CBSE Important Questions Class 7 Maths Chapter 3 A Peek Beyond the Point
CBSE Important Questions Class 7 Maths Chapter 3 – A Peek Beyond the Point
Class 7 Maths Chapter 3 A Peek Beyond the Point is a chapter of the new NCERT (National Council of Educational Research and Training) textbook Ganita Prakash. It introduces tenths and hundredths, the parts of a number that come after the decimal point. Students learn to read such numbers in words, to add and subtract them, to compare and arrange lengths, and to read measurements on a ruler or scale.
These CBSE important questions for Class 7 Maths Chapter 3 include writing numbers in words, sums and differences of tenths and hundredths, comparing fish lengths, number sequences, marking lengths on a scale and reading a ruler. Every question has a clear step-by-step answer in simple English, so students can practise on their own, parents can check the working, and teachers can use them in class. A free printable PDF of these important questions is also available on this page.
CBSE Important Questions Class 7 Maths Chapter 3 A Peek Beyond the Point
Q1. Write the given lengths in words.
(i) 2 + 1/10 + 5/100
(ii) 6 + 19/100
(iii) 118/100
Answer: (i) 2 + 1/10 + 5/100: Two and one-tenth and five-hundredths
(ii) 6 + 19/100: Six and nineteen-hundredths
(iii) 118/100: One hundred and eighteen-hundredths
Q2. Find the sum and the difference for the following.
(i) 15 + 6/10 + 15/100 and 13 + 2/10 + 1/100 (sum)
(ii) 8 + 15/100 and 2 + 3/100 (difference)
Answer: (i) Add the whole numbers, the tenths and the hundredths separately:
(15 + 13) + (6/10 + 2/10) + (15/100 + 1/100) = 28 + 8/10 + 16/100
16/100 = 10/100 + 6/100 = 1/10 + 6/100, so the sum = 28 + 9/10 + 6/100
(ii) (8 − 2) + (15/100 − 3/100) = 6 + 12/100
The difference = 6 + 12/100
Q3. A Celestial Pearl Danio's length is 3 6/10 cm, and the length of the Philippine Goby is 7/10 cm. What is the difference in their lengths?
Answer: Length of the Celestial Pearl Danio = 3 6/10 cm = 36/10 cm
Length of the Philippine Goby = 7/10 cm
Difference = 36/10 − 7/10 = 29/10 cm = 2 9/10 cm
Q4. Arrange the given lengths in ascending order.
7 6/10, 6 7/10, 130/10, 13 1/10
Answer: Write all the lengths as fractions with denominator 10:
7 6/10 = 76/10, 6 7/10 = 67/10, 130/10 stays 130/10, 13 1/10 = 131/10
When the denominators are the same, the fraction with the smaller numerator is smaller. Arranging the numerators: 67, 76, 130, 131.
So the ascending order is 6 7/10, 7 6/10, 130/10, 13 1/10.
Q5. In the given sequence, identify the change after each term and write the next three terms.
13 6/10, 12 5/10, 11 4/10
Answer: 13 6/10 − 12 5/10 = 1 1/10 and 12 5/10 − 11 4/10 = 1 1/10. So each term is 1 1/10 less than the term before it (the whole number goes down by 1 and the tenths go down by 1).
The next three terms are 10 3/10, 9 2/10 and 8 1/10.
The sequence is 13 6/10, 12 5/10, 11 4/10, 10 3/10, 9 2/10, 8 1/10.
Q6. We can ask similar questions about fractional parts:
(a) How many hundredths make one unit?
(b) How many tenths make one unit?
(c) How many tenths make one hundredth?
(d) How many hundredths make one tenth?
Answer: (a) Since 1/100 × 100 = 1, 100 hundredths make one unit.
(b) Since 1/10 × 10 = 1, 10 tenths make one unit.
(c) Since 1/10 × 1/10 = 1/100, one-tenth of a tenth makes one hundredth.
(d) Since 1/100 × 10 = 1/10, 10 hundredths make one tenth.
Q7. Find the sums and differences:
(a) 3/10 + 3 + 4/100
(b) (9 + 5/10 + 7/100) + (2 + 1/10 + 3/100)
(c) (15 + 6/10 + 4/100) + (14 + 3/10 + 6/100)
(d) (7 + 7/100) − (4 + 4/100)
(e) (8 + 6/100) − (5 + 3/100)
(f) (12 + 6/10 + 2/100) − (9/10 + 9/100)
Answer: (a) 3/10 = 30/100, so 3 + 30/100 + 4/100 = 3 + 34/100 = 3 34/100
(b) (9 + 2) + (5/10 + 1/10) + (7/100 + 3/100) = 11 + 6/10 + 10/100 = 11 + 6/10 + 1/10 = 11 + 7/10 = 11 7/10
(c) (15 + 14) + (6/10 + 3/10) + (4/100 + 6/100) = 29 + 9/10 + 10/100 = 29 + 9/10 + 1/10 = 29 + 1 = 30
(d) (7 − 4) + (7/100 − 4/100) = 3 + 3/100 = 3 3/100
(e) (8 − 5) + (6/100 − 3/100) = 3 + 3/100 = 3 3/100
(f) 12 + 6/10 + 2/100 = 12 + 62/100 and 9/10 + 9/100 = 99/100. Since 62/100 is less than 99/100, borrow 1 from 12: 11 + 162/100 − 99/100 = 11 + 63/100 = 11 63/100
Q8. Solve this by converting to hundredths: 25 9/10 − (6 + 4/10 + 7/100)
Answer: 25 9/10 = 25 90/100 (because 9/10 = 90/100)
6 + 4/10 + 7/100 = 6 + 40/100 + 7/100 = 6 47/100
25 90/100 − 6 47/100 = (25 − 6) + (90/100 − 47/100) = 19 + 43/100 = 19 43/100
Q9. (a) What will be the sum of 15 + 3/10 + 4/100 and 2 + 6/10 + 8/100?
(b) Are both these methods different?
Answer: (a) Method 1 (adding the parts separately):
(15 + 2) + (3/10 + 6/10) + (4/100 + 8/100) = 17 + 9/10 + 12/100
12/100 = 10/100 + 2/100 = 1/10 + 2/100, so the sum = 17 + 9/10 + 1/10 + 2/100 = 17 + 10/10 + 2/100 = 17 + 1 + 2/100 = 18 2/100
Method 2 (column method): write the numbers one below the other, add hundredths (4 + 8 = 12), tenths (3 + 6 = 9) and units (15 + 2 = 17). Carry 10 hundredths to the tenths (10 tenths), then carry 10 tenths to the units: 18 2/100.
(b) No, both methods use the same logic and give the same answer. They differ only in layout: Method 1 is written across the page, Method 2 is written in columns like we do for whole numbers.
Q10. In each group, identify the longest and the shortest lengths. Mark each length on the scale.
(a) 3/10, 3/100, 33/100
(b) 3 1/10, 30/10, 1 3/10
(c) 45/100, 54/100, 5/10, 4/10
(d) 3 6/10, 3 6/100, 3 6/10 6/100
(e) 8/10 2/100, 9/100, 1 8/100
(f) 7 3/10 5/100, 7 5/10, 7 41/100
(g) 65/10 15/100, 5 87/100, 5 7/100
(a) 3/10, 3/100, 33/100 3/10 = 30/100. Comparing 30/100, 3/100 and 33/100: longest length = 33/100, shortest length = 3/100.

(b) 3 1/10, 30/10, 1 3/10 3 1/10 = 31/10, 30/10 = 3, 1 3/10 = 13/10. Longest length = 3 1/10, shortest length = 1 3/10.

(c) 45/100, 54/100, 5/10, 4/10 5/10 = 50/100 and 4/10 = 40/100. Longest length = 54/100, shortest length = 40/100 (that is, 4/10).

(d) 3 6/10, 3 6/100, 3 6/10 6/100 3 6/10 = 36/10, 3 6/100 = 30/10 + 6/100, 3 6/10 6/100 = 36/10 + 6/100. Largest number = 3 6/10 6/100, smallest number = 3 6/100.

(e) 8/10 2/100, 9/100, 1 8/100 8/10 + 2/100 = 82/100, 9/100 and 1 8/100 = 108/100. Largest number = 1 8/100, smallest number = 9/100.

(f) 7 3/10 5/100, 7 5/10, 7 41/100 7 3/10 5/100 = 7 35/100, 7 5/10 = 7 50/100, 7 41/100. Largest number = 7 5/10, smallest number = 7 3/10 5/100.

(g) 65/10 15/100, 5 87/100, 5 7/100 65/10 + 15/100 = 66/10 + 5/100 = 6 65/100, 5 87/100 and 5 7/100. Largest number = 65/10 15/100, smallest number = 5 7/100.

Q11. For the lengths shown below, write the measurements and read out the measures in words.


Answer: First ruler: 5 3/10 7/100 = Five and three-tenths and seven-hundredths.
Second ruler: 15 3/100 = Fifteen and three-hundredths.
Q12. Observe the figure below. Notice the markings and the corresponding lengths written in the boxes when measured from 0. Fill the lengths in the empty boxes.

Answer: The distance from 0 to 1 is divided into 100 small parts, so each small mark is 1/100. Counting the marks from 0, the empty boxes are 55/100, 155/100, 164/100, 202/100 and 240/100.

Related Study Material on Extramarks
- CBSE Important Questions Class 7 Maths Chapter 1
- CBSE Important Questions Class 7 Maths Chapter 2
- CBSE Important Questions Class 7 Maths Chapter 4
- CBSE Important Questions Class 7 Maths Chapter 5
- CBSE Important Questions Class 7 Maths Chapter 6
- CBSE Important Questions Class 7 Maths Chapter 7
- CBSE Important Questions Class 7 Maths Chapter 8
- CBSE Important Questions Class 7 Maths Chapter 9
- CBSE Important Questions Class 7 Maths Chapter 10
- CBSE Important Questions Class 7 Maths Chapter 11
Q.1 Probability of any event always lies between _____________.
Marks:1
1. 0 and 1
2. 1 and 2
3. 2 and 3
4. 3 and 4
Ans
1. 0 and 1
Explanation
Probability always lies between 0 and 1.
Q.2 When a coin is thrown, then the probability of getting head is ______________.
Marks:1
0
1
Ans 3.
Explanation
When s coin is thrown, it has two possible outcomes i.e. Head or Tail
Therefore, the required probability =
Q.3 The heights of 15 students are given below:
165, 155, 168, 160, 163, 162, 165, 168, 156, 159, 160, 164, 163, 165, 160.
Find:
(i) the range of heights
(ii) the mode
(iii) the median
Marks:3
Ans
Arranging in ascending order 155, 156, 159, 160, 160, 160, 162, 163, 163, 164, 165, 165, 165, 168, 168
(i) Range = 168 – 155 = 13
(ii) Mode = 160 and 165 occurs 3 times. Therefore, modes are 160 and 165 both.
(iii) Median = 163 (central value)
