Sample Paper 1

In the number 6524783, the digit 5 lies at the lakhs place.

Place value of 5 = 5 × 1,00,000 = 5,00,000 = 5 lakh.

Hence, option 1 is the correct option.

The whole numbers are 0, 1, 2, 3, 4, … They begin from 0, so the smallest whole number is 0. (-1 is not a whole number.)

Hence, option 2 is the correct option.

The additive inverse of a number is the number which when added to it gives 0.

Since -18 + 18 = 0, the additive inverse of -18 is 18.

Hence, option 1 is the correct option.

1 is a factor of every number. 0 cannot be considered a factor because division by 0 is not defined.

Hence, option 2 is the correct option.

A set is well-defined if its members can be definitely identified. “Tall students”, “good books” and “beautiful flowers” depend on personal opinion and are not well-defined. The set of prime numbers less than 20 is well-defined.

Hence, option 3 is the correct option.

In an improper fraction the numerator is greater than or equal to the denominator. In $\frac{9}{7}$, the numerator 9 is greater than the denominator 7.

Hence, option 3 is the correct option.

In 45.672, the digit 7 is the second digit after the decimal point, i.e., it is at the hundredths place.

Place value of 7 = 7 hundredths.

Hence, option 2 is the correct option.

Converting to the same unit: 3 km = 3 × 1000 m = 3000 m.

Ratio of 3 km to 750 m = $\frac{3000}{750} = \frac{4}{1}$ = 4 : 1.

Hence, option 2 is the correct option.

A variable is a symbol that can take different values. Here, x is a variable, while 5, 3 + 4 and 8 are constants.

Hence, option 2 is the correct option.

A ray has exactly one end point; it starts at a fixed point and extends without end in one direction.

Hence, option 2 is the correct option.

Parallel lines never meet, however far they are extended in either direction.

Hence, option 3 is the correct option.

An equilateral triangle has 3 lines of symmetry, one through each vertex and the mid-point of the opposite side.

Hence, option 3 is the correct option.

Perimeter of a square = 4 × side = 4 × 8 = 32 cm.

Hence, option 3 is the correct option.

Arranging the data 4, 7, 3, 9, 5 in ascending order: 3, 4, 5, 7, 9.

The number of observations = 5 (odd), so the median is the middle term = $\frac{5 + 1}{2}\text{ term} = \frac{6}{2}\text{ term}$ = 3rd term = 5.

Hence, option 2 is the correct option.

The first five even numbers are 2, 4, 6, 8, 10.

Mean = $\frac{2 + 4 + 6 + 8 + 10}{5} = \frac{30}{5} = 6$.

Hence, option 1 is the correct option.

Given,

Number = 846793

(a) Indian system: Starting from the right, commas are placed after every two digits after the first three digits.

⇒ 8,46,793 — Eight lakh forty-six thousand seven hundred and ninety-three.

(b) International system: Starting from the right, commas are placed after every three digits.

⇒ 846,793 — Eight hundred forty-six thousand seven hundred and ninety-three.

The absolute value of an integer is its distance from 0 on the number line, and is never negative.

(a) |-15| = 15

(b) |12| = 12

(c) |0| = 0

Steps of construction:

(i) Mark a point O as the centre.

(ii) Open the compass to a suitable radius. Place the compass needle at O and draw a complete circle.

(iii) Mark any point A on the circle and join OA. Then OA is a radius.

(iv) Extend AO to meet the circle again at B and join AB. Then AB is a diameter passing through the centre O.

Given,

Observations = 8, 12, 15, 9, 16

By formula,

Mean = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

= $\frac{8 + 12 + 15 + 9 + 16}{5}$

= $\frac{60}{5}$

= 12.

Hence, the mean = 12.

(a) To form the greatest number, arrange the digits in descending order:

8 > 5 > 3 > 0, so the greatest 4-digit number = 8530.

(b) To form the smallest number, arrange the digits in ascending order. The thousands place cannot be 0, so we place the smallest non-zero digit (3) first:

Smallest 4-digit number = 3058.

Given,

Numbers = 24 and 32

Prime factorisation of 24 and 32:

$$\begin{array}{l|r} 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$$

⇒ 24 = 2 × 2 × 2 × 3 = 2³ × 3

⇒ 32 = 2 × 2 × 2 × 2 × 2 = 2⁵

HCF = product of the common prime factors with the lowest powers = 2³ = 8

LCM = product of all prime factors with the highest powers = 2⁵ × 3 = 32 × 3 = 96

Verification:

HCF × LCM = 8 × 96 = 768

Product of the two numbers = 24 × 32 = 768

Since 768 = 768, HCF × LCM = product of the two numbers is verified.

Given,

Length = 25 m, Width = 15 m.

(a) Perimeter = 2 × (length + width)

⇒ Perimeter = 2 × (25 + 15) = 2 × 40 = 80 m.

(b) Area = length × width

⇒ Area = 25 × 15 = 375 m².

(a) Given,

Number = 6,847

The tens digit is 4 (less than 5), so we round down.

⇒ 6,847 ≈ 6,800

(b) Given,

Number = 94,562

The hundreds digit is 5, so we round up.

⇒ 94,562 ≈ 95,000

(c) Given,

Number = 7,83,456

The thousands digit is 3 (less than 5), so we round down.

⇒ 7,83,456 ≈ 7,80,000

Given,

a = 8, b = 5 and c = 3.

The distributive property of multiplication over addition states that:

a × (b + c) = (a × b) + (a × c)

LHS = 8 × (5 + 3) = 8 × 8 = 64

RHS = (8 × 5) + (8 × 3) = 40 + 24 = 64

Since LHS = RHS = 64, multiplication is distributive over addition for the whole numbers 8, 5 and 3.

(a) “3°C below zero” is represented as the integer -3°C (morning temperature).

A rise of 8°C is represented as the integer +8°C (the change).

(b) Afternoon temperature = morning temperature + change

⇒ Afternoon temperature = -3 + 8

⇒ Afternoon temperature = 5°C.

Given,

Even numbers between 1 and 15 are 2, 4, 6, 8, 10, 12, 14.

(a) Roster form: {2, 4, 6, 8, 10, 12, 14}

(b) Set-builder form: {x : x is an even number between 1 and 15}

Given,

Fractions = $\frac{48}{72}$ and $\frac{5}{8}$, $\frac{7}{12}$.

Simplifying $\frac{48}{72}$:

HCF of 48 and 72 :

$$\begin{array}{l|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

⇒ 72 = 2³ × 3²

⇒ 48 = 2⁴ × 3

⇒ HCF = 2³ × 3 = 8 × 3 = 24

The HCF of 48 and 72 is 24.

$\frac{48}{72} = \frac{48 \div 24}{72 \div 24} = \frac{2}{3}$

Comparing $\frac{5}{8}$ and $\frac{7}{12}$ by cross multiplication:

Cross multiply: 5 × 12 = 60 and 7 × 8 = 56.

Since 60 > 56, we have $\frac{5}{8} \gt \frac{7}{12}$.

Given,

Cost of 3.5 kg of apples = ₹ 210.

(a) Cost per kg = $\frac{210}{3.5}$ = ₹ 60.

(b) Cost of 7.2 kg of apples = 7.2 × 60 = ₹ 432.

Given,

Number of chocolates = 45, number of toffees = 30, total items = 45 + 30 = 75.

(a) Ratio of chocolates to toffees = $\frac{45}{30} = \frac{3}{2}$ = 3 : 2.

(b) Ratio of toffees to total items = $\frac{30}{75} = \frac{2}{5}$ = 2 : 5.

Given,

Expression = 5x + 12.

Variable term = 5x

Constant term = 12

Coefficient of the variable term = 5

Solving 4x + 7 = 23:

⇒ 4x + 7 = 23

⇒ 4x = 23 - 7 [transposing 7]

⇒ 4x = 16

⇒ x = $\frac{16}{4}$ = 4.

Steps of construction:

(i) Draw a line segment AB = 7.5 cm using a ruler.

(ii) With A as centre and a convenient radius, draw an arc cutting AB at a point P.

(iii) With P as centre and the same radius, draw an arc cutting the first arc at Q; with Q as centre, draw another arc cutting it at R.

(iv) With Q and R as centres and a radius more than half of QR, draw two arcs meeting at S.

(v) Join AS and produce it. Then AS is perpendicular to AB at A, i.e., ∠SAB = 90°.

Recall: an acute angle is less than 90°, a right angle is exactly 90°, an obtuse angle is between 90° and 180°, a straight angle is exactly 180°, and a reflex angle is between 180° and 360°.

(a) 35° → Acute angle (since 35° < 90°)

(b) 90° → Right angle

(c) 125° → Obtuse angle (since 90° < 125° < 180°)

(d) 180° → Straight angle

(e) 280° → Reflex angle (since 180° < 280° < 360°)

Given,

Number of students who liked Social Studies = 40 - (15 + 12 + 8) = 40 - 35 = 5.

The data to be represented is:

Subject	Number of students
Mathematics	15
Science	12
English	8
Social Studies	5

Taking the subjects on the horizontal axis and the number of students on the vertical axis (scale: 1 unit = 1 student), the bar graph is drawn with bars of equal width and equal spacing.

Given,

Red balloons = 36, Yellow balloons = 48.

The greatest number of equal bunches with no balloons left over is the HCF of 36 and 48.

Prime factorisation of 36 and 48:

$$\begin{array}{l|r} 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

⇒ 36 = 2² × 3²

⇒ 48 = 2⁴ × 3

⇒ HCF = 2² × 3 = 4 × 3 = 12

(a) The greatest number of bunches she can make = 12.

(b) Number of red balloons in each bunch = $\frac{36}{12}$ = 3

Number of yellow balloons in each bunch = $\frac{48}{12}$ = 4

Hence, each bunch will have 3 red and 4 yellow balloons.

Given,

Floor dimensions = 6 m × 4 m, Tile size = 30 cm × 20 cm.

(a) Area of the floor = length × width

⇒ Area = 6 × 4 = 24 m².

(b) Area of one tile = 30 × 20 = 600 cm².

(c) First convert the floor area into cm²:

⇒ 24 m² = 24 × 10000 cm² = 2,40,000 cm² [since 1 m² = 10000 cm²]

Number of tiles required = $\frac{\text{Area of floor}}{\text{Area of one tile}} = \frac{2,40,000}{600}$ = 400 tiles.

Taking the days of the week on the horizontal axis and the number of books on the vertical axis (scale: 1 unit = 1 book), the bar chart is drawn with bars of equal width.

(a) Total number of books read over the week

⇒ 2 + 3 + 1 + 4 + 2 + 5 + 3

⇒ 20 books.

(b) Mean = $\frac{\text{Total number of books}}{\text{Number of days}} = \frac{20}{7} = 2\frac{6}{7}$ books per day.

(c) Arjun read the maximum number of books (5) on Saturday.

A triangle whose three sides are of different lengths has no line of symmetry. Such a triangle is a scalene triangle.

The capital English letters having exactly one vertical line of symmetry include A, M, T.

(a) Given,

⇒ 17 : 100 = $\frac{17}{100}$

⇒ $\frac{17}{100} \times 100%$ = 17%.

(b) Given,

⇒ 3 : 25 = $\frac{3}{25}$

$\Rightarrow \frac{3}{25} \times 100% = 3 \times 4$ = 12%

(c) Given,

Students passed = 18 out of 20.

Percentage passed = $\frac{18}{20} \times 100%$ = 90%.

Steps of construction:

(i) Draw a ray BC. With the help of a protractor, place its centre at B and mark a point at 75°. Join B to this point to obtain ray BA, so that ∠ABC = 75°.

(ii) With B as centre and a convenient radius, draw an arc cutting BA at P and BC at Q.

(iii) With P and Q as centres and a radius more than half of PQ, draw two arcs meeting at R.

(iv) Join BR. Then BR bisects ∠ABC.

Each of the two angles formed = $\frac{75°}{2}$ = 37.5°.

On measuring with the protractor, ∠ABR = ∠RBC = 37.5°, and 37.5° + 37.5° = 75° = ∠ABC, which verifies the bisection.

From the given figure (with lines l, m, n, p and points A, B, C, D, E):

(a) Pairs of parallel lines: l and m (l ∥ m).

(b) Pairs of intersecting lines: n and p; n and l; n and m; p and l; p and m.

(c) Lines whose point of intersection is E: l and p.

(d) Collinear points: A, B, C and A, E, D.