# Mixed Problems Using Unitary Method | Unitary Method Questions and Answers

Mixed Problems using Unitary Method includes questions on direct variation and inverse variation. Solve the Mixed Problems using Unitary Method available here on a regular basis and learn how to approach. Make the most out of the Mixed Questions using Unitary Method and get a grip on the concept. Enhance your problem-solving skills and become proficient in the concept of Unitary Method by availing the Mixed Problems using Unitary Method available. If you have any difficulty you can check our solutions and assess where you went wrong.

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## 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.