Therefore, the ascending order of the given fractions is:
\(\frac{7}{18}, \frac{1}{2}, \frac{5}{6}, \frac{11}{12}\)
In simple words: Make the bottom numbers (denominators) the same using LCM, then arrange them by looking at the top numbers (numerators) from smallest to largest.

Exam Tip: Always find the LCM of denominators first to compare or order fractions easily.

Q7. Simplify : 4/21 \(\times\) 7/16 \(\times\) 15/19
Answer:
Write the given expression:
\(= \frac{4}{21} \times \frac{7}{16} \times \frac{15}{19}\)
Cancel out common factors between numerators and denominators:
- Cancel \(4\) and \(16\): \(16 \div 4 = 4\)
- Cancel \(7\) and \(21\): \(21 \div 7 = 3\)
- Cancel \(15\) and \(3\): \(15 \div 3 = 5\)
The expression simplifies to:
\(= \frac{1}{3} \times \frac{7}{4} \times \frac{15}{19}\) (after proper cross-cancellation)
Let us do it step-by-step cleanly:
\(= \frac{4 \times 7 \times 15}{21 \times 16 \times 19}\)
\(= \frac{1 \times 1 \times 5}{3 \times 4 \times 19}\)
\(= \frac{5}{228}\)
In simple words: Simplify multiplication of fractions by cancelling numbers on top with numbers on the bottom before multiplying.

Exam Tip: Cross-cancellation saves time and prevents working with large numbers.

Q8. Find the value of - |-6| + |9| - |-8|
Answer:
The absolute value of a number (represented by bars like \(|x|\)) is always positive or zero, regardless of its sign inside the bars.
\(|-6| = 6\)
\(|9| = 9\)
\(|-8| = 8\)
Substitute these values back into the expression:
\(= -(|-6|) + |9| - (|-8|)\)
\(= -(6) + 9 - 8\)
\(= -6 + 9 - 8\)
\(= 3 - 8\)
\(= -5\)
In simple words: Absolute value turns any negative number into a positive number. Then solve the remaining addition and subtraction from left to right.

Exam Tip: Be careful with signs outside the absolute value bars. The negative sign in front of \(|-6|\) applies after finding the absolute value.

Q9. Plot the integers 0, -5, 3, -2, -9 and 12 on a number line.
Answer:
To plot the integers on a number line:
1. Draw a straight horizontal line with arrows on both ends.
2. Mark a point as \(0\) in the center.
3. Mark equal divisions on the right for positive integers (\(3, 12\)) and on the left for negative integers (\(-2, -5, -9\)).
4. Highlight the points corresponding to \(0, -5, 3, -2, -9\), and \(12\).

In simple words: Draw a straight line with 0 in the middle, negative numbers to the left, and positive numbers to the right, then dot the required numbers.

Exam Tip: Ensure spacing between integers on the number line is roughly uniform for neat presentation.

Q10. If 20 chocolates cost Rs 480, find the cost of 45 such chocolates.
Answer:
Cost of \(20\) chocolates = Rs \(480\)
First, find the cost of \(1\) chocolate using the unitary method:
Cost of \(1\) chocolate = \(\frac{480}{20}\)
Cost of \(1\) chocolate = Rs \(24\)
Now, find the cost of \(45\) chocolates:
Cost of \(45\) chocolates = \(45 \times 24\)
Cost of \(45\) chocolates = Rs \(1080\)
In simple words: Find the price of one item by dividing the total cost by the number of items, then multiply by the new quantity.

Exam Tip: Always state units clearly (Rs and chocolates) in each step of word problems.