CBSE Class 7 Mathematics Practice Assignment Set 01

Section - A

Question 1. Find the product of (-3) x (0) x (-6)
Answer: (-3) \times 0 \times (-6) = 0
In simple words: Multiplying any number by zero always gives zero.

Exam Tip: Look for a zero in a long product chain; the entire result is immediately zero without needing further multiplication.

Question 2. Evaluate 50 \div (-5)
Answer: 50 \div (-5) = -10
In simple words: A positive number divided by a negative number results in a negative quotient.

Exam Tip: Remember the sign rules for division: positive divided by negative is negative.

Question 3. Find the mean of 5, 1, 0, 4, 6, 3
Answer: Sum of observations = \(5 + 1 + 0 + 4 + 6 + 3 = 19\)
Total number of observations = 6
Mean = \(\frac{\text{Sum of observations}}{\text{Number of observations}} = \frac{19}{6} = 3.17\)
In simple words: Mean is the average found by adding all the given numbers and dividing by how many numbers there are.

Exam Tip: Double check your addition of the terms before dividing.

Question 4. Solve x - 3 = 7
Answer: Given equation is \(x - 3 = 7\)
Transposing -3 to the right side:
\(x = 7 + 3\)
\(x = 10\)
In simple words: To find the unknown variable, move the constant to the other side by changing its sign.

Exam Tip: Always perform the inverse operation to isolate the variable.

Question 5. Find the complement of 35°.
Answer: Complementary angles add up to 90°.
Complement = \(90^\circ - 35^\circ = 55^\circ\)
In simple words: The complement is what you need to add to 35 degrees to make 90 degrees.

Exam Tip: Complement means 90 degrees, while supplement means 180 degrees.

Question 6. Define median of a triangle.
Answer: A line segment joining a vertex of a triangle to the midpoint of the opposite side is called a median of the triangle.
In simple words: A median connects a corner of a triangle to the exact middle of the opposite side.

Exam Tip: A triangle always has exactly three medians that intersect at a single point called the centroid.

Question 7. Solve 1\(\frac{1}{2}\) + \(\frac{3}{4}\)
Answer: Convert mixed fraction to improper fraction:
\(1\frac{1}{2} = \frac{3}{2}\)
Find a common denominator for \(\frac{3}{2}\) and \(\frac{3}{4}\), which is 4:
\(\frac{3}{2} = \frac{6}{4}\)
Now add the fractions:
\(\frac{6}{4} + \frac{3}{4} = \frac{9}{4} = 2\frac{1}{4}\)
In simple words: Convert mixed numbers to improper fractions, find a common denominator, and add the numerators.

Question 8. 7 Rupees 7 paise = ____________ rupees.
Answer: 7.07 rupees
In simple words: Since 1 rupee equals 100 paise, 7 paise is written as 0.07 rupees.
Exam Tip: Be careful with place values when converting paise to rupees.

Question 9. Is it possible to have a triangle with sides 3 cm, 6 cm, 7 cm?
Answer: Yes, it is possible.
Verification using the triangle inequality theorem (sum of any two sides must be greater than the third side):
\(3 + 6 = 9 > 7\)
\(6 + 7 = 13 > 3\)
\(3 + 7 = 10 > 6\)
Since all conditions are satisfied, a triangle can be formed.
In simple words: A triangle is possible because the sum of any two shorter sides is greater than the longest side.

Exam Tip: Always check if the sum of the two smaller sides is strictly greater than the largest side.

Question 10. Find the value of x in the given figure
Answer: From the given right-angled triangle figure with an exterior angle of 125°, using the exterior angle property:
The exterior angle of a triangle is equal to the sum of its two interior opposite angles.
One interior opposite angle is 90° (indicated by the right-angle symbol).
\(90^\circ + x = 125^\circ\)
\(x = 125^\circ - 90^\circ = 35^\circ\)
In simple words: The exterior angle equals the sum of the two opposite interior angles.

Exam Tip: Identify the right angle (90°) correctly before applying the exterior angle theorem.

Section - B

Question 11. Find a) \(\frac{4}{9} \times \frac{3}{8}\) b) \(\frac{3}{16} \div \frac{7}{20}\)
Answer:
a) \(\frac{4}{9} \times \frac{3}{8} = \frac{4 \times 3}{9 \times 8} = \frac{12}{72} = \frac{1}{6}\)
b) \(\frac{3}{16} \div \frac{7}{20} = \frac{3}{16} \times \frac{20}{7} = \frac{3 \times 20}{16 \times 7} = \frac{60}{112} = \frac{15}{28}\)
In simple words: Multiply numerators and denominators for fraction multiplication. For division, multiply by the reciprocal of the divisor.

Exam Tip: Simplify fractions by canceling common factors before multiplying to save time.

Question 12. Write down a pair of integers whose : a) Sum is -4 b) Difference is -3
Answer:
a) Sum is -4: Example pair is -6 and 2 (since \(-6 + 2 = -4\))
b) Difference is -3: Example pair is 2 and 5 (since \(2 - 5 = -3\))
In simple words: Choose any two integers that fit the required arithmetic operation. Many correct pairs exist.

Question 13. Shaili finished her home work in \(\frac{7}{12}\) hour. Vijay finished the same homework in \(\frac{3}{4}\) hour. Who worked longer ? By what fraction was it longer ?
Answer:
Time taken by Shaili = \(\frac{7}{12}\) hour
Time taken by Vijay = \(\frac{3}{4}\) hour = \(\frac{3 \times 3}{4 \times 3} = \frac{9}{12}\) hour
Comparing \(\frac{7}{12}\) and \(\frac{9}{12}\), since \(\frac{9}{12} > \frac{7}{12}\), Vijay worked longer.
Difference in time = \(\frac{9}{12} - \frac{7}{12} = \frac{2}{12} = \frac{1}{6}\) hour.
In simple words: Convert fractions to a common denominator to compare them, then subtract to find the difference.
Exam Tip: Always reduce the final fractional difference to its simplest form.

Question 14. PQR is a triangle, right - angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Answer:
Using Pythagoras theorem in right-angled triangle PQR:
\(QR^2 = PQ^2 + PR^2\)
\(QR^2 = 10^2 + 24^2\)
\(QR^2 = 100 + 576\)
\(QR^2 = 676\)
\(QR = \sqrt{676} = 26 \text{ cm}\)
In simple words: In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

Exam Tip: Memorize squares of common numbers up to 30 for faster calculation.

Question 15. You have to show that \(\triangle ABC \cong \triangle ADC\). Write the missing reasons in the following proof :
(i) BC = DC [________________]
(ii) \(\angle BCA = \angle DCA\) [________________]
(iii) AC = AC [________________]
(iv) \(\triangle ABC \cong \triangle ADC\) [________________]
Answer:
(i) BC = DC [Given]
(ii) \(\angle BCA = \angle DCA\) [Given / 90° each]
(iii) AC = AC [Common side]
(iv) \(\triangle ABC \cong \triangle ADC\) [SAS congruence criterion]
In simple words: State the geometric reason for each matching part between the two triangles.

Exam Tip: Check the order of sides and angles to identify the correct congruence criterion (SSS, SAS, ASA, RHS).

Section - C

Question 16. Evaluate :-
1. \(\frac{3}{5} + \frac{2}{7}\)
2. \(\frac{9}{11} - \frac{4}{15}\)
3. \(\frac{4}{5} \times \frac{12}{7}\)
4. \(\frac{4}{9} \div \frac{2}{3}\)
Answer:
1. \(\frac{3}{5} + \frac{2}{7} = \frac{3 \times 7 + 2 \times 5}{35} = \frac{21 + 10}{35} = \frac{31}{35}\)
2. \(\frac{9}{11} - \frac{4}{15} = \frac{9 \times 15 - 4 \times 11}{165} = \frac{135 - 44}{165} = \frac{91}{165}\)
3. \(\frac{4}{5} \times \frac{12}{7} = \frac{4 \times 12}{5 \times 7} = \frac{48}{35}\)
4. \(\frac{4}{9} \div \frac{2}{3} = \frac{4}{9} \times \frac{3}{2} = \frac{12}{18} = \frac{2}{3}\)
In simple words: Apply standard fraction operations: find common denominators for addition/subtraction, multiply straight across for multiplication, and multiply by the reciprocal for division.

Exam Tip: Always double check LCM calculations when adding or subtracting fractions with different denominators.

Question 17. Fill in the following Blanks.
1. -69 + ____ = -69
2. -89 x _____ = -89
3. -13 + \{17 + (-40)\} = ____ + _____ + _____
4. -609 \div ___ = 1
Answer:
1. 0 (Additive identity property)
2. 1 (Multiplicative identity property)
3. -13 + 17 + (-40) (Associative property of addition)
4. -609 (Division of a number by itself)
In simple words: Use basic properties of integers such as identity elements and associative laws.

Exam Tip: Remember that adding zero leaves a number unchanged, and multiplying by one leaves it unchanged.

Question 18. Find x
Answer:
From the given right-angled triangle DEF (where E is the right angle, DE = 5, EF = 12, and hypotenuse DF = x):
Using Pythagoras theorem:
\(x^2 = DE^2 + EF^2\)
\(x^2 = 5^2 + 12^2\)
\(x^2 = 25 + 144\)
\(x^2 = 169\)
\(x = \sqrt{169} = 13\)
In simple words: Use the Pythagorean theorem to find the missing hypotenuse of the right triangle.

Exam Tip: The triplet (5, 12, 13) is a well-known Pythagorean triplet.

Question 19. If \(\triangle PQR \cong \triangle LMN\) then find
a) PR
b) QR
c) LR
d) LQ
Answer:
Based on corresponding parts of congruent triangles (CPCTC) for \(\triangle PQR \cong \triangle LMN\):
a) PR corresponds to LN
b) QR corresponds to MN
c) LR corresponds to KP (or PK)
d) LQ corresponds to MP (or PM)
In simple words: Corresponding vertices match in the exact same order given in the congruence statement.

Exam Tip: Write down the letter correspondence order (\(P \leftrightarrow L\), \(Q \leftrightarrow M\), \(R \leftrightarrow N\)) before answering.

Section - D

Question 20. Define RHS congruence criterion. In the given figure, \(DA \perp AB\), \(CB \perp AB\) and \(AC = BD\). State the three pairs of equal parts in \(\triangle ABC\) and \(\triangle DAB\). Which of the following statements is meaningful?
i) \(\triangle ABC \cong \triangle BAD\)
ii) \(\triangle ABC \cong \triangle ABD\)
Answer:
Definition: If in two right-angled triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.
Three pairs of equal parts in \(\triangle ABC\) and \(\triangle DAB$\):
1. \(\angle ABC = \angle BAD = 90^\circ\) (given)
2. Hypotenuse \(AC = BD\) (given)
3. Side \(AB = BA\) (common side)
Meaningful statement: (i) \(\triangle ABC \cong \triangle BAD\)
In simple words: RHS states that right triangles with equal hypotenuses and matching side lengths are congruent.

Exam Tip: Ensure corresponding vertices align correctly when writing triangle congruence statements.

Question 21. a) A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
b) Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Answer:
a) Height above ground where tree broke = 5 m
Distance from base to top on ground = 12 m
Length of the broken part (hypotenuse) = \(\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ m}\)
Original height of the tree = Breaking height + Length of broken part = \(5 + 13 = 18 \text{ m}\)
b) Length of rectangle (\(l\)) = 40 cm
Diagonal (\(d\)) = 41 cm
Breadth (\(b\)) = \(\sqrt{d^2 - l^2} = \sqrt{41^2 - 40^2} = \sqrt{(41 - 40)(41 + 40)} = \sqrt{1 \times 81} = 9 \text{ cm}\)
Perimeter of rectangle = \(2(l + b) = 2(40 + 9) = 2(49) = 98 \text{ cm}\)
In simple words: Use the Pythagorean theorem to find missing lengths in both real-world tree and rectangle problems.

Exam Tip: For the rectangle breadth calculation, using the difference of squares identity (\(a^2 - b^2 = (a-b)(a+b)\)) makes arithmetic much faster.

Question 22. Solve:
a) \(2y + \frac{5}{2} = \frac{37}{2}\)
b) \(7m - \frac{19}{2} = 13\)
c) \(10 = 4 + 3(t - 2)\)
d) When you toss a coin, what is the probability of getting a HEAD?
Answer:
a) \(2y + \frac{5}{2} = \frac{37}{2}\)
\(2y = \frac{37}{2} - \frac{5}{2}\)
\(2y = \frac{32}{2} = 16\)
\(y = 8\)
b) \(7m - \frac{19}{2} = 13\)
\(7m = 13 + \frac{19}{2}\)
\(7m = \frac{26 + 19}{2} = \frac{45}{2}\)
\(m = \frac{45}{14}\)
c) \(10 = 4 + 3(t - 2)\)
\(10 - 4 = 3(t - 2)\)
\(6 = 3(t - 2)\)
\(2 = t - 2\)
\(t = 4\)
d) Probability of getting a HEAD = \(\frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{1}{2}\)
In simple words: Isolate the variable step-by-step by transposing constants and dividing by coefficients.

Exam Tip: Always verify your solution by substituting the value back into the original equation.

Question 23. Find:
a) \(\frac{4}{9} \div 5\)
b) \(2\frac{1}{3} + \frac{3}{5}\)
c) \(2\frac{2}{3} \times 3\)
d) Give an example of each - like and unlike fraction.
Consider this data collected from a survey of a colony:
[Table inserted below]
i) Draw a double bar graph choosing an appropriate scale.
ii) Which sport is most popular?
iii) Which is more preferred, watching or participating in sports?
Answer:
a) \(\frac{4}{9} \div 5 = \frac{4}{9} \times \frac{1}{5} = \frac{4}{45}\)
b) \(2\frac{1}{3} + \frac{3}{5} = \frac{7}{3} + \frac{3}{5} = \frac{35 + 9}{15} = \frac{44}{15} = 2\frac{14}{15}\)
c) \(2\frac{2}{3} \times 3 = \frac{8}{3} \times 3 = 8\)
d) Example of like fractions: \(\frac{2}{7}\) and \(\frac{5}{7}\). Example of unlike fractions: \(\frac{1}{3}\) and \(\frac{1}{4}\).
 

i) Double bar graph construction representation:
ii) Cricket is the most popular sport (total 1240 + 620 = 1860 votes).
iii) Watching sports is more preferred compared to participating in sports.
In simple words: Like fractions share denominators, while unlike fractions have different denominators. Bar graphs visually compare categorical data frequencies.

Exam Tip: Always include a clear legend and proper scale labeling when drawing bar graphs.

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