CBSE Important Questions Class 7 Maths Chapter 11 Another Peek Beyond the Point

CBSE Important Questions Class 7 Maths Chapter 11 – Another Peek Beyond the Point

Class 7 Maths Chapter 11 Another Peek Beyond the Point is a chapter of the new NCERT (National Council of Educational Research and Training) textbook Ganita Prakash. It takes students beyond whole numbers into tenths and hundredths, which are 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 them, and to convert tenths into hundredths.

These CBSE important questions for Class 7 Maths Chapter 11 include reading numbers in words, sums and differences of tenths and hundredths, comparing lengths of fish, arranging lengths in order, number sequences and unit conversions. 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 11 Another 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

Table showing the lengths in units, tenths and 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

Bar graph comparing the lengths of the two fish

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.

Number line showing the four lengths in ascending order

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.

Sequence decreasing by 1 1/10 each time

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

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