ML Aggarwal CBSE Solutions Class 8 Math Eighth Chapter Direct and Inverse Variation Exercise 8.2

ML Aggarwal CBSE Solutions Class 8 Math 8th Chapter Direct and Inverse Variation Exercise 8.2

(1) (i) Inverse variation

(ii) Inverse variation

(iv) Inverse variation

(2) (i) 90 X 10 = 900

60 X 15 = 900

45 X 20 = 900

30 X 30 = 900

20 X 45 = 900

∴ xy = 900

∵ xy is constant

∴ x ∝ 1/y

∴ In the following table all pair of variable are in inverse variation.

(ii) 75 X 10 = 750

45 X 30 = 1350

30 X 25 = 750

20 X 35 = 700

10 X 65 = 650

xy is not constant.

∴ x & y are not inverse.

(3) Given, v ∝ 1/p

Where, volume = v

Pressure = P

Temperature = T

∴ VP = T. [Where, T = constant]

Now, given data –

V₁ = 630 cc, v₂ = 720 cc

P₁ = 360 mm, p₂ =?

We know, V₁ P₁ = = v₂ p₂ = T

Or, v₁ p₁ = v₂ p₂

Or, 630 X 360 = 720 X p₂

Or, P₂ = 630X360/720

Or, p₂ = 215 mm (Ans)

(6) Given, M₁ = 12, D₁ = 8 days

M₂ =?, d₂ = 6 days

∴ We know, M₁ d₁ = M₂ d₂

Or, M₂ = M₁ d₁/d₂ = 12X8/6

= 16 (Ans)

(7) No of taps = 8

Time = 27 min

Again no of tapes = 8 – 2 = 6

Let, x time to need

We know, x X 6 = 8 X 27

Or, x = 8X27/6 = 36 min (Ans)

(8) M₁ = 560 person, D₁ = 9 month

M₂ =? D₂ = 5 months.

We know, M₂ X d₂ = M₁ X d₁

Or, M₂ X 5 = 9 X 560

Or, M₂ = 9X560/5 = 1008

∴ Person required = 1008 – 560 = 448 (Ans)

(9) Let, no of x bottles would be filled.

∴ We know that,

x X 12 = 30 X 10

Or, x = 30X10/12

= 100/4

= 25

∴ 25 bottles would be field (Ans)

(10) Given, s₁ = 5km/h, T₁ = 24 min

S₂ =?, T₂ = 20 min

We know,

S₂ T₂ = S₁ T₁

Or, S₂ X 20 = 5 X 24

Or, S₂ = 5X24/20 = 6 km/hr (Ans)

(11) Firstly, total class time = 8X40 = 320 min.

Total school time = (20+20+320) min.

Let, class period time would be x.

Now, According to question –

20+20+320 = 40+40+x

Or, 360 = 80 + x

Or, 360 – 80 = x

Or, x = 280

∴ Each period time = 280/8 = 35 min (Ans)

(12) M₁ = 80, D₁ = 60 – 15 = 45

M₂ = (80+20), d₂ =?

∴ We know, M₁ d₁ = M₂ d₂

Or, d₂ = M₁ d₁/M₂ = 80X45/100 = 36 days

∴ D₂ = 36 days (Ans)

(13) M₁= 1200,

d₁ = 28-4

= 24 days

Let, x soldiers are left.

∴ M₂ = (1200 – k) d₁ = 32 days

∴ We know, M₁ d₁ = M₂ d₂

Or, (1200X24) = (1200 – x) X 32

Or, 1200 – x = 1200X24/32 = 900

Or, 1200 – 900 = x

Or, x = 300

∴ 300 soldiers are left (Ans)