Worksheet on Word Problems on Unitary Method | Printable Unitary Method Worksheet PDF with Answers

Worksheet on Word Problems on Unitary Method is prepared by experts keeping in mind the concept and rules associated with it. The Problems available in the Worksheet covers mixed questions on both direct and indirect variation.

Use the Unitary Method Word Problems Worksheet over here as a quick guide and learn the problem-solving approach used. Practice the Question and Answers in the Worksheet for Unitary Method Word Problems on a daily basis and improve your mathematics knowledge.  and you can access them free. Download the handy Worksheet on Unitary Method Word Problems and prepare as and when you want.

Do Refer:

Unitary Method Questions with Solutions

I. Jaya types 530 words in half an hour. How many words would she type in 5 minutes?

Solution:

Given that,
Jaya types 530 words in half an hour.
By the unitary method,
In 1 minute, Jaya types 530/30 = 17 words
In 5 minutes Jaya types 17 x 5 = 85 words
Hence, Jaya types 85 words in five minutes.


II. A worker is paid Rs.1250 for 5 days of work. If he works for 28 days, how much money will he get?

Solution:

Given that,
A worker is paid money for 5 days of work=Rs 1250
A worker will get money for one day work=1250/5=Rs 250
A worker will get the money for 28 days of work=250 ×28=Rs 7000
Therefore, A Worker will get the money of Rs 7000.


III. 3 men or 5 women can earn Rs 500 in a day. Find how much 7 men and 10 women will earn in a day?

Solution:

Given that,
3 men or 5 women can earn money in a day=Rs 500
Let one man earn=x
one woman earn=y
3x=500
x=500/3=166
5y=500
y=500/5=100
7 men and 10 women=7(166) + 10(100)
=1162 + 1000
=Rs 2162
Hence, 7 men and 10 women will earn money in a day is Rs 2162.


IV. In a camp, there are provisions for 130 persons for 20 days. If 50 more persons join the camp, find the number of days the provision will last?

Solution:

Given that,
There are provisions for 130 persons for 20 days.
If 50 more students join the camp, the number becomes 180.
The more number of persons, the sooner would the provisions exhaust. Therefore this is the case of inverse proportion.
So, Number of students x Camp days = constant
130 x 20 = 180 x n
n = 14
So, the provisions will last for 14 days.


V. A can do a piece of work in 10 days while B can do it in 20 days. With the help of C, they finish the work in 4 days. At what time would C alone do it?

Solution:

Given that,
A can do a piece of work in no.of days=10
B can do a piece of work in no.of days=20
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 = 1/20
C’s 1 day work = 1/x
(A + B + C)’s 1 day work = 1/10 + 1/20 + 1/x
L.C.M of (20, 15, x) = 20x.
Then, we have
(A + B + C)’s 1 day work = (2x/20x) + (x/20x) + (20/20x)
(A + B + C)’s 1 day work = (2x + x + 20)/20x
(A + B + C)’s 1 day work = (3x + 20)/20x ——(1)
Also Given that A, B and C together can do the work in 4 days.
So, we have
(A + B + C)’s 1 day work = 1/4 ——(2)
From (1) and (2), we get
(3x + 20)/20x = 1/4
4(3x + 20) = 1 ⋅ 20x
12x + 80 = 20x
80 = 8x
10 = x
So, C alone can complete the work in 10 Days.


VI. A and B together can do a piece of work in 15 days, while A alone can do it in 30 days. How long would B alone take to do it?

Solution:
Given that,
A and B together can do a piece of work in 15 days.
A alone can do it in 30 days
Let “x” be the no. of days taken by B to complete the work.
A’s 1 day work = 1/30
B’s 1 day work = 1/x
(A + B)’s 1 day work = 1/30 + 1/x
L.C.M of (30, x) = 30x.
Then, we have
(A + B)’s 1 day work = (x/30x) + (30/30x)
(A + B)’s 1 day work = (x + 30)/30x ——(1)
Also given, A and B together can do the work in 15 days.
So, we have
(A + B)’s 1-day work = 1/15 ——(2)
From (1) and (2), we get
(x + 30)/30x = 1/15
15(x + 30) = 1 ⋅ 30x
15x + 450 = 30x
450 = 15x
30= x
Hence, B alone can complete the work in 30 Days.

VII. A can do 1/3 part of a work in 4 days, while B can do 1/2 part of the work in 3 days. In how many days can both do it together?

Solution:
Given that,
A can-do 1/3 part of a work in 4 days while B can do 1/2 part of the work in 3 days.
A’s 1 day work = (1/3) / 4 = 1/12
B’s 1 day work = (1/2) / 3 = 1/6
(A + B)’s 1 day work = 1/12 + 1/6
L.C.M of (12, 6) = 12.
Then, we have
(A + B)’s 1 day work = 1/12 + 2/12
(A + B)’s 1 day work = 3/12
(A + B)’s 1 day work = 1/4
Hence, A, and B can together finish the work in 4 days.

VII. Rakesh goes to the shop to buy some books. The shopkeeper told him that 2 books would cost Rs 50. Can you find the cost of 8 books with the help of the unitary method?

Solution:

Given that,
The cost of two books is=Rs 50
First, we will find the cost of 1 book.
Cost of 1 book =Total cost of books/Total number of books
=50/2
=25.
Now, we will find the cost of 5 books.
Cost of 5 books = Cost of 1 book × Number of books
=25×8=200.
Therefore, the cost of 8 books is Rs 200.


VIII. 8 farmers harvest the crops in the field in 16 hours. How many workers are required to do the same amount of work in 12 hours?

Solution:

Given that,
8 farmers harvest the crops in the field in 16 hours.
First, we have to find no. of farmers required to harvest the crop in one hour.
16 hours = 8 farmers
1 hour = 16 x 8 =128 farmers
Now, we have to find the no. of farmers required to complete the work in 12 hours.
1 hour = 128 farmers
12 hours = 128 ÷ 12 =10 farmers.
Hence, 10 farmers are needed to complete the work in 12 hours.


IX. The cost of 5 apples is Rs150. Find the number of apples that can be purchased with Rs500.

Solution:

Given that,
The cost of 5 apples =Rs 150.
Hence the cost of 1 apple =Rs 150/5=Rs 30.
Now, the number of apples that can be purchased is Rs 30=1.
The number of apples that can be purchased with Rs 1=1/30.
The number of apples that can be purchased with Rs 500=1/30×500=16 apples.
Hence, the number of apples that can be purchased with Rs 500 is 16 apples.


X. A car traveling at a speed of 110kmph covers 330km. How much time will it take to cover 220km?

Solution:

Given that,
A car traveling at a speed of 110kmph covers 330km.
First, we need to find the time required to cover 330km.
Speed=Distance/Time
110=330/T
110 T=330
T=3 hours
Applying the unitary method, we get,
To travel 330km, the time required is 3 hours
So, to travel 1km, the time required is 3/330 hours
To travel 220km, the time required is =3/330×220=2 hours
Hence, to travel 220km by car, the time required is 2 hours.


Leave a Comment