ICSE Class 6 Mathematics Sample Paper with Solutions Set 01

SECTION A (40 marks)

Question 1

(a) Write the following numbers in the descending order of their values and represent them on a number line.
i. 8, -6, 2, -12, 0, 3, 15 and -1
ii. Integers greater than -6 and less than 2. [2 Marks]

Answer:
i. Descending order: \( 15, 8, 3, 2, 0, -1, -6, -12 \)
ii. Integers: \( -5, -4, -3, -2, -1, 0, 1 \)

Teacher's Note:
a) Descending order means arranging numbers from the largest to the smallest value, keeping in mind that negative numbers with larger absolute values are smaller.
b) Students must ensure that when listing integers between two given numbers, the boundary numbers are excluded unless specified as inclusive.

(b) Aruna covered \( \frac{1}{2} \) the distance to school by walking and \( \frac{1}{3} \)^rd the distance by bus and the rest by train. Find out the fraction of distance covered in the train? [2 Marks]

Answer:
Let the total distance be \( 1 \).
Fraction covered by walking and bus = \( \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6} \)
Fraction of distance covered in the train = \( 1 - \frac{5}{6} = \frac{1}{6} \)

Teacher's Note:
a) To add fractions with different denominators, first find the least common multiple (LCM) of the denominators.
b) A common error is subtracting numerators directly without finding a common denominator first.

(c) Evaluate: \( 5x - 14 = x - (24 + 4x) \) [3 Marks]

Answer:
\( 5x - 14 = x - 24 - 4x \)
\( 5x - 14 = -3x - 24 \)
\( 5x + 3x = -24 + 14 \)
\( 8x = -10 \)
\( x = \frac{-10}{8} \)
\( x = -\frac{5}{4} = -1\frac{1}{4} \)

Teacher's Note:
a) Always expand brackets carefully by distributing negative signs correctly to all terms inside.
b) Remember to transpose variable terms to one side and constant terms to the other before simplifying.

(d) State whether true or false: [3 Marks]
i. \( \{3, 5, 7, \dots\} \) is a finite set.
ii. A line has infinite number of points on it.
iii. \( 0.45 = 45\% \)

Answer:
i. False
ii. True
iii. True

Teacher's Note:
a) An infinite set has elements that cannot be counted to an end, indicated by the ellipsis (\( \dots \)).
b) To convert a decimal to a percentage, multiply it by \( 100\% \).

Question 2

(a) What are the greatest and smallest possible numbers which can be formed using the digits 9, 8, 7 and 4 without repetition and with 7 always at the ones place? [2 Marks]

Answer:
Greatest number: \( 9847 \)
Smallest number: \( 4897 \)

Teacher's Note:
a) Since digit 7 is fixed at the ones place, arrange the remaining digits (\( 9, 8, 4 \)) in descending order for the greatest number.
b) For the smallest number, arrange the remaining digits in ascending order after fixing 7 at the units place.

(b) Express each of the following as an algebraic expression. [2 Marks]
i. Sum of y and 7.
ii. number m divided by 23 and added to 5

Answer:
i. \( y + 7 \)
ii. \( 5 + \frac{m}{23} \)

Teacher's Note:
a) Translate words like "sum" into addition and "divided by" into fractions or division symbols.
b) Ensure proper ordering of operations as stated in the verbal phrase.

(c) The sides of a triangle are in the ratio \( 3 : 2 : 4 \). If the perimeter of the triangle is \( 27\text{ cm} \), find the length of each side. [3 Marks]

Answer:
Let the common multiple be \( x \).
The sides are \( 3x \), \( 2x \), and \( 4x \).
Perimeter = \( 3x + 2x + 4x = 27 \)
\( 9x = 27 \)
\( x = \frac{27}{9} = 3 \)
First side = \( 3(3) = 9\text{ cm} \)
Second side = \( 2(3) = 6\text{ cm} \)
Third side = \( 4(3) = 12\text{ cm} \)

Teacher's Note:
a) Use a variable multiplier to represent quantities given in a ratio.
b) The perimeter of a triangle is the sum of all its three sides.

(d) The H.C.F. and L.C.M. of two numbers are 144 and 6480, respectively. If one of the numbers is 720, find the other number. [3 Marks]

Answer:
We know that: Product of two numbers = \( \text{H.C.F.} \times \text{L.C.M.} \)
\( 720 \times \text{Second number} = 144 \times 6480 \)
\( \text{Second number} = \frac{144 \times 6480}{720} \)
\( \text{Second number} = 1296 \)

Teacher's Note:
a) The fundamental property relating two numbers to their HCF and LCM is Product = HCF \( \times \) LCM.
b) Simplify the fraction carefully before final multiplication to avoid calculation errors.

Question 3

(a) Name the following: [3 Marks]
[Figure: Two circles with labelled parts. First circle shows a circle with a shaded sector and a horizontal line segment representing a chord. Second circle shows a circle with centre O, diameter AB passing through O, and two points X and Y on the circumference.]
i. Name the two parts of the circle in blue.
ii. What is AB and how does it divide the circle.

Answer:
i. A sector and a chord.
ii. AB is the diameter and it divides the circle into two equal halves (semicircles).

Teacher's Note:
a) A sector is enclosed by two radii and an arc, while a chord connects any two points on a circle.
b) The diameter is the longest chord that passes through the centre of the circle.

(b) Use the divisibility tests to determine whether the number 505 is divisible by 2, 3, 4, 5, 6 and 9. [3 Marks]

Answer:
1. Divisibility by 2: The units digit of 505 is 5 (not even), so it is not divisible by 2.
2. Divisibility by 3: The sum of digits = \( 5 + 0 + 5 = 10 \), which is not divisible by 3, so not divisible by 3.
3. Divisibility by 4: The number formed by the last two digits is 05, which is not divisible by 4.
4. Divisibility by 5: The units digit is 5, so it is divisible by 5.
5. Divisibility by 6: Since it is not divisible by 2 and 3, it is not divisible by 6.
6. Divisibility by 9: The sum of digits is 10, which is not divisible by 9.
Conclusion: 505 is divisible only by 5.

Teacher's Note:
a) Memorize standard divisibility rules for quick verification.
b) A number is divisible by 6 only if it satisfies divisibility tests for both 2 and 3 simultaneously.

(c) Construct an angle of \( 60^{\circ} \), using a ruler and compass. [4 Marks]

Answer:
Steps of Construction:
1. Draw a ray AB.
2. With A as centre and any convenient radius, draw an arc intersecting AB at point C.
3. With C as the centre and the same radius, draw a small arc intersecting the previous arc at point D.
4. Join ray AD through D. Angle DAB is the required \( 60^{\circ} \) angle.

Teacher's Note:
a) Do not change the compass radius once set when cutting the arc for a \( 60^{\circ} \) angle.
b) Always use a sharp pencil and ensure all construction arcs are clearly visible.

Question 4

(a) Add the expressions: \( -17x^2 - 2xy + 23y^2 \), \( -9y^2 + 15x^2 + 7xy \) and \( 13x^2 + 3y^2 - 4xy \) [3 Marks]

Answer:
\( (-17x^2 - 2xy + 23y^2) + (-9y^2 + 15x^2 + 7xy) + (13x^2 + 3y^2 - 4xy) \)
= \( (-17 + 15 + 13)x^2 + (-2 + 7 - 4)xy + (23 - 9 + 3)y^2 \)
= \( 11x^2 + xy + 17y^2 \)

Teacher's Note:
a) Group like terms together by combining their coefficients carefully.
b) Pay close attention to positive and negative signs during addition.

(b) Identify the 3D shapes which form the following nets: [3 Marks]
[Figure: Three nets shown. Net 1 is a circle attached to a rectangle and another circle; Net 2 is a square base surrounded by four triangles forming a pyramid; Net 3 is a cross-shaped net of six rectangles forming a rectangular box.]

Answer:
Net 1: Cylinder
Net 2: Tetrahedron (or Square Pyramid based on standard school terminology)
Net 3: Cuboid

Teacher's Note:
a) Visualize folding the flat 2D nets along their edges to identify the resulting 3D solid.
b) Count the number and shapes of faces in each net to verify.

(c) Find the H.C.F. of 780 and 462 by the division method. [4 Marks]

Answer:
\( 780 \div 462 = 1 \) with remainder \( 318 \)
\( 462 \div 318 = 1 \) with remainder \( 144 \)
\( 318 \div 144 = 2 \) with remainder \( 30 \)
\( 144 \div 30 = 4 \) with remainder \( 24 \)
\( 30 \div 24 = 1 \) with remainder \( 6 \)
\( 24 \div 6 = 4 \) with remainder \( 0 \)
H.C.F. = \( 6 \)

Teacher's Note:
a) In the division method, continue dividing the divisor by the remainder until the remainder becomes zero.
b) The last non-zero divisor is the H.C.F. of the given numbers.

SECTION B (40 marks)

Question 5

(a) Write each statement below in algebraic form: [3 Marks]
i. 28 more than twice of x is equal to 45.
ii. 3y reduced by 5z is greater than 8x.
iii. 6x divided by 13y is less than 17.

Answer:
i. \( 2x + 28 = 45 \)
ii. \( 3y - 5z > 8x \)
iii. \( \frac{6x}{13y} < 17 \)

Teacher's Note:
a) Translate words like "more than" as addition and "reduced by" as subtraction.
b) Use inequality symbols correctly for phrases like "greater than" (\( > \)) and "less than" (\( < \)).

(b) Mark the lines of symmetry in the given figure? [2 Marks]
[Figure: A regular hexagram (Star of David) with dotted lines indicating lines of symmetry passing through vertices and opposite sides.]

Answer:
The given figure has 6 lines of symmetry (3 passing through opposite vertices and 3 passing through opposite indentations).

[Figure: A Star of David showing 6 dashed lines of symmetry passing through opposite vertices and opposite inner points.]

Teacher's Note:
a) A line of symmetry divides a figure into two identical mirror-image halves.
b) Regular polygons and symmetric stars possess multiple lines of symmetry.

(c) The area of a rectangular plot is \( 340\text{ m}^2 \). If its breadth is \( 17\text{ m} \), find its length and perimeter. [3 Marks]

Answer:
Area = \( \text{length} \times \text{breadth} \)
\( 340 = l \times 17 \)
\( l = \frac{340}{17} = 20\text{ m} \)
Perimeter = \( 2(l + b) = 2(20 + 17) = 2(37) = 74\text{ m} \)

Teacher's Note:
a) Use the area formula to find the missing dimension before calculating the perimeter.
b) Always include proper units in the final answer.

(d) The following pie-chart shows the percentage distribution of the expenditure incurred in publishing a book. Study the pie-chart and the answer the questions based on it. [3 Marks]
[Figure: A pie chart showing Publishing Expenditure: Printing Cost 20%, Paper Cost 25%, Transportation Cost 10%, Promotion Cost 10%, Royalty 15%, Binding 20%.]
i. If for a certain quantity of books, the publisher has to pay Rs. 30,600 as printing cost, what will be the amount of royalty to be paid for these books?
ii. What is the central angle of the sector corresponding to the expenditure incurred on Royalty?

Answer:
i. Let the amount of royalty be Rs. \( r \).
Printing cost percentage = \( 20\% \), Royalty percentage = \( 15\% \)
\( \frac{30600}{20} = \frac{r}{15} \)
\( r = \frac{30600 \times 15}{20} = \text{Rs. } 22,950 \)
ii. Central angle = \( 15\% \text{ of } 360^{\circ} = \frac{15}{100} \times 360^{\circ} = 54^{\circ} \)

Teacher's Note:
a) Use proportions to relate percentages to actual monetary values in pie charts.
b) To find the central angle, multiply the percentage fraction by \( 360^{\circ} \).

Question 6

(a) From the sum of \( x + y - 2z \) and \( 2x - y + z \) subtract \( x + y + z \) [3 Marks]

Answer:
Sum = \( (x + y - 2z) + (2x - y + z) = 3x - z \)
Subtracting \( x + y + z \) from sum:
\( (3x - z) - (x + y + z) = 3x - z - x - y - z = 2x - y - 2z \)

Teacher's Note:
a) First perform addition on the first two expressions, then subtract the third expression.
b) Distribute the negative sign correctly across all terms inside the parentheses during subtraction.

(b) Divide 81 into three parts in the ratio \( 2 : 3 : 4 \) [3 Marks]

Answer:
Sum of ratio terms = \( 2 + 3 + 4 = 9 \)
1st part = \( \frac{2}{9} \times 81 = 18 \)
2nd part = \( \frac{3}{9} \times 81 = 27 \)
3rd part = \( \frac{4}{9} \times 81 = 36 \)

Teacher's Note:
a) Divide a quantity in a given ratio by dividing it by the sum of ratio parts and multiplying by each individual ratio term.
b) Verify the answer by adding the three parts to ensure their sum equals the original number.

(c) A vegetable trader buys some tomatoes and onions for Rs. 420 such that the ratio of the total weight of tomatoes to the total weight of onions is \( 2 : 3 \). The total weight of the tomatoes and onions is 60 kg. If the ratio of the total price of tomatoes to the total price of onions is \( 8 : 27 \), then what is the cost of 5 kg of tomatoes and 5 kg of onions. [4 Marks]

Answer:
Total weight = 60 kg, ratio \( 2 : 3 \)
Weight of tomatoes = \( \frac{2}{5} \times 60 = 24\text{ kg} \)
Weight of onions = \( \frac{3}{5} \times 60 = 36\text{ kg} \)
Total price = Rs. 420, price ratio \( 8 : 27 \)
Price of tomatoes = \( \frac{8}{35} \times 420 = \text{Rs. } 96 \)
Price of onions = \( \frac{27}{35} \times 420 = \text{Rs. } 324 \)
Cost of 1 kg of tomatoes = \( \frac{96}{24} = \text{Rs. } 4 \)
Cost of 1 kg of onions = \( \frac{324}{36} = \text{Rs. } 9 \)
Cost of 5 kg of tomatoes and 5 kg of onions = \( (5 \times 4) + (5 \times 9) = 20 + 45 = \text{Rs. } 65 \)

Teacher's Note:
a) Break down multi-step word problems by calculating weights and total prices separately first.
b) Find unit rates per kilogram before calculating costs for specific quantities.

Question 7

(a) Simplify: \( 3(a + b) - 2(2a - b) + 4a - 7 \) [3 Marks]

Answer:
\( = 3a + 3b - 4a + 2b + 4a - 7 \)
\( = (3a - 4a + 4a) + (3b + 2b) - 7 \)
\( = 3a + 5b - 7 \)

Teacher's Note:
a) Expand parentheses by multiplying each term inside by the outer coefficient.
b) Group like terms together to simplify the expression efficiently.

(b) Find the cost of fencing a rectangular park of length \( 250\text{ m} \) and breadth \( 175\text{ m} \) at the rate of Rs. 12 per metre. [3 Marks]

Answer:
Perimeter of the park = \( 2(l + b) = 2(250 + 175) = 2(425) = 850\text{ m} \)
Cost of fencing = \( 850 \times \text{Rs. } 12 = \text{Rs. } 10,200 \)

Teacher's Note:
a) Fencing requires calculating the perimeter of the given shape, not its area.
b) Multiply the total perimeter by the given rate per metre to get the total cost.

(c) Write each of the following sets in roster form as well as in set builder form:
1) Set of factors of 48
2) Set of integers between -3 and 8 [4 Marks]

Answer:
1) Factors of 48:
Roster form: \( \{1, 2, 3, 4, 6, 8, 12, 16, 24, 48\} \)
Set builder form: \( \{x : x \text{ is a factor of } 48\} \)
2) Integers between -3 and 8:
Roster form: \( \{-2, -1, 0, 1, 2, 3, 4, 5, 6, 7\} \)
Set builder form: \( \{x : x \text{ is an integer and } -3 < x < 8\} \)

Teacher's Note:
a) Roster form lists all elements inside curly brackets separated by commas.
b) Set builder form defines properties that elements must satisfy.

Question 8

(a) A school has 1625 students out of which 750 are girls and the rest are boys. Find the ratio between the number of boys to the number of girls in the school. [3 Marks]

Answer:
Total students = 1625
Number of girls = 750
Number of boys = \( 1625 - 750 = 875 \)
Ratio of boys to girls = \( \frac{875}{750} = \frac{35}{30} = \frac{7}{6} \) or \( 7 : 6 \)

Teacher's Note:
a) Subtract the number of girls from total students to find the number of boys.
b) Always reduce ratios to their simplest lowest terms.

(b) Construct a perpendicular line from a point not on the line. [4 Marks]

Answer:
Steps of Construction:
1. Let AB be the given line and C be a point outside it.
2. With C as centre and a convenient radius, draw an arc intersecting line AB at two points P and Q.
3. With P and Q as centres and equal radii, draw two arcs intersecting each other at point D on the opposite side of AB.
4. Join C and D cutting AB at M. CM is the required perpendicular.
[Figure: A line AB with an external point C above it, an arc cutting AB at P and Q, intersecting arcs below AB at D, and a perpendicular line segment CM joining C to AB.]

Teacher's Note:
a) Ensure the compass radius is large enough to intersect the line at two distinct points.
b) Keep the same compass radius when drawing intersecting arcs from points P and Q.

(c) Find the value of: \( 3x^3 - 4x^2 + 5x - 6, \text{ when } x = -1 \) [3 Marks]

Answer:
Substituting \( x = -1 \) into the expression:
\( = 3(-1)^3 - 4(-1)^2 + 5(-1) - 6 \)
\( = 3(-1) - 4(1) - 5 - 6 \)
\( = -3 - 4 - 5 - 6 \)
\( = -18 \)

Teacher's Note:
a) Pay close attention to exponents of negative numbers (odd powers remain negative, even powers become positive).
b) Perform additions of negative numbers carefully to avoid sign errors.

Question 9

(a) The following data gives total marks (out of 500) obtained by 5 students of class VII. Represent the data by a bar graph. [3 Marks]

Answer:
[Figure: A bar graph with student names (ajay, raj, desu, deepak, sonal) on the horizontal axis and marks from 0 to 600 on the vertical axis, showing vertical rectangular bars of heights 350, 375, 400, 450, and 485 respectively.]

Teacher's Note:
a) Choose an appropriate scale on the vertical axis (e.g., 1 unit = 100 marks) to accommodate all data values clearly.
b) Maintain equal width for all bars and equal spacing between them.

(b) Name the types of the following sets: [3 Marks]
i. Set of even numbers which are not divisible by 2.
ii. {Number of people in India}
iii. Set of odd numbers between 7 and 19.

Answer:
i. Empty set (Null set)
ii. Finite set
iii. Finite set

Teacher's Note:
a) An empty set contains no elements since no even number fails to be divisible by 2.
b) A finite set has a countable and definite number of elements.

(c) Simplify: \( \left(\frac{4}{5} - \frac{1}{4}\right) \div \left(1\frac{9}{20} + 1\frac{3}{10}\right) \) [4 Marks]

Answer:
\( = \left(\frac{16 - 5}{20}\right) \div \left(\frac{29}{20} + \frac{13}{10}\right) \)
\( = \frac{11}{20} \div \left(\frac{29 + 26}{20}\right) \)
\( = \frac{11}{20} \div \frac{55}{20} \)
\( = \frac{11}{20} \times \frac{20}{55} \)
\( = \frac{1}{5} \)

Teacher's Note:
a) Convert mixed fractions to improper fractions and simplify expressions inside parentheses first following BODMAS.
b) Division of fractions is performed by multiplying the dividend by the reciprocal of the divisor.