Answer: 4

Explanation:
Let the numbers be x, y.
x(x+y) = 510, y(x+y) = 390
x2 + xy = 510 …… (1)
y2 + xy = 390 …… (2)
Adding (1) and (2)
x2 + xy + xy + y2 = 510 + 390 = 900
(x+y) 2 = 900 = (30) 2
(x+y) = 30
Subtracting (2) from (1)
x2 + xy – xy - y2 = 510 - 390 = 120
x2 - y2 = 120
(x+y) (x-y) = 120
(x-y) = 4.

Answer: yes

Explanation:
n2 – 1 = (n + 1) (n – 1)
Every prime number greater than 3 can be written in the form of
(6K + 1) or (6K – 1), where K is a positive integer. Let n = 6K + 1,
then (n + 1) (n – 1) = (6K + 1 + 1) (6K + 1 - 1)
= (6K + 2) (6K)
= 12 K (3K + 1)
Either K or (3K + 1) is even. K (3K + 1) is even.
Since K (3K + 1) is even, it is divisible by 2. Hence, 12 × K(3K + 1) is Divisible by 12 × 2 = 24

Answer: 60

Explanation:
Kishor erases all numbers and leaves only numbers divisibly by 3. So on the board we have 3, 6, 9, 12, … 99. A total of 33 numbers are left out. Basu erases all but multiples of 4. From 33 numbers only multiples of 12 will remain, i.e, 12, 24, 36,… 96. A total of 8 members are left out. Komal erases all numbers that are not multiples of 5. The numbers that are multiples of 5 will be left out. Only 60 is a multiple of 5. The numbers erased are 12, 24, 36, 48, 72, 84 and 96. A total of 7 numbers will be erased and one number, i.e 60, remains on the board.
Alternately, the numbers that are left on the board = Multiples of LCM of (3, 4, 5).
This equals to 60. Hence only one number remains on the board .

Answer: 0

Explanation:
 17 = -2 mod 19
1725 = -225 mod 19
21 = 2 mod 19
2125 = 225 mod 19
Using the concept, “If a = b mod c and p = q mod c, then a + p = b + q
mod c”. 1725 + 2125 = -225 + 225 mod 19
= 0 mod 19 , i.e 0 is the remainder.   

Answer: 166

Explanation:
Step 1: A. Let the required number and the remainder in each case be
N. Using Rule 3, N + 2 is divisible by 6, 7 and 8.
B. N + 2 must be a multiple of 6, 7 and 8
C. The smallest possible value of N + 2 must be the LCM of (6, 7, 8). Any such value of N + 2 can be written as follows.
N + 2 = K × LCM(6, 7, 8) where K = 1, 2, 3
Step 2: × LCM(6, 7, 8) = K × 23 × 3 × 7 = 168K
Step 3: To find out the least possible value of N + 2, Substituting K = 1in (1), we get N + 2 = 168. N = 166

Answer: 39

Explanation:
 The maximum number of swords or spears that can be kept in each box such that they are equal in number = HCF of (378, 675) = 27. The minimum number of boxes required = (378 + 675) / 27 = 39

Answer: 35

Explanation:
 The smallest time after which all the bells toll together again = K × LCM of (20, 25, 28, 50), where K = 1, 2, 3…
20 = 2 × 10 = 22 × 5; 25 = 52
28 = 2 × 2 × 7 = 22 × 7; 50 = 5 × 10 = 2 × 52
Time = K × LCM of (20, 25, 28, 50) = K × 22 × 52 × 7 = 700k sec
When K – 1, Time = 700 sec, which is not an integral minute.
When K = 3, Time = 3 × 700 sec = 2100 sec = 35 minutes.

Answer: 44

Explanation:
 252 × k = 6x x364
6x x364 is divisible by 4 × 7 × 9
Using test of divisibility by 9, we get 6 + x + x + 3 + 6 + 4 is divisible by 9
19 + 2x is divisible by 9
x has to be 4, the number is 644364.

Answer: 30

Explanation:
The bulb glows every 20 sec and the bell chimes every 24 sec. The time after which the bulb glows and the bell chime is = LCM (24,20) =120 second or 2 minutes. In 60 minutes, they will glow and chime together 30 times. 

Answer: 2

Explanation:
(8 + 7 + H + 4) - (2 + 6 + 8 + 5) = H – 2
=0 or multiple of 11
∴only digit at H that makes the number divisible by 11 is 2.