RBSE Class 7 Mathematics Chapter 12 Solutions in English — Another Peek Beyond the Point

Another Peek Beyond the Point Class 7 Solutions RBSE Maths Ganita Prakash Part 2 Chapter 4

PracticingGanita Prakash Class 7 Solutionsand RBSE Class 7 Maths Part 2 Chapter 4 Another Peek Beyond the Point Solutions Question Answer helps develop logical thinking and accuracy.

Ganita Prakash Class 7 Part 2 Chapter 4 Solutions

Class 7 Ganita Prakash Part 2 Chapter 4 Solutions

RBSE Class 7 Maths Ganita Prakash Part 2 Chapter 4 Solutions

In-text Questions
Page 67

Question 1.
Jonali and Pallabi play a game. Jonali says a fraction and Pallabi gives the equivalent decimal. Write Pallabi’s answer in the blank spaces.

Solution:

Question 2.
Jonali goes to the market to buy spices. She purchases 50 g of Cinnamon, 100 g of Cumin seeds, 25 g of Cardamom and 250 g of Pepper. Express each of the quantities in kilograms by writing them in terms of fractions as well as decimals. The fractions Jonali gave Pallabi have denominators 10, 100, 1000, and so on.
Solution:

Page 68

Question 1.
Write the following fractions as a sum of fractions and also as decimals:

Solution:

Page 70

Question 1.
Can the product of two decimals be a natural number?
Solution:
Yes, e g., 2.5 × 0.4 = 1

Question 2.
Can the product of a decimal and a natural number be a natural number?
Solution:
Yes, e g., 0.5 × 2 = 1.

Page 71

Question 1.
Suppose we know that 596 × 248 = 147808, can you immediately write down the product of 5.96 × 24.8?
Solution:
5.96 has 2 decimal places and 24.8 has 1 decimal place.
Total decimal places = 2 + 1
= 3 decimal places.
596 × 248 = 147808
Therefore, 5.96 × 24.8 = 147.808

Page 75

Question 1.
What is 0.039 m in centimetres and millimetres?
Solution:
0.039 m = 3.9 cm = 39 mm
Since, 1 m = 100 cm and 1 cm = 10 mm

Question 2.
Complete the following table.

Solution:

Question 3.
How do we convert $\frac {1}{2}$ into a decimal?
Solution:
It is easy to express a fraction as a decimal if the denominator is 1, 10, 100, 1000, etc.
$\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}$(multiplying the numerator and denominator by 5).
We know that the fraction 5/10 can be represented as a decimal 0.5.

Page 76

Question 1.
Now, what if the ribbon of length 29 m was shared between four friends instead of 2?
Solution:
So, each will get 29 ÷ 4 m, that is, $\frac {29}{4}$ m
Now, the denominator of the fraction is 4. To convert a fraction to a decimal, it helps if the denominator is of the form 1, 10, 100, 1000, and so on.
So we can get an equivalent fraction of $\frac {29}{4}$ by multiplying the numerator and denominator by 25.
$\frac{29 \times 25}{4 \times 25}=\frac{725}{100}=7.25$
So, each of the 4 friends will get 7.25 m of ribbon.

Page 85

Question 1.
Can you find the quotients of 10 ÷ 9, and 100 ÷ 11?
Solution:
10 ÷ 9 = 1.111… (never ending)
100 ÷ 11 = 9.0909… (never ending)

Question 2.
Now divide 1 by 7 (1 ÷ 7). Will this end?
Solution:

Note all the remainders we get. It starts with 1, then 3, then 2, then 6, and so on. Let us represent this as a chain.

Not only do the remainders repeat in a cycle, the digits of the quotient also repeat in a cycle! 0.142857 142857 14…
This division never ends.

Pages 85-86

Question 1.
Let us consider the number 142857 that arose when dividing 1 by 7. Multiply 142857 by numbers from 1 to 6.
What are the products? What do you notice?
Solution:
142857 × 1 = 142857
142857 × 2 = 285714
142857 × 3 = 428571
142857 × 4 = 571428
142857 × 5 = 714285
142857 × 6 = 857142
Observation : You get the same number back, but with the digits cycled around!

Question 2.
Multiply 142857 by 7. What do you observe?
Solution:
142857 × 7 = 999999. We get the repeating blocks.

Question 3.
Are there other such numbers?
Solution:
Yes, to find one such number, you can find 1 ÷ 17 in decimal, and use the repeating block of digits.
1 ÷ 17 = 0.0588235294117647… (repeating block)
The repeating block is 0588235294117647.

Page 87

Question 1.

What pattern do you observe? Why are 2 and 5 related in this way?
Solution:
$\frac{1}{2 \times 2 \times 2 \times 2 \times 2}=0.03125$
and $\frac{1}{5 \times 5 \times 5 \times 5 \times 5}=0.00032$

• To find the first value we calculate 2 × 2 × 2 × 2 × 2 = 32, then 1 ÷ 32 = 0.03125
• To find the second value we calculate 5 × 5 × 5 × 5 × 5 = 3125 then 1 ÷ 3125 = 0.00032

We can see the pattern that number of decimal places in the result is equal to the exponent of the denominator.
When a fraction’s denominator contains only powers of 2 or 5, it can be easily written as a terminating decimal. This is because 2 and 5 are the prime factors of 10 and the decimal number system is based on powers of 10.

Page 90

Question 1.
Can you write an expression for the number of days in 100 calendar years with this new adjustment?
Solution:
There are 25 years divisible by 4 in 100 years.
100 is also divisible by 4, but we have to exclude it. So only 24 years have 366 days, and the rest (76 years) have 365 days.
Expression : (24 × 366) + (76 × 365) = 8784 + 27740 = 36,524 days.

Page 93

Question 1.
Do you wonder how people figured out that the Earth completes one revolution around the Sun in exactly 364.2422 days?
Solution:

People discovered how long the Earth takes to go around the Sun through centuries of careful observation and measurement of the sky.
Ancient astronomers watched the repeated pattern of seasons, the position of the Sun among the stars, and events like solstices and equinoxes. By recording the time between two identical events (such as one spring equinox to the next), they could estimate the length of a year. Over many years, averaging these measurements gave a very accurate value. Modem scientists now use precise instruments, telescopes, atomic clocks, and space satellites to measure Earth’s motion, which is how we know the length of a year so accurately (about 364.2422 days).

Class 7 Maths Another Peek Beyond the Point Solutions

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 = 2.4
(b) 0.3 = 3 tenths = 7 × 3 tenths = 21 tenths = 2.1
(c) 9 × 5 hundredths = 45 hundredths = 0.45

Question 2.
Find the products :
(a) 27.34 × 6
(b) 4.23 × 3.7
(c) 0.432 × 0.23
Solution:
(a) 2734 × 6 = 16404, with 2 decimal places = 164.04
(b) 423 × 37 = 15651, with 2+ 1 = 3 decimal places = 15.651
(c) 432 × 23 = 9936, with 3 + 2 = 5 decimal places = 0.09936

Question 3.
Thejus needs 1.65 m of cloth for a shirt. How many metres of cloth are needed for 3 shirts?
Solution:
Cloth needed for 1 shirt = 1.65 m
3 shirts need to be made.
Cloth needed = 1.65 × 3 -4.95 m

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 4 notebooks = 4 × 15.50
= ₹ 62.00
Cost of 3 erasers = 3 × 2.75
= ₹ 8.25
Total spent = 62.00+ 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 centimeters.
Solution:
Total height = 36 × 1.45
= 52.2 mm
1 cm = 10 mm
Therefore, 52.2 mm = 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 = 56.50 × 2.250 = ₹ 127.125
Yes, we can write 56.50 as 56.5 and 2.250 as 2.25 because trailing zeroes after the decimal don’t change the value.
The product we get will also be the same : 56.5 × 2.25 = 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:
Cost price of 50 notebooks = 50 × 23.6 = ₹ 1180
Selling price of 50 notebooks
= 50 × 30 = ₹ 1500
Profit earned = 1500 – 1180 = ₹ 320 .

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:
(a) 18 × 1.2 = 21.6 (1 decimal place)
(b) 18 × 0.12 = 2.16 (2 decimal places)
(c) 1.8 × 1.2 = 2.16 (1 + 1 = 2 decimal places)
(d) 0.18 × 0.12 = 0.0216 (2 + 2 = 4 decimal places)
(e) 0.018 × 0.012 =0.000216 (3 + 3 =6 decimal places)
(f) 1.8 × 12 = 21.6 (1 decimal place)
Cases (d) and (e) have products 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:
(a) 7 × 0.6 = 4.2, not less than 1.
(b) 0.7 × 0.6 = 0.42, less than 1.
(c) 0.7 × 6 = 4.2, not less than 1.
(d) 0.07 × 0.06 = 0.0042, less than 1.

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

Solution:

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) Make denominator 10:
$18 \div 5=\frac{18 \times 2}{5 \times 2}=\frac{36}{10}=3.6$
Verification: 18 ÷ 5
Dividing I 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 remains. 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
therefore 18 ÷ 5 = 3.6

Hence verified

(b) Make denominator 100:
$415 \div 4=\frac{415 \times 25}{4 \times 25}=\frac{10375}{100}=103.75$
Verifications:
By following the steps:

∴ 415 ÷ 4 = 103.75
Hence verified

(c) Make denominator 10:
$1217 \div 2=\frac{12.17 \times 5}{2 \times 5}=\frac{6085}{10}=608.5$
By following the steps:

∴ 1217 ÷ 2 = 608.5
Hence verified

(d) Make denominator 1000:
$4827 \div 8=\frac{4827 \times 125}{8 \times 125}$
$=\frac{603375}{1000}=603.375$
By following the steps:

∴ 4827 ÷ 8 = 603.375
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:
(iii) 381.5
Explanation : Multiply numerator and denominator by 25.
$\frac{1526}{4}=\frac{1526 \times 25}{4 \times 25}=\frac{38150}{100}$
= 381.50

(b) $\frac {3567}{8}$ =
(i) 4458.75
(ii) 44.5875
(iii) 445.875
(iv) 4458.75
Solution:
(iii) 445.875
Explanation : Multiply numerator and denominator by 125.
$\frac{3567}{8}=\frac{3567 \times 125}{8 \times 125}=\frac{445875}{1000}$
= 445.875

Question 3.
What ¡s the quotient?
(a) 132 ÷ 4 =
(b)13.2 ÷ 4 =
(c) 1.32 ÷ 4 =
(d) 0.132 ÷ 4=
Solution:
We solve each by making the. denominator 100(convert ÷ 4 into division by 100 by multiplying numerator and denominator by 25).
(a) 132 ÷ 4 = $\frac{132}{4}=\frac{132 \times 25}{100}=\frac{3300}{100}=33$

(b)13.2 ÷ 4 = $\frac{13.2}{4}=\frac{13.2 \times 25}{100}=\frac{330}{100}=3.3$

(c) 1.32 ÷ 4 = $\frac{1.32}{4}=\frac{1.32 \times 25}{100}=\frac{33}{100}=0.33$

(d) 0.132 ÷ 4 = $\frac{0.132}{4}=\frac{0.132 \times 25}{100}=\frac{3.3}{100}=0.033$

Question 4.
What is the quotient?
(a) 126 ÷ 8 =
(b) 12.6 ÷ 8 =
(c) 1.26 ÷ 8 =
(d) 0.126 ÷ 8 =
Solution:
Since 8 × 125 = 1000
(a) $\frac{126}{8}=\frac{126 \times 125}{8 \times 125}=\frac{15750}{1000}=15.75$
Dividing by 10 moves the decimal one place
(b) 12.6 ÷ 8 = 1.575
(c) 1.26 — 8 = 0.1575
(d) 0.126 ÷ 8 = 0.01575
(e) 0.0126 ÷ 8 = 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) Make denominator 10:
$\frac{2}{5}=\frac{2 \times 2}{5 \times 2}=\frac{4}{10}=0.4$

(b) Make denominator 100:
$\frac{13}{4}=\frac{13 \times 25}{4 \times 25}=\frac{325}{100}=3.25$

(c) Make denominator 100:
$\frac{4}{50}=\frac{4 \times 2}{50 \times 2}=\frac{8}{100}=0.08$

(d) Make denominator 1000:
$\frac{5}{8}=\frac{5 \times 125}{8 \times 125}=\frac{625}{1000}=0.625$

Question 2.
Find the quotients:
(a) 24.86 ÷ 1.2
(b) 5.728 ÷ 1.52
Solution:
(a) $\frac{24.86 \times 10}{1.2 \times 10}=\frac{248.6}{12}=20.716 \ldots \ldots$
(b) $\frac{24.86 \times 10}{1.2 \times 10}=\frac{248.6}{12}=20.716 \ldots \ldots$

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:
(a) Given : 156 × 12 = 1872.
1 + 1 = 2 decimal places, so 18.72

(b) $\frac{187.2 \times 10}{1.2 \times 10}=\frac{1872}{12}=156$

(c) $\frac{18.72 \times 100}{15.6 \times 100}=\frac{1872}{1560}=1.2$

(d) 3 + 2 = 5 decimal places, so 156 × 12 = 1872, answer = 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:
(a) 25 ÷ 1000 = 0.025, so answer 1000
(b) 25 ÷ 0,1 = 250, so answer = 0.1
(c) 25 ÷ 1o = 2.5, so answer = 1o
(d) 25 ÷ 10 = 25 × 0.1, so answer 0.1
(e)25 ÷ 0.10 = 25 × 10, so answer 10
(f) 25 ÷ 0.01 = 25 × 100, so answer = 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) 2.46 ÷ 1.5 = $\frac{2.46 \times 10}{1.5 \times 10}$
= $\frac{24.6}{15}$
= 1.64

(b) 2.45 ÷ 0.15 = $\frac{2.46 \times 100}{0.15 \times 100}$
= $\frac{246}{15}$
= 16.4

(c) 2.46 ÷ 0.015 = $\frac{2.46 \times 1000}{0.015 \times 1000}$
= $\frac{2460}{15}$
= 164
24.6 ÷ 1.5 = 16.4 and 2.46 ÷ 0.15 = 16.4. Yes, they are the same

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:
Length of each piece = 4 ÷ 5 = $\frac{4}{5}$
$=\frac{4 \times 2}{5 \times 2}=\frac{8}{10}=0.8 \mathrm{~m}$

Question 7.
If the perimeter of a regular polygon with 12 sides is 208.8 cm, what is the length of its side?
Solution:
In regular polygon all sides are equal.
Length of each side = 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:
Juice per person = 3 ÷ 8 = $\frac{3}{8}$
$=\frac{3 \times 125}{8 \times 125}=\frac{375}{1000}$
= 0.375 litres
= 375 millilitres

Question 9.
A car covers 234.45 km using 12.6 litres of petrol. What is the distance travelled per litre?
Solution:
Distance per litre = 234.45 ÷ 12.6
= $\frac{234.45 \times 10}{12.6 \times 10}$
= $\frac{234.45 \times 10}{12.6 \times 10}$
= 18.6km/litre

Question 10.
13.5 kg of flour (aata) was distributed equally among 15 students. How much flour did each student receive?
Solution:
Flour per student = 13.5 ÷ 15
= 0.9 kg

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:
Solution : Price per gram of peanut chikki
= 70.5 ÷ 210 = 0.3357 per gram
Price per gram of potato chips
= 33.25 ÷ 110 = 0.3023 per gram
∴ Potato chips are cheaper per gram.

Question 2.
Write the decimal number at the arrow mark:

Solution:
Arrow between 3.1 and 3.2, so answer could be 3.16 (or any value between 3.1 and 3.2).

Arrow between 2.15 and 2.17, so answer could be 2.162 (or any value between 2.15 and 2.17).

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:
Cost price = 3 × 30 = ₹ 90
Selling price = 35 ÷ 5 = ₹ 175
Profit = 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:
Space needed for 80 books
= 80 × 2.5 = 200 cm
Bookshelf length = 160 cm
Number of books that can fit
= 160 ÷ 2.5 = 64 books
Space left = 160 – (64 × 2.5)
= 160 – 160 = 0 cm (no space left)

Question 5.
Fill in the following blanks appropriately : 1 cm = 10 mm 1 m = 100 cm 1 km = 1000 m

Solution:

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:
6$\frac {1}{4}$ = 6 + 0.25 = 6.25 4
and 2$\frac {1}{2}$ = 2 + 0.5 = 2.5
∴ 6.25 ÷ 2.5 =2.5 (Quotient)
and 60$\frac {1}{4}$ = 60 + 0.25 = 60.25
3$\frac {1}{2}$ = 3 + 0.5 = 3.5
∴ 60.25 ÷ 3.5 = 17.21 (Quotient)

Question 7.
Fill the boxes in at least 2 different ways :
(a) ☐ × ☐ = 2.4
(b) ☐ × ☐ = 14.5
Solution:
(a) 1.2 × 2 = 2.4, or 0.6 × 4 = 2.4, or 2.4 × 1 = 2.4
(b) 29 × 0.5 = 14.5, or 5 × 2.9 = 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:
(a) Both have 1 decimal place, so quotient = 21
(b) Both have 2 decimal places, so quotient = 21
(c) (756 × 100)/(0.36 × 100) = 75600/36 = 2100
(d) 75.6 ÷ 360 = 0.21
(e) (7560 × 10)/(3.6 × 10) = 75600/36 = 2100
(f) (7.56 × 100)/(0.36 × 100) = 756/36 = 21

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

Solution:

Question 10.
Using the digits 2, 4, 5, 8, and 0 fill the boxes get the : ☐☐ . ☐ × ☐ . ☐ 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:
(a) 42.0 × 8.5 = 357
(b) 45.8 × 0.2 = 9.16
(c) 85.4 × 2.0 = 170.8
(d) 20.5 × 4.8 = 98.4
(e) 45.8 × 0.2 = 9.16

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:
(a) 245.05 × 0.942368
Value ≈ 230.93
(b) 245.05 × 7.9682
Value ≈ 1952.32
(c) 245.05 ÷ 7.9682
Value ≈ 30.75
(d) 245.05 ÷ 0.942368
Value ≈ 260.02
(e) 245.05
Value = 245.05
(f) 7.9682
Value = 7.9682
Increasing oder: (f) < (c) < (a) < (e) < (d) < (b)

It’s Puzzle Time
Hidato

Puzzle

Solution:

In Hidato, a grid of cells is given. It is usually square-shaped, like Sudoku or Kakuro, but it can also include hexagons or any shape that forms a tessellation. It can have inner holes (like a disc), but it is made of only one piece. Usually, in every Hidato puzzle the lowest and the highest numbers are given on the grid. Your task is to fill the grid such that there is a continuous path of consecutive numbers from the lowest to the highest number. The next number must be in any one of the adjacent cells, including diagonally adjacent cells.

The grid comes pre-filled with some numbers (with values between the smallest and the highest) to ensure that these puzzles have a single solution.
Try solving the following Hidato puzzles.

Solution: