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**Frank Learning Maths Class 8 Solutions Chapter 11 Direct and Inverse Variation**

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 Book, Chapter 11 Direct and Inverse Variation. Here students can easily find step by step solutions of all the problems for Direct and Inverse Variation. Here students will find solutions for Exercise 11.1, 11.2 and 11.3. 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 in this post all Mathematic solutions are based on the CBSE latest curriculum.

**Direct and Inverse Variation Exercise 11.1 Solution**

**Question no – (1) **

**Solution : **

As per the given question,

x_{1} = 17,

x_{2} = 13,

y_{1} = 212.50,

y_{2} = ?

**∴** x_{1}/y_{1} = x_{2}/y_{2}

= 17/212.50 = 13/y_{2}

= y_{2} = 13 × 212.50/17

= y_{2} = 162.5

Therefore, the cost of 13 such notebooks are 162.5

**Question no – (2) **

**Solution : **

According to the given question,

x_{1} = 1h,

y_{1} = 70 km,

x_{2} = 24,

y_{2} = ?

= 60 min

**∴** x_{1}/y_{1} = x_{2}/y_{2}

⇒ 60/70 = 24/y_{2}

= y_{2} = 24 × 70/60

= 28 km

Therefore, the car will travel 28 km in 24 minutes.

**Question no – (3) **

**Solution : **

As per the question,

x_{1} = 375,

y_{1} = 150m,

x_{2} = 562.5,

y_{2} = ?

**∴** 375/150 = 562.5/y_{2}

= y_{2} = 562.5 × 150/375 × 10

= 225 km

Therefore, the taxi can be travelled 225 km.

**Question no – (4) **

**Solution : **

According to the question,

x_{1} = 2070,

y_{1} = 1 ½

h = 3/2 h,

x_{2} = 3450,

y_{2} = ?

**∴** 2070/3/2 = 3450/y_{2}

Or y_{2} = 3450 × 3/2 / 2070

= 2.5 h

Given, x_{2} = 3450,

y_{2 }= 2.5 h,

x_{3 }= ? ,

y_{3} = 3h

**∴** 3450/x_{3} = 2.5/3

Or x_{3} = 3450 × 3 × 10/ 2.5

= 4140 words

Therefore, Rita will type 4140 words in 3 hours.

**Question no – (5) **

**Solution : **

As per the given question,

x_{1} = 210 m,

y_{1} = 90 min,

x_{2} = 560 m,

y_{2} = ?

Or 210/90 = 560/y_{2}

Or y_{2} = 560 × 90/210

= 240 min

= 4h

Thus, Karan take 4h to cover a distance of 560 m.

**Question no – (6) **

**Solution : **

Given in the question,

x_{1} = 8,

y_{1} = 680,

x_{2} = 26,

y_{2} = ?

**∴** 8/680 = 26/y_{2}

= y_{2} = 26 × 680/8

= 2210 rupees

Thus, his salary for that month would be 2210 rupees.

**Question no – (7) **

**Solution : **

1^{st} washing powder,

= (18 × 1.5) kg

= 27 kg

2^{nd} washing powder,

= (14.2)

= 28 kg

Let, x_{1} = 27kg, y_{1} = 1242, x_{2} = 28kg, y_{2} = ?

or 27/1242 = 28/y_{2}

y_{2} = 28 × 12 × 2 / 27

= 1288 rupees.

Therefore, the cost of washing powder will be 1288 rupees.

**Question no – (8) **

**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 = ?

Step by Step Solution :

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 – (9) **

**Solution : **

Length of the bacteria,

= 5/50,000

= 0.00001 cm

**∴** x_{1} = 5 cm, y_{1} = 50000, x_{2} = ? , y_{2} = 20000

5/50000 = x_{2}/20000

x_{2} = 2 cm

Therefore, its enlarged length would be 2 cm.

**Question no – (10) **

**Solution : **

Let, x_{1} = 12 cm,

y_{1} = 28m,

x_{2} = 9 cm,

y_{2} = ?

**∴** 12/28 = 9/y_{2}

y_{2} = 9 × 28/12

= 21 cm

Therefore, the model ship will be 21 cm in length.

**Question no – (11) **

**Solution : **

**(a) 1 : 4,00,00,000**

distance between two cities

= 11cm

= 11/100000 km

**∴** actual distance,

= 11/100000 × 400,00,000

= 4400 km.

**(b) x _{1} = 1, y_{1} = 40000000, x_{2} = ?, y_{2} = 250**

1/40000000 = x_{2}/250

x_{2} = 250/40000000

x_{2} = 1/160000 km

= 1/160000 × 100000 cm

= 10/16 cm

= 0.625 cm

**Question no – (12) **

**Solution : **

**(a) Given, x _{1} = 12 cm, y_{1} = 15.6m, x_{2} = 18m, y_{2} = ?**

12/18 = 15.6m/y_{2}

y_{2} = 15.6 × 18/12

= 23.4 m

**(b) Given, x _{1} = 12m, y_{1} = 15.6m, x_{3} = ?, y_{3} = 11.7m**

**∴** 12/15.6 = x_{3}/11.7

x_{3} = 12 × 11.7/15.6

= 9 m

**Question no – (13) **

**Solution : **

**(a) Given, x _{1} = 2kg, y_{1} = 9 × 10^{6}, x_{2} = 5 kg, y_{2} = ?**

2/9 × 10^{6 }= 5/y_{2}

y_{2} = 5 × 9 × 10^{6} / 2

= 5 × 9 × 10 × 10/2

= 225 × 10^{5} crystals

**(b) Given, x _{1} = 2kg, y_{1} = 9 × 10^{6}, x_{3} = 1.2 kg, y_{3} = ?**

2/9 × 10^{6} = 1.2/y_{3}

y_{3} = 1.2 × 9 × 10 / 2 × 10

= 54 × 10^{5} crystals

**Question no – (14) **

**Solution : **

**(a) Given, x _{1} = 1h = 60 min, y_{1} = 75 km, x_{2} = 20min, y_{2} = ?**

60/75 = 20/y_{2}

y_{2} = 20 × 75/60

= 25 km

**(b) Given, x _{1} = 60min, y_{1} = 75 km, x_{3} = ?, y_{3} = 250 km**

60/75 = x_{3}/2.50

x_{3} = 60 × 250/75

= 200 min

= 3 hours 20 min

**Direct and Inverse Variation Exercise 11.2 Solution**

**Question no – (1) **

**Solution : **

Let’s, x_{1} = 18 men,

y_{1} = 25 day,

x_{2} = 15 men,

y_{2} =?

**∴** Inverse variation, x_{1}y_{1} = x_{2}y_{2}

18 × 25 = 15 × y_{2}

y_{2} = 18 × 25/15 = 30 days.

Therefore, 15 men will need 30 days to reap the field.

**Question no – (2) **

**Solution : **

Let, x_{1} = 45 cow,

y_{1} = 22 days,

x_{2} = ?, y_{2} = 15days

Inverse, = 45 × 22/15 = x_{2}

= x_{2 }= 66 cows.

Therefore, 66 cows will graze the same field in 15 days.

**Question no – (3) **

**Solution : **

As per the question we get,

x_{1} = 39 men, y_{1} = 15days, x_{2} = 45 men, y_{2} = ?

= 39 × 15/ 45 = y_{2}

= y_{2} = 13 days

Therefore, (13 × 2) = 26 days because the length is double.

**Question no – (4) **

**Solution : **

Let, x_{1} = 1200 men, y_{1} = 36 day,

And, x_{2} = (1200 + 400) = 1600 men , y_{2} = ?

**∴ **1200 × 36/1600 = y_{2}

= 27 = y_{2}

= y_{2} = 27 days

Hence, the provisions will the last 27 days

**Question no – (5) **

**Solution : **

x_{1} = 270, y_{1} = 14day, x_{2} = ?, y_{2} = 21 day

**∴** 270 × 14/21 = x_{2}

= x_{2} = 270 × 14/21

= 180 men

**∴** (270 – 180) = 90 men fell sick.

Thus, 90 men will fell sick.

**Question no – (6) **

**Solution : **

Cost of suitcases

= (500 × 42)

= 21000

**∴** x_{1} = 42, y_{1} = 500, x_{2} = ?, y_{2} = 600

= 42 × 500/600 = x_{2}

= x_{2} = 35 suitcases.

Thus, the person will able to buy 35 such suitcases.

**Question no – (7) **

**Solution : **

Let, x_{1} = 18kg, y_{1} = 25, x_{2} = ?, y_{2} = 20

**∴** 18 × 25/20 = x_{2}

= x_{2} = 45/2 kg

= x_{2} = 22 1/2 kg

Therefore, the person now can buy 22 1/2 kg sugar.

**Question no – (8) **

**Solution : **

Let, x^{1} = 15 pipes, y^{1} = 4h, x^{2} = 12, y^{2} = ?

= 15 × 4/12 = y^{2}

**∴** y^{2} = 5h

x^{1} = 15 pipes, y^{1} = 4h, x^{2} = ?, y^{2} = 3h

= 15 × 4/3 = x^{2}

**∴** x^{2} = 20 pipes

Therefore, 20 pipes will required.

**Question no – (9) **

**Solution : **

Let, x_{1} = 60 km/h, y_{1} = 15h,

And, x_{2} = 90 km/h, y_{2} = ?

**∴** 60 × 15/90 = y_{2}

= y_{2} = 10 h

Hence, it take 10 hours to complete the same journey.

**Question no – (10) **

**Solution : **

Let, x_{1} = 100 km/h, y_{1} = 18 m,

And, x_{2} = ?, y_{2} = 15 min

**∴** 100 × 18/15 = x_{2}

= x_{2} = 120 km/h

Therefore, the speed should be 120 km/h.

**Question no – (11) **

**Solution : **

x_{1} = 8 periods, y_{1} = 45 min,

x_{2} = 9 period, y_{2} = ?

**∴** 8 × 45/9 = y_{2}

= y_{2} = 40 min

Therefore, each period will be 40 min long.

**Direct and Inverse Variation Exercise 11.3 Solution**

**Question no – (1) **

**Solution : **

A will complete = 1/16 days

B will complete = 1/18 days

C will complete = 1/24 days

**∴** 1/16 + 1/18 + 1/24

= 18 + 16 + 12/288

= 46/288

**∴** Complete the word,

= 288/46

**∴** 6 6/23 days.

Therefore, they will take 6 6/23 days to complete the work.

**Question no – (2) **

**Solution : **

Work done by A in 1 day = 1/24

Work done B in 1 day = 1/30

Work done C in 1 day = 1/18

**∴** Them in 1 day 1/24 + 1/30 + 1/18

= 15 + 12 + 20/360

= 47/360

In 4 day = 4 × 47/360

= 47/90

Work let = 1 – 47/90

= 43/90

**∴** Time taken to done 43/90 of the work,

= 43/90 × (24 + 18)

= 43/90 × 42

= 43 × 7/15

= 301/15

= 20 1/15

Therefore, A and C can complete the work in 20 1/15 days.

**Question no – (3) **

**Solution : **

1/15 + 1/24 + 1/30 …(As per the question)

= 24 + 15 + 12 / 360

= 51 /360

**∴** They together do,

= 120/17

= 7 1/17 h

Therefore, they will take 7 1/17 h to fill the tank.

**Question no – (6) **

**Solution : **

A is 50% more efficient than B

B is 50% less efficient time than A to complete the work,

**∴** A can complete work = 30/2 – 15 day

C can complete the work 30 × 4 = 120 days

**∴** A, B, C, is one work = 1/15, 1/30, 1/120

**∴** Together, 1/15 + 1/30 + 1/120

= 13/120

**∴** They taken,

= 120/13 ÷ 120

= 3 3/13

Therefore, 3 days for complete the work.

**Question no – (7) **

**Solution :**

M^{1} + 49, W^{1} = 28m, T^{1} = 52 days

M^{2} = ? W^{2} = 40m, T^{2} = 35 day

M^{1}T^{1}/W^{1} = M^{2} × T^{2}/W^{1} × T^{2}

= 49 × 52 × 4/28 × 35

= 104 men

Therefore, 104 men will be required to dig the well in 35 days.

**Previous Chapter Solution : **