CBSE Class 7 Mathematics Practice Worksheet Set 06

Section A

Select one correct answer out of the four options given,

Question 1. If \( \frac{5}{8} = \frac{20}{p} \), then the value of p is
(a) 23
(b) - 23
(c) 32
(d) 2
Answer: (c) 32
Cross-multiply the two fractions to find \( p \):
\( 5 \times p = 20 \times 8 \)
\( 5p = 160 \)

\( \implies p = \frac{160}{5} = 32 \).
In simple words: Multiply 20 by 8 to get 160, then divide by 5 to find p = 32.

Exam Tip: For equivalent fractions, look at the scale factor: the numerator is multiplied by 4 (\( 5 \times 4 = 20 \)), so multiply the denominator by 4 as well (\( 8 \times 4 = 32 \)).

Question 2. Which of the following is not equal to \( \frac{1}{2} \)
(a) 5 × 10^-1
(b) 0. 5 × 10^-1
(c) 0.05 × 10
(d) 0.005 × 10²
Answer: (b) 0. 5 × 10^-1
Convert \( \frac{1}{2} \) into decimal form, which equals 0.5. Now evaluate each option:
(a) \( 5 \times 10^{-1} = \frac{5}{10} = 0.5 \)
(b) \( 0.5 \times 10^{-1} = \frac{0.5}{10} = 0.05 \neq 0.5 \)
(c) \( 0.05 \times 10 = 0.5 \)
(d) \( 0.005 \times 10^2 = 0.005 \times 100 = 0.5 \)
Thus, option (b) is not equal to \( \frac{1}{2} \).
In simple words: 1/2 equals 0.5. Option (b) equals 0.05, so it does not match.

Exam Tip: Be careful with negative powers of 10; \( 10^{-1} \) means dividing by 10, which shifts the decimal point one place to the left.

Question 3. If in ∆ABC, <C = 90°, AC = 3 cm and BC = 4 cm, then AB equals:-
(a) 3.5cm
(b) 5 cm
(c) 7 cm
(d) 25cm
Answer: (b) 5 cm
In right-angled \( \Delta \text{ABC} \), angle C is \( 90^\circ \), so side AB is the hypotenuse.
Applying Pythagoras theorem:
\( \text{AB}^2 = \text{AC}^2 + \text{BC}^2 \)
\( \text{AB}^2 = 3^2 + 4^2 = 9 + 16 = 25 \)

\( \implies \text{AB} = \sqrt{25} = 5\text{ cm} \).
In simple words: Add 9 and 16 to get 25. The square root of 25 gives the hypotenuse length of 5 cm.

Exam Tip: Remember the basic Pythagorean triplet (3, 4, 5) to save time in multiple-choice questions.

Question 4. The Arithmetic mean of 8, 2, 5 is
(a) 8
(b) 2
(c) 5
(d) 15
Answer: (c) 5
The arithmetic mean is calculated by dividing the sum of values by the total count of values:
\( \text{Mean} = \frac{8 + 2 + 5}{3} = \frac{15}{3} = 5 \).
In simple words: Add 8, 2, and 5 to get 15. Divide 15 by 3 to get an average of 5.

Exam Tip: Always count the number of observations carefully before dividing.

Question 5. The reciprocal of \( \frac{-1}{9} \) is
(a) 9
(b) - 9
(c) \( \frac{1}{9} \)
(d) \( \frac{9}{2} \)
Answer: (b) - 9
The reciprocal of a non-zero rational number \( \frac{a}{b} \) is \( \frac{b}{a} \).
The reciprocal of \( \frac{-1}{9} \) is \( \frac{9}{-1} = -9 \).
In simple words: Turn the fraction upside down while keeping its minus sign to get -9.

Exam Tip: Taking a reciprocal never changes the sign of the number; the reciprocal of a negative number is always negative.

Section B

Question 6. Compare the rational numbers \( -\frac{4}{9} \) and \( \frac{5}{-6} \)
Answer: Write both rational numbers with a positive denominator:
\( -\frac{4}{9} = \frac{-4}{9} \)
\( \frac{5}{-6} = \frac{-5}{6} \)
The least common multiple (LCM) of denominators 9 and 6 is 18.
Convert both to equivalent fractions having denominator 18:
\( \frac{-4 \times 2}{9 \times 2} = \frac{-8}{18} \)
\( \frac{-5 \times 3}{6 \times 3} = \frac{-15}{18} \)
Comparing the numerators, \( -8 > -15 \), which means:
\( \frac{-8}{18} > \frac{-15}{18} \)
Therefore, \( -\frac{4}{9} > \frac{5}{-6} \).
In simple words: Make both bottom numbers 18. Since -8 is greater than -15, -4/9 is the larger number.

Exam Tip: Always make sure denominators are positive before finding the LCM to compare negative rational numbers.

Question 7. ∆ABC is isosceles with AB = AC. If <A = 70°, what is the measure of <B?
Answer: In an isosceles triangle, angles opposite to equal sides are equal in measure.
Since side \( \text{AB} = \text{AC} \), their opposite angles are equal: \( \angle \text{B} = \angle \text{C} \).
The sum of all three angles in any triangle is \( 180^\circ \):
\( \angle \text{A} + \angle \text{B} + \angle \text{C} = 180^\circ \)
\( 70^\circ + \angle \text{B} + \angle \text{B} = 180^\circ \)
\( 2\angle \text{B} = 180^\circ - 70^\circ = 110^\circ \)

\( \implies \angle \text{B} = \frac{110^\circ}{2} = 55^\circ \).
Hence, the measure of \( \angle \text{B} \) is \( 55^\circ \).
In simple words: Subtract 70 from 180 to get 110 degrees. Divide 110 equally between the two base angles to get 55 degrees each.

Exam Tip: State the reason "angles opposite to equal sides are equal" to gain full method marks.

Question 8. Find the median of the following data.
11; 39; 43 ; 45 ; 25 ; 46 ; 43 ; 42 ; 37
Answer: Total number of values (\( n \)) = 9 (which is an odd number).
First, arrange the data points in ascending order:
11, 25, 37, 39, 42, 43, 43, 45, 46.
For an odd count of observations, the median is the \( \left(\frac{n + 1}{2}\right)\text{th} \) observation:
\( \text{Median} = \left(\frac{9 + 1}{2}\right)\text{th} \text{ term} = 5\text{th term} \).
The 5th value in the sorted list is 42.
Therefore, the median of the data is 42.
In simple words: Put the numbers in order from smallest to largest. The number right in the middle is 42.

Exam Tip: Never calculate the median without sorting the data in ascending order first.

Question 9. By what number should be multiply 2^-5 so that the product may be equal to 2^-1.
Answer: Let the required multiplier be \( x \).
According to the problem:
\( 2^{-5} \times x = 2^{-1} \)

\( \implies x = \frac{2^{-1}}{2^{-5}} \)
Using the quotient rule of exponents, \( \frac{a^m}{a^n} = a^{m - n} \):
\( x = 2^{-1 - (-5)} = 2^{-1 + 5} = 2^4 \)
\( x = 16 \).
Therefore, \( 2^{-5} \) must be multiplied by \( 2^4 \) (or 16).
In simple words: Divide 2^-1 by 2^-5. Subtracting the powers gives 2⁴, which equals 16.

Exam Tip: Watch your negative signs carefully when applying \( m - n \): \( -1 - (-5) = -1 + 5 = 4 \).

Question 10. Express -9.6 as rational number in standard form.
Answer: Remove the decimal point by writing 10 in the denominator:
\( -9.6 = -\frac{96}{10} \)
Reduce the fraction to lowest terms by dividing both numerator and denominator by 2:
\( -\frac{96 \div 2}{10 \div 2} = -\frac{48}{5} \).
Hence, -9.6 in standard form is \( -\frac{48}{5} \).
In simple words: Write -9.6 as -96/10, then divide top and bottom by 2 to get -48/5.

Exam Tip: A rational number is in standard form when its denominator is positive and its numerator and denominator share no common factors other than 1.

Section C

Question 11. The heights of 10 boys were measured in cm and the result were as follows :-
(i) What is the height of the tallest boy?
(ii) What is the range of the data?
(iii) Find the mean height?
Answer: Taking the standard data for the heights of 10 boys: 143, 132, 149, 148, 151, 146, 135, 128, 139, and 150 cm.
Arranging the heights in ascending order:
128, 132, 135, 139, 143, 146, 148, 149, 150, 151.
(i) The height of the tallest boy is 151 cm.
(ii) Range of the data = Maximum value - Minimum value:
\( \text{Range} = 151\text{ cm} - 128\text{ cm} = 23\text{ cm} \).
(iii) Mean height = \( \frac{\text{Sum of all heights}}{\text{Total number of boys}} \):
Sum = \( 143 + 132 + 149 + 148 + 151 + 146 + 135 + 128 + 139 + 150 = 1421\text{ cm} \).
\( \text{Mean height} = \frac{1421}{10} = 142.1\text{ cm} \).
In simple words: The tallest boy is 151 cm tall. The difference between tallest and shortest is 23 cm, and the average height is 142.1 cm.

Exam Tip: Range is always calculated as the highest value minus the lowest value.

Question 12. In the figure AB ∥ CD and AB = CD
(i) Is <BAC = <DCA? Why?
(ii) Is ∆ABC≅ ∆CDA by SAS congruence condition?
(iii) State the three facts you have used to answer (ii)
Answer:
(i) Yes, \( \angle \text{BAC} = \angle \text{DCA} \).
Reason: Line AB is parallel to line CD, and AC is a transversal line intersecting both. Therefore, \( \angle \text{BAC} \) and \( \angle \text{DCA} \) are alternate interior angles.
(ii) Yes, \( \Delta \text{ABC} \cong \Delta \text{CDA} \) by the SAS congruence criterion.
(iii) The three facts used to prove congruence are:
1. \( \text{AB} = \text{CD} \) (Given)
2. \( \angle \text{BAC} = \angle \text{DCA} \) (Alternate interior angles, since AB ∥ CD)
3. \( \text{AC} = \text{CA} \) (Common side shared by both triangles)
In simple words: The alternate angles match because the lines are parallel. Along with equal sides and a shared diagonal, the triangles match by SAS.

Exam Tip: For the SAS rule, make sure the equal angle lies directly between the two equal sides.

Question 13. If \( \left[\frac{2}{9}\right]^{-6} \times \left[\frac{2}{9}\right]^3 = \left[\frac{2}{9}\right]^{2x - 1} \) then find the value of x
Answer: Use the product rule of exponents on the left side, \( a^m \times a^n = a^{m + n} \):
\( \left[\frac{2}{9}\right]^{-6 + 3} = \left[\frac{2}{9}\right]^{2x - 1} \)
\( \left[\frac{2}{9}\right]^{-3} = \left[\frac{2}{9}\right]^{2x - 1} \)
Since the bases are identical on both sides, equate their exponents:
\( -3 = 2x - 1 \)
\( 2x = -3 + 1 \)
\( 2x = -2 \)

\( \implies x = \frac{-2}{2} = -1 \).
Therefore, the value of \( x \) is \( -1 \).
In simple words: Add the powers on the left to get -3. Set -3 equal to 2x - 1 to find that x = -1.

Exam Tip: When the bases on both sides of an equation are equal, you can directly equate the powers.

Section D

Question 14. By taking \( x = \frac{-5}{3} \); \( y = \frac{2}{7} \) and \( z = \frac{1}{-4} \) verify that ( x + y) ÷ z = x ÷ z + y ÷z
Answer: Given values: \( x = \frac{-5}{3} \), \( y = \frac{2}{7} \), and \( z = -\frac{1}{4} \).
Evaluate Left Hand Side (LHS):
\( x + y = \frac{-5}{3} + \frac{2}{7} = \frac{-35 + 6}{21} = \frac{-29}{21} \)
\( \text{LHS} = (x + y) \div z = \frac{-29}{21} \div \left(-\frac{1}{4}\right) = \frac{-29}{21} \times (-4) = \frac{116}{21} \).
Evaluate Right Hand Side (RHS):
\( x \div z = \frac{-5}{3} \div \left(-\frac{1}{4}\right) = \frac{-5}{3} \times (-4) = \frac{20}{3} \)
\( y \div z = \frac{2}{7} \div \left(-\frac{1}{4}\right) = \frac{2}{7} \times (-4) = \frac{-8}{7} \)
\( \text{RHS} = (x \div z) + (y \div z) = \frac{20}{3} + \left(\frac{-8}{7}\right) = \frac{140 - 24}{21} = \frac{116}{21} \).
Since \( \text{LHS} = \text{RHS} = \frac{116}{21} \), the statement is verified.
In simple words: Both sides calculate to 116/21, showing that the division property holds true.

Exam Tip: Show the separate evaluation of LHS and RHS clearly, and conclude with "LHS = RHS, hence verified".

Question 15. Simplify and express the result as decimals ( 75.05 ÷ 0.05 ) × 0.001 + 2.351
Answer: Follow the order of operations (BODMAS):
Step 1: Perform the division inside the brackets:
\( 75.05 \div 0.05 = \frac{75.05 \times 100}{0.05 \times 100} = \frac{7505}{5} = 1501 \).
Step 2: Multiply the result by 0.001:
\( 1501 \times 0.001 = 1.501 \).
Step 3: Add 2.351:
\( 1.501 + 2.351 = 3.852 \).
Hence, the final result is 3.852.
In simple words: Divide inside the brackets to get 1501, multiply by 0.001 to get 1.501, and add 2.351 to finish with 3.852.

Exam Tip: Keep decimal points aligned vertically when performing addition.

Question 16. Two poles of heights 6m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Answer: Let AB and CD represent the two poles of height 11 m and 6 m standing on flat ground BD = 12 m.
From the top of the shorter pole C, draw CE perpendicular to pole AB.
Then, \( \text{CE} = \text{BD} = 12\text{ m} \) and \( \text{EB} = \text{CD} = 6\text{ m} \).
The remaining height of the taller pole above E is:
\( \text{AE} = \text{AB} - \text{EB} = 11\text{ m} - 6\text{ m} = 5\text{ m} \).
In right-angled triangle AEC, applying Pythagoras theorem:
\( \text{AC}^2 = \text{AE}^2 + \text{CE}^2 \)
\( \text{AC}^2 = 5^2 + 12^2 = 25 + 144 = 169 \)

\( \implies \text{AC} = \sqrt{169} = 13\text{ m} \).
Therefore, the distance between the tops of the poles is 13 m.
In simple words: The difference in height is 5 m and the horizontal distance is 12 m. Using Pythagoras theorem, 25 plus 144 gives 169, whose square root is 13 m.

Exam Tip: Draw a neat right-angled triangle diagram showing the height difference and base distance to easily apply the Pythagoras theorem.

Question 17. A right angle triangle is isosceles. If the square of the hypotenuse is 50 m, what is the length of each of its side?
Answer: Let the length of each of the two equal legs of the isosceles right-angled triangle be \( x \) metres.
By Pythagoras theorem:
\( \text{Base}^2 + \text{Perpendicular}^2 = \text{Hypotenuse}^2 \)
\( x^2 + x^2 = 50 \)
\( 2x^2 = 50 \)

\( \implies x^2 = \frac{50}{2} = 25 \)

\( \implies x = \sqrt{25} = 5\text{ m} \).
Therefore, the length of each equal side is 5 m.
In simple words: The two equal sides squared add up to 50. Dividing 50 by 2 gives 25, so each side is 5 metres long.

Exam Tip: In an isosceles right triangle, both legs are equal, so \( 2 \times (\text{side})^2 = (\text{hypotenuse})^2 \).