CBSE Class 7 Mathematics Practice Worksheet Set 05

Question 1. Find five rational numbers between – 3 and – 2 .
Answer: To find five rational numbers, write \( -3 \) and \( -2 \) as fractions with a common denominator of \( 5 + 1 = 6 \):
\( -3 = -\frac{3 \times 6}{6} = -\frac{18}{6} \)
\( -2 = -\frac{2 \times 6}{6} = -\frac{12}{6} \)
Now, pick five rational numbers between \( -\frac{18}{6} \) and \( -\frac{12}{6} \):
\( -\frac{17}{6}, -\frac{16}{6}, -\frac{15}{6}, -\frac{14}{6}, -\frac{13}{6} \)
Simplifying these fractions gives:
\( -\frac{17}{6}, -\frac{8}{3}, -\frac{5}{2}, -\frac{7}{3}, -\frac{13}{6} \).
In simple words: Change both integers to fractions over 6. Then pick five fractions that sit between them.

Exam Tip: To find \( n \) rational numbers between two integers, multiply and divide both by \( n + 1 \).

Question 2. Represent \( -\frac{3}{4} \) on the number line.
Answer: The number \( -\frac{3}{4} \) is a negative fraction lying between 0 and \( -1 \).
Divide the unit distance between 0 and \( -1 \) into 4 equal segments.
Starting from 0, count 3 marks to the left.
Point P on the number line represents \( -\frac{3}{4} \).In simple words: Split the gap between 0 and -1 into 4 equal steps. Move 3 steps left from 0 to place -3/4.

Exam Tip: Since the denominator is 4, divide the space between 0 and -1 into 4 equal sections using 3 tick marks.

Question 3. Find the cost of \( 3\frac{1}{3} \) metres of cloth at Rs \( 40\frac{1}{2} \) per metre.
Answer: Convert the mixed fractions into improper fractions:
Length of cloth = \( 3\frac{1}{3}\text{ m} = \frac{10}{3}\text{ m} \).
Cost per metre = Rs \( 40\frac{1}{2} = \text{Rs } \frac{81}{2} \).
Total cost = Length of cloth \( \times \) Cost per metre:
\( \text{Total cost} = \frac{10}{3} \times \frac{81}{2} \)
Simplify by cancelling common factors:
\( = \frac{10}{2} \times \frac{81}{3} = 5 \times 27 = \text{Rs } 135 \).
Hence, the total cost of the cloth is Rs 135.
In simple words: Turn both mixed numbers into improper fractions and multiply them. The total cost is Rs 135.

Exam Tip: Change mixed numbers to improper fractions before multiplying to avoid errors.

Question 4. By what number should \( 3^{-3} \) be multiplied to obtain 4?
Answer: Let the required multiplier be \( x \).
According to the given condition:
\( 3^{-3} \times x = 4 \)
Write \( 3^{-3} \) with a positive exponent:
\( \frac{1}{3^3} \times x = 4 \)
\( \frac{1}{27} \times x = 4 \)

\( \implies x = 4 \times 27 = 108 \).
Therefore, the required number is 108.
In simple words: 3 to the power -3 is 1/27. Multiply 4 by 27 to find the answer, which is 108.

Exam Tip: Remember that a negative exponent means taking the reciprocal: \( a^{-m} = \frac{1}{a^m} \).

Question 5. Find the value of n when: \( 5^{2n} \times 5^3 = 5^9 \)
Answer: Apply the product rule of exponents on the left side, \( a^m \times a^n = a^{m+n} \):
\( 5^{2n + 3} = 5^9 \)
Since the bases on both sides are identical, equate their powers:
\( 2n + 3 = 9 \)
\( 2n = 9 - 3 \)
\( 2n = 6 \)

\( \implies n = \frac{6}{2} = 3 \).
Thus, the value of \( n \) is 3.
In simple words: Add the powers on the left to get 2n + 3. Setting that equal to 9 gives n = 3.

Exam Tip: When bases are the same across an equals sign, set the exponents equal to solve for the unknown.

Question 6. Write the following numbers in standard form:
(a) 684502
(b) 0.000987
Answer: Standard form expresses a number as \( k \times 10^n \), where \( 1 \le k < 10 \).
(a) Place the decimal point after the first non-zero digit by shifting it 5 places left:
\( 684502 = 6.84502 \times 10^5 \).
(b) Shift the decimal point 4 places right to place it after the first non-zero digit:
\( 0.000987 = 9.87 \times 10^{-4} \).
In simple words: Move the dot so only one non-zero number stays on the left. Count the jumps to find the power of 10.

Exam Tip: Shifting the decimal to the left gives a positive power, while moving it right gives a negative power.

Question 7. Subtract (2a – 3b + 4c) from the sum of (a + 3b – 4c) & (4a – b + 9c).
Answer: First, add the two expressions together:
\( \text{Sum} = (a + 3b - 4c) + (4a - b + 9c) \)
Group like terms:
\( = (a + 4a) + (3b - b) + (-4c + 9c) \)
\( = 5a + 2b + 5c \).
Now, subtract \( (2a - 3b + 4c) \) from this sum:
\( (5a + 2b + 5c) - (2a - 3b + 4c) \)
Change the signs of each term being subtracted:
\( = 5a + 2b + 5c - 2a + 3b - 4c \)
Combine like terms:
\( = (5a - 2a) + (2b + 3b) + (5c - 4c) \)
\( = 3a + 5b + c \).
In simple words: Add the first two groups to get 5a + 2b + 5c. Then subtract the third group to reach 3a + 5b + c.

Exam Tip: Be careful with signs when subtracting an algebraic expression: reverse the sign of every term inside the brackets.

Question 8. Multiply \( -\frac{8}{21}x^2y^3 \) by \( -\frac{7}{16}xy^2 \) and verify your result for x = 3 and y = 2.
Answer: Multiply the two monomials:
\( \text{Product} = \left(-\frac{8}{21}x^2y^3\right) \times \left(-\frac{7}{16}xy^2\right) \)
Multiply the numerical coefficients and variables separately:
\( = \left(-\frac{8}{21} \times -\frac{7}{16}\right) \times (x^2 \times x) \times (y^3 \times y^2) \)
\( = \left(\frac{8 \times 7}{21 \times 16}\right) x^3 y^5 \)
Cancel common factors (\( 8 \) into \( 16 \) twice, and \( 7 \) into \( 21 \) thrice):
\( = \frac{1}{3 \times 2} x^3 y^5 = \frac{1}{6}x^3y^5 \).
Verification for \( x = 3 \) and \( y = 2 \):
Left Hand Side (LHS):
\( \text{LHS} = \left(-\frac{8}{21} \times 3^2 \times 2^3\right) \times \left(-\frac{7}{16} \times 3 \times 2^2\right) \)
\( = \left(-\frac{8}{21} \times 9 \times 8\right) \times \left(-\frac{7}{16} \times 3 \times 4\right) \)
\( = \left(-\frac{192}{7}\right) \times \left(-\frac{21}{4}\right) = \frac{192 \times 21}{7 \times 4} = 48 \times 3 = 144 \).
Right Hand Side (RHS):
\( \text{RHS} = \frac{1}{6} \times 3^3 \times 2^5 = \frac{1}{6} \times 27 \times 32 = \frac{864}{6} = 144 \).
Since \( \text{LHS} = \text{RHS} = 144 \), the result is verified.
In simple words: Multiply numbers and add the powers of matching letters to get (1/6)x³y⁵. Putting in 3 and 2 gives 144 on both sides.

Exam Tip: Calculate both LHS and RHS separately during verification, and conclude with "LHS = RHS, hence verified".

Question 9. Simplify: (3x + 4) (2x – 3) + (5x – 4)(x + 2)
Answer: Expand the first product:
\( (3x + 4)(2x - 3) = 3x(2x - 3) + 4(2x - 3) \)
\( = 6x^2 - 9x + 8x - 12 = 6x^2 - x - 12 \).
Expand the second product:
\( (5x - 4)(x + 2) = 5x(x + 2) - 4(x + 2) \)
\( = 5x^2 + 10x - 4x - 8 = 5x^2 + 6x - 8 \).
Add both expressions together:
\( (6x^2 - x - 12) + (5x^2 + 6x - 8) \)
Combine like terms:
\( = (6x^2 + 5x^2) + (-x + 6x) + (-12 - 8) \)
\( = 11x^2 + 5x - 20 \).
In simple words: Multiply out both sets of brackets, then combine like terms to get 11x² + 5x - 20.

Exam Tip: Expand each binomial pair separately first to avoid sign errors when combining terms.

Question 10. Solve:
(a) t – (2t + 5 ) – 5(1 – 2t) = 2(3 + 4t) – 3(t – 4)
(b) \( \frac{3x - 1}{5} - \frac{x}{7} = 3 \)
Answer:
(a) Remove brackets on both sides:
Left side: \( t - 2t - 5 - 5 + 10t = 9t - 10 \).
Right side: \( 6 + 8t - 3t + 12 = 5t + 18 \).
Equate both sides:
\( 9t - 10 = 5t + 18 \)
Collect terms with \( t \) on the left and numbers on the right:
\( 9t - 5t = 18 + 10 \)
\( 4t = 28 \)

\( \implies t = \frac{28}{4} = 7 \).

(b) Find the LCM of denominators 5 and 7, which is 35.
Multiply the entire equation by 35 to clear fractions:
\( 35 \times \left(\frac{3x - 1}{5}\right) - 35 \times \left(\frac{x}{7}\right) = 35 \times 3 \)
\( 7(3x - 1) - 5x = 105 \)
\( 21x - 7 - 5x = 105 \)
\( 16x - 7 = 105 \)
\( 16x = 105 + 7 \)
\( 16x = 112 \)

\( \implies x = \frac{112}{16} = 7 \).
In simple words: Clear the brackets in (a) to find t = 7. Clear the fractions in (b) by multiplying by 35 to find x = 7.

Exam Tip: Multiplying every term by the LCM of the denominators is the cleanest way to clear fractions in linear equations.

Question 11. In an examination, a student requires 40% of the total marks to pass. If Rupa gets 185 marks and fails by 15 marks, find the total marks.
Answer: Marks obtained by Rupa = 185.
Shortfall to pass = 15 marks.
Minimum marks needed to pass = \( 185 + 15 = 200 \).
Let the maximum (total) marks be \( M \).
Since 40% is required to pass:
\( 40\% \text{ of } M = 200 \)
\( \frac{40}{100} \times M = 200 \)

\( \implies M = \frac{200 \times 100}{40} = 5 \times 100 = 500 \).
Therefore, the total marks for the examination are 500.
In simple words: Rupa needed 200 marks to pass. Since 200 is 40% of the total, the full test had 500 marks.

Exam Tip: Find the passing score by adding the student's score and the failing margin before setting up the percentage equation.

Question 12. Two lines AB and CD intersect at a point O. If ∠AOC = 40°, find the measure of each of the angles ∠AOD , ∠BOD and ∠BOC .
Answer: Given that lines AB and CD intersect at O, and \( \angle \text{AOC} = 40^\circ \).
1. Vertically opposite angles are equal:
\( \angle \text{BOD} = \angle \text{AOC} = 40^\circ \).
2. Angles on a straight line form a linear pair and sum to \( 180^\circ \):
\( \angle \text{AOC} + \angle \text{AOD} = 180^\circ \)
\( 40^\circ + \angle \text{AOD} = 180^\circ \)

\( \implies \angle \text{AOD} = 180^\circ - 40^\circ = 140^\circ \).
3. Angles \( \angle \text{BOC} \) and \( \angle \text{AOD} \) are vertically opposite angles:
\( \angle \text{BOC} = \angle \text{AOD} = 140^\circ \).
Hence, the required measures are \( \angle \text{AOD} = 140^\circ \), \( \angle \text{BOD} = 40^\circ \), and \( \angle \text{BOC} = 140^\circ \).In simple words: The angle opposite to 40 degrees is also 40 degrees. The other two angles along the straight line are 140 degrees each.

Exam Tip: State the geometric reasons clearly: "vertically opposite angles are equal" and "linear pair angles sum to 180°".

Question 13. Two angles of a triangle are equal and the third angle measures 70°. Find the measure of each of the unknown angles.
Answer: Let the measure of each of the two equal angles be \( x \).
The sum of the three interior angles of a triangle is always \( 180^\circ \):
\( x + x + 70^\circ = 180^\circ \)
\( 2x + 70^\circ = 180^\circ \)
\( 2x = 180^\circ - 70^\circ \)
\( 2x = 110^\circ \)

\( \implies x = \frac{110^\circ}{2} = 55^\circ \).
Therefore, each of the unknown angles measures \( 55^\circ \).
In simple words: Subtract 70 from 180 to get 110 degrees. Split 110 equally between the two unknown angles to get 55 degrees each.

Exam Tip: Use the angle sum property of a triangle (\( \angle A + \angle B + \angle C = 180^\circ \)) to set up your equation.

Question 14. Two sides of a triangle are 6cm and 8cm long. What can be the length of its third side?
Answer: In any triangle, two fundamental rules govern side lengths:
1. The sum of lengths of any two sides is greater than the third side:
\( \text{Third side} < 8\text{ cm} + 6\text{ cm} = 14\text{ cm} \).
2. The difference between lengths of any two sides is less than the third side:
\( \text{Third side} > 8\text{ cm} - 6\text{ cm} = 2\text{ cm} \).
Hence, the length of the third side can be any value strictly between 2 cm and 14 cm.
In simple words: The third side must be longer than 8 - 6 = 2 cm, and shorter than 8 + 6 = 14 cm.

Exam Tip: State the range clearly: the third side must be greater than the difference and less than the sum of the other two sides.

Question 15. A 15m long ladder is placed against a wall in such a way that the foot of the ladder is 9m away from the wall. Up to what height does the ladder reach the wall?
Answer: The wall, the ground, and the ladder form a right-angled triangle.
Hypotenuse (length of ladder, \( c \)) = 15 m.
Base (distance from the wall, \( b \)) = 9 m.
Let the height reached on the wall be \( h \).
By Pythagoras theorem:
\( h^2 + b^2 = c^2 \)
\( h^2 + 9^2 = 15^2 \)
\( h^2 + 81 = 225 \)
\( h^2 = 225 - 81 = 144 \)

\( \implies h = \sqrt{144} = 12\text{ m} \).
Thus, the ladder reaches a height of 12 m on the wall.In simple words: Using Pythagoras theorem, 15 squared minus 9 squared gives 144. The square root of 144 is 12 metres.

Exam Tip: Identify the ladder as the hypotenuse opposite the right angle formed between the vertical wall and horizontal ground.