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Important Questions: A Peek Beyond the Point | Mathematics (Ganita Prakash) Class 7 - New NCERT PDF Download

Q1: Why is a ruler divided into 10 equal parts in centimeters useful for measuring small objects like screws?

Sol: Dividing a centimeter into 10 equal parts (tenths) allows for precise measurements of small objects, as it captures differences smaller than a centimeter. This is crucial for objects like screws, where lengths may differ by fractions of a centimeter.

Q2: Write the length 3.56 cm in words, explaining the meaning of each digit.

Sol: Three and fifty-six hundredths centimeters.

  • 3: Three whole centimeters.

  • 5: Five tenths of a centimeter.

  • 6: Six hundredths of a centimeter.

Q3: You are given a ruler with markings divided into 10 equal parts between each centimeter. Measure the length of a key and a pencil. Write their lengths in centimeters, including tenths and hundredths.

Sol: Let’s assume example measurements; actual objects may vary.
Key: 5.40 cm (5 cm + 4 tenths + 0 hundredth)
Pencil: 7.20 cm (7 cm + 2 tenths + 0 hundredth)

Q4: Convert the following quantities to decimal form:
a) 345 hundredths
b) 67 tenths

Sol:
a) 345 hundredths = 345/100 = 3.45
b) 67 tenths = 67/10 = 6.7

Q5: Arrange the following lengths in ascending order: 2.34 cm, 2.3 cm, 2.345 cm, 2.304 cm.

Sol: Compare digits:

  • 2.34 = 2.340

  • 2.3 = 2.300

  • 2.345 = 2.345

  • 2.304 = 2.304
    Ascending order: 2.3 < 2.304 < 2.34 < 2.345

Q6: Which is closer to 5: 4.89, 5.12, or 5.03?

Sol: Calculate differences:

  • |5 - 4.89| = 0.11

  • |5 - 5.12| = 0.12

  • |5 - 5.03| = 0.03

  • Closest: 5.03

Q7: Find the total length of three ribbons: 1.45 m, 2.3 m, and 0.78 m.

Sol: Total length = 1.45 + 2.3 + 0.78 = 4.53 m

Q8: A fish is 3.2 cm long, and another is 1.8 cm long. What is the difference in their lengths?

Sol: length of one fish = 3.2 cm
length of the other fish = 1.8 cm
difference in lengths = 3.2 − 1.8
= 1.4 cm

Q9: Identify the pattern and find the next three terms: 7.2, 7.5, 7.9, 8.4, 9.0.

Sol: Differences: +0.3, +0.4, +0.5, +0.6 (increasing by 0.1 each step).
Next: +0.7, +0.8, +0.9
Terms: 9.7, 10.5, 11.4

Q10: Create a sequence of five decimal numbers with a pattern and challenge a friend to find the next term.

Sol: Sequence: 1.5, 2.0, 2.6, 3.4, 4.4
Pattern: +0.5, +0.6, +0.8, +1.0 (differences increase by 0.1, 0.2, 0.2, ...)
Next term: 4.4 + 1.2 = 5.6

Q11: How many millimeters are in 2.5 cm? How many centimeters are in 1.2 m?

Sol: millimeters in 2.5 cm 
= 2.5 × 10 
= 25 mm
centimeters in 1.2 m 
= 1.2 × 100 
= 120 cm

Q12: Convert 345 paise to rupees and express in decimal form.

Sol: Since 1 paisa = 1/100 rupee 
345 paise = 345 × (1/100) rupee
= ₹3.45

Q13: A shop sells 2.5 kg of apples, 1.75 kg of oranges, and 0.9 kg of bananas. Calculate the total weight.

Sol: Weight of apples = 2.5 kg
Weight of oranges = 1.75 kg
Weight of bananas = 0.9 kg
Total weight = 2.5 + 1.75 + 0.9 = 5.15 kg

Q14: A snail climbs 4.5 cm up a wall each day and slips back 2.3 cm at night. The wall is 15 cm high. How many days will it take to reach the top?

Sol: Daily climb = 4.5 cm

Nightly slip = 2.3 cm

Net gain per full day = 4.5 − 2.3 = 2.2 cm

Height needed before final day’s climb = 15 − 4.5 = 10.5 cm
Full days needed = 10.5 ÷ 2.2 = 4.77… → 5 full days
Total days = 5 + 1 = 6 days

Q15: Write 12.345 as a sum of tenths, hundredths, and thousandths.

Sol: Whole parts = 12
Tenths = 3/10 (= 0.3)
Hundredths = 4/100 (= 0.04)
Thousandths = 5/1000 (= 0.005)
As sum = Important Questions: A Peek Beyond the Point | Mathematics (Ganita Prakash) Class 7 - New NCERT= 12.345

Q16: For two decimal numbers 17.892 and 6.345, verify if their sum is greater than the sum of their whole number parts and less than the sum of their whole number parts plus 2.

Sol: Whole parts = 17 and 6
Sum of whole parts = 17 + 6 = 23
Sum of decimals = 17.892 + 6.345 = 24.237
Sum of whole parts + 2 = 23 + 2 = 25
Check = 24.237 > 23 and 24.237 < 25
Conclusion = Yes, 23 < 24.237 < 25

The document Important Questions: A Peek Beyond the Point | Mathematics (Ganita Prakash) Class 7 - New NCERT is a part of the Class 7 Course Mathematics (Ganita Prakash) Class 7 - New NCERT.
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FAQs on Important Questions: A Peek Beyond the Point - Mathematics (Ganita Prakash) Class 7 - New NCERT

1. What is the significance of the title "A Peek Beyond the Point"?
Ans. The title "A Peek Beyond the Point" suggests exploring concepts or ideas that extend beyond a certain limit or understanding. It implies a deeper investigation into subjects that may initially seem simple but have greater depth and complexity when examined closely.
2. What are some key themes discussed in "A Peek Beyond the Point"?
Ans. Key themes in "A Peek Beyond the Point" may include the exploration of perspective, the importance of looking beyond the obvious, and the value of critical thinking. These themes encourage readers to consider different viewpoints and to seek deeper meanings in various contexts.
3. How does the article encourage critical thinking among students?
Ans. The article encourages critical thinking by prompting students to question their assumptions, analyze information critically, and consider multiple perspectives. It may provide examples or scenarios that challenge students to think beyond surface-level interpretations and engage in deeper analysis.
4. What are some practical applications of the concepts discussed in the article?
Ans. The concepts discussed in "A Peek Beyond the Point" can be applied in various areas such as academic research, problem-solving in everyday life, and decision-making processes. By applying critical thinking and exploring beyond the obvious, students can enhance their understanding and approach to various subjects.
5. How can teachers incorporate the ideas from "A Peek Beyond the Point" into their lessons?
Ans. Teachers can incorporate ideas from the article by designing activities that promote inquiry-based learning, encouraging discussions that require students to defend their viewpoints, and using real-world examples that illustrate the importance of looking beyond initial impressions. This can help foster a classroom environment that values exploration and critical analysis.
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