 # NCERT Solutions for Class 8 Maths Chapter 13 :Direct and Inverse Proportion CBSE Part 1

In this page we have NCERT book Solutions for Class 8th Maths Chapter 13 :Direct and Inverse Proportion for EXERCISE 1 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
Following are the car parking charges near a railway station upto
4 hours Rs 60
8 hours Rs 100
12 hours Rs 140
24 hours Rs 180
Check if the parking charges are in direct proportion to the parking time.
We know that two quantities are in direct proportion if whenever the values of one quantity increase, then the value of another quantity increase in such a way that ratio of the quantities remains same
Here The charges are not increasing in direct proportion to the parking time because
4/60 ≠ 8/100 ≠ 12/140 ≠ 24/180
Question 2
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.
 Parts of red pigments 1 4 7 12 20 Part of base 8
The ratio of Parts  of red pigments and part of base = 1/8
 Parts of red pigments 1 4 7 12 20 Part of base 8 a b c d
Now Parts of red pigments and part of base are in direct proportion
Now according to law of direct proportion
4/a =1/8
a=32
7/b= 1/8
b=56
12/c=1/8
c=96
20/d=1/8
d= 160
So results is
 Parts of red pigments 1 4 7 12 20 Part of base 8 32 56 96 160
Question 3
In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Let a be the red pigment part required with 1800ml of base,the as per law of direct proportion
1/75=   a/1800
a=24
Question 4
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Let  x bottles will be filled in 5 hours,
Now  hours  and amount of bottles are in  direct proportion
Then as per law of direct proportion
840/6= a/5
a=700

Question 5
A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Let x be the actual length of bacteria
Now length of bacteria and enlargement are in direct proportion, so
5/50000= x/1
Or x=1/10000 cm
Now let us assume y be the length when it is enlarged 20,000
Again as per law of direct proportion
y/20000= 5/50000
y= 2cm
Question 6
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?
The model of the ship and actual ship are in direct proportion
Let x be the length of the model of the ship, then
9/12= x/28
X= 21 cm
Question 7
Suppose 2 kg of sugar contains 9 × 106 crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?
Amount of sugar in kg is direct proportion to amount of crystals
1. Let x be the crystals in 5k,then 2/(9×106)=   5/x
x=45 × 54 crystals
2. Let y be the crystals in 1.2k,then 2/(9×106)=   1.2/x
x=10.8 × 53 crystals
Question 8
Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Answer: Length of scale and distance covered are in direct proportion
Let x be the scale in map
1/18-=x/72
x=4 cm
Question 9
A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high
(ii) the height of a pole which casts a shadow 5m long.
Answer: Length of the pole and length of shadow are in direct proportion
Also let us convert everything in meter
1. Let x be the length of the shadow cast by another pole 10 m 50 cm high 6.5/3.2= 10.5/x
Or x =6 m
2. Let x be the length of the shadow cast by another pole 10 m 50 cm high 6.5/3.2= 5/x
Or x =8.75 m
Question 10
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
As speed is same, then distance travelled and time taken will be in direct proportion
So let us assume x be the distance travelled in 5 hours or 300 minutes
14/25= x/300
x=168 km

Reference Books for class 8 Math

Given below are the links of some of the reference books for class 8 Math.

1. Mathematics Foundation Course for JEE/Olympiad : Class 8 This book can take students maths skills further. Only buy if child is interested in Olympiad/JEE foundation courses.
2. Mathematics for Class 8 by R S Aggarwal Detailed Mathematics book to clear basics and concepts. I would say it is a must have book for class 8 student.
3. Pearson Foundation Series (IIT -JEE / NEET) Physics, Chemistry, Maths & Biology for Class 8 (Main Books) | PCMB Combo : These set of books could help your child if he aims to get extra knowledge of science and maths. These would be helpful if child wants to prepare for competitive exams like JEE/NEET. Only buy if you can provide help to the child while studying.
4. Reasoning Olympiad Workbook - Class 8 :- Reasoning helps sharpen the mind of child. I would recommend students practicing reasoning even though they are not appearing for Olympiad.

You can use above books for extra knowledge and practicing different questions.

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