Another Peek Beyond Class 7 Solutions Maths Ganita Prakash Part 2 Chapter 4 - #NCSOLVE 📚

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Class 7 Maths Ganita Prakash Part 2 Chapter 4 Solutions

Ganita Prakash Class 7 Chapter 4 Solutions Another Peek Beyond

Class 7 Maths Ganita Prakash Part 2 Chapter 4 Another Peek Beyond Solutions Question Answer

4.1 A Quick Recap of Decimals, 4.2 Decimal Multiplication

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) Here, 6 × 4 tenths = 24 tenths.
(b) 7 × 0.3 = 7 × 3 tenths = 21 tenths
(c) Here, 9 × 5 hundredths = 45 hundredths

Question 2.
Find the products:
(a) 27.34 × 6
(b) 4.23 × 3.7
(c) 0.432 × 0.23
Solution:

Question 3.
Thejus needs 1.65 m of cloth for a shirt. How many metres of cloth are needed for 3 shirts?
Solution:
Given: Thejus needs 1.65 m of cloth for a shirt.
For 3 shirts, the total cloth needed = 1.65 × 3
= \(\frac {165}{100}\) × 3
= \(\frac {495}{100}\)
= 4.95

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:
Here cost of 1 notebook = ₹ 15.50
∴ Cost of 4 notebooks = 4 × 15.50
= \(\frac{4 \times 1550}{100}\)
= \(\frac {6200}{100}\)
= ₹ 62
and cost of 1 eraser = ₹ 2.75
∴ Cost of 3 erasers = 3 × ₹ 2.75
= \(\frac{3 \times 275}{100}\)
= \(\frac {825}{100}\)
= ₹ 8.25
∴ Total amount spent = 62 + 8.25 = ₹ 70.25

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 centimetres.
Solution:
Thickness of 1 coin = 1.45 mm
Total thickness of 36 coins = 36 × 1.45
= \(\frac{36 \times 145}{100}\)
= \(\frac {5220}{100}\)
= 52.2 mm
Now 10 mm = 1 cm
1 mm = \(\frac {1}{10}\) cm
∴ 52.2 mm = \(\frac {52.2}{10}\) = 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:
Price of 1 kg of oranges = ₹ 56.50
Price of 2.250 kg of oranges = 56.50 × 2.250
= \(\frac{5650 \times 2250}{100 \times 1000}\)
= \(\frac {12712500}{100000}\)
= ₹ 127.125
Now 56.5 × 2.25 = 127.125
Hence, we will get the same product.
The zeroes at the end of a decimal do not change its value.
As we saw, 56.50 is the same as 56.5, and 2.250 is the same as 2.25.
Hence, the product of the two numbers will be the same.

Question 6.
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:
Profit per notebook = Selling price – wholesale price
= 30 – 23.6
= ₹ 6.4
Total profit = Profit per notebook × No. of notebooks
= 6.4 × 50
= ₹ 320

Question 7.
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:
(a) Here 18 × 12 = 216 …..(i)
Now 18 × 1.2 = \(\frac{18 \times 12}{10}\) [Using (i)]
= \(\frac {216}{10}\)
= 21.6 (1 decimal place)

(b) 18 × 0.12 = \(\frac{18 \times 12}{100}\) (Using (i))
= \(\frac {216}{100}\)
= 2.16 (2 decimal places)

(c) 1.8 × 1.2 = \(\frac{18}{10} \times \frac{12}{10}=\frac{216}{100}\) [Using (i)]
= 2.16 (2 decimal places)

(d) 0.18 × 0.12 = \(\frac{18}{100} \times \frac{12}{100}=\frac{216}{100 \times 100}\) [Using (i)]
= 0.0216 (4 decimal places)

(e) 1.8 × 12 = \(\frac{18 \times 12}{10}=\frac{216}{10}\) [Using (i)]
= 21.6 (1 decimal place)
When multiplying two numbers positive if both numbers are less than 1, their product will also be less than 1.
In (d) and (e) product is less than 1.

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:
Yes, we can find the answer without actual multiplication, only by using decimal place values.
Multiplying by a number greater than 1: The product is greater than the original number.
Multiplying by a number between 0 and 1: The product is less than the original number.
Multiplying two numbers between 0 and 1: The product will be less than both factors, and therefore definitely less than 1.
(a) Greater than 1.
(b) Less than 1.
(c) Greater than 1.
(d) Less than 1.

Question 10.
Multiplying the following numbers by 10, 100, and 1000 to complete the table.

Solution:

4.3 Decimal Division

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) Given \(\frac {18}{5}\)
To convert the denominator 5 into 10, multiply both the numerator and Dr by 2.
\(\frac{18 \times 2}{5 \times 2}=\frac{36}{10}\) = 3.6
Verification
18 ÷ 5
Dividing 1 ten and 8 ones into 5 equal parts.
1 < 5
It means we need to regroup 1 ten as 10 ones,
i.e., 10 + 8 = 18 ones
18 ones ÷ 5
3 ones remain.
To divide 3 ones into 5 equal parts.
Regroup the 3 ones as 30 Tenths. (Place a decimal while regrouping ones into tenths).
30 Tenths ÷ 5 = 6
Then, 18 ÷ 5 = 3.6

Hence verified.

(b) Given \(\frac {415}{4}\)
To convert the denominator 4 into 100, multiply both the Nr and Dr by 25.
\(\frac{415 \times 25}{4 \times 25}=\frac{10375}{100}\) = 103.75
Verification
By following the steps

∴ 414 ÷ 4 = 103.75
Hence verified.

(c) Given \(\frac {1217}{2}\)
To convert the denominator 2 into 10, multiply both the Nr and Dr by 5.
\(\frac{1217 \times 5}{2 \times 5}=\frac{6085}{10}\) = 608.5
Verification
By following the steps:

∴ 1217 ÷ 2 = 608.5
Hence verified.

(d) Given \(\frac {4827}{8}\)
To convert the denominator 8 into 1000, multiply both the Nr and Dr by 125.
\(\frac{4827 \times 125}{8 \times 125}=\frac{603375}{1000}\) = 603.375
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
(b) \(\frac {3567}{8}\) = _____
(i) 4458.75
(ii) 44.5875
(iii) 445.875
(iv) 4458.75
Solution:
(a) \(\frac {1526}{4}\)
By using the Long Division Method:

Hence, option (iii) is correct.

(b) Given \(\frac {3567}{8}\)
By using the Long Division Method:

Hence, option (iii) is correct.

Question 3.
What is the quotient?
(a) 132 ÷ 4 = _____
(b) 13.2 ÷ 4 = _____
(c) 1.32 ÷ 4 = _____
(d) 0.132 ÷ 4 = _____
Solution:
(a) \(\frac {132}{4}\)

∴ Quotient = 33

(b) \(\frac {13.2}{4}\) = \(\frac {132}{40}\)

∴ Quotient = 3.3

(c) Here \(\frac{1.32}{4}=\frac{132}{400}\)

∴ Quotient = 0.33

(d) Here \(\frac{0.132}{4}=\frac{132}{4000}\)

∴ Quotient = 0.033

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 (Pages 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) \(\frac {2}{5}\)
Multiply both the Nr and Dr by 2.
\(\frac{2}{5} \times \frac{2}{2}=\frac{4}{10}\)
Now, put decimal \(\frac {4}{10}\) = 0.4
Hence \(\frac {2}{5}\) in decimal form is 0.4.

(b) \(\frac {13}{4}\)
Multiply both the Nr and Dr by 25.
\(\frac{13}{4} \times \frac{25}{25}=\frac{325}{100}\)
Now, put a decimal
\(\frac {325}{100}\) = 3.25
Hence \(\frac {13}{4}\) in decimal form is 3.25.

(c) \(\frac {4}{50}\)
Multiply both the Nr and Dr by 2.
\(\frac{4}{50} \times \frac{2}{2}=\frac{8}{100}\)
Now, put the decimal
\(\frac {8}{100}\) = 0.08
Hence \(\frac {4}{50}\) in decimal form is 0.08.

(d) \(\frac {5}{8}\)
Multiply both the Nr and Dr by 125.
\(\frac{5}{8} \times \frac{125}{125}=\frac{625}{1000}\)
Now, put the decimal
\(\frac {625}{1000}\) = 0.625
Hence \(\frac {5}{8}\) in decimal form is 0.625.

Question 2.
Find the quotients:
(a) 24.86 ÷ 1.2
(b) 5.728 ÷ 1.52
Solution:
(a) Here 24.86 ÷ 1.2
Converting division into a fraction, we get
\(\frac{24.86}{1.2}=\frac{2486 \times 10}{12 \times 100}=\frac{2486}{120}\)
Now

∴ Quotient = 20.7166…

(b) Converting division into a fraction, we get

∴ Quotient = 3.76

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 ……(i)
⇒ 156 = \(\frac {1872}{12}\) …….(ii)
⇒ 12 = \(\frac {1872}{156}\) …….(iii)
(a) Now converting division into a fraction
15.6 × 1.2 = \(\frac{156 \times 12}{10 \times 10}=\frac{1872}{100}\) = 18.72 [using (i)]

(b) Converting division into a fraction 187.2 ÷ 1.2
187.2 ÷ 1.2 = \(\frac{187.2}{1.2}=\frac{1872}{12}\) = 156 [using (ii)]

(c) Converting division into a fraction 18.72 ÷ 15.6, we get
18.72 ÷ 15.6 = \(\frac{18.72}{15.6}=\frac{1872 \times 10}{156 \times 100}\) = \(\frac {12}{10}\) = 1.2 [Using (iii)]

(d) Here 0.156 × 0.12 = \(\frac{156 \times 12}{1000 \times 100}\) [Using (i)]
= \(\frac{1872}{1000 \times 100}\)
= 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:

Question 5.
Find the quotients:
(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:
Converting 2.46 ÷ 1.5 into fraction
\(\frac{2.46}{1.5}=\frac{246 \times 10}{15 \times 100}=\frac{246}{150}\)

∴ Quotient = 1.64

(b) Converting 2.46 ÷ 0.15 into a fraction, we get
\(\frac{2.46}{0.15}=\frac{246 \times 100}{15 \times 100}=\frac{246}{15}\)

∴ Quotient = 16.4

(c) Converting 2.46 ÷ 0.015 into a fraction, we get
\(\frac{2.46}{0.015}=\frac{246 \times 1000}{15 \times 100}=\frac{2460}{15}\)

∴ Quotient = 164
Now \(\frac{24.6}{1.5}=\frac{246 \times 10}{15 \times 10}=\frac{246}{15}\)
and \(\frac{2.46}{0.15}=\frac{246 \times 100}{15 \times 100}=\frac{246}{15}\)
Both are the same.
Hence quotient obtained in 24.6 ÷ 1.5 is the same as the quotient obtained in 2.46 ÷ 0.15.

Question 6.
A 4 m long wooden block has to be cut into 5pieces of equal length. What is the length of each piece?
Solution:
Here total length = 4 m
No. of pieces = 5
Length of each piece = \(\frac{\text { Total length }}{\text { No. of pieces }}\)
= \(\frac {4}{5}\)
= 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:
Here Perimeter = 208.8 cm
No. of sides = 12
Length 0f a side = \(\frac{\text { Perimeter }}{\text { No. of sides }}\)
= \(\frac {208.8}{12}\)
= 17.4 cm

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:
Here total quantity of juice = 3 litres
No. of friends = 8
∴ Juice per friend = \(\frac {3}{8}\) litre
= \(\frac {3}{8}\) × 1000 ml
= \(\frac {3000}{8}\)
= 375 ml

Question 9.
A car covers 234.45 km using 12.6 litres of petrol. What is the distance travelled per litre?

Solution:
Given total distance = 234.45 km
Total petrol = 12.6 litres


Hence, the total distance travelled per litre of petrol is 18.607 km.

Question 10.
13.5 kg of flour (aata) was distributed equally among 15 students. How much flour did each student receive?
Solution:
Total quantity of flour = 13.5 kg
No. of students = 15
Flour per student = \(\frac {13.5}{15}\) = 0.9 kg
Now

Each student receives = 0.9 kg.

4.4 Look Before You Leap!

Figure It Out (Pages 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 a 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 the 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 of 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:
Given that Shyamala bought 3 kg of bananas at ₹ 30 per kg.
Total cost = 3 × 30 = ₹ 90
She sold 35 bananas for ₹ 5 each.
Total revenue = 35 × 5 = ₹ 175
Profit = Total revenue – Total cost
= 175 – 90
= ₹ 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:
Given that, the teacher wanted to place 80 textbooks, each is 2.5 cm thick.
Total thickness required = 80 × 2.5 = 200 cm
The bookshelf is 160 cm long.
The thickness of one book is 2.5 cm.
Number of books that can fit = \(\frac{160 \mathrm{~cm}}{2.5 \mathrm{~cm}}\) = 64
64 books could be placed on the shelf.
The total thickness of these books = 64 × 2.5 = 160 cm, which is the full length of the shelf.
The teacher wanted to place 80 textbooks, but only 64 textbooks can fit.
Therefore, there is no space left after placing the maximum number of books.

Question 5.
Fill in the following blanks appropriately:

Solution:
Here, (i) 1 km = 1000 m
∴ 5.5 km = 5.5 × 1000 = 5500 m

(ii) 1 m = 100 cm
∴ 35 cm = \(\frac {35}{100}\) = 0.35 m

(iii) 1 cm = 10 mm
∴ 14.5 cm = 14.5 × 10 = 145 mm

(iv) 1 kg = 1000 g
∴ 68 g = \(\frac {68}{1000}\) = 0068 kg

(v) 1 m = 1000 mm
∴ 9.02 m = 9.02 × 1000 = 9020 mm

(vi) 1 l = 1000 ml
∴ 125.5 ml = \(\frac {125.5}{1000}\) = 0.1255 l

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 by converting the fractions into decimals?
Solution:


∴ Quotient = 17.21

Question 7.
Fill the boxes in at least 2 different ways:

Solution:
(a) Here 1.2 × 2 = 2.4 and 0.4 × 6 = 2.4
(b) Here 2.9 × 5 = 14.5 and 14.5 × 1 = 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:

Question 9.
Find the missing cells if each cell represents a ÷ b:

Solution:
Here table given here represents division, where the value in each cell is the result of dividing the number in the corresponding ‘a’ column b the number in the corresponding row ‘b’.
Given 1517 ÷ 37 = 41
\(\frac {151.7}{3.7}\) = 41
\(\frac {15.17}{0.37}\) = 41
\(\frac {1.517}{37}\) = 0.041

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
Solution:
Let A = 245.05, B = 0.942368, C = 7.9682
We note that B < 1 and C > 1
(a) Now A × B = 245.05 × 0.942368 < 245.05
(∴ Multiplying a number by a value less than 1 results in a smaller number)

(b) A × C = 245.05 × 7.9682 > 245.05
(Multiplying a number by a value greater than 1 results in a larger number)

(c) A ÷ C = 2.4505 ÷ 7.9682 < 245.05
(Dividing a number by a value greater than 1 results in a smaller number)

(d) A ÷ B = 245.05 ÷ 0.942368 > 245.05
(Dividing a number by a value greater than 1 results in a larger number)

(e) Now, expression less than 245.05
∴ 0.942368 is closer to 1 than 7.9682 is to 1.
0.942368 will result in a value closer to 245.05 than dividing by 7.9682.
∴ (c) < (a)
Again, 0.942368 is closer to 1 than 7.9682 is to 1.
∴ Dividing by 0.942368 will result in a value closer to 245.05 than multiplying by 7.9682.
∴ (d) < (b)
Also, (f) 7.9682 is significantly smaller than 245.05; it will be the smallest value.

(f) ∴ 7.9682 < (e) 245.05 ÷ 7.9682
Combining all, we get (f), (c), (a), (e),(d), (b).

The post Another Peek Beyond Class 7 Solutions Maths Ganita Prakash Part 2 Chapter 4 appeared first on Learn CBSE.

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