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

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.

Scale showing 3/100, 3/10 and 33/100

(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.

Scale showing 1 3/10, 30/10 and 3 1/10

(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).

Scale showing 4/10, 45/100, 5/10 and 54/100

(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.

Scale from 30/10 to 40/10 showing the three lengths

(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.

Scale showing 9/100, 8/10 2/100 and 1 8/100

(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.

Scale from 68/10 showing the three lengths near 7

(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.

Scale showing 5 7/100, 6, 64/10 and 65/10 15/100

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

Ruler showing a length between 5 and 6

Ruler showing a length near 15

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.

Number line with some lengths marked and empty boxes to fill

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.

Number line with all the lengths filled in

Related Study Material on Extramarks

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

14

12

1

Ans  3.

12

Explanation

When s coin is thrown, it has two possible outcomes i.e. Head or Tail

Therefore, the required probability =

12

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)

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