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Frank ICSE Mathematics Class 8 Solutions Chapter 7 Direct and Inverse Variation
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 7, Direct and Inverse Variation. Here students can easily find step by step solutions of all the problems for Direct and Inverse Variation, Exercise 7.1, 7.2 and 7.3 Also our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 7 solutions. Here in this post all the solutions are based on latest Syllabus.
Direct and Inverse Variation Exercise 7.1 Solution :
Question no – (1)
Solution :
(a) The number of articles and their cost.
= Directly
(b) The number of hours of work and the wages received.
= Directly
(c) The preparation done by a student for an examination and his or her performance.
= Directly
(d) Distance travelled and time taken by a vehicle when its speed is constant.
= Directly
Question no – (2)
Solution :
(a) Area of land and its cost
= Definite direct.
(b) Length of the side of a square and its area b.
= Direct ratio with square.
(c) Height and weight of a human being
= Indefinite.
(d) Length of a book and number of pages in it
= Indefinite.
Question no – (3)
Solution :
(a)
| X | 5 | 7 | 9 |
| Y | 40 | 56 | 72 |
(b)
| X | 2.5 | 4.5 | 6.5 |
| Y | 10 | 18 | 26 |
(c)
| X | 7 | 9 | 18 |
| Y | 84 | 108 | 216 |
(d)
| X | 7.5 | 21 | 33 |
| Y | 10 | 28 | 44 |
Question no – (4)
Solution :
According to the question,
The cost of 17 notebooks is 212.50.
The cost of 12 such notebooks will be = ?
Let the cost 12 notebooks be x rupees
∴ 17/212.50 = 12/x
or, 17x = 212.50 × 12
or, x = 212.50 × 12/17
= x = 150
Therefore, the cost of 12 such notebooks will be Rs 150.
Question no – (5)
Solution :
According to the given question,
A car is travelling at a speed of 70 km/h.
The distance travelled by the car in 24 minutes = ?
Let, the distance be x kilometres,
∴ 60/70 = 24/x
or, 60x = 24 × 70
or, x = 24 × 70/60
= x = 28 km
Therefore, the car will travel 28 km in 24 minutes.
Question no – (6)
Solution :
In the given question,
A tourist taxi charges ₹3750 for travelling a distance of = 150 km.
Distance that can be travelled for ₹5650 by the taxi = ?
Let, the distance be x kilometers,
∴ 3750/150 = 5650/x
or, 3750x = 150 × 5650
or, x = 150 × 5650/3750
= x = 226 km
Therefore, 226 km can be travel by taxi.
Question no – (7)
Solution :
As per the question,
Rita types in 1 h = 2070 words
She will type 3450 words =?
Words she will she type in 3 hours = ?
= 1 1/2 hour = 90 minutes
∴ 2070/90 = 3450/x
or, 2070 = 90 × 3450
or, x = 90 × 3450/2070 = 150 minutes
∴ 150 minutes 2 1/2 hours
Now,
Let, she write x words in 3 hours (180 minutes)
∴ 2070/90 = x/180
or, 90x = 2070 × 180
or, x = 070 ×180/90
= x = 5140 words
Therefore, In 3 hours she shall type 5140 words.
Question no – (8)
Solution :
In the given question,
Karan can walk a distance of 2.1 m in = 30 minutes
He will cover a distance of 5.6 km = ?
Let, he takes x minutes
∴ 21/30 = 5.6/x
or, 2.1x = 30 × 5.6/2.1
= x = 80 minutes
Therefore, Karan will cover 5.6 km in 80 minutes.
Question no – (9)
Solution :
According to the question,
A worker is paid for 8 days of work = ₹680
He works for 26 days in a month
So, what would be his salary for that month = ?
Let, his salary be x rupees
∴ 8/680 = 6/x
or, 8x = 680 × 26
or, x = 680 × 26/8
= x = 2210 rupees
Therefore, his salary for that month will be Rs. 2210
Question no – (10)
Solution :
Total amount of washing powder of 18 packets each weighing 1.5 kg,
= (18 × 1.5)
= 27 kg
Total amount of powder of 14 packets each weighing 2 kg,
= (14 × 2)
= 28 kg
Let, the cost be x rupees
∴ 27/1242 = 28/x
or, 27x = 28 × 1242
or, x = 28 × 1242/27
∴ x = 28 × 46
= x = 1288 rupees
Therefore, the cost of 14 packets washing powder will be 1288 Rs.
Question no – (11)
Solution :
According to the given question,
25 men can dig a canal of length = 62.5 m
Men will be required to dig a similar canal of length 155 m = ?
Let, x number of men are required
∴ 25/62.5 = x/155
or, 62.5x = 25 × 155 × 10/62
or, x = 31 × 2
= x = 62
Therefore, 62 men will be required to dig a similar canal.
Question no – (12)
Solution :
(a) The actual distance between the two cities.
∴ Actual distance,
= (11 × 4,00,00,000) cm
= (11 × 4,00) km
= 4400 km
(b) The distance in cm between these two cities on this map,
∴ Distance on map,
= 250/4,00,00,000 km
= 250,000,00/4,00,00,000
= 25/40 cm
= 250/40 mm
= 6.25 mm
Question no – (13)
Solution :
(a) The length of the shadow cast by another pole of length 18 m.
Let the length be x m
∴ 12/15.6 = 18/x
or, 12x = 15.6 × 18
or, x = 15.6 × 18/12
∴ x = 23.4 m
(b) The height of a pole which casts a shadow of length 11.7 m.
Let, the height be x m
∴ 12/15.6 = x/11.7
or, 15.6 = 12 × 11.7
or, 12 × 117/15.6
∴ x = 9 m
Direct and Inverse Variation Exercise 7.2 Solution :
Question no – (1)
Solution :
The correct option – (b)
Time taken to travel a certain distance and speed – in this case variation is inverse.
Question no – (2)
Solution :
(a)
| x | 48 | 36 | 2 | 6 |
| y | 3 | 4 | 72 | 24 |
(b)
| x | 26 | 4 | 65 | 13 |
| y | 20 | 130 | 8 | 40 |
(c)
| x | 3 | 12 | 64 | 8 |
| y | 128 | 32 | 6 | 48 |
(d)
| x | 18 | 24 | 54 | 9 |
| y | 36 | 27 | 12 | 72 |
Question no – (3)
Solution :
According to the question,
18 men can reap a wheat field in = 25 days,
15 men take to reap the same field = ? days
Let, they take x days
∴ 18 × 25 = 15x
or, x = 18 × 25/15
= x = 30 days
Therefore, 15 men will take 30 days to reap the same field.
Question no – (4)
Solution :
Let, x cows will graze the filled in 15 cow
= 45 × 22 = 15x
or, x = 45 × 22/15
= 66 cow
Therefore, 66 cow will graze the same field in 15 days
Question no – (5)
Solution :
Let, x days will be needed
= 2 × 39 × 15 = 45 × x ……[2 is multiplied because length of canal is doubled]
or, x = 2 × 39 × 15/45
= 26
∴ 26 days
Therefore, 45 men will dig the canal in 26 days
Question no – (6)
Solution :
Let, the provisions will last x days
= 1200 × 36 = (1600)x
or, x = 1200 × 36/1600
= 27
The provisions will last 27 days.
Therefore, The provisions will last 27 days.
Question no – (7)
Solution :
Let, men did the work
∴ 270
∴ 270 × 14 = 21 × x
or, x = 270 × 4/21
= 180 men
No of men fell sick is = 270 – 180
= 90 men
Therefore, 90 men fell sick
Question no – (8)
Solution :
Let, he can buy x suitcases
∴ 42 × 500 = 600x
or, = 42 × 500/60
= 35
∴ He is able to buy 35 suitcases
Therefore, He is able to buy 35 suitcases
Question no – (9)
Solution :
Let, x kg sugar can be bought now
∴ 18 × 25 = 20 × x
or, x = 18 × 25/20 = 22.5 kg
∴ 22.5 kg sugar can be bought
Therefore, He can buy now 22.5 kg sugar.
Question no – (10)
Solution :
Let, x hours is needed
∴ 15 × 4 × x
= y = 15 × 4/3 = 10 hours
Now, Let y pipes are needed to fill the tank in 3 hours
∴ 15 × 4 = 3 × y or, y = 20
or, y = 15 × 4/3 or, y = 20
Therefore, 25 are pipes
Question no – (11)
Solution :
Let x hours is needed,
∴ 60 × 15 = 90 × x
or, x = 60 × 15/90 = 10 hours
Therefore, The train will complete the same journey in 10 hours
Question no – (12)
Solution :
Let, the speed should be x km/hours
∴ 100 × 18 = 15 × x
or, x = 100 × 18/15 = 120 km/hour
Therefore, the required speed of the train will be 120 km/hour.
Direct and Inverse Variation Exercise 7.3 Solution :
Question no – (1)
Solution :
In a day, A can do 1/6 work
In a day B can do 1/18 work
In a day C can do 1/9 work
∴ In a day A, B, C together can be do
= (1/6 + 1/18 + 1/9)
= 3 + 1 + 2/18
= 6/18
= 1/3 work
In a day they do 1/3 work
Question no – (2)
Solution :
In a hour tap A can fill 1/12 of the tank
In a hour tap B can fill 1/20 of the tank
In a hour tap C can fill 1/30 of the tank
In a hour, they together can fill
= (1/12 + 1/20 + 1/30)
= 5 + 3 + 2/60
= 11/60 of the tank
∴ To fully fill the tank, needed time is 60/111, 5 5/11 hours
Therefore, to fully fill the tank, needed time is 60/111, 5 5/11 hours
Question number – (3)
Solution :
In a hour tap A can fill 1/6 of the tank
In a hour top B can fill 1/3 of the tank
In a hour tap C can fill EMPTY 1/4 of the tank
∴ If all the taps are open then in 1 hour
= 1/6 + 1/3 – 1/4
= 2 + 4 – 3/12
= 3/12
= 1/4 part of the tank will be filled
∴ To Fully fill the tank 4 hours is needed
Therefore, to Fully fill the tank 4 hours is needed
Question no – (4)
Solution :
In a day A can do 1/20 of the work
In a day B can do 1/30 of the work
In a day they together can work
= (1/20 + 1/30)
= 3 + 2/60
= 5/60 part of the work
So, in 3 days they can complete = 5/60 × 3
= 1/4 part of the work
∴ So, after 3 days, the remaining work will be
= (1 – 1/4)
= 3/4 part
Now, A alone will do 3/54 part of the work
So, A needs 3/4/1/20
= 3/4 × 20/1
= 15 days
So, in 15 days A can complete the remaining work.
Therefore, A will finish the remaining work in 15 days
Question no – (5)
Solution :
A can do 1/12 work in 1 day
B can do 1/20 work in 1 day
∴ In 1 day A and B can do
= (1/12 + 1/20)
= 8/60
= 2/15 part of the work.
So, in 3 days, A and B can do 2/15 × 3
= 2/5 part of the work.
So, the remaining work will be
= (1 – 2/5)
= 3/5 part
So, to do 3/5 part of work, A needs
= 3/5/1/12
= 3/5 × 12/1
= 36/5
= 7 1/5 days
Therefore, A will complete the work in 7 1/5 days
Question no – (6)
Solution :
In 1 day A can do 1/9 part of the work
In 1 day B can do 1/12 part of the work
In 1 day C can do 1/18 part of the work
So, In 1 day A, B, C together can do
= (1/9 + 1/12 + 1/18)
= 4 + 3 + 2/36
= 9/36
= 1/4 part of the work
∴ So, the remaining work is (1 – 1/4)
= 3/4 part
Now,
In a day, A, C together can do (1/9 + 1/18)
= 2 + 1/18
= 1/6
Part of the work,
∴ To do the remaining work A, C need 3/4/1/6
= 3/4 × 6/1
= 9/2
= 4 1/2 days
Therefore, A and C will take 4 1/2 days to complete the work
Question no – (7)
Solution :
In 1 day, A, B, C can do 1/3 part of the work
In 1 day A can do 1/6 part of the work
In 1 day B can do 1/8 part of the work
In 1 day A,B can do (1/6 + 1/8)
= 7/24 part of the work
∴ In day C can do 1/3 – 7/24
= 8 – 7/24
= 1/24 part of the work
∴ To complete 1 unit of work alone needs 1/1/
= 1 × 24/1
= 24 days
Therefore, C will take 24 days to do it alone
Question no – (8)
Solution :
In 1 day, A, B, C can do 1/3 part of the work
In 1 day A can do 1/6 part of the work
In 1 day B can do 1/8 part of the work
In 1 day A,B can do (1/6 + 1/8)
= 7/24 part of the work
∴ In day C can do 1/3 – 7/24
= 8 – 7/24
= 1/24 part of the work
∴ To complete 1 unit of work alone needs 1/1/
= 1 × 24/1
= 24 days
Therefore, C will take 24 days to do it alone
Question no – (9)
Solution :
In 1 day Sheetal can do 1/6 part of work
In 1 day Ray can do 1/8 part of work
∴ In 1 day they together can do (1/6 + 1/8)
= 7/24 part of work
∴ In 2 days they together can do 7/24 × 2
= 7/12 part of work
So, the remaining work is
= (1 – 7/12)
= 5/12 part of work
To complete this amount of work Ray needs
= 5/12/1/8
= 5/12 × 8/1
= 10/3 days 3 1/3 days
Therefore, Raj will take 10/3 days 3 1/3 days to complete the work
Question no – (10)
Solution :
In 1 day Altaf can do 1/12 part of work
In 1 day Balraj 1/8 part of work
In 1 day they together can do
= (1/12 + 1/8)
= 5/24 part of the work
So, the remaining work will be (1 – 5/24)
= 19/24 part of the work
Altaf does this work alone then he needs 19/24/1/12
= 19/24 × 12/1
= 19/2
= 9 1/2 days
Therefore, Altaf will compete the work in 9 1/2 days
Question no – (11)
Solution :
In a day Ayush can do,
= 1/(2/5)
= 1/10 part of work
In 1 day Ravi can do 1/5/3 part of work
In 1 day together can do,
= (1/10 + 1/15)
= 5/30
= 1/6 part of work
∴ To complete the work they need,
= 1/(1/6)
= 6 days
Therefore, Ayush will complete the work in 6 days
Question no – (12)
Solution :
In 1 hour tap A can fill,
= 1/4 × 12
= 1/48 part of the tank
In 1 hour tap B can fill,
= 1/6 × 6
= 1/36 part of the tank
In 1 hour A,B together can fill,
= 1/48 + 1/36
= 7/144 part of the work
So, to completely fill the tank A,B together need,
= 1/7/144
= 144/7
= 20 4/7 hours
Therefore, it will take 20 4/7 hours 2 fill the tank completely
Question no – (13)
Solution :
In 1 day A,B can part 1/8 part of the fence
In 1 day B,C can part 1/12 part of the fence
In 1 day C, A can part 1/6 part of the fence
A, B, C can paint (1/8 +1/12 + 1/6)/2 part of the fence 1 day that is
= 9/24 × 1/2
= 9/48 part
∴ A can do in 1 day
= (9/48 – 1/12)
= 9 – 4/48
= 5/48 parts
B can do in 1 day
= (9/48 – 1/6)
= 9 – 8/48
= 1/48 parts
C can do in 1 day
= (9/48 – 1/8)
= 9 – 6/48
= 3/48
= 1/10 Parts
Question no – (14)
Solution :
In 1 day A, B can do 1/10 part of the work
In 1 day B, C can do 1/15 part of the work
In 1 day C, A can do 1/12 part of the work
In 1 day A, B, C can do,
= (1/10 + 1/15 + 1/12)/2
= (6 + 5 + 4/60)/2
= 1/(4/2)
= 1/8 part of the work,
In 1 day A can do
= 1/8 – 1/15
= 7/120 part of the work
In 1 day B can do
= 1/8 – 1/12
= 1/24 part of the work
In 1 day C can do
= 1/8 – 1/10
= 1/40 part of the work
A can do the whole work in,
= 120/7
= 17 1/7 days
B can do the whole work,
= 24/1
= 24 days
C can do the whole work,
= 40/1
= 40 days
Therefore, A will take 17 1/7 days B will take 24 days and C will take 40 days
Question no – (15)
Solution :
In 1 hour A can do 1/6 part of the work
In 1 hour B can do 1/8 part of the work
In 1 hour C can do 1/12 part of the work
In hour A, B, C can do
= (1/6 + 1/8 + 1/12)
= 4 + 3 + 2/24
= 9/24
= 3/8 part of the work
To complete the whole work, they need 8/3 2 2/3days
Now, Ratio of efficiency of A, B, C is 1/6 : 1/8 : 1/12
= 4 : 3 : 2
In 3600, A will get
= 3600 × 4/9
= 1600 rupees
In 3600, b will get
= 3600 × 3/9
= 1200 rupees
In 36,00 C will get
= 3600 × 2/9
= 800 rupees
Question no – (16)
Solution :
In 1 day, the man can do 1/16 part the work
In 1 day the man and woman do 1/12 part of the work
In 1 day the woman can do
= (1/12 – 1/16)
= 1/48 part of the work
So, the ratio of their efficiency
= 1/16 : 1/48
= 3 : 1
(i) In 2400 the man will get
= 2400 × 3/4
The woman will get
= 2400 × 1/4
= 600
(ii) The woman does 1/48 part of the work in 1 day
The woman days 1 part of the work in
= 1/(1/48)
= 48 days
Question no – (17)
Solution :
Ratio of efficiency of peter and Mihir
= 100 : 150
= 2 : 3
Ratio of time of peter and Mihir
= 1/2 : 1/3
= 3 : 2
Now, Let, Mihir takes x days to complete the work alone
= 3 : 2 : : 12 : x
or, 3 × x = 2 × 12
or, x = 2 × 12/3
or, x = 8 days
Therefore, Mihir will complete the work 8 days.
Revision Exercise Questions Solution :
Question no – (1)
Solution :
(a)
| x | 5 | 7 | 9 |
| y | 40 | 56 | 72 |
(b)
| x | 2.5 | 4.5 | 6.5 |
| y | 10 | 18 | 26 |
(c)
| x | 7 | 9 | 18 |
| y | 84 | 108 | 216 |
(d)
| x | 7.5 | 21 | 33 |
| y | 10 | 28 | 44 |
Question no – (2)
Solution :
(a)
| x | 48 | 36 | 2 | 6 |
| y | 3 | 4 | 72 | 24 |
(b)
| x | 26 | 4 | 65 | 13 |
| y | 20 | 130 | 8 | 40 |
(c)
| x | 3 | 12 | 64 | 8 |
| y | 128 | 32 | 6 | 48 |
(d)
| x | 18 | 24 | 54 | 9 |
| y | 36 | 27 | 12 | 72 |
Question no – (3)
Solution :
Suppose 20 men will take x days,
So, 25 × x
or, x = 25 × 40/20
= 50 days
Therefore, 20 men will take 50 days to complete the same work
Question no – (5)
Solution :
In 1 day A can do 1/5 of the work
In 1 day B can do 1/10 of the work
In 1 day A, B can do
= 1/5 + 1/10
= 3/10 part of the work
In 2 day A, B can do 3/10 × 2
= 3/5 part of the work
The remaining work is
= (1 – 3/5)
= 2/5 part
Now, In 1 day C can do 1/10 par of the work
Now, In 1 day ABC can do,
= 1/5 + 1/10 + 1/10
= 2/5 part of the work
If A, B, C work together then the work will be finished in
= 2/5/(2/5)
= 1 day
Therefore, They all three will take 1 day to complete the work together.
Question no – (6)
Solution :
A : B = 50 : 100
B : C = 50 : 100
= 100 : 200
A : B : C = 50 : 100 : 200
= 1 : 2 : 4
A will get = 1/7 × 6300
= 900 rupees
B will get = 2/7 × 6300
= 1800
C will get = 4/7 × 6300
= 3600
Therefore, The ratio will be 1 : 2 : 4 and A will get Rs 900 B will get Rs 1800 C will get Rs 3600
Question no – (7)
Solution :
In 1 hour the first pipe can fill 1/2 part of the tank
In 1 hour the second pipe can empty 1/3 part of the tank
In 1 hour both pipes can full,
= (1/2 – 1/3)
= 1/6 part of the pipe
If both pipes are open, then to fill the tank the needed time will be
= 1/(1/6)
= 6 hours
Therefore, the tank will fill in 6 hours if both taps are open together.
Next Chapter Solution :
👉 Chapter 8 👈
