VITEEE PCME Mock Test - 1 - JEE MCQ

Solution:

• Explanation:
  
  
  • We know that the cofactor of a matrix is obtained by taking the determinant of the matrix formed by removing the row and column containing the element for which the cofactor is being calculated.
  
  
  • Given that A1, B1, C1, etc are cofactors of a1, b1, c1, it implies that A1 is the cofactor of a1, B1 is the cofactor of b1, and C1 is the cofactor of c1.
  
  
  • Therefore, A1 = Δ, B1 = Δ2, C1 = Δ3 (where Δ represents the determinant of the original matrix).
  
  
   


• Conclusion:
  
  
  • So, the relationship between the cofactors A1, B1, C1 and the determinant Δ is A: Δ, B: Δ2, C: Δ3.
  
  
  • Hence, the correct answer is b. Δ².
  
  
   


 

The total number of units manufactured by companies Q and R together in the year 2001 = 6.5 + 6.5 = 13 thousands
The total number of units manufactured by companies S and T together in the year 2003 = 16.4 + 12.5 = 28.9 thousands
Required ratio = 13 : 28.9
= 130 : 289

Total number of hard disks produced by Company 'X' over all the years together = 6 + 10 + 18 + 15 + 30 + 28 + 38 + 35 = 180 (in thousand) = 1,80,000

Required average = 1, 80, 000/8 = 22,500

Total number of hard disks and mouses produced by Company 'X' in the year 2010 = 6 + 30 = 36 thousand = 36,000
Number of keyboards and pen drives produced by Company 'X' in the year 2015 = 55 + 36 = 91 thousand = 91,000
Required difference = 91,000 - 36,000 = 55,000

The number of pen drives produced by Company 'X' in years 2010, 2011, 2014 and 2015 = 10 + 15 + 22 + 36 = 83 (in thousands)
The number of hard disks produced by Company 'X' in years 2011, 2012, 2015 and 2017 = 10 + 18 + 28 + 35 = 91 (in thousands)
Required ratio = 83 : 91

Let us assume that the number with which Anita has to perform the multiplication is 'x'.

Instead of finding 35x, she calculated 53x.

The difference = 53x - 35x = 18x = 540

Therefore, x = 540/18 = 30

So, the new product = 30 x 53 = 1590.

A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So red light fashes after every 1/3 min and green light flashes every 2/5 min. LCM of both the fractions is 2 min.
Hence, they flash together after every 2 min. So in an hour they flash together 30 times.

For number of zeroes we must count number of 2 and 5 in prime numbers below 100.
We have just 1 such pair of 2 and 5.
Hence we have only 1 zero.

Since P to R is double the distance of P to Q,
Therefore, it is evident that the time taken from P to R and back would be double the time taken from P to Q and back (i.e. double of 6.5 hours = 13 hours).

Since going from P to R takes 9 hours, coming back from R to P would take 4 hours i.e. 13- 9 = 4

So Option A is correct

A and B complete a work in = 15 days
⇒ One day's work of (A + B) = 1/ 15

B complete the work in = 20 days
⇒ One day's work of B = 1/20

⇒ A's one day's work = 1/15 − 1/20 = (4−3)/6 = 1/60

Thus, A can complete the work in = 60 days.

So Option A is correct

Given:

A and B together can do a piece of work in 50 days.

A is 40% less efficient than B

Concept used:

Total work = Efficiency of the workers × time taken by them

Calculation:

Let the efficiency of B be 5a

So, efficiency of A = 5a × 60%

⇒ 3a

So, total efficiency of them = 8a

Total work = 8a × 50

⇒ 400a

Now,

60% of the work = 400a × 60%

⇒ 240a

Now,

Required time = 240a/3a

⇒ 80 days

∴ A can complete 60% of the work working alone in 80 days.