Kendriya Vidyalaya Sangathan
Periodic Test 1 (PT-1) — Sample Paper 2 (2026–27)
Class: VII
Subject: Mathematics (Ganita Prakash)
Time Allowed: 90 Minutes
Maximum Marks: 40
SECTION A: Multiple Choice Questions (1 Mark Each)
Q1. What is the face value of the digit $7$ in the number $5,73,82,109$? (Chapter 1)
(a) $70,00,000$
(b) $7,00,000$
(c) $7$
(d) $70,000$
Q2. Which of the following fractions represents $0.05$ in its simplest form? (Chapter 3)
(a) $\dfrac{1}{5}$
(b) $\dfrac{1}{20}$
(c) $\dfrac{5}{10}$
(d) $\dfrac{1}{50}$
Q3. What is the value of $30 - 5 \times 4 + 2$? (Chapter 2)
(a) $102$
(b) $12$
(c) $10$
(d) $20$
Q4. Place value of $3$ in the number $12.435$ is: (Chapter 3)
(a) $3 \text{ tens}$
(b) $3 \text{ tenths}$
(c) $3 \text{ hundredths}$
(d) $3 \text{ thousandths}$
Q5. According to the International System of Numeration, the number $6,504,321$ is read as: (Chapter 1)
(a) Sixty-five lakh four thousand three hundred twenty-one
(b) Six million five hundred four thousand three hundred twenty-one
(c) Sixty-five million four thousand three hundred twenty-one
(d) Six hundred fifty thousand four hundred thirty-one
SECTION B: Very Short Answer Questions (2 Marks Each)
Q6. Form the largest and smallest 7-digit numbers using the digits $4, 0, 9, 1, 6, 2, 8$ without repetition. (Chapter 1)
Q7. Simplify using the BODMAS rule: (Chapter 2)
Q8. Write the following fractional sums as decimals: (Chapter 3)
(a) $200 + 30 + 5 + \dfrac{6}{10} + \dfrac{2}{100}$
(b) $7 + \dfrac{9}{100}$
Q9. Insert brackets in the expression $8 + 12 \div 4 - 2$ so that its value becomes $3$. (Chapter 2)
SECTION C: Short Answer Questions (3 Marks Each)
Q10. A sugar factory produced $38,45,200\text{ kg}$ of sugar in Year 1, $42,10,800\text{ kg}$ in Year 2, and $35,90,500\text{ kg}$ in Year 3.
(a) Find the total production over 3 years.
(b) Round off the Year 2 production to the nearest lakh. (Chapter 1)
Q11. Solve the following nested bracket expression: (Chapter 2)
Q12. Arrange these decimals in descending (decreasing) order: (Chapter 3)
Q13. A piece of wire measuring $15 \dfrac{5}{10}\text{ cm}$ is bent into three segments: two segments measure $4 \dfrac{2}{10}\text{ cm}$ and $5 \dfrac{7}{10}\text{ cm}$ respectively. Find the length of the third segment in decimal form. (Chapter 3)
SECTION D: Long Answer / Case-Based Questions (5 Marks Each)
Q14. (Case Study - Large Numbers & Estimation) (Chapter 1)
A stadium held three major events in a month. Event A sold $62,840$ tickets, Event B sold $84,190$ tickets, and Event C sold $45,610$ tickets.
(a) Estimate the total ticket sales by rounding off each event's sales to the nearest thousand. (2 Marks)
(b) Calculate the actual total ticket sales and find the difference between the actual total and your estimated total. (2 Marks)
(c) Write the actual total in expanded notation using powers of 10 or place values. (1 Mark)
Q15. (Expression Evaluation & Problem Creation) (Chapter 2)
(a) Evaluate:
(3 Marks)
(b) Write a word problem that corresponds to the mathematical expression $(15 + 25) \div 5$. (2 Marks)
Q16. (Measurement, Decimals & Fractions) (Chapter 3)
A chef is mixing ingredients for a recipe:
Flour = $2 \text{ kg and } 4 \text{ tenths}$
Sugar = $1 \text{ kg and } 75 \text{ hundredths}$
Butter = $8 \text{ tenths of a kg}$
(a) Write the quantity of each ingredient as a fraction or mixed fraction in tenths/hundredths, and convert each to decimal kilograms. (3 Marks)
(b) Calculate the total combined mass of all three ingredients in kilograms. (2 Marks)
Here is the step-by-step answer key and marking scheme for Sample Paper 2.
SECTION A: MCQs (1 Mark Each)
Q1. Answer: (c) $7$
Solution: Face value of a digit is the digit itself, regardless of its place in the number. Therefore, the face value of $7$ is $7$. (Place value would be $70,00,000$).
Q2. Answer: (b) $\dfrac{1}{20}$
Solution: $0.05 = \dfrac{5}{100}$. Dividing numerator and denominator by $5$ gives $\dfrac{1}{20}$.
Q3. Answer: (b) $12$
Solution: By BODMAS, perform multiplication first:
$$30 - (5 \times 4) + 2 = 30 - 20 + 2$$Then addition/subtraction left-to-right: $10 + 2 = 12$.
Q4. Answer: (c) $3\text{ hundredths}$
Solution: In $12.435$, the digit $4$ is in the tenths place, $3$ is in the hundredths place ($\frac{3}{100}$), and $5$ is in the thousandths place.
Q5. Answer: (b) Six million five hundred four thousand three hundred twenty-one
Solution: In the International System, commas group digits in threes ($6,504,321 \rightarrow \text{Millions, Thousands, Units}$).
SECTION B: Very Short Answer (2 Marks Each)
Q6.
Largest 7-digit number: Arrange digits in descending order $\rightarrow$ $98,64,210$ (1 Mark)
Smallest 7-digit number: Arrange digits in ascending order (cannot start with $0$) $\rightarrow$ $10,24,689$ (1 Mark)
Q7.
Bracket first: $48 \div 8 + 3 \times 2$ (1 Mark)
Division & Multiplication: $6 + 6 = 12$ (1 Mark)
Q8.
(a) $200 + 30 + 5 + 0.6 + 0.02 = \mathbf{235.62}$ (1 Mark)
(b) $7 + 0.09 = \mathbf{7.09}$ (1 Mark)
Q9.
Given expression: $8 + 12 \div 4 - 2$
Target value: $3$
Correct bracket placement: $(8 + 12) \div (4 - 2)$ (1 Mark)
Verification: $20 \div 2 = 10$ (Wait, let's re-group):
$$(8 + 12 \div 4) - 2 \rightarrow (8 + 3) - 2 = 9$$$$(8 + 12) \div 4 - 2 \rightarrow 20 \div 4 - 2 = 5 - 2 = 3 \quad \checkmark$$Final Placement: $(8 + 12) \div 4 - 2 = 3$ (1 Mark)
SECTION C: Short Answer (3 Marks Each)
Q10.
(a) Total Production:
$$38,45,200 + 42,10,800 + 35,90,500 = \mathbf{1,16,46,500\text{ kg}}$$(2 Marks)
(b) Round off Year 2 ($42,10,800$) to nearest lakh:
Look at ten-thousands digit ($1 < 5$), so round down $\rightarrow$ $42,00,000\text{ kg}$ (1 Mark)
Q11.
Innermost parenthesis $(9 - 3)$: $50 - \{15 + [18 \div 6]\}$ (1 Mark)
Square brackets $[18 \div 6]$: $50 - \{15 + 3\}$ (1 Mark)
Curly braces $\{18\}$: $50 - 18 = \mathbf{32}$ (1 Mark)
Q12.
Write in aligned decimal places:
$$3.44, \quad 3.40, \quad 3.04, \quad 3.004, \quad 0.34$$Descending Order (Largest to Smallest):
$$\mathbf{3.44 > 3.40 > 3.04 > 3.004 > 0.34}$$(3 Marks)
Q13.
Total wire: $15 \dfrac{5}{10} = 15.5\text{ cm}$
Segment 1: $4 \dfrac{2}{10} = 4.2\text{ cm}$
Segment 2: $5 \dfrac{7}{10} = 5.7\text{ cm}$ (1 Mark)
Combined length of 2 segments: $4.2 + 5.7 = 9.9\text{ cm}$ (1 Mark)
Third segment: $15.5 - 9.9 = \mathbf{5.6\text{ cm}} \quad \left(\text{or } 5 \dfrac{6}{10}\text{ cm}\right)$ (1 Mark)
SECTION D: Long Answer / Case-Based (5 Marks Each)
Q14.
(a) Estimated total:
Event A: $62,840 \rightarrow 63,000$
Event B: $84,190 \rightarrow 84,000$
Event C: $45,610 \rightarrow 46,000$
Estimated Sum: $63,000 + 84,000 + 46,000 = \mathbf{1,93,000\text{ tickets}}$ (2 Marks)
(b) Actual total & Difference:
Actual Sum: $62,840 + 84,190 + 45,610 = \mathbf{1,92,640\text{ tickets}}$ (1 Mark)
Difference: $1,93,000 - 1,92,640 = \mathbf{360\text{ tickets}}$ (1 Mark)
(c) Expanded form of $1,92,640$:
$$(1 \times 1,00,000) + (9 \times 10,000) + (2 \times 1,000) + (6 \times 100) + (4 \times 10)$$(1 Mark)
Q15.
(a) Evaluate: $120 \div [2 \times \{18 - (10 - 2)\}]$
Step 1: $(10 - 2) = 8 \implies 120 \div [2 \times \{18 - 8\}]$
Step 2: $\{18 - 8\} = 10 \implies 120 \div [2 \times 10]$
Step 3: $[2 \times 10] = 20 \implies 120 \div 20 = \mathbf{6}$ (3 Marks)
(b) Sample Word Problem:
"A group of 15 boys and 25 girls participate in a school quiz. If they are divided equally into 5 teams, how many students are in each team?" (2 Marks)
Q16.
(a) Conversion:
Flour: $2 \dfrac{4}{10}\text{ kg} = \mathbf{2.4\text{ kg}}$
Sugar: $1 \dfrac{75}{100}\text{ kg} = \mathbf{1.75\text{ kg}}$
Butter: $\dfrac{8}{10}\text{ kg} = \mathbf{0.8\text{ kg}}$ (3 Marks)
(b) Total combined mass:
$$\begin{array}{rl} 2.40 \\ 1.75 \\ + 0.80 \\ \hline \mathbf{4.95\text{ kg}} \end{array}$$Total Mass: $\mathbf{4.95\text{ kg}}$ (or $4 \dfrac{95}{100}\text{ kg}$) (2 Marks)
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