VITEEE PCBE Mock Test - 12 - JEE MCQ

Total number of Mining Engineering students in 2014 at all the IITs = 600
Total number of Mining Engineering students at IIT Madras and IIT Kanpur = 600 – (95 + 155 + 65 + 90 + 140) = 55
So, all the Mining Engineering students at IIT Madras and IIT Kanpur received a scholarship of Rs. 2000 each. (As the number of students at each of the two IITs is less than 55)
Then, total scholarship amount awarded to Mining Engineering students = 55  6  2000 + (40 + 45 + 10 + 35 + 45)1200 = 6,60,000 + 2,10,000 = Rs. 8,70,000

Suppose the total numbers of Mechanical Engineering students at IIT Delhi and IIT Roorkee in 2014 are y and z, respectively.
So, y - z = 29 ...(1)
Then, total number of students at IIT Delhi in 2014 = y + 174 + 225 + 95 + 80 = y + 574
And total number of students at IIT Delhi in 2015 = 1.1y + 174 + 225  1.08 + 95 + 80  1.1
= 1.1y + 174 + 243 + 95 + 88 = 1.1y + 600 ...(2)
Total number of Mechanical Engineering students at all the seven IITs in 2014 = y + z + 100 + 150
+ 220 + 190 + 220 = y + z  + 880 ...(3)
Then, from equations (2) and (3),
1.5(1.1y + 600) = y + z + 880
⇒ 1.65y + 900 = y + z + 880 
⇒ 0.65y - z = -20
⇒ 0.65y - y + 29 = -20 (from equation 1)
⇒ - 0.35y = -49
⇒ y = 140 and z = 111
Hence, ratio of the total number of students in Mechanical Engineering at IIT Delhi and IIT Roorkee in 2014 = 140 : 111

 (i) Average runs scored in the 1st innings = 67.5
Total score = 67.5 × 4 = 270.0
(ii) Average runs of Sachin = 72.5
Total runs of Sachin = 72.5 × 4 = 290
So, the runs made by Sachin in 3rd innings = 290 - 80 - 70 - 80 = 60
(iii) Total score in the 4th innings = 270 (from 7)
Now, let Paramjeet`s, Dhooni`s and Jeetu`s scores in 4th innings be x, x and y.
=> 2x + y + 80 = 270
=> 2x + y = 190 ..…(i)
Let Jeetu`s and Dhooni`s scores in 1st innings be p and q, respectively.
=> p + q + 80 + 60 = 270
=> p + q = 130 ..…(ii)
From 2:
i.e. 6p - x = 170 ..….(iii)
From 5: Jeetu`s and Dhooni`s scores in 2nd innings are p - 10 and q - 10, respectively.
=> Jeetu`s total runs = 220
=> p + p - 10 + 60 + y = 220
2p + y = 170 ..….(iv)
Solving (i) and (iv), we get 2x - 2p = 20 ……(v)
Solving (iii) and (v), we get x = 46 and p = 36
From (i) and (ii),
y = 98 and q = 94
Thus,

In 2nd innings, Jeetu Malik scored the lowest.

(i) Average runs scored in the 1st innings = 67.5
Total score = 67.5 × 4 = 270.0
(ii) Average runs of Sachin = 72.5
Total runs of Sachin = 72.5 × 4 = 290
So, the runs made by Sachin in 3rd innings = 290 - 80 - 70 - 80 = 60
(iii) Total score in the 4th innings = 270 (from 7)
Now, let Paramjeet`s, Dhooni`s and Jeetu`s scores in 4th innings be x, x and y.
=> 2x + y + 80 = 270
=> 2x + y = 190 ..…(i)
Let Jeetu`s and Dhooni`s scores in 1st innings be p and q, respectively.
=> p + q + 80 + 60 = 270
=> p + q = 130 ..…(ii)
From 2:
i.e. 6p - x = 170 ..….(iii)
From 5: Jeetu`s and Dhooni`s scores in 2nd innings are p - 10 and q - 10, respectively.
=> Jeetu`s total runs = 220
=> p + p - 10 + 60 + y = 220
2p + y = 170 ..….(iv)
Solving (i) and (iv), we get 2x - 2p = 20 ……(v)
Solving (iii) and (v), we get x = 46 and p = 36
From (i) and (ii),
y = 98 and q = 94
Thus,

Lowest runs were scored in 1st innings by Jeetu Malik.

However you arrange the digits, you will observe that the sum of digits is always equal to (100)(1+2+3) = 600.

Notice that 600 is divisible by 3 but not by 9.

Hence the original number is divisible by 3 but not by 3^2 which is 9.

ORIGINAL NUMBER = 3(p) where p is an integer does not have the prime factor 3. But 3 is not a perfect square!

Hence there will be no number which has 100 1’s, 100 2’s and 100 3’s which will be a perfect square.

Given:

73 × 75 × 78 × 57 × 197 × 37 is divided by 34

Calculation:

73 × 75 × 78 × 57 × 197 × 3734

We have taken individual remainder like

When 73 is divided by 34 gives remainder is 5

Similarly

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Calculation:

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Now we have to take LCM of time interval

⇒ LCM of (12, 15, 20, 30) = 60

Total seconds in 8 hours = 8 × 3600 = 28800

Number of times bell rings = 28800/60

⇒ Number of times bell rings = 480

If four bells ring together in starting

⇒ 480 + 1 

∴ The bell ringing 481 times in 8 hours. So the correct option is B

Note: In these type of question we assume that we have started counting the time after first ringing. Due to this when we calculate the LCM it gives us the ringing at 2nd time not the first time. So, we needed to add 1.

A number is repeated. Then it is reduced by 13 and the result is repeated.

► The next number is 27-13=14.

Pattern:

► 4 × 2 + 3 = 11,
► 11 × 2 – 3 = 19,
► 19 × 2 + 3 = 41,
► 41 × 2 – 3 = 79,
► 79 × 2 + 3 = 161.