MCQ: Unitary Method - 2 - SSC CGL MCQ

# MCQ: Unitary Method - 2 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Unitary Method - 2

MCQ: Unitary Method - 2 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Unitary Method - 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Unitary Method - 2 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Unitary Method - 2 below.
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MCQ: Unitary Method - 2 - Question 1

### In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk?

Detailed Solution for MCQ: Unitary Method - 2 - Question 1

According to question,
40 cows eat 40 bags of husk in 40 days.
1 cows eat 40 bags of husk in 40 x 40 days.
1 cows eat 1 bags of husk in 40 x 40 / 40 days.
1 cows eat 1 bags of husk in 40 days.

MCQ: Unitary Method - 2 - Question 2

### If 8 men can reap 80 hectares in 24 days, then how many hectares can 36 men reap in 30 days?

Detailed Solution for MCQ: Unitary Method - 2 - Question 2

Since 8 men in 24 days can reap 80 hectares.
∴ 1 men in 24 days can reap 80 / 8 hectares.
∴ 1 men in 1 days can reap 80 / 8 x 24 hectares.
∴ 36 men in 1 days can reap 80 x 36 / 8 x 24 hectares.
∴ 36 men in 30 days can reap 80 x 36 x 30 / 8 x 24 hectares.
∴ 36 men in 30 days can reap 10 x 36 x 30 / 24 hectares.
∴ 36 men in 30 days can reap 10 x 3 x 30 / 2 hectares.
∴ 36 men in 30 days can reap 10 x 3 x 15 hectares.
∴ 36 men in 30 days can reap 450 hectares.

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MCQ: Unitary Method - 2 - Question 3

### If the price of 23 toys is Rs. 276, then what will the price (Rs.) of 12 toys?

Detailed Solution for MCQ: Unitary Method - 2 - Question 3

According to given question,
Price of 23 toys = Rs. 276.

∴ Price of 12 toys = 12 x 12 = 144

MCQ: Unitary Method - 2 - Question 4

An industrial loom weaves 0.128 meters of cloth every second. Approximately, how many seconds will it take for the loom to weave 25 meter of cloth?

Detailed Solution for MCQ: Unitary Method - 2 - Question 4

According to given question,
Since an industrial loom weaves 0.128 meters of cloth in 1 seconds.
∴ an industrial loom weaves 1 meters of cloth in 1 / 0.128 seconds.
∴ an industrial loom weaves 25 meters of cloth in 1 x 25 / 0.128 seconds.
∴ an industrial loom weaves 25 meters of cloth in 25 / 0.128 seconds.
∴ an industrial loom weaves 25 meters of cloth in 195 seconds.

MCQ: Unitary Method - 2 - Question 5

If the cost of x meters of wire is d rupees, then what is the cost of y meters of wire at the same rate?

Detailed Solution for MCQ: Unitary Method - 2 - Question 5

According to given question,
Since Cost of x meters of wire is d rupees.

MCQ: Unitary Method - 2 - Question 6

A certain industrial loom weaves 0.128 meters of cloth every second. Approximately how many seconds will it take for the loom to weave 25 meters of cloth?

Detailed Solution for MCQ: Unitary Method - 2 - Question 6

A certain industrial loom weaves 0.128 meters of cloth in 1 second.
A certain industrial loom weaves 1 meters of cloth in 1/0.128 second.
A certain industrial loom weaves 25 meters of cloth in 25 x 1/ 0.128 second.
A certain industrial loom weaves 25 meters of cloth in 195.312 second.

MCQ: Unitary Method - 2 - Question 7

If the Price of 6 toys is Rs. 264.37, What will be the approximate price (Rs) of 5 toys?

Detailed Solution for MCQ: Unitary Method - 2 - Question 7

Given in the question,
Price of 6 toys is Rs. 264.37.
∴ Price of 1 toys is Rs. 264.37/ 6 .
∴ Price of 5 toys is Rs. 264.37 x 5 / 6 .
∴ Price of 5 toys is Rs. 44 x 5 .
∴ Price of 5 toys is Rs. 220.

MCQ: Unitary Method - 2 - Question 8

A rope can make 70 rounds of the circumference of a cylinder whose radius of the base is 14 cm. how many times can it go round a cylinder having radius 20 cm?

Detailed Solution for MCQ: Unitary Method - 2 - Question 8

Given in the question,
If radius is 14 cm then rope can make 70 rounds.
If radius is 1 cm then rope can make 70 x 14 rounds.
If radius is 20 cm then rope can make 70 x 14 / 20 rounds.
If radius is 20 cm then rope can make 7 x 7 rounds.
If radius is 20 cm then rope can make 49 rounds.

MCQ: Unitary Method - 2 - Question 9

If 18 binders bind 900 books in 10 days, How many binders will be required to bind 660 books in 12 days?

Detailed Solution for MCQ: Unitary Method - 2 - Question 9

According to given question,
Since 900 books are bonded in 10 days by 18 binders.
∴ 1 books are bonded in 10 days by 18 / 900 binders.
∴ 1 books are bonded in 1 days by 18 x 10 / 900 binders.
∴ 660 books are bonded in 1 days by 18 x 10 x 660 / 900 binders.
∴ 660 books are bonded in 12 days by 18 x 10 x 660 / 12 x 900 binders.
∴ 660 books are bonded in 12 days by 18 x 10 x 55 / 900 binders.
∴ 660 books are bonded in 12 days by 18 x 55 / 90 binders.
∴ 660 books are bonded in 12 days by 55 / 5 binders.
∴ 660 books are bonded in 12 days by 11 binders.

MCQ: Unitary Method - 2 - Question 10

A person works on a project and completes 5/8 of the job in 10 days. At this rate, how many more days will he it take to finish the job?

Detailed Solution for MCQ: Unitary Method - 2 - Question 10

A person completes 5/8 part of the project in 10 days.
A person completes the project in 10/5/8 days.
A person completes the project in 10 x 8 /5 days.
A person completes the project in 2 x 8 days.
A person completes the project in 16 days.

MCQ: Unitary Method - 2 - Question 11

A garrison had provisions for a certain number of days. After 10 days, 1 / 5 of the men desert and it is found that the provisions will now last just as long as before. How long was that?

Detailed Solution for MCQ: Unitary Method - 2 - Question 11

Suppose that y men had provisions for D days .Then y men had provisions for (D - 10) days .
[y - ( y/5)] men had provisions for D days .
According to question ,
∴ y( D - 10) = (4y/5) x D ⇒ 5Dy - 50y = 4Dy
⇒ Dy - 50y = 0 ⇒ y(D - 50) = 0 ⇒ D = 50 days
Hence required answer is 50 days .

MCQ: Unitary Method - 2 - Question 12

If 7 spiders make 7 webs in 7 days, then 1 spider will make 1 web in how many days?

Detailed Solution for MCQ: Unitary Method - 2 - Question 12

Given in the question,
M1 = 7 spiders , D1 = 7 days , W1 = 7 webs
M2 = 1 spider , D2 = ? , W2 = 1 web
We know that ,

⇒ D2 = 7 x 1 = 7 days .
Hence , 1 spider will make 1 web in 7 days.

MCQ: Unitary Method - 2 - Question 13

Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?

Detailed Solution for MCQ: Unitary Method - 2 - Question 13

Since 6 identical machines in 1 minute can produce a total of 270 bottles.
∴ 1 identical machines in 1 minute can produce a total of 270 / 6 bottles.
∴ 10 identical machines in 1 minute can produce a total of 270 x 10 / 6 bottles.
∴ 10 identical machines in 4 minute can produce a total of 270 x 10 x 4 / 6 bottles.
∴ 10 identical machines in 4 minute can produce a total of 45 x 10 x 4 bottles.
∴ 10 identical machines in 4 minute can produce a total of 1800 bottles.

MCQ: Unitary Method - 2 - Question 14

A certain number of men can finish a piece of work in 100 days. If there were 10 men less, it would take 10 days more for the work to be finished. How many men were there originally?

Detailed Solution for MCQ: Unitary Method - 2 - Question 14

Given in the question,
Let k be the men there originally .
Work done by y men can finish = 100 days
If there were 10 men less, it would take more days finish the work in 10 days
i.e. ( k - 10 ) = 110
According to question ,
k × 100 = ( k - 10 ) × 110
⇒ 100k = 110k - 1100 ⇒ 10k = 1100 ⇒ k = 110
Therefore , 110 men were there originally .

MCQ: Unitary Method - 2 - Question 15

A flagstaff 17.5 meters high casts a shadow of length 40.25 meters. What will be the height of a building, which casts a shadow of length 28.75 meters under similar conditions?

Detailed Solution for MCQ: Unitary Method - 2 - Question 15

According to question,
A shadow of length 40.25 meters is build for a height of 17.5 meters.
A shadow of length 1 meters is build for a height of 17.5 /40.25 meters.
A shadow of length 28.75 meters is build for a height of 17.5 x 28.75 /40.25 meters.
A shadow of length 28.75 meters is build for a height of 17.5 x 2875 /4025 meters.
A shadow of length 28.75 meters is build for a height of 17.5 x 575 /805 meters.
A shadow of length 28.75 meters is build for a height of 17.5 x 115/161 meters.
A shadow of length 28.75 meters is build for a height of 12.5 meters.

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## Quantitative Aptitude for SSC CGL

314 videos|170 docs|185 tests