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See More:

- Worksheet on Word Problems on Unitary Method
- Worksheet on Inverse Variation Using Unitary Method
- Worksheet on Direct Variation using Unitary Method

## Mixed Questions using Unitary Method with Solutions PDF

**I. **The cost of 10 pens is 200 Rs. Find:

i.The number of pens that can be purchased with 320 Rs.

ii. The cost of 5 pens?

**Solution:**

Given that,

The cost of 10 pens is= Rs 200

The cost of 1 pen=200/10=20

The no. of pens that can be purchased with Rs 20 is=1

The no. of pens that can be purchased with Rs 1 is=1/20

The no. of pens that can be purchased with Rs 320=1/20 × 320=16

The cost of one pen=Rs 20

The cost of 5 pens=5 × Rs 20=Rs 100.

Hence, no. of pens that can be purchased with Rs 320 is 16 and the cost of 5 pens is Rs 100.

**II. **10 potters can make 130 pots in 7 days. How many potters will be required to make 160 pots in 5 days?

**Solution:**

10 potters can make 130 pots in 7 days.

1 potter can make 130 pots in 7 × 10 days.

1 potter can make 1 pot in (7 × 10)/130 days.

Let the number of potters required be x, then;

x potters can make 1 pot in (7 × 10)/( 130 × x) days

x potters can make 160 pots in (7 × 10 × 160)/(130 × x ) days

But the number of days given = 5

According to the problem;

(7 × 10 × 160)/(130 × x ) = 5

11200/130x = 5

650x = 11200

x = 11200/650

x = 17

Therefore, 17 potters are required to make 160 pots in 5 days.

**III. **The cost of 12 bananas is Rs 48. Find the cost of 20 bananas?

**Solution:**

Given that,

The cost of 12 bananas is= Rs 48

The cost of 1 banana is=Rs 48/12=Rs 4

The Cost of 20 bananas=20 ×Rs 4=Rs 80

Therefore, the cost of 20 bananas is Rs 80.

**IV. **If 30 students can consume a stock of food in 60 days, then for how many days the same stock of food will last for 20 students?

**Solution:**

Given that,

30 students can consume a stock of food in =60 days

Let the food be consumed by 20 students in x days.

By inverse Proportion,

30 × 60= 20 × x

1800=20x

x=1800/20

x=90

Hence, The food is consumed by 20 students in 90 days.

**V. **A car runs 120 km on 10 litres of fuel, how many kilometres will it run on 15 litres of fuel?

**Solution:**

No. of km a car travel on 10 litres = 120 km

No. of km a car travel on 1 litre = 120/10 = 12 km

No. of km a car travel on 15 litres = 15 x 10 = 150 km

Therefore, the car will run 150 kilometres on 15 litres of fuel.

**VI. **If 4 buses can carry 260 people, find out the total number of people which 10 buses can carry?

**Solution:**

Given that,

No. of people carried by 4 buses= 260

No. of people carried by 1 bus=260/4=65

No. of people carried by 10 buses=10 x 65=650

Hence, 10 buses can carry 650 people.

**VII. **A truck covers a particular distance in 2 hours with a speed of 70 miles per hour. If the speed is increased by 10 miles per hour, find the time taken by the truck to cover the same distance?

**Solution:**

This is a situation of inverse variation.

Because more speed —–> less time

Given that, Time = 2 hours and Speed = 70 mph

We know that Distance = Time ⋅ Speed

Distance = 2 ⋅ 70 = 140 miles

If the given speed of 70 mph is increased by 10 mph, then the new speed will be 80 mph.

We also know that Time = Distance / Speed

Time = 140 / 80

Time = 1.75 hours

If the speed is increased by 10 mph, the time taken by the truck is 1.75 hours.

**VIII. **If 5 men can complete the work in 12 hours, how many men will be able to do the work in 5 hours?

**Solution:**

This is a situation of inverse variation.

Because fewer hours —–> more men

In 12 hours, Work can be completed by 5 men

No. of hours taken by 1 man to complete the work is

= No. of hours ⋅ No. of men

= 12 ⋅ 5

= 60 hours

No. of men required to complete the work in 5 hours is= 60 /5

= 12 men

Therefore, 12 men will be able to complete the work in 10 hours.

**IX. **A man has enough money to buy 12 kg of apples at Rs 50 per kg. How much can he buy, if the price is increased by Rs 2 per kg?

**Solution:**

This is a situation of inverse variation.

Because

more price —–> fewer kgs of apples

The cost of 12 kg of apples at Rs 50 per kg is

= 12 ⋅ 50

= Rs 600

So, the person has Rs 600

If the price is increased by Rs 2 per kg, then the new price per kg is

= Rs 52

No. of kgs of apples he can buy with Rs 600 is

= 600 / 52

= 11.53 kg

If the price is increased by Rs 2 per kg, the person can buy 11.53 kg of apples.

**X. **A can do a piece of work in 10 days while B can do it in 15 days. With the help of C, they finish the work in 5 days. At what time would C alone do it?

**Solution:**

Let “x” be the no. of days taken by C to complete the work.

A’s 1 day work = 1/10

B’s 1 day work = 1/15

C’s 1 day work = 1/x

(A + B + C)’s 1 day work = 1/10 + 1/15 + 1/x

L.C.M of (10, 15, x) = 30x.

Then, we have (A + B + C)’s 1 day work =(3x/30x)+(2x/30x)+(30/30x)

(A + B + C)’s 1 day work =(5x+30)/30x —>(1)

Also given A, B and C together can do the work in 5 days.

So, we have (A + B + C)’s 1-day work = 1/5 ——(2)

From (1) and (2), we get

(5x+30)/30x=1/5

5(5x+30)=30x

25x+150=30x

5x=150

x=150/5=30

Therefore, C alone can complete the work in 30 days.

**XI. **A can do a piece of work in 10 days and B alone can do it in 20 days. They worked together on it for 3 days and then A left. How long would B take to finish the remaining work?

**Solution:**

A’s 1 day work = 1/10

B’s 1 day work = 1/20

(A + B)’s 1 day work = 1/10 + 1/20

L.C.M of (10, 20) = 20.

Then, we have

(A + B)’s 1 day work = 2/20 + 1/20

(A + B)’s 1 day work = 3/20

Then, the amount of work completed by A and B together in 3 days is= 3 ⋅ 3/20

= 9/20

The amount of work left for B to complete is= 11/20

Number of days that B will take to finish the work is

= Amount of work/part of the work done in 1 day

= (11/20) / (1/20)

= (11/20) ⋅ (20/1)

= 11

So, no. of days taken by B to finish the remaining work is 11 days.

**XI. **A book contains 100 pages and each page has 20 lines. How many pages will the book contain if every page has 10 lines per page?

**Solution:**

This is a situation of inverse variation.

Because,

less lines ——–> more pages

20 lines ——–> 100 pages

1 line ——–> 20 ⋅ 100 = 2000 pages

10 lines ——–> 2000 / 10 = 200 pages

Therefore, If the book has 10 lines per page, then it will contain 200 pages.

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