**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 –

V_{1} = 630 cc, v_{2} = 720 cc

P_{1} = 360 mm, p_{2} =?

We know, V_{1} P_{1} = = v_{2} p_{2} = T

Or, v_{1} p_{1} = v_{2} p_{2}

Or, 630 X 360 = 720 X p_{2}

Or, P_{2} = 630X360/720

Or, p_{2} = 215 mm (Ans)

**(6) **Given, M_{1} = 12, D_{1} = 8 days

M_{2} =?, d_{2} = 6 days

∴ We know, M_{1} d_{1} = M_{2} d_{2}

Or, M_{2} = M_{1} d_{1}/d_{2} = 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) **M_{1} = 560 person, D_{1} = 9 month

M_{2} =? D_{2} = 5 months.

We know, M_{2} X d_{2} = M_{1} X d_{1}

Or, M_{2} X 5 = 9 X 560

Or, M_{2} = 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, s_{1} = 5km/h, T_{1} = 24 min

S_{2} =?, T_{2} = 20 min

We know,

S_{2} T_{2} = S_{1} T_{1}

Or, S_{2} X 20 = 5 X 24

Or, S_{2} = 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) M_{1} = 80, D_{1} = 60 – 15 = 45

M_{2} = (80+20), d_{2} =?

∴ We know, M_{1} d_{1} = M_{2} d_{2}

Or, d_{2} = M_{1} d_{1}/M_{2} = 80X45/100 = 36 days

∴ D_{2} = 36 days (Ans)

**(13) **M_{1}= 1200,

d_{1} = 28-4

= 24 days

Let, x soldiers are left.

∴ M_{2} = (1200 – k) d_{1} = 32 days

∴ We know, M_{1} d_{1} = M_{2} d_{2}

Or, (1200X24) = (1200 – x) X 32

Or, 1200 – x = 1200X24/32 = 900

Or, 1200 – 900 = x

Or, x = 300

∴ 300 soldiers are left (Ans)