MCQ: Unitary Method - 2 - SSC CGL MCQ

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.

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.

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

∴ Price of 12 toys = 12 x 12 = 144

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.

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

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.

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.

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.

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.

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.

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 .

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.

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.

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 .

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.