A Peek Beyond the Point | IT

Question 15

Observe the given sequences of numbers. Identify the change after each term and extend the pattern:

(a) 44, 43104\dfrac{3}{10}, 46104\dfrac{6}{10}, ______, ______, ______

(b) 82108\dfrac{2}{10}, 87108\dfrac{7}{10}, 92109\dfrac{2}{10}, ______, ______, ______

(c) 76107\dfrac{6}{10}, 87108\dfrac{7}{10}, ______, ______, ______, ______

(d) 57105\dfrac{7}{10}, 53105\dfrac{3}{10}, ______, ______, ______, ______

(e) 1351013\dfrac{5}{10}, 1313, 1251012\dfrac{5}{10}, ______, ______, ______, ______

(f) 1151011\dfrac{5}{10}, 1041010\dfrac{4}{10}, 93109\dfrac{3}{10}, ______, ______, ______

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Solution
Understand the Question
  • To extend a numerical sequence, determine the rule of change (the common difference) between consecutive terms by subtracting the earlier term from the succeeding term.
  • If the difference is positive, the sequence increases (add the value each time).
  • If the difference is negative, the sequence decreases (subtract the value each time).
  • When working with mixed fractions having tenths, regroup 1010 tenths as 11 whole when adding, and regroup 11 whole as 1010 tenths when subtracting.

(a) 44, 43104\dfrac{3}{10}, 46104\dfrac{6}{10}, ______, ______, ______

Step 1 · Find the Common Difference

Find the difference between consecutive terms:

4310−4=3104\dfrac{3}{10} - 4 = \dfrac{3}{10} 4610−4310=3104\dfrac{6}{10} - 4\dfrac{3}{10} = \dfrac{3}{10}

Rule: Add 310\text{Rule: Add } \dfrac{3}{10}

Step 2 · Calculate the Next Three Terms

Fourth term:

4610+310=49104\dfrac{6}{10} + \dfrac{3}{10} = 4\dfrac{9}{10}

Fifth term:

4910+310=41210=4+1210=5210\begin{aligned} 4\dfrac{9}{10} + \dfrac{3}{10} &= 4\dfrac{12}{10} \\[0.6em] &= 4 + 1\dfrac{2}{10} \\[0.6em] &= 5\dfrac{2}{10} \end{aligned}

Sixth term:

5210+310=55105\dfrac{2}{10} + \dfrac{3}{10} = 5\dfrac{5}{10}
Answer

(a) Pattern: Add 310\dfrac{3}{10}; Next terms: 49104\dfrac{9}{10}, 52105\dfrac{2}{10}, 55105\dfrac{5}{10}

(b) 82108\dfrac{2}{10}, 87108\dfrac{7}{10}, 92109\dfrac{2}{10}, ______, ______, ______

Step 1 · Find the Common Difference

Find the difference between consecutive terms:

8710−8210=5108\dfrac{7}{10} - 8\dfrac{2}{10} = \dfrac{5}{10} 9210−8710=81210−8710=510\begin{aligned} 9\dfrac{2}{10} - 8\dfrac{7}{10} &= 8\dfrac{12}{10} - 8\dfrac{7}{10} \\[0.6em] &= \dfrac{5}{10} \end{aligned}

Rule: Add 510\text{Rule: Add } \dfrac{5}{10}

Step 2 · Calculate the Next Three Terms

Fourth term:

9210+510=97109\dfrac{2}{10} + \dfrac{5}{10} = 9\dfrac{7}{10}

Fifth term:

9710+510=91210=10210\begin{aligned} 9\dfrac{7}{10} + \dfrac{5}{10} &= 9\dfrac{12}{10} \\[0.6em] &= 10\dfrac{2}{10} \end{aligned}

Sixth term:

10210+510=1071010\dfrac{2}{10} + \dfrac{5}{10} = 10\dfrac{7}{10}
Answer

(b) Pattern: Add 510\dfrac{5}{10}; Next terms: 97109\dfrac{7}{10}, 1021010\dfrac{2}{10}, 1071010\dfrac{7}{10}

(c) 76107\dfrac{6}{10}, 87108\dfrac{7}{10}, ______, ______, ______, ______

Step 1 · Find the Common Difference

Find the difference between the given terms:

8710−7610=(8−7)+(710−610)=1+110=1110\begin{aligned} 8\dfrac{7}{10} - 7\dfrac{6}{10} &= (8 - 7) + \left(\dfrac{7}{10} - \dfrac{6}{10}\right) \\[0.6em] &= 1 + \dfrac{1}{10} \\[0.6em] &= 1\dfrac{1}{10} \end{aligned}

Rule: Add 1110\text{Rule: Add } 1\dfrac{1}{10}

Step 2 · Calculate the Next Four Terms

Third term:

8710+1110=(8+1)+(710+110)=9810\begin{aligned} 8\dfrac{7}{10} + 1\dfrac{1}{10} &= (8 + 1) + \left(\dfrac{7}{10} + \dfrac{1}{10}\right) \\[0.6em] &= 9\dfrac{8}{10} \end{aligned}

Fourth term:

9810+1110=(9+1)+(810+110)=10910\begin{aligned} 9\dfrac{8}{10} + 1\dfrac{1}{10} &= (9 + 1) + \left(\dfrac{8}{10} + \dfrac{1}{10}\right) \\[0.6em] &= 10\dfrac{9}{10} \end{aligned}

Fifth term:

10910+1110=(10+1)+(910+110)=11+1010=11+1=12\begin{aligned} 10\dfrac{9}{10} + 1\dfrac{1}{10} &= (10 + 1) + \left(\dfrac{9}{10} + \dfrac{1}{10}\right) \\[0.6em] &= 11 + \dfrac{10}{10} \\[0.6em] &= 11 + 1 \\[0.6em] &= 12 \end{aligned}

Sixth term:

12+1110=1311012 + 1\dfrac{1}{10} = 13\dfrac{1}{10}
Answer

(c) Pattern: Add 11101\dfrac{1}{10}; Next terms: 98109\dfrac{8}{10}, 1091010\dfrac{9}{10}, 1212, 1311013\dfrac{1}{10}

(d) 57105\dfrac{7}{10}, 53105\dfrac{3}{10}, ______, ______, ______, ______

Step 1 · Find the Common Difference

Since the sequence decreases, find the amount subtracted:

5710−5310=4105\dfrac{7}{10} - 5\dfrac{3}{10} = \dfrac{4}{10}

Rule: Subtract 410\text{Rule: Subtract } \dfrac{4}{10}

Step 2 · Calculate the Next Four Terms

Third term:

5310−410=41310−410=4910\begin{aligned} 5\dfrac{3}{10} - \dfrac{4}{10} &= 4\dfrac{13}{10} - \dfrac{4}{10} \\[0.6em] &= 4\dfrac{9}{10} \end{aligned}

Fourth term:

4910−410=45104\dfrac{9}{10} - \dfrac{4}{10} = 4\dfrac{5}{10}

Fifth term:

4510−410=41104\dfrac{5}{10} - \dfrac{4}{10} = 4\dfrac{1}{10}

Sixth term:

4110−410=31110−410=3710\begin{aligned} 4\dfrac{1}{10} - \dfrac{4}{10} &= 3\dfrac{11}{10} - \dfrac{4}{10} \\[0.6em] &= 3\dfrac{7}{10} \end{aligned}
Answer

(d) Pattern: Subtract 410\dfrac{4}{10}; Next terms: 49104\dfrac{9}{10}, 45104\dfrac{5}{10}, 41104\dfrac{1}{10}, 37103\dfrac{7}{10}

(e) 1351013\dfrac{5}{10}, 1313, 1251012\dfrac{5}{10}, ______, ______, ______, ______

Step 1 · Find the Common Difference

Find the amount subtracted between consecutive terms:

13510−13=51013\dfrac{5}{10} - 13 = \dfrac{5}{10} 13−12510=121010−12510=510\begin{aligned} 13 - 12\dfrac{5}{10} &= 12\dfrac{10}{10} - 12\dfrac{5}{10} \\[0.6em] &= \dfrac{5}{10} \end{aligned}

Rule: Subtract 510\text{Rule: Subtract } \dfrac{5}{10}

Step 2 · Calculate the Next Four Terms

Fourth term:

12510−510=1212\dfrac{5}{10} - \dfrac{5}{10} = 12

Fifth term:

12−510=111010−510=11510\begin{aligned} 12 - \dfrac{5}{10} &= 11\dfrac{10}{10} - \dfrac{5}{10} \\[0.6em] &= 11\dfrac{5}{10} \end{aligned}

Sixth term:

11510−510=1111\dfrac{5}{10} - \dfrac{5}{10} = 11

Seventh term:

11−510=101010−510=10510\begin{aligned} 11 - \dfrac{5}{10} &= 10\dfrac{10}{10} - \dfrac{5}{10} \\[0.6em] &= 10\dfrac{5}{10} \end{aligned}
Answer

(e) Pattern: Subtract 510\dfrac{5}{10}; Next terms: 1212, 1151011\dfrac{5}{10}, 1111, 1051010\dfrac{5}{10}

(f) 1151011\dfrac{5}{10}, 1041010\dfrac{4}{10}, 93109\dfrac{3}{10}, ______, ______, ______

Step 1 · Find the Common Difference

Find the amount subtracted between consecutive terms:

11510−10410=(11−10)+(510−410)=1+110=1110\begin{aligned} 11\dfrac{5}{10} - 10\dfrac{4}{10} &= (11 - 10) + \left(\dfrac{5}{10} - \dfrac{4}{10}\right) \\[0.6em] &= 1 + \dfrac{1}{10} \\[0.6em] &= 1\dfrac{1}{10} \end{aligned} 10410−9310=(10−9)+(410−310)=1+110=1110\begin{aligned} 10\dfrac{4}{10} - 9\dfrac{3}{10} &= (10 - 9) + \left(\dfrac{4}{10} - \dfrac{3}{10}\right) \\[0.6em] &= 1 + \dfrac{1}{10} \\[0.6em] &= 1\dfrac{1}{10} \end{aligned}

Rule: Subtract 1110\text{Rule: Subtract } 1\dfrac{1}{10}

Step 2 · Calculate the Next Three Terms

Fourth term:

9310−1110=(9−1)+(310−110)=8210\begin{aligned} 9\dfrac{3}{10} - 1\dfrac{1}{10} &= (9 - 1) + \left(\dfrac{3}{10} - \dfrac{1}{10}\right) \\[0.6em] &= 8\dfrac{2}{10} \end{aligned}

Fifth term:

8210−1110=(8−1)+(210−110)=7110\begin{aligned} 8\dfrac{2}{10} - 1\dfrac{1}{10} &= (8 - 1) + \left(\dfrac{2}{10} - \dfrac{1}{10}\right) \\[0.6em] &= 7\dfrac{1}{10} \end{aligned}

Sixth term:

7110−1110=(7−1)+(110−110)=6\begin{aligned} 7\dfrac{1}{10} - 1\dfrac{1}{10} &= (7 - 1) + \left(\dfrac{1}{10} - \dfrac{1}{10}\right) \\[0.6em] &= 6 \end{aligned}
Answer

(f) Pattern: Subtract 11101\dfrac{1}{10}; Next terms: 82108\dfrac{2}{10}, 71107\dfrac{1}{10}, 66

Common Mistakes
  • Borrowing/Regrouping Errors: When subtracting 410\dfrac{4}{10} from 53105\dfrac{3}{10}, remember to borrow 11 whole as 1010\dfrac{10}{10} to make 413104\dfrac{13}{10} before subtracting.
  • Whole Number Form: Forgetting that 1010=1\dfrac{10}{10} = 1, so 11+1010=1211 + \dfrac{10}{10} = 12, and 7010=6+1=67\dfrac{0}{10} = 6 + 1 = 6 (or 7−1=67-1=6).

More questions in IT

Q1

In the following figure, screws are placed above a scale. Measure them and write their length in the space provided.

Q2

Which scale helped you measure the length of the screws accurately? Why?

Q3

What is the meaning of 2710 cm2 \dfrac{7}{10} \text{ cm} (the length of the first screw)?

Q4

Context: As seen on the ruler, the unit length between two consecutive numbers is divided into 10 equal parts. To get the length 2710 cm2\dfrac{7}{10}\text{ cm}, we go from 0 to 2 and then take seven parts of 110\dfrac{1}{10}. The length of the screw is 2 cm2\text{ cm} and 710 cm\dfrac{7}{10}\text{ cm}. Similarly, we can make sense of the length 3210 cm3\dfrac{2}{10}\text{ cm}.

Q. Can you explain why the unit was divided into smaller parts to measure the screws?

Q5

Measure the following objects using a scale and write their measurements in centimeters (as shown earlier for the lengths of the screws): pen, sharpener, and any other object of your choice.

Q6

Write the measurements of the objects shown in the picture:

Q7

For the objects shown below, write their lengths in two ways and read them aloud. An example is given for the USB cable. (Note that the unit length used in each diagram is not the same).

Q8

Arrange these lengths in increasing order:

(a) 910\dfrac{9}{10}

(b) 17101 \dfrac{7}{10}

(c) 13010\dfrac{130}{10}

(d) 1311013 \dfrac{1}{10}

(e) 1051010 \dfrac{5}{10}

(f) 76107 \dfrac{6}{10}

(g) 67106 \dfrac{7}{10}

(h) 410\dfrac{4}{10}

Q9

Arrange the following lengths in increasing order: 41104\dfrac{1}{10}, 410\dfrac{4}{10}, 4110\dfrac{41}{10}, 4111041\dfrac{1}{10}.

Q10

The lengths of the body parts of a honeybee are given. Find its total length.

Head: 23102\dfrac{3}{10} units Thorax: 54105\dfrac{4}{10} units Abdomen: 75107\dfrac{5}{10} units

Q11

The length of Shylaja's hand is 1241012\dfrac{4}{10} units, and her palm is 67106\dfrac{7}{10} units, as shown in the picture. What is the length of the longest (middle) finger?

Q12

Context: The length of Shylaja's hand is 1241012 \dfrac{4}{10} units, and her palm is 67106 \dfrac{7}{10} units.

Q. Try computing the difference by converting both lengths to tenths.

Q13

A Celestial Pearl Danio's length is 2410 cm2\dfrac{4}{10} \text{ cm}, and the length of a Philippine Goby is 910 cm\dfrac{9}{10} \text{ cm}. What is the difference in their lengths?

Q14

How big are these fish compared to your finger?

Q15

Observe the given sequences of numbers. Identify the change after each term and extend the pattern:

(a) 44, 43104\dfrac{3}{10}, 46104\dfrac{6}{10}, ______, ______, ______

(b) 82108\dfrac{2}{10}, 87108\dfrac{7}{10}, 92109\dfrac{2}{10}, ______, ______, ______

(c) 76107\dfrac{6}{10}, 87108\dfrac{7}{10}, ______, ______, ______, ______

(d) 57105\dfrac{7}{10}, 53105\dfrac{3}{10}, ______, ______, ______, ______

(e) 1351013\dfrac{5}{10}, 1313, 1251012\dfrac{5}{10}, ______, ______, ______, ______

(f) 1151011\dfrac{5}{10}, 1041010\dfrac{4}{10}, 93109\dfrac{3}{10}, ______, ______, ______

Q16

What is the length of this smaller part? How many such smaller parts make a unit length?

Q17

How many one-hundredths make one-tenth? Can we also say that the length is 4 units and 45 one-hundredths?

Q18

Observe the figure below. Notice the markings and the corresponding lengths written in the boxes when measured from 00. Fill the lengths in the empty boxes.

Q19

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

Q20

In each group, identify the longest and the shortest lengths. Mark each length on the scale.

(a) 310,3100,33100\dfrac{3}{10}, \dfrac{3}{100}, \dfrac{33}{100}

(b) 3110,3010,13103\dfrac{1}{10}, \dfrac{30}{10}, 1\dfrac{3}{10}

(c) 45100,54100,510,410\dfrac{45}{100}, \dfrac{54}{100}, \dfrac{5}{10}, \dfrac{4}{10}

(d) 3610,36100,361061003\dfrac{6}{10}, 3\dfrac{6}{100}, 3\dfrac{6}{10}\dfrac{6}{100}

(e) 810,2100,9100,18100\dfrac{8}{10}, \dfrac{2}{100}, \dfrac{9}{100}, 1\dfrac{8}{100}

(f) 73105100,7510,7411007\dfrac{3}{10}\dfrac{5}{100}, 7\dfrac{5}{10}, 7\dfrac{41}{100}

(g) 651015100,587100,57100\dfrac{65}{10}\dfrac{15}{100}, 5\dfrac{87}{100}, 5\dfrac{7}{100}

Q21

Context: What will be the sum of 15 310 410015\ \dfrac{3}{10}\ \dfrac{4}{100} and 2 610 81002\ \dfrac{6}{10}\ \dfrac{8}{100}? This can be solved in different ways. Some are shown below.

Are both these methods different?

Q22

Observe the addition done below for 483+268483 + 268. Do you see any similarities between the methods shown above?

(400+80+3)+(200+60+8)=(400+200)+(80+60)+(3+8)=600+140+11=600+150+1=700+50+1=751\begin{aligned} &(400 + 80 + 3) + (200 + 60 + 8) \\ &= (400 + 200) + (80 + 60) + (3 + 8) \\ &= 600 + 140 + 11 \\ &= 600 + 150 + 1 \\ &= 700 + 50 + 1 \\ &= 751 \end{aligned}
Q25

Can we extend this further?

Q26

We can ask similar questions about fractional parts:

(a) How many thousandths make one unit? (b) How many thousandths make one tenth? (c) How many thousandths make one hundredth? (d) How many tenths make one ten? (e) How many hundredths make one ten?

Q27

Make a few more questions of this kind and answer them.

Q28

Can the quantity 42104 \dfrac{2}{10} be written as 42 (skipping the 110\dfrac{1}{10} in 2×1102 \times \dfrac{1}{10})?

If yes, how would we know if 42 means 4 tens and 2 units or it means 4 units and 2 tenths?

Q29

Make a place value table similar to the one above. Write each quantity in decimal form and in terms of place value, and read the number:

(a) 2 ones, 3 tenths and 5 hundredths

(b) 1 ten and 5 tenths

(c) 4 ones and 6 hundredths

(d) 1 hundred, 1 one and 1 hundredth

(e) 8100\dfrac{8}{100} and 910\dfrac{9}{10}

(f) 5100\dfrac{5}{100}

(g) 110\dfrac{1}{10}

(h) 211002\dfrac{1}{100}, 41104\dfrac{1}{10} and 7710007\dfrac{7}{1000}

Q30

Write these quantities in decimal form:

(a) 234 hundredths

(b) 105 tenths

Q31

Fill in the blanks below (mm↔cm\text{mm} \leftrightarrow \text{cm})

Q32

Fill in the blanks below (cm↔m\text{cm} \leftrightarrow \text{m}):

Q33

How many mm does 1 meter have?

Q34

Can we write 1 mm=11000 m1\text{ mm} = \dfrac{1}{1000}\text{ m}?

Q35

Fill in the blanks below (g⇄kg\text{g} \rightleftarrows \text{kg})

Q36

Fill in the blanks below (rupee ↔\leftrightarrow paise)

Q37

Name all the divisions between 11 and 1.11.1 on the number line.

Q38

Identify and write the decimal numbers against the letters.

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