By using Ganita Prakash Class 7 Solutions and Part 2 Chapter 4 Another Peek Beyond the Point Class 7 Question Answer, students can improve their problem-solving skills.
Class 7 Maths Ganita Prakash Part 2 Chapter 4 Solutions
Class 7 Maths Another Peek Beyond the Point Solutions
Class 7 Ganita Prakash Part 2 Chapter 4 Solutions Another Peek Beyond the Point
Figure it Out (Pages: 73-74)
Question 1.
Recall that a tenth is 0.1, a hundredth is 0.01, and so on. Find the following products in tenths, hundredths and so on:
(a) 6 × 4 tenths = 24 tenths
(b) 7 × 0.3
(c) 9 × 5 hundredths;
Solution:
(a) 6 × 4 tenths = 24 tenths
⇒ 6 × 4 × 10 = 24 × 10
⇒ 240 = 240 = 24 tenths = 2.4
(b) 7 × 0.3
⇒ 7 × 3 = 21 (0.3 is equivalent to 3 tenths)
⇒ 21 tenths = In decimal form 2.1
(c) 9 × 5 hundredths
⇒ 9 × 5 = 45 hundredths
In decimal form this is 0.45
Question 2.
Find the products :
(a) 27.34 × 6
(b) 4.23 × 3.7
(c) 0.432 × 0.23
Solution:
(a) 27.34 × 6
⇒ 2734 × 6 = 16404
Since there are two digits after the decimal point in 27.34
∴ 164.04
(b) 4.23 × 3.7
⇒ 423 × 37 = 15651
Since we count total number of digits after the decimal points in the factors (two in 4.23 and one in 3.7, total of three)
∴ 4.23 × 3.7 = 15.651
(c) 0.432 × 0.23
⇒ 432 × 23 = 9936
We count the total number of digits after the decimal points (three is 0.432 and two in 0.23)
⇒ 0.432 × 0.23 = 0.9936
Question 3.
Thejus needs 1.65 m of cloth for a shirt. How many metres of cloth are needed for 3 shirts?
Solution:
For 1 shirt Thejus needs = 1.65 m of cloth
For 3 shirts Thejus needs = 3 × 1.65 m
= 3 × 165
= 495
Since 1.65 has two digits after the decimal point.
∴ 1.65 × 3 = 4.95
∴ Thejus needs 4.95 meters of cloth for 3 shirts
Question 4.
Meenu bought 4 notebooks and 3 erasers. The cost of each book was ₹15.50 and each eraser was ₹2.75. How much did she spend in all?
Solution:
Cost of 1 notebook = ₹15.50
Cost of 4 notebooks = 4 × 15.50 = ₹ 62.00
Meenu bought 4 notebooks
Cost of 1 eraser = ₹ 2.75
Cost of 3 erasers as she bought 3 = 2.75 × 3
= 8.25
So, total money spent = Cost of notebook + Cost of erasers
Total amount spent = 62.00, 8.25 = 70.25
∴ Meenu has spent a total of ₹70.25 in all
Question 5.
The thickness of a rupee coin is 1.45 mm. What is the total height of the cylinder formed by placing 36 rupee coins one over the other? Write the answer in centimeters.
Solution:
Total height (mm) = Number of coins × Thickness per coin
⇒ 36 × 1.45 = 52.2 mm
To convert the height to centimeters
We know that 1 cm = 10 mm
To convert mm’s to cm’s, we divide the value in mm’s by 10
∴ Total height (cm) = \(\frac{\text { Total height }(\mathrm{mm})}{10}\)
⇒\(\frac{52.2}{10}\) = 5.22 cm
So, the total height of the cylinder formed by the coins is 5.22 cm
Question 6.
The price of 1 kg of oranges is ₹56.50. What is the price of 2.250 kg of oranges? Can we write 56.50 as 56.5 and 2.250 as 2.25 and multiply? Will we get the same product? Why?
Solution:
Cost of 1 orange = ₹ 56.50
We need to find the price of 2.250 oranges
∴ Total price = 56.50 × 2.250
Total price = ₹ 127.125
Yes, we can write 56.50 as 56.5 and 2.250 as 2.25 and multiply them
⇒ 56.50 = 56 + \(\frac{50}{100}\)
⇒ 56 + \(\frac{5}{10}\) = 56.5
⇒ 2.250 = 2 + \(\frac{250}{100}\)
⇒ 2 + \(\frac{25}{10}\) = 2.20
Since the actual values being multiplied are identical the result of the multiplication will also be identical, because trailing zeros in a decimal number do not affect its value.
The price of 2.250 kg of oranges is ₹127.125
Question 7. Dwarakanath purchases notebooks at a wholesale price of ₹23.6 per piece and sells each notebook at ₹30/-. How much profit does he make if he sells 50 books in a week?
Solution:
We know
the C.P. (Cost Price) of 1 notebook = ₹ 23.6 and the S.P. (Selling Price) of 1 notebook
= ₹ 30
∴ Profit per notebook = SP – CP
= 30 – 23.6 = ₹ 6.4
₹6.4 is profit for per notebook
To calculate total profit for 50 notebooks
Total profit = Profit per notebook × Number of notebook sold
= 6.4 × 50 = ₹ 320
Dwarkanath makes a total profit of ₹320, if he sell 50 books in a week.
Question 8.
Given that 18 × 12 = 216, find the products:
(a) 18 × 1.2
(b) 18 × 0.12
(c) 1.8 × 1.2
(d) 0.18 × 0.12
(e) 0.018 × 0.012
(f) 1.8 × 12
In which of the cases above is the product less than 1?
Solution:
Firstly calculate for (a) and (f) as there is one digit after the decimal.
Given 18 × 12 = 216
(a) 18 × 1.2 = 21.6
(f) 1.8 × 12 = 21.6
Secondly, let’s calculate for two digits after the decimal point.
(b) 18 × 0.12=2.16
(c) 1.8 × 1.2=2.16
Thirdly, calculate (d) and (e) as there are two digits in each factor after the decimal point.
(d) 0.18 × 0.12=0.0216
(e) 0.018 × 0.012=0.000216
Question 9.
In which of the following multiplications is the product less than 1? Can you find the answer without actually doing the multiplications?
(a) 7 × 0.6
(b) 0.7 × 0.6
(c) 0.7 × 6
(d) 0.07 × 0.06
Solution:
(a) 7 × 0 . 6
Here, 7 is multiplied by 0.6 (which is between 0 and 1). The product will be less than 7. Since 0.6 > 17, the product is greater than 1 (4.2>1).
(b) 0 . 7 × 0.6
Both factors (0.7 and 0.6) are less than 1. The product must be less than 1(0.42 < 1). (c) 0 . 7 × 6 Here, 6 is multiplied by 0.7 (between 0 and 1). The product will be less than 6. Since 0.7 >16, the product is greater than 1(4.2 > 1).
(d) 0 . 0 7 × 0 . 0 6
Both factors are much less than 1. The product will be very small and less than 1 (0.0042 < 1).
The products less than 1 are found in the following multiplications :
(b) 0.7 × 0.6
(d) 0.07 × 0.06
Question 10.
Multiply the following numbers by 10, 100 and 1000 to complete the table.
Solution:
Multiplying a decimal number by 10,100, or 1000 involves shifting the decimal point to the right by the number of zeros in the multiplier (one place for 10, two for 100, and three for 1000).
Step 2: Calculate the products for all numbers
Using the rule from Step 1, the products for each number are calculated :
For 5.7 :
5.7 × 10 = 57
5.7 × 100 = 570
5.7 × 1000 = 5700
For 23.02 :
23.02 × 10 = 230.2 ;
23.02 × 100 = 2302 ;
23.02 × 1000 = 23020
For 0.92 :
0.92 × 10 = 9.2 ;
0.92 × 100 = 92 ;
0.92 × 1000 = 920
For 0.306 :
0.306 × 10 = 3.06 ;
0.306 × 100 = 30.6 ;
0.306 × 1000 = 306
For 24.67 :
24.67 × 10 = 246.7 ;
24.67 × 100 = 2467 ;
24.67 × 1000 = 24670
The completed table with the products is :
Figure it Out (Page: 83)
Question 1.
Find the quotient by converting the denominator into 1, 10, 100 or 1000 and verify the solution by the long division method (division by place value).
(a) \(\frac{18}{5}\)
(b) \(\frac{415}{4}\)
(c) \(\frac{1217}{2}\)
(d) \(\frac{4827}{8}\)
Solution:
(a) \(\frac{18}{5}\)
Converting denominator method :
To convert the denominator to 10, multiply both the numerator and the denominator by 2
\(\frac{18}{5}\) × \(\frac{2}{2}\) = \(\frac{36}{10}\) = 3.6
(As dividing by 10 shifts the decimal point one place to the left.)
Long Division Method : It will also give the same result
The quotient is 3.6
(b) \(\frac{415}{4}\)
Converting denominator method :
To convert the denominator to 100, multiply both the numerator and denominator by 25
\(\frac{415}{4}\) × \(\frac{25}{250}\) = \(\frac{10375}{100}\) = 103.75
(As dividing by 100 shifts two decimal points)
Long Division Method :
Quotient = 103.75
(c) \(\frac{1217}{2}\)
Converting denominator method :
To convert denominator to 10, multiply both numerator and denominator by 5
\(\frac{1217}{2}\) × \(\frac{5}{5}\) = \(\frac{6085}{10}\) = 608.5
(Dividing by 10 shifts decimal point one place to the left)
Long Division Method :
Quotient = 608.5
(d) Given \(\frac{4827}{8}\)
To convert the denominator 8 into 1000 , multiply both the N’ and D’ by 125.
\(\frac{4827 \times 125}{8 \times 125}\) = \(\frac{603375}{1000}\)
Verification
By following the steps:
We get 4827 ÷ 8 = 0603.375
Hence verified.
Question 2.
Choose the correct answer :
(a) \(\frac{1526}{4}\) =
(i) 38.15
(ii) 380.15
(iii) 381.5
(iv) 381.05
Solution:
Correct answer is (iii) 381.5
Explanation :
Converting denominator
To convert denominator (4) to 100 , we multiply both the numerator and the denominator by 25
\(\frac{1526}{4}\) × \(\frac{25}{25}\) = \(\frac{38150}{100}\) = 381.5
(Dividing by 100 shifts the decimal point two places to the left)
(b) \(\frac{3567}{8}\) =
(i) 4458.75
(ii) 44.5875
(iii) 445.875
(iv) 4458.75
Solution:
Correct option (iii) 445.875
Question 3.
What is the quotient ?
(a) 132 ÷ 4 =
(b) 13.2 ÷ 4 =
(c) 1.32 ÷ 4 =
(d) 0.132 ÷ 4 =
Solution:
(a) 132 ÷ 4 = quotient = 33
(b) 13.2 ÷ 4 = quotient = 3.3 (Decimal is placed after one digit)
(c) 1.32 ÷ 4 = quotient = 0.33 (Decimal is placed after two digits)
(d) 0.132 ÷ 4 = quotient = 0.033 (Decimal is placed after three digits)
Question 4.
What is the quotient ?
(a) 126 ÷ 8 =
(b) 12.6 ÷ 8 =
(c) 1.26 ÷ 8 =
(d) 0.126 ÷ 8 =
(e) 0.0126 ÷ 8 =
Solution:
(a) \(\frac{126}{8}\)
Hence quotient = 15.75
(b) \(\frac{12.6}{8}\)
Hence quotient = 1.575
(c) Here 1.26 ÷ 8
Hence quotient = 0.1575
(d) Here 0.126 ÷ 8 =
Hence quotient = 0.01575
(e) Here 0.0126 ÷ 8 =
Hence quotient = 0.001575
Figure it Out (Page : 86-87)
Question 1.
Express the following fractions in decimal form :
(a) \(\frac{2}{5}\)
(b) \(\frac{13}{4}\)
(c) \(\frac{4}{50}\)
(d) \(\frac{5}{8}\)
Solution:
(a) \(\frac{2}{5}\)
\(\frac{2}{5}\) = 0.4 (Division method)
\(\frac{2}{5}\) = \(\frac{2 × 2}{5 × 2}\) = \(\frac{4}{10}\) = 0.4
(b) \(\frac{13}{4}\)
\(\frac{13}{4}\) = 3.25 (Division method)
\(\frac{13}{4}\) = \(\frac{13×25}{4×25}\) = \(\frac{325}{100}\) = 3.25.
(c) \(\frac{4}{50}\)
\(\frac{4}{50}\) = 0.08 (Division method)
\(\frac{4}{50}\) = \(\frac{4×2}{50×2}\) = \(\frac{8}{100}\) = 0.08
(d) \(\frac{5}{8}\)
\(\frac{5}{8}\) = 0.625 (Division method)
\(\frac{5}{8}\) = \(\frac{5×125}{8×125}\) = \(\frac{625}{1000}\) = 0.625
Question 2.
Find the quotients :
(a) 24.86 ÷ 1.2
(b) 5.728 ÷ 1.52
Solution:
(a) 24.86 ÷ 1.2
To remove the decimal multiply both numbers by 10
(24.86 × 10) ÷ (1.2 × 10) = 248.6 ÷ 12
Performing the division (e.g., long division), the exact result is 20.716, a repeating decimal.
\(\frac{24.86}{1.2}\) ≈ 20.72
(b) 5.728 ÷ 1.52
To remove the decimal from the divisor, multiply both numbers by 100
(5.728 × 100) ÷ (1.52 × 100) = 572.8 ÷ 152
Performing the division
\(\frac{5.728}{1.52}\) ≈ 3.77
Question 3.
Evaluate the following using the information 156 × 12 = 1872.
(a) 15.6 × 1.2= ______
(b) 187.2 ÷ 1.2= ______
(c) 18.72 ÷ 15.6= ______
(d) 0.156 × 0.12= ______
Solution:
Given 156 × 12=1872. This relationship forms the basis for solving all parts of the problem by adjusting for the position of the decimal points or using the inverse operation of division.
(a) 15.6 × 1.2 = ______
Rewrite the expression :
= \(\frac{156}{10}\) × \(\frac{12}{10}\) = \(\frac{1872}{100}\) = 18.72
(b) 187.2 ÷ 1.2 = ______ We know that if 156 × 12=1872, or 1872 ÷ 12=156
Rewrite the expression :
= \(\frac{1872 / 10}{12 / 10}\) = \(\frac{1872}{12}\) = 156
(c) 18.72 ÷ 15.6 = ______
We use the fact that 1872 ÷ 156 = 12
Rewrite the expression :
= \(\frac{1872 / 10}{12 / 10}\) = \(\frac{1872}{156}\) × \(\frac{10}{100}\) = 12 × \(\frac{1}{2}\) = 1.2
(d) 0.156 × 0.12 = ______
Rewrite the expression :
=\(\frac{156}{1000}\) × \(\frac{12}{100}\)
=\(\frac{156 × 12}{100000}\) = \(\frac{1872}{1000000}\) = 0.01872
Question 4.
Evaluate the following :
(a) 25 ÷ ______ = 0.025
(b) 25 ÷ ______ = 250
(c) 25 ÷ ______ = 2.5
(d) 25 ÷ 10 = 25 × ______
(e) 25 ÷ 0.10 = 25 × ______
(f) 25 ÷ 0.01 = 25 × ______
Solution:
The missing values for the given expressions are: (a) 1000, (b) 0.1, (c) 10, (d) 0.1, (e) 10 To solve let’s put x as missing number
(a) 25 ÷ ….. = 0.025
25 ÷ x = 0.025
x = 25 ÷ 0.025 = 1000
(b) 25 ÷ …… = 250
25 ÷ x = 250
x = 25 ÷ 250 = 0.1
(c) 25 ÷ ….. = 2.5
25 ÷ x = 2.5
x =25 ÷ 2.5=10
(d) 25 ÷ 10 = 25 × ……
25 ÷ 10 = 25 × x
25 ÷ 10 = 25 x
2.5 =25 x
x = 2.5 ÷ 25 = 0.1 or \(\frac{1}{10}\)
(e) 25 ÷ 0.10 = 25 × …….
25 ÷ 0.10 = 25 × x
250 = 25 x
x = 250 ÷ 25 = 10
(f) 25 ÷ 0.01 = 25 × ……
25 ÷ 0.01 = 25 × x
2500 = 25 x
x = 2500 ÷ 25 = 100
Division by 0.01 is equivalent to multipication by 100.
Question 5.
Find the quotient :
(a) 2.46 ÷ 1.5=
(b) 2.46 ÷ 0.15=
(c) 2.46 ÷ 0.015=
Is the quotient obtained in 24.6 ÷ 1.5 the same as the quotient obtained in 2.46 ÷ 0.15?
Solution:
(a) To evaluate 2.46 ÷ 1.5 =
We can multiply both numbers by 10
= 1.64
(By division method)
(b) 2.46 ÷ 0.15 =
We can multiply both numbers by 100
= 16.4
(By division method)
(c) 2.46 ÷ 0.015 =
We can multiply both numbers by 1000
= 164
(By division method)
Question 6.
A 4 m long wooden block has to be cut into 5 pieces of equal length. What is the length of each piece?
Solution:
To find the length of each of the 5 equal piece
= \(\frac{4 m}{5}\) = 0.08 m
∴ Length of each piece is 0.8 m
Question 7.
If the perimeter of a regular polygon with 12 sides is 208.8 cm , what is the length of its side?
Solution:
To find the perimeter
Use this formula P = n × S
Rearrange to find the length
S = \(\frac{P}{n}\) = \(\frac{208.8 cm}{12}\)
The length its side is S = 17.4 m
Question 8.
3 litres of watermelon juice is shared among 8 friends equally. How much watermelon juice will each get? Express the quantity of juice in millilitres.
Solution:
To find the amount of juice
= \(\frac{3L}{8}\) = 0.375 L
To express quantity in mL (milliliters)
Multiply the volume by 1000 , since
1 L = 1000 mL
Volume per friend in mL = 0.375 L × 1000 = 375 mL
Each friend will get 375 mL of watermelon juice
Question 9.
A car covers 234.45 km using 12.6 litres of petrol. What is the distance travelled per litre?
Solution:
To calculate distance per litre :
= \(\frac{234.45 km}{12.6L}\)
= 18.607 km / L
= 18.61 km / L (Rounding off)
Distance travelled per litre =18.61 km / L
Question 10.
13.5 kg of flour (aata) was distributed equally among 15 students. How much flour did each student receive?
] Solution:We know that
Total amount of flour is 13.5 kg
Number of students =15
Flour per student =\(\frac{13.5 kg}{15}\) = 0.9 kg
Each student will receive 0.9 kg of flour
Completion of Numerical sequences
12 × 2 × 2 × 2 × 2 = 0.03125
15 × 5 × 5 × 5 × 5 = 0.00032
Figure it Out (Page : 93-95)
Question 1.
A 210 gram packet of peanut chikki costs ₹70.5, while a 110 gram packet of potato chips costs ₹33.25. Which is cheaper?
Solution:
Given cost of peanut is ₹ 70.5 per 210 grams.
∴ Cost per gram = \(\frac{70.5}{210}\) = 0.3357
and cost of potato chips is ₹ 33.25 for 110 grams.
∴ Cost per gram = \(\frac{33.25}{110}\) = 0.3023
∵ 0.3023<0.3357.
Hence, potato chips are cheaper.
Question 2.
Write the decimal number at the arrow mark :
Solution:
(i) Here number line is divided into 10 equal parts.
Difference between 3.2 and 3.1 = 3.2 – 3.1 = 0.1
Value of each mark = \(\frac{0.1}{10}\) = 0.01
Now arrow is on the sixth mark after 3.1
∴ Decimal number at the arrow mark
= 3.1 + 6 × 0.01
= 3.1 + 0.06
= 3.16
(ii) Here number line is divided into 10 equal parts between 2.15 and 2.17.
∴ Difference between 2.17 and 2.15
= 2.17 – 2.15 = 0.02
Value of each mark = \(\frac{0.02}{10}\) = 0.002
Arrow is on the sixth mark after 2.15
∴ Decimal number at the arrow mark
= 2.15 + 6 × 0.002 = 2.150 + 0.012
= 2.162
Question 3.
Shyamala bought 3 kg bananas at ₹30/- per kg. She counted 35 bananas in all. She sells each banana for ₹5/-. How much profit does she make selling all the bananas?
Solution:
Shyamala bought 3 kg of bananas at ₹30/-per kg.
Total Cost = (Price per kg) × (Kilograms bought)
Total Cost = ₹30/- per kg × 3 kg
Total Cost = ₹90/-
Calculate the total revenue (selling price)
She counted 35 bananas in total and sold each banana for ₹5/-.
Total Revenue = (Price per banana) × (Number of bananas sold)
Total Revenue = ₹ 5 /- per banana × 35 bananas
Total Revenue = ₹175/-
Calculate the profit
Profit is the difference between the total revenue and the total cost.
Profit = Total Revenue – Total Cost
Profit =₹ 175 /- ₹ 90 /-
Profit =₹ 85 /-
Question 4.
A teacher placed textbooks that are 2.5 cm thick on a bookshelf. The teacher wanted to place 80 textbooks on the shelf. The bookshelf is 160 cm long. How many books could be placed on the shelf? Was there any space left? If yes, how much?
Solution:
To Calculate the required shelf length for all 80 books
We know that-
Textbook thickness: 2.5 cm
Number of textbooks: 80
[-] Required length = 80 books × 2.5 cm / book
Required length = 200 cm
The bookshelf is only 160 cm long, which is less than the required 200 cm . The teacher could not place all 80 books.
To find out how many books could be placed, divide the shelf length by the book thickness:
Bookshelf length: 160 cm
Textbook thickness: 2.5 cm
Number of books that fit =160 cm / 2.5 cm / book
Number of books that fit =64 books
Calculate the remaining space
After placing 64 books, all the available space is filled. The amount of space left on the 160 cm shelf is 0 cm if the maximum number of books are placed snugly.
However, if the question asks if there was space left after attempting to place the total requested number of books (80 books), the answer is that the shelf was too short.
Question 5.
Fill in the following blanks appropriately:
Solution:
5.5 k m = 5500 m (Multiply by 1000)
5.5 × 1000 = 5500
14.5 cm = 145 mm (Multiply by 10)
14.5 × 10 = 145
9.02 m = 9020 mm (Multiply by 1000
as 1 m=100 cm = 1000 mm)
.02 × 1000 = 9020
Conversions based on division (smaller unit to larger unit) :
35 cm = 0.35 m (Divide by } 100)
35 ÷ 100 = 0.35
68 g} =0.068 kg (Divide by 1000)
68 ÷ 1000 =0.068
125.5 ml =0.1255 l (Divide by 1000)
125.5 ÷ 1000 = 0.1255
Question 6.
The following problem was set by Sridharacharya in his book, Patiganita. ” 6 \(\frac{1}{4}\) is divided by 2 \(\frac{1}{2}\), and 60 \(\frac{1}{4}\) is divided by 3 \(\frac{1}{2}\). Tell the quotients separately.” Can you try to solve it by converting the fractions into decimals?
Solution:
Given fractions 6 \(\frac{1}{4}\), 2 \(\frac{1}{2}\), 60 \(\frac{1}{4}\) and 3 \(\frac{1}{2}\) To convert them to decimals
6 \(\frac{1}{4}\) = 6 + \(\frac{1}{4}\) = 6+0.25=6.25
2 \(\frac{1}{2}\) = 2 + \(\frac{1}{2}\) = 2+0.5=2.5
60 \(\frac{1}{4}\) = 60 + \(\frac{1}{4}\) = 60+0.25=60.25
3 \(\frac{1}{2}\) = 3 + \(\frac{1}{2}\) = 3+0.5=3.5
First division
6 \(\frac{1}{4}\) by 2 \(\frac{1}{2}\) (using their decimal equivalents)
\(\frac{6.25}{2.25}\) = 2.5
The first quotient is 2.5
Second division
60 \(\frac{1}{4}\) by 3 \(\frac{1}{2}\) (using their decimal equivalents)
\(\frac{60.25}{3.5}\) = 17.21428571
The decimal is non-terminating
∴ fractional form is \(\frac{241}{14}\)
The quotients are 2.5 and \(\frac{241}{14}\) (or approximately 17.214)
Question 7.
Fill the boxes in at least 2 different ways:
(a) [ ] × [ ] = 2.4
(b) [ ] × [ ] = 14.5
Solution:
(a) [ ] × [ ] = 2.4
=1 × 2.4 = 2.4
or 2 × 1.2 = 2.4
(b) [ ] × [ ] = 14.5
= 1 × 14.5 = 14.5
or 2 × 7.25 = 14.5
Question 8.
Find the following quotients given that 756 ÷ 36 = 21 :
(a) 75.6 ÷ 3.6
(b) 7.56 ÷ 0.36
(c) 756 ÷ 0.36
(d) 75.6 ÷ 360
(e) 7560 ÷ 3.6
(f) 7.56 ÷ 0.36
Solution:
Fundamental relationship is 756 ÷ 36 = 21 We will use the properties of division and decimal points to solve. The principle is that multiplying or dividing both the dividend and the divisor by the same power of 10 keeps the quotient unchanged.
To calculate quotient for (a), (b) and (f)
We can manipulate the decimals so the numbers became 756 and 36.
Question 9.
Find the missing cells if each cell represents a ÷ b :
Solution:
The problem is based on the given fundamental division fact :
1517 ÷ 37 = 41
All other missing cells represent quotients where the dividend (a) and divisor (b) are scaled versions of 1517 and 37, respectively. The results will all be variations of 41, differing only by the position of the decimal point.
To calculate all quotients
We calculate the value for each cell using the formula a ÷ b, applying the rules of decimal division :
The missing cells are filled as shown in the table in Step 2. Each value is calculated using the division a ÷ b
Question 10.
Using the digits 2,4,5,8, and 0 fill the boxes [ ][ ].[ ]× [ ].[ ] to get the :
(a) maximum product
(b) minimum product
(c) product greater than 150
(d) product nearest to 100
(e) product nearest to 5
Solution:
Question 11.
Sort the following expressions in increasing order :
(a) 245.05 × 0.942368
(b) 245.05 × 7.9682
(c) 245.05 ÷ 7.9682
(d) 245.05 ÷ 0.942368
(e) 245.05
(f) 7.9682
Answer:
The expressions in increasing order are :
(f) 7.9682 , (c) 245.05 ÷ 7.9682, (a) 245.05 × 0.942368, (e) 245.05, (d) 245.05 ÷ 0.942368, and (b) 245.05 × 7.9682
The sorting is based on understanding how operations with numbers greater than 1 and less than 1 affect the magnitude of the result
Numbers less than 1 (0.942368): Multiplying by a number less than 1 decreases the value; dividing by a number less than 1 increases the value.
Numbers greater than 1 (7.9682,245.05): Multiplying by a number greater than 1 increases the value; dividing by a number greater than 1 decreases the value.