Number System - Online Test
| Let the required fraction be x. Then | 1 | - x = | 9 |
| x | 20 |
 | 1 - x2 | = | 9 |
| x | 20 |
20 - 20x2 = 9x
20x2 + 9x - 20 = 0
20x2 + 25x - 16x - 20 = 0
5x(4x + 5) - 4(4x + 5) = 0
(4x + 5)(5x - 4) = 0
| x = | 4 |
| 5 |
Formula: (Divisor*Quotient) + Remainder = Dividend.
Soln:
(56*Q)+29 = D -------(1)
D%8 = R -------------(2)
From equation(2),
((56*Q)+29)%8 = R.
=> Assume Q = 1.
=> (56+29)%8 = R.
=> 85%8 = R
=> 5 = R.
| 107 x 107 + 93 x 93 | = (107)2 + (93)2 |
| = (100 + 7)2 + (100 - 7)2 | |
| = 2 x [(100)2 + 72] [Ref: (a + b)2 + (a - b)2 = 2(a2+ b2)] | |
| = 20098 |
(xn + 1) will be divisible by (x + 1) only when n is odd.
(6767 + 1) will be divisible by (67 + 1)
(6767 + 1) + 66, when divided by 68 will give 66 as remainder.
Let the number be x and on dividing x by 5, we get k as quotient and 3 as remainder.
x = 5k + 3
x2 = (5k + 3)2
= (25k2 + 30k + 9)
= 5(5k2 + 6k + 1) + 4
On dividing x2 by 5, we get 4 as remainder.
3-digit number divisible by 6 are: 102, 108, 114,... , 996
This is an A.P. in which a = 102, d = 6 and l = 996
Let the number of terms be n. Then tn = 996.
a + (n - 1)d = 996
102 + (n - 1) x 6 = 996
6 x (n - 1) = 894
(n - 1) = 149
n = 150
Number of terms = 150.
Required numbers are 24, 30, 36, 42, ..., 96
This is an A.P. in which a = 24, d = 6 and l = 96
Let the number of terms in it be n.
Then tn = 96 a + (n - 1)d = 96
24 + (n - 1) x 6 = 96
(n - 1) x 6 = 72
(n - 1) = 12
n = 13
Required number of numbers = 13.
Marking (/) those which are are divisible by 3 by not by 9 and the others by (X), by taking the sum of digits, we get:s
2133 
9 (X)
2343 12 (/)
3474 18 (X)
4131 9 (X)
5286 21 (/)
5340 12 (/)
6336 18 (X)
7347 21 (/)
8115 15 (/)
9276 24 (/)
Required number of numbers = 6.
| (963 + 476)2 + (963 - 476)2 | = ? |
| (963 x 963 + 476 x 476) |
| Given Exp. = | (a + b)2 + (a - b)2 | = | 2(a2 + b2) | = 2 |
| (a2 + b2) | (a2 + b2) |
