Warning: Undefined array key "https://nctbsolution.com/joy-of-mathematics-class-8-solutions/" in /home/862143.cloudwaysapps.com/hpawmczmfj/public_html/wp-content/plugins/wpa-seo-auto-linker/wpa-seo-auto-linker.php on line 192
Joy of Mathematics Class 8 Solutions Chapter 7 Direct and Inverse Variation
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 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 wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 7.1 and 7.2
Direct and Inverse Variation Exercise 7.1 Solution
Question no – (1)
Solution :
7/8 quintal of sugar costs = 245
1 quintal of sugar costs = 245/7/8
0.8 quintal of sugar costs = 245 × 8/7 × 0.8/10
= 7 × 32
= 224
∴ The cost of 0.8 quintal of sugar is 224
Question no – (2)
Solution :
15 men can do a job in = 24 days
1 men can do a job in = 24 × 15 days
40 men can do a job in = 24 × 15/40 days
= 9 days
∴ 40 men can do a job in 9 days.
Question no – (3)
Solution :
After 10 days; the provisions would be sufficient for = (40 – 10)
460 soldiers = 30 days
For 1 soldiers the provisions would be sufficient for = 30 × 460
(460 + 140) = 600 soldiers the provisions would be sufficient = 30 × 460/600
= 23 days
∴ Total days will the provisions last at in the same rate = (23 + 10) = 33 days.
Question no – (4)
Solution :
After 4 days; the provisions would be sufficient for 200 soldiers
= 12 – 4
= 8 days
For 1 soldiers; the provisions would be sufficient for
= 8 × 200
(200 – 40) = 160 soldiers; the provisions would sufficient for
= 8 × 200/160
= 10 days
Hence, the provisions will least for 10 days.
Question no – (5)
Solution :
90m2 space is accommodated for 20 girls.
For 1m2 space is accommodated for = 20/90 girls
22.5m2 space is accommodated for
= 20/90 × 22.5/10 girls
= 5 girls
Therefore 22.5m2 space is accommodated for 5 girls.
Question no – (6)
Solution :
In 12 weeks Jaya earns = 1,60,000
In 1 weeks Jaya earns = 1,60000/12
In 1 year = 52 weeks Jaya earns = 160000 × 52/12
= 693333.33
∴ She earn Rs 693333.33 in a year.
Question no – (7)
Solution :
45 acres of plot are dug by = 40 men
1 acres of plot dug by = 40/45 men
752 acres of plot are dug by = 40/45 × 72 men
= 64 men
∴ 64 men will be required to dig 72 acres of the plot in a day.
Question no – (8)
Solution :
Here,
the work of 5 men = work of 8 women
the work of 1 men = work of 8/5 women
the work of 4 men = work of 8/5 × 4 women
= 32/5 women
(a) 5 men earn = 4160 per day
1 men earn = 4160/5 per day
= 832 per day.
(b) 8 women earn = 4160 per day
1 women earn = 4160/8 per day
= 520 per day.
(c) 4 men and 5 women
= 32/5 women + 5 women
= 32 + 25/5 women
= 57/5 women
Now, 8 women earn = 4160 per day
1 women earn = 4160 per day
57/5 women earn = 4160/8 × 57/5 per day
= 5928 per day
∴ 4 men and 5 women earn in 3 days
= 3 × 5928
= 17784.
Question no – (9)
Solution :
14 men, working 8 hours a day complete the work in
= 11 days.
14 men working 1 hours a day complete the work in
= 11 × 8 days
1 men working 1 hours a day complete the work in
= 11 × 8 × 14 days
22 men working 1 hours a day complete the work in
= 11 × 8 × 14/22 days
22 men working 7 hours a day complete the work in
= 11 × 8 × 14/22 × 7
= 8 days.
Question no – (10)
Solution :
In 20 days typing 8 hours 6 typists
In 20 days typing 1 hours 6 × 8 typists
In 1 days typing 1 hours 6 × 8 × 20 typists
Now,
In 12 days typing 1 hours 6 × 8 × 20/12 typists
In 12 days typing 5 hours 6 × 8 × 20/12 × 5 typists
= 16 typists.
∴ Typing should be employed
= (16 – 6)
= 10
Hence, 10 typist works for 5 hours a day.
Question no – (11)
Solution :
In 15 days the work can be finished by 2 men and 8 women.
In 1 days the work can be finished by 15 × (2men + 8 women) = 30 men + 120 women
Again,
In 6 days the work can be finished by 8 men and 16 women
In 1 days the work can be finished by 6(8 men + 15 women) = 48 men + 96 women
According to the amount of work
30 men + 120 women = 48 men + 96 women
= 120 women – 96 women = 48 men – 30 men
= 24 women = 18 men
= 4 women = 3 men
(a) 3 man are equivalent to 4 women.
(b) 2 men + 8 women
= 2men + 6 men
= 8 men
8 men can finished a work in 15 days
1 men can finished a work in = 15 × 8 days
(9 men + 4women) = (9 men + 3 men) = 12 men can finished a work in = 15 × 8/12 days
= 10 days
Question no – (12)
Solution :
In 12 days a piece of work can be finished by 3 men and 4 women.
In 1 days a piece of work can be finished by 12 (3 men + 4 women) = 36 men + 48 women
Again,
In 9 days the work can be finished by = 5 men and 4 women
In 1 days the work can be finished by = 9 (5 men + 4 women)
= 45 men + 36 women
∴ 36 men + 48 women = 45 men + 36 women
= 48 women – 36 women = 45 men – 36 men
= 12 women = 9 men
= men/women = 12/9
= 4/3
= 4 : 3
Question no – (13)
Solution :
In 10 days earn 17500 by 5 workers
In 10 days earn 1 by = 5/17500 workers
In 1 days earn 1 by = 5 × 10/17500 Workers
Again, In 7 days earn 29400 by
= 5 × 10 × 29400/17500 × 7
= 16 workers
∴ 16 workers will be earn 29400.
Question no – (14)
Solution :
Here,
The work of 5 men = the work of 6 women
The work of 1 men = the work of 6/5 women
Now, 8 men and 9women
= 8 × 6/5 women + 9 men
= (48 + 45/5) women
= 93/5 women.
Again,
6 women earn = 5400 per day
1 women earn = 5400/6 per day
51/5 women earn = 5400/6 × 93/5 per day
= (180 × 93) per day
= 16740 per day
∴ Rs 16740 earn per day.
Question no – (15)
Solution :
In Rs 90 bought = 9 mangoes
In Rs 1 bought = 9/90 mangoes
In Rs 1160 bought
= 9/90 × 1160 mangoes
= 116 mangoes
∴ In Rs 1160 bought 116 mangoes.
Direct and Inverse Variation Exercise 7.2 Solution
Question no – (1)
Solution :
A can do a piece of work in 10 days
A’s one day’s work = 1/10
B can do a piece of work in = 15 days
B’s one day’s work = 1/15
(A + B)’s one day’s work
= 1/10 + 1/15
= 3 + 6/30
= 9/30
= 3/10
∴ Therefore the time taken by A and B to finish the work = 10/3 days
∴ Thus A and B together can finish the work in 3 1/3 days.
Question no – (2)
Solution :
A’s one day’s work = 1/20
B’s one day’s work = 1/30
(A + B)’s one day’s work = 1/20 + 1/30
= 3 + 2/60
= 5/60
= 1/12
∴ Therefore the time taken by A and B to finish the work together
= 12/1 days
= 12 days
∴ Thus A and B together finish the work in 12 days.
Question no – (3)
Solution :
A and B can finish the work = 4 days
(A + B)’s one day’s work = 1/4
A can alone finish the work = 6 days
A’s one day’s work = 1/6
So, work done by B in one day
= 1/4 – 1/6
= 3 – 2/12
= 1/12
∴ Therefore B can finish the work in 12 days.
Question no – (4)
Solution :
A and B together can complete the work in 12 days
(A + B)’s one day’s work = 1/12
B alone can finish the work in = 30 day’s
B’s one day’s work = 1/30
So, work done by A in one day
= 1/12 – 1/30
= 5 – 2/60
= 3/60
= 1/20
Therefore, A alone can finish the work in 20 days.
Question no – (5)
Solution :
A and B can complete some work in 20 days
(A + B)’s one day’s work = 1/20 …… (i)
B and C can complete some work in 15 days
(B + C)’s one day’s work = 1/5 …….(ii)
C and A can complete some work in 12 day
(C + A)’s one day’s work = 1/12 ……. (iii)
Adding (i), (ii), (iii) we get,
2(A + B + C)’s one day’s work = 1/20 + 1/15 + 1/12
= 3 + 4 + 5/60
= 12/60
= 1/5
∴ (A + B + C)’s one day’s work
= 1/5 × 1/2
= 1/10 ……. (iv)
(a) Therefore A, B and C working all together can finish the work in 10 days.
(b) Subtracting (i) from (iv) we have
C’s one day’s work = 1/10 – 1/20
= 2 – 1/20
= 1/20
C alone can do the work in 20 days
Again, subtracting (ii) from (iv) we have
A’s one day’s work = 1/10 – 1/15
= 3 – 2/30
= 1/30
A alone can do the work in 30 days.
Now, Subtracting (iii) from (iv) we have,
B’s one day’s work = 1/10 – 1/12
= 6 – 5/60
= 1/60
∴ B alone can do the work in 60 days.
Question no – (6)
Solution :
A’s one day’s work = 1/8
B’s one day’s work = 1/12
C’s one day’s work = 1/24
(A + B + C)’s one day’s work = 1/8 + 1/12 + 1/24
= 3 + 2 + 1/24
= 6/24
= 1/4
A, B and C together can complete the work in 4 days.
Question no – (7)
Solution :
A’s one day’s work = 1/15
B’s one day’s work = 1/20
(A + B)’s one day’s work
= 1/15 + 1/20
= 4 + 3/60
= 7/60
∴ C’s one day’s work = 1/5 – 7/60
= 12 – 7/60
= 5/60
= 1/12
∴ C alone can finish the work in 12 days.
Question no – (8)
Solution :
A’s one day’s work = 1/25
B’s one day’s work = 1/20
(A + B)’s one day’s work = 1/25 + 1/20
= 4 + 5/100
= 9/100
(A + B)’s 5 day’s work = 9/100 × 5 = 9/20
∴ Remaining work = 1 – 9/20
= 20 – 9/20
= 11/20
Number of days taken by B to finish the remaining work = work to be done/B’s one day’s work
= 11/20 / 1/20
= 11/20 × 20/1
= 11
∴ B finish the remaining work in 11 days.
Question no – (9)
Solution :
(A + B + C)’s one day’s work = 1/4
A’s one day’s work = 1/12
C’s one day’s work = 1/10
B’s one day’s work = (A + B + C)’s one day’s work = (A + C)’s one day’s work
= 1/4 – {1/12 + 1/10}
= 1/4 – {5 + 6/60}
= 1/4 – 11/60
= 15 – 11/60
= 4/60
= 1/15
Therefore, B alone can do the work in 15 days.
Question no – (10)
Solution :
(A + B + C)’s one day’s work = 1/8
A’s one day’s work = 1/20
(A + B)’s one day’s work = 1/14
Therefore, C’s one day’s work = (A + B + C)’s one day’s work (A + B)’s one day’s work
= 1/8 – 1/14
= 7 – 4/56
= 3/56
A and C one day’s work
= 1/20 + 3/56
= 14 + 15/280
= 29/280
Therefore A and C together can do the work
Question no – (11)
Solution :
(A + B)’s one day’s work = 1/20
∴ B alone can do 1/5th of the work in 12 days
∴ B’s one day’s work = 1/12 × 1/5
= 1/60
∴ A’s one day’s work = 1/20 – 1/60
= 3 – 1/60
= 2/60
= 1/30
Therefore, A alone can do the work in 30 days.
Question no – (12)
Solution :
A’s one day’s work = 1/6
B’s one day’s work = 1/4
(A + B)’s one day’s work = 1/6 + 1/4
= 2 + 3/12
= 5/12
A started the worked, after 2 days he was joined B
= 1/6 × 2
= 1/3
∴ Remaining work,
= 1 – 1/3
= 2/3
∴ The time taken to finish the remaining work
= 2/3 / 5/12
= 2/3 × 12/5
= 8/5
Question no – (13)
Solution :
In one minutes, tap A alone can fill = 1/20 of the tank.
In one minutes tap B alone can fill 1/30 of the tank.
In one minutes tap C alone can empty = 1/15 of the tank.
Therefore, in one minutes all the three pipes are used simultaneously
= 1/20 + 1/30 – 1/15
= 3 + 2 – 4/60
= 5 – 4/60
= 1/60
∴ The tank will be filled in 60 minutes.
Question no – (14)
Solution :
In one minutes, tap A alone can fill = 1/12 of the tank.
In one minutes, tap B alone can fill = 1/16 of the tank.
In one minutes, tap C alone can empty = 1/8 of the tank.
If all the three pipes are turned on simultaneously
= 1/12 + 1/16 – 1/8
= 4 + 3 – 6/48
= 7 – 6/48
= 1/48
Therefore the tank will be empty in 48 minutes.
Question no – (15)
Solution :
In one hour, tap A can fill 1/8 of the tank
In one hour, from a waste pipe in the bottom, the tank is filled = 1/12 of the tank.
∴ Time taken to empty it
= 1/8 – 1/12
= 3 – 2/24
= 1/24
∴ Time taken to empty it 24 hours.
Previous Chapter Solution :
