Proportional Reasoning 2 Class 8 Extra Questions Maths Part 2 Chapter 3 - #NCSOLVE 📚

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Class 8 Proportional Reasoning 2 Extra Questions

Class 8 Maths Chapter 3 Proportional Reasoning 2 Extra Questions

Proportional Reasoning 2 Extra Questions Class 8

Question 1.
If the cost of 8 notebooks is ₹ 280, find the cost of 15 such notebooks.
Solution:
Let the cost of 15 notebooks be ₹ x.

The more notebooks, the higher the cost.
This is the case of direct proportion.
So, \(\frac{8}{280}=\frac{15}{x}\)
⇒ 8 × x = 15 × 280
⇒ x = \(\frac{15 \times 280}{8}\)
x = 525
Thus, the cost of 15 notebooks is ₹ 525.

Question 2.
Check whether x andy are inversely proportional to each other in each of the following tables:


Solution:
We find the product xy for the corresponding values of x and y and compare them.
(a) We have 4 × 60 = 240
8 × 30 = 240
15 × 20 = 300
10 × 24 = 240
30 × 12 = 360
36 × 5 = 180
Since the products of the values of x and the corresponding values of y are not the same or fixed.
So, x and y do not vary inversely.

(b) We have 8 × 18 = 144
12 × 12 = 144
9 × 16 = 144
24 × 6 = 144
36 × 4 = 144
3 × 48 = 144
Clearly, xy = 144
Here, the products of the values of x and the corresponding values of y are the same or fixed, i.e., 144.
So, x and y vary inversely.

Question 3.
Shabnam takes 20 minutes to reach her school if she goes at a speed of 6 km/h. If she wants to reach school in 24 minutes, what should be her speed?
Solution:
Given, speed = 6 km/h = \(\frac{6 \times 1000}{60}\) m/min = 100 m/min.
Let the required speed be x m/min.
Thus, we have the following table:

For a constant or fixed distance, speed and time are inversely proportional.
Therefore, 100 × 20 = x × 24
⇒ x = \(\frac{100 \times 20}{24}=\frac{250}{3}\)
Thus, Shabnam’s speed should be \(\frac {250}{3}\) m/min = \(\frac{250 \times 60}{3 \times 1000}\) = 5 km/h

Question 4.
In a scout camp, there is food provision for 350 cadets for 36 days. If 50 cadets leave the camp, for how many days will the provision last?
Solution:
Let the required number of days be x.
Thus, we have the following table:

The fewer cadets, the longer the provision will last.
So, this is the case of inverse proportion.
Therefore, 350 × 36 = 300 × x
⇒ x = \(\frac{350 \times 36}{300}\) = 42
Thus, the provision will last 42 days when 50 cadets leave the camp.

Question 5.
If 24 workers can build a wall in 80 hours, how many extra workers will be required to finish the same work in 60 hours?
Solution:
To finish the work in fewer hours, more workers will be needed.
Therefore, this is the case of inverse proportion.
Let the required number of extra workers be x.
Thus, we have the following table:

Then, 24 × 80 = (24 + x) × 60
⇒ 24 + x = \(\frac{24 \times 80}{60}\) = 32
⇒ x = 32 – 24 = 8
Thus, the number of extra workers required is 8.

Question 6.
A cricket coach schedules practice sessions that include different activities in the following ratio:
time for warm-up/cool-down : time for batting : time for bowling : time for fielding :: 3 : 4 : 3 : 5.
If each session is 150 minutes long, how much time is spent on each activity?
Solution:
The session is 150 minutes long, so the time spent on each activity is:
time for warm-up/cool-down = 150 × \(\frac{3}{3+4+3+5}\)
= 150 × \(\frac {3}{15}\)
= 30 minutes
time for batting = 150 × \(\frac{4}{3+4+3+5}\)
= 150 × \(\frac {4}{15}\)
= 40 minutes
time for bowling = 150 × \(\frac{3}{3+4+3+5}\)
= 150 × \(\frac {3}{15}\)
= 30 minutes
time for fielding = 150 × \(\frac{5}{3+4+3+5}\)
= 150 × \(\frac {5}{15}\)
= 50 minutes

Question 7.
Show that the numbers 22, 33, 42, and 63 are in proportion.
Solution:
Numbers are 22, 33, 42, 63

We have product of extremes = 22 × 63 = 1386
and product of means = 33 × 42 = 1386
∵ Product of extremes = Product of means.
∴ Hence, 22, 33, 42, and 63 are in proportion.

Question 8.
Show that the numbers 36, 49, 6, and 7 are not in proportion.
Solution:
Numbers are 36, 49, 6, 7
We have product of extremes = 36 × 7 = 252
and product of means = 49 × 6 = 294
∵ Product of extremes ≠ Product of means.
∴ The numbers 36, 49, 6, and 7 are not in proportion.

Question 9.
Find the fourth proportional of 4, 8, and 12.
Solution:
Let the fourth proportional of 4, 8, and 12 be x.
∴ By definition, the ratios 4 : 8 and 12 : x are in proportion.
∴ Product of extremes = Product of means
⇒ 4 × x = 8 × 12
⇒ x = \(\frac{8 \times 12}{4}\)
⇒ x = 2 × 12
⇒ x = 24
∴ The required fourth proportional is 24.

Question 10.
Find the mean proportional of 6 and 96.
Solution:
We know that mean proportional b of a = 6 and c = 96 is \(\sqrt{a c}\)
⇒ b = \(\sqrt{6 \times 96}=\sqrt{576}\) …(1)
∴ From (1), the mean proportional
b = \(\sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3}\)
= 2 × 2 × 2 × 3
= 24

Question 11.
A person spends his salary in the following manner:
Total salary = ₹ 12000
Food = ₹ 1800
Clothing = ₹ 1200
House rent = ₹ 3000
Education of children = ₹ 1200
Misc. expenditure = ₹ 2400
Savings = ₹ 2400
Make a pie chart exhibiting the above information.
Solution:
Let us prepare a table to find the sector angles corresponding to different components (of expenditure) by the formula:

Steps of construction:
1. Draw a circle of any convenient radius (a little larger).
2. Draw a horizontal radius.
3. Starting with this horizontal radius, draw sectors whose central (sector) angles are 54°, 36°, 90°, 36°, 72°, 12°, respectively, as shown in the figure below.

4. Shade the six sectors with different patterns and designs and label (name) each of them.
Also, give a heading to the pie chart. Thus, we obtain the required pie chart as shown in the figure.

Question 12.
Megha’s report card states her marks as follows:

Make a pie chart (circle graph) exhibiting her marks in various subjects.
Solution:
Let us prepare a table to find the sector angles corresponding to different components (or marks) by the formula:

Steps of construction:
1. Draw a circle of any convenient radius (a little larger).
2. Draw any line as a radius (see figure).
3. Starting with this horizontal radius, draw sectors whose central (sector) angles are 72°, 66°, 102°, 66°, 54° as shown in the figure.
(We can take the sector, clockwise or anti-clockwise, as per your convenience.)
4. Shade the five sectors with different patterns and designs and label (name) each of them.
Also, give a heading to the circle graph. Thus, we obtain the required circle as shown in the figure.

Question 13.
Observe the following tables and find if x and y are directly proportional (i.e., are in direct proportion), and if so, also write the constant of direct proportion.

Solution:
(i) The ratios of the corresponding values of x and y in the given table are:
\(\frac{x}{y}: \frac{20}{40}=\frac{1}{2}, \frac{17}{34}=\frac{1}{2}, \frac{14}{28}=\frac{1}{2}, \frac{11}{22}=\frac{1}{2},\) \(\frac{8}{16}=\frac{1}{2}, \frac{5}{10}=\frac{1}{2}, \frac{2}{4}=\frac{1}{2}\)
Since all these ratios are equal (\(\frac {1}{2}\))
∴ (Variables) x and y are in direct proportion, i.e., are directly proportional, and the constant of (direct) proportion is \(\frac {1}{2}\)

(ii) The ratios of the corresponding values of x and y in the given table are
\(\frac{x}{y}: \frac{6}{4}=\frac{3}{2}, \frac{10}{8}=\frac{5}{4}, \frac{14}{12}=\frac{7}{6}, \frac{18}{16}=\frac{9}{8},\) \(\frac{22}{20}=\frac{11}{10}, \frac{26}{24}=\frac{13}{12} \text { and } \frac{30}{28}=\frac{15}{14}\)
Since these ratios are not all equal, variables x and y are not in direct proportion, i.e., are not directly proportional.

(iii) The ratios of the corresponding values of x and y in the given table are:
\(\frac{5}{15}=\frac{1}{3}, \frac{8}{24}=\frac{1}{3}, \frac{12}{36}=\frac{1}{3}, \frac{15}{60}=\frac{1}{4},\) \(\frac{18}{72}=\frac{1}{4} \text { and } \frac{20}{100}=\frac{1}{5}\)
Since these ratios are not all equal, variables x and y are not in direct proportion, i.e., are not directly proportional.

Question 14.
The following are the car parking charges near a railway station upto
4 hours – ₹ 60
8 hours – ₹ 100
12 hours – ₹ 140
24 hours – ₹ 180
Check if the parking charges are in direct proportion to the parking time.
Solution:
Let x denote the number of parking hours, and y denote the parking charges in ₹.
The ratios of the corresponding values of x and y in the given table are:
\(\frac{x}{y}: \frac{4}{60}=\frac{1}{15}, \frac{8}{100}=\frac{2}{25}, \frac{12}{140}=\frac{3}{35},\) \(\frac{24}{180}=\frac{12 \times 2}{12 \times 15}=\frac{2}{15}\)
Since these ratios are not all equal, variables x and y are not in direct proportion, i.e., are not directly proportional.

Question 15.
(a) A mixture of paint is prepared by mixing I part of red pigments with 8 parts of base.
In the following table, find the parts of the base that need to be added.

(b) In the above question, if 1 part of a red pigment requires 75 ml of base, how much red pigment should we mix with 1800 ml of base?
Solution:
(a) Given: 1 part of red pigment is mixed with 8 parts of base to prepare the mixture.
∴ The variables x (parts of red pigments) and y (parts of base) are in direct proportion. …(1)
Let the missing entries be a, b, c, and d.
Given table becomes:

Because it is a case of direct proportion [By (1)], the ratios of corresponding entries of x and y are equal.
\(\frac{1}{8}=\frac{4}{a}, \frac{1}{8}=\frac{7}{b}, \frac{1}{8}=\frac{12}{c}, \frac{1}{8}=\frac{20}{d}\)
Cross-multiplying
a = 8 × 4 = 32
b = 8 × 7 = 56
c = 8 × 12 = 96
and d = 8 × 20 = 160
∴ Parts of the base that need to be added are as follows:

(b) Given: x = 1, y = 75ml
x = ?, y = 1800 ml
Because x and y are in direct proportion [By (1)],
therefore \(\frac{1}{x}=\frac{75}{1800}\) (Ratio of corresponding values is same)
Cross-multiplying,
75x = 1800
⇒ x = \(\frac{1800}{75}\)
⇒ x = 24
∴ 24 parts of red pigments should be mixed with 1800 ml of base.

Question 16.
In which of the following tables x and y vary inversely, i.e., x and y are inversely proportional:

Solution:
(i) x₁y₁ = 6 × 24 = 144
x₂y₂ = 3 × 48 = 144
x₃y₃ = 36 × 4 = 144
x₄y₄ = 72 × 2 = 144
x₅y₅ = 16 × 9 = 144
∴ x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄ = x₅y₅
∴ xy = k = 144 for all corresponding values of x and y.
∴ x andy are inversely proportional.

(ii) x₁y₁ = 9 × 8 = 72
x₂y₂ = 18 × 4 = 72
x₃y₃ = 2 × 30 = 60
x₄y₄ = 12 × 6 = 72
We can see that all values of xy are not equal.
∴ x and y are not inversely proportional.

Question 17.
If x and y vary inversely, fill in the blanks:

Solution:
(i) Because x and y vary inversely (given), therefore xy remains constant and
8 × 10 (from the first column of the given table)
= 80
= k
As xy = k = 80
∴ y = \(\frac {k}{x}\) and x = \(\frac {k}{y}\)
∴ First missing entry in row of y = \(\frac {k}{x}\)
= \(\frac {k}{2}\)
= \(\frac {80}{2}\)
= 40
Again first missing entry in row of x = \(\frac {k}{y}\)
= \(\frac {80}{20}\)
= 4
Now second missing entry in row of y = \(\frac {k}{x}\)
= \(\frac {80}{5}\)
= 16

(ii) Because x and y vary inversely (given), therefore xy remains constant = 16 × 4 = 64
and 128 × 0.5 = 64
∴ First missing in row of y = \(\frac {64}{32}\) = 2
and second missing entry in row of y = \(\frac {64}{8}\) = 8

Question 18.
If x varies inversely as y and x = 6 when y = 6, find x when y = 9.
Solution:
Because x and y vary inversely (given), xy is constant
i.e., x₁y₁ = x₂y₂
⇒ 6 × 6 = x × 9
⇒ 36 = 9x
⇒ 9x = 36 [∵ If a = b, then b = a]
Dividing both sides by 9,
⇒ \(\frac{9 x}{9}=\frac{36}{9}\)
⇒ x = 4

Question 19.
In a television game show, the prize money of ₹ 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners.

Solution:
Since the total prize money is fixed to be ₹ 1,00,000 and all winners are to get equal prize, the more the number (x) of winners, the less the prize (y) for each winner.
∴ x and y are in inverse proportion.
Hence, xy = constant
= 1 × 1,00,000 (from the first row of the given table)
= ₹ 100,000
= k
Also, 2 × 50,000 = ₹ 1,00,000
∴ First missing entry in column of y = \(\frac {k}{x}\)
= \(\frac {100000}{4}\)
= ₹ 25,000
Second missing entry in column of y = \(\frac {100000}{5}\) = ₹ 20,000
Third missing entry in column of y = \(\frac {100000}{8}\) = ₹ 12,500
Fourth missing entry in column of y = \(\frac {100000}{10}\) = ₹ 10,000
Fifth missing entry in column of y = \(\frac {100000}{20}\) = ₹ 5,000

The post Proportional Reasoning 2 Class 8 Extra Questions Maths Part 2 Chapter 3 appeared first on Learn CBSE.

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