1. (c)
The change in the amount of chemical in each tank after every minute is as follows:
A:−20–10+90=60
B:−100+110+20=30
C:−50−90+100=−40
D:−110+10+50=−50
Since tank D loses the maximum amount of chemical in a minute, it will be emptied first.
Let n minutes be the time taken by tank D to get empty. ∴1000–50n=0 ∴n=20minutes
Hence, option 3.

2. (c)
If we consider the third term to be 'x"
The 15th term will be (x + 12d)
6th term will be (x + 3d)
11th term will be (x + 8d) and 13th term will be (x + 10d)
Thus, as per the given condition, 2x + 12d = 3x + 21d.
Or x + 9d = 0
x + 9d will be the 12th term.

3. (d)
Let the original length, breadth and height of the room be 3x, 2x and x respectively.
∴ The new length, breadth and height are 6x, x and x/2 respectively.
Area of four walls = (2 × length × height) + (2 × breadth × height)
Original area of four walls = 6x2 + 4x2 = 10x2
New area of four walls = 6x2 + x2 = 7x2
∴ Area of wall decreases by [(10x2 − 7x2)/10x2] × 100 = 30%


4. (c)
The number of goats remains the same.
If the percentage that is added every time is equal to the percentage that is sold, then there should be a net decrease. The same will be the case if the percentage added is less than the percentage sold. 

5. (a)
Let there be g girls and b boys.
Number of games between two girls = gC2
Number of games between two boys = bC2
∴ g(g – 1)/2 = 45 ∴ g2 – g – 90 = 0 ∴ (g – 10)(g + 9) = 0
∴ g = 10 Also, b(b – 1)/2 = 190 ∴ b2 – b – 380 = 0
∴ (b + 19)(b – 20) = 0 ∴ b = 20
∴ Total number of games = (g + b)C2 = 30C2 = 435
∴ Number of games in which one player is a boy and the other is a girl = 435 – 45 – 190 = 200 Hence, option a.

6. (d)
Let there be n rows and a students in the first row.
∴ Number of students in the second row = a + 3
∴ Number of students in the third row = a + 6 and so on.
∴ the number of students in each row forms an arithmetic progression with common difference = 3 the total number of students = The sum of all terms in the arithmetic progression= n[IMG]file:///C:/Users/Mansi/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]/2= 630
Now consider options,
If n=3, a= 207
IF n=4, a= 153
If n=5, a=120
If n=6, a is not an integer
If n=7, a= 81
Therefore, only 6 is not an integer.

7. (d)
1/m + 4/n = 1/12
∴ 1/m = 1/12 – 4/n
∴ m = 12n / (n – 48)
As, m is a positive integer, n should be greater than 48 and moreover since n is a positive odd integer lesser than 60, n can take values 49, 51, 53, 55, 57 and 59.
If n = 49, 51, 57 then m is a positive integer.
If n = 53, 55, 59 then m is not an integer.
∴ 3 pairs of values of m and n satisfy the given equation.

Questions 8 and 9

Ram and Shyam run a race between points A and B, 5 km apart. Ram starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Shyam starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed.
Ram starts at 9:00 am and Shyam starts at 9:45 am from A.
Ram reaches B at 10:00 am (as his speed is 5km/hr and the distance between A and B is 5km)
When Ram reaches B, Shyam is 15/60 × 10 = 2.5 km away from A. Ram meets Shyam (2.5 × 60)/(10 + 5) minutes after 10:00 a.m. i.e., at 10:10 a.m. Shyam reaches B at 10:15 a.m.
At 10:15 a.m., Ram is (15/60) × 5 = 1.25 km away from B. Shyam overtakes Ram in 1.25/(10 – 5) = 0.25 hrs = 15 minutes after 10:15 am i.e. at 10:30 a.m.

8. (b)
9. (b)

10. (c) 

The four consecutive two-digit odd numbers will have (1, 3, 5, 7) or (3, 5, 7, 9) or (5, 7, 9, 1) or (7, 9, 1, 3) as units digits.
As the sum divided by 10 yields a perfect square, the sum is a multiple of 10.
∴ The unit's digits have to be (7, 9, 1, 3).
Thus the four numbers will be (10x + 7), (10x + 9), (10x + 11) and (10x + 13), where 0 < x < 9 (as each of these numbers is a two digit number)
Sum of these numbers = 40x + 40 = 40(x + 1)
Now, 40(x + 1)/10 = 4(x + 1) is a perfect square
As 4 is a perfect square (x + 1) is some perfect square < 10
x + 1 = 4, x = 3, and the four numbers are 37, 39, 41 and 43
x + 1 = 9, x = 8, and the four numbers are 87, 89, 91 and 93