Rd Sharma Solutions Class 8 Chapter 10 Direct and Inverse Variations
Welcome to NCTB Solution. Here with this post we are going to help 8th Class students for the Solutions of RD Sharma Class 8 Mathematics, Chapter 10, Direct and Inverse Variations. Here students can easily find Exercise wise solution for chapter 10, Direct and Inverse Variations. Students will find proper solutions for Exercise 10.1 and 10.2 Our teacher’s solved every problem with easily understandable methods so that every students can understand easily.
Direct and Inverse Variations Exercise 10.1 Solution :
Question no – (1)
Solution :
Direct variation :
If two quantities are linked in such a way that an increase in are quantity leads to a corresponding increase in the other and vice versa then such a variation in called a direct variation.
Example :
x ∝ y [where x , y = variable; k = constant]
∴ x/y = k
[∴ x = y k ]
Question no – (2)
Solution :
(i) Number of articles (x) and their price (y).
= x ∝ y
(ii) Weight of articles (x) and their cost (y).
= x ∝ y
(iii) Distance x and time y, speed remaining the same.
= x ∝ 1/y
(iv) Wages (y) and number of hours (x) of work.
= x ∝ y
(v) Speed (x) and time (y).
= x ∝ 1/y
(vi) Area of a land (x) and its cost (y).
= x ∝ y
Question no – (3)
Solution :
(i) a/b 7/21 = 9/27 = 13/39 = 21/63 = 25/75
∴ a/b = 1/3 [a = b/3] ∴ a ∝ b [Direct]
(ii) a/b = 10/5 = 20/10 = 30/15 = 40/20 = 46/23
∴ a/b = 2 ∴ a ∝ b [Direct]
(iii) a/b = 2/6 = 3/9 = 4/12 = 5/17 = 6/20
∴ It is nothing directly
(iv) a/b = 1²/1³ = 2²/2³ = 3²/3³ = 4²/4³ = 5²/5³
∴ It is not vary directly.
Question no – (4)
Solution :
(i) Two quantities are said to vary Directly with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same.
(ii) x and y are said to vary directly with each other if for some positive number k, x/y = k.
Question no – (5)
Solution :
(i)
| x | 2.5 | 4.0 | 6.0 | 15 |
| y | 5 | 8 | 12 | 30 |
= 0.5
(ii)
| x | 5 | 15/2 | 10 | 35 | 25 | 20 |
| y | 8 | 12 | 16 | 56 | 40 | 32 |
(iii)
| X | 6 | 8 | 10 | 16 | 20 |
| y | 15 | 20 | 25 | 40 | 50 |
= 3/5
(iv)
| X | 4 | 9 | 12 | 9 | 3 | 1 |
| y | 16 | 36 | 48 | 36 | 12 | 4 |
(v)
| x | 3 | 5 | 7 | 9 |
| y | 12 | 20 | 28 | 36 |
= 1/4
Question no – (7)
Solution :
Cost of 12 registers Rs 156
∴ Cost of 7 registers,
= 156 × 7/12
= 91
Therefore, the cost of 7 such registers is 91.
Question no – (8)
Solution :
| Time (t) | Dis (x) |
| 125 | 100 |
| 315 | ? |
∴ Distance = 315 × 100/125
= 252 meter.
Therefore, she will cover 252 meters.
Question no – (9)
Solution :
| Length of plastic sheet (m) | Price of plastic sheet (Rs) |
| 93 | 1395 |
| 105 | ? |
∴ Price of plastic sheet,
= 105 × 1395/93
= 1575
Therefore, the cost to buy 105 m of such plastic sheet will be Rs. 1575.
Question no – (10)
Solution :
We know, 1 hour = 60 minute
| Time | GWAM |
| 60 | 1080 |
| 105 | ? |
∴ GWAM = 1080/60
= 18 words
Therefore, Suneeta’s gross word a minute rate will be 18 words.
Question no – (11)
Solution :
We know,
1 hr = 60 min
| Time(min) | Speed |
| 60 | 50 |
| 12 | ? |
∴ Speed = 50 × 12/60
= 10 km/min
Therefore, it would travel 10 km in 12 minutes.
Question no – (12)
Solution :
| Shelf-length(m) | Number of boxes |
| 13.6 | 68 |
| 20.4 | ? |
∴ Number of Boxes,
= 20.4 × 68/13.6
= 102 Boxes
Therefore, 102 Boxes are required.
Question no – (13)
Solution :
| Shelf-length | copies |
| 3.4 | 136 |
| 5.1 | ? |
∴ Copies,
= 5.1 × 136/3.4
= 204
Therefore, there are 204 boxes can be occupy.
Question no – (14)
Solution :
| Distance (km) | Price |
| 240 | 15.00 |
| 5.1 | ? |
∴ Price = 139.2 × 15/240
= 8.70
Therefore, fare for the journey of 139.2 km is Rs. 8.70
Question no – (15)
Solution :
| Pile of cardboards | Thickness |
| 12 | 35 |
| 294 | ? |
∴ Thickness,
= 294 × 35 /12
= 857.5 mm
Therefore, thickness of a pile of 294 cardboards will be 857.5 mm.
Question no – (16)
Solution :
| Cost of cloth | Length of cloth |
| 242.50 | 97 |
| 30. 2.50 | ? |
∴ Length of cloth,
= 97 × 302.50/242.50
= 121 m
Therefore, 121 m of this can be purchased for Rs 302.50.
Question no – (18)
Solution :
| Paid cost | Days |
| 210 | 6 |
| 875 | ? |
∴ Days,
= 875 × 6/210
= 25
Therefore, he work for 25 days.
Question no – (21)
Solution :
| Weight | Amount of extension |
| 200 | 3.5 |
| 700 | ? |
∴ Amount of extension,
= 700 × 3.5/205
= 9.8 cm
Therefore, the extension produced by the weight of 700 gm is 9.8 cm.
Question no – (22)
Solution :
| Days | Dust quantity |
| 10 | 2.6 × 10⁸ |
| 45 | ? |
∴ Dust quantity,
= 45 × 2.6 × 10⁸/10
= 11.7 × 10⁸ pound
Therefore, Earth will pick up 11.7 × 10⁸ pound dust in 45 days.
Question no – (23)
Solution :
| Dust quantity | Days |
| 1.2 × 10⁸ | 15 |
| 4.8 × 10⁸ | ? |
∴ Days,
= 15 × 4.8 × 10⁸/1.2 × 10⁸
= 60 Days
Therefore, in 60 days pick up 4.8 × 10⁸ kg of dust.
Direct and Inverse Variations Exercise 10.2 Solution :
Question no – (1)
Solution :
(i) xy = 4×6 = 3×8 = 12×2 = 1× 24 = 24
∴ x ∝ 1/y = xy = k = 24
∴ It is vary inversely
(ii) xy = 5×20 = 20× 5 = 10×10= 4×25 = 100
∴ x ∝ 1/y = xy = 100
∴ It is vary inversely
(iii) xy = 4×9 = 3×12 = 6×8 = 1×36
∴ It is not vary inversely
(iv) xy = 9×8 = 24×3 = 15×4 = 3×25
∴ It is not vary inversely
Question no – (2)
Solution :
(i)
| x | 12 | 16 | 12 | 8 | 384 |
| y | 8 | 6 | 4 | 12 | 0.25 |
(ii)
| x | 16 | 32 | * | 8 | 128 |
| y | 4 | 2 | * | 8 | 0.25 |
(iii)
| x | 9 | 27 | 81 | * | 243 |
| y | 27 | 9 | 3 | * | 1 |
Question no – (3)
Solution :
(i) The number of x men hired to construct a wall and the time y taken to finish the job.
= x ∝ 1/y = xy = k (vary inversely)
(ii) The length x of a journey by bus and price y of the ticket.
= Not vary inversely.
(iii) Journey (x km) undertaken by a car and the petrol (y litres) consumed by it.
= x ∝ 1/y = x y = k (Vary inversely)
Question no – (4)
Solution :
| V (in cm ³) | 45 | 48 | 60 | 90 | 100 | 180 | 200 |
| P (in atmospheres) | 2 | 15/8 | 2/3 | 1 | 0.9 | 1/2 | 9/20 |
Question no – (5)
Solution :
Let, d₂ = 25 days, m₁ = 36 men,
m₂ = 15 days, d₂ = ?
We know,
m₁ d₁ = m₂d₂
or, d₂ = m₁d₁/m₂ = 36 × 25 /15
= 60 days
Therefore, 15 men will do it in 60 days.
Question no – (6)
Solution :
Let, m₁ = 50 men, d₁ = 5 month,
m₂ = 125 men, d₂ = ?
We know
m₁d₁ = m₂d₁
or, d₂ = m₁d₁/m₂
= 50 × 5/125
= 2 month
Therefore, in 2 month same work can be completed by 125 men.
Question no – (7)
Solution :
Let, m₁ = 420 men, d₁ = month
And, m₂ = ? d₂ = 7 month
We know that,
m₁d₁ = m₂d₂
or, 420 × 9/7
= 540 men
∴ Extra men need,
= (540 – 420)
= 120 men
Therefore, to complete the job in 7 months 120 extra men will needed.
Question no – (8)
Solution :
Let, m₁ = 1200 men, d₁ = 35 days,
m₂ = 25 days, d₂ = ?
We know that
m₁d₁ = m₂d₂
or, d₂ = m₁d₁/m₂
= 1022 × 35/25
= 1680 men
∴ Men should join,
= (1680 – 1200)
= 480 men
Therefore, more 480 men should be join to complete the work.
Question no – (9)
Solution :
Number of girl = g₁ = 50, d₁ = 40 days
Number of girl = g₂ = (50 + 30) = 80, d₂ = ?
We know that,
m₁d₁ = m₂d₂
or, d₂ = m₁d₁/m₂
= 50 × 40/80
= 25 days
Therefore, these provisions will last 25 days.
Question no – (10)
Solution :
Let, s₁ = 48 km/hr, h₁ = 10 hr,
And, s₂ = ? h₂ = 8 hr
We know that,
s₁h₁ = s₂h₂
or, s₂ = s₁h₁/h₁
= 48 × 10/8
= 60 km/hr
Therefore, Speed should be increased = (60 – 48) = 12 km/hr.
Question no – (11)
Solution :
Let, x soldiers are left,
∴ After 4 days, (28 – 4) = 24 days x soldiers left now soldiers are,
= (120 – x)
Now, m₁ = 1200, d₁ = 24 days,
m₂ = (1200 – x), d₂ = 32 days
According to question –
m₁d₁ = m₂d₂
or, m₁ = m₁d₁/d₂
or, (1200 – x) = 1200 × 24/32 = 900
or, x = 1200 – 900 = 300
Therefore, the 300 soldiers left the fort.
Question no – (12)
Solution :
We know that,
No of spray machine₁ × T₁ = no of spray machine₂ × T₂
or, 3 × 60 = 5 × T₂
or, T₂ = 3 × 60/5
= 36 min
Therefore, it will take 36 minutes to do the same job.
Question no – (13)
Solution :
Let, x members are there,
m₁ = 3, d₁ = 1 month= 30 days, m₂ = x, d₂ = 18 days
We know that,
m₁d₁ = m₂d₂
or, 3 × 30 = x × 18
or, x = 3 × 30/18
= 5
∴ New member
= (5 – 3)
= 2 members.
Therefore, there are 2 new member.
Question no – (14)
Solution :
Let, c₁ = 55, d₁ = 16,
And, d₂ = 10, c₂ = ?
We know that,
= c₁d₁ = c₂d₂
or, d₂ = c₁d₁/c₂
= 55 × 16/10
= 88 cow
Therefore, 88 cow will graze the same field in 10 days.
Question no – (15)
Solution :
Let, m₁ = 18 men d₁ = 35 days,
And, m₂ =?, d₂ = 15 days
We know that,
m₁d₁ = m₂d₂
or, 18 × 35/15 = d₂
or, d₂ = 42 men
Therefore, 42 men will be required to reaping the field.
Question no – (16)
Solution :
Price of 1 cycles = 500
∴ Price of 25 cycles = (500 × 25) = 12500
∴ Price of 1 new cycles = (500 + 125) = 625
∴ Numbers of cycles,
= 12500/625
= 20
Therefore, he will be able to buy 20 cycles.
Question no – (17)
Solution :
Cost price of 1 machines = 200
∴ Cost price of 75 machines,
= 200 × 75
= 10500
Cost price of 1 machines after discount,
= 200 – 50
= 150
∴ No of machines,
= 1500/150
= 100 machines
Therefore, he can buy 100 machines.
Question no – (18)
Solution :
(i) x = 3 when y = 8, find y when x = 4
x y = x₂y₂
or, 3 × 8 = 4 × y₂
or, y₂ = 3 × 8/4 = 6
∴ y₂ = 6
(ii) x = 5 when y = 15, find x when y = 12
x₁y₁ = x₂ y₂
or, 5 × 15 = x₂12
or, x₂ = 5 × 15/12 = 25/4
∴ x₂ = 25/4
(iii) x = 30, find y when constant of variation = 900.
xy = k
or, 30y = 900
or, y =900/30
or, y = 30
(iv) y = 35, find x when constant of variation = 7
xy = k
or, x 35 = 7
or, x = 7/35
or, x = 1/5
Next Chapter Solution :
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