10, 14, 16, 18, 21, 24, 26
[A].
[B].
[C].
[D].
Answer: Option C
Explanation:
Each of the numbers except 21 is an even number.
[B].
[C].
[D].
Answer: Option C
Explanation:
Each of the numbers except 21 is an even number.
Explanation:
Each of the numbers is a prime number except 12.
| 3 |
| 5 |
[B].
| 3 |
| 10 |
[C].
| 4 |
| 5 |
[D].
| 4 |
| 3 |
Answer: Option C
Explanation:
| Let the required fraction be x. Then | 1 | – x = | 9 |
| x | 20 |
| 1 – x2 | = | 9 | |
| x | 20 |
20 – 20×2 = 9x
20×2 + 9x – 20 = 0
20×2 + 25x – 16x – 20 = 0
5x(4x + 5) – 4(4x + 5) = 0
(4x + 5)(5x – 4) = 0
| x = | 4 |
| 5 |
| 12 |
| 35 |
[B].
| 1 |
| 35 |
[C].
| 35 |
| 8 |
[D].
| 7 |
| 32 |
Answer: Option A
Explanation:
Let the numbers be a and b. Then, a + b = 12 and ab = 35.
| a + b | = | 12 | 1 | + | 1 | = | 12 | ||||
| ab | 35 | b | a | 35 |
| Sum of reciprocals of given numbers = | 12 |
| 35 |
[B].
[C].
[D].
Answer: Option D
Explanation:
Number = (12 x 35)
Correct Quotient = 420 21 = 20
[B].
[C].
[D].
Answer: Option B
Explanation:
Let the smaller number be x. Then larger number = (x + 1365).
x + 1365 = 6x + 15
5x = 1350
x = 270
Smaller number = 270.
[B].
[C].
[D].
Answer: Option D
Explanation:
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.
[B].
[C].
[D].
Answer: Option C
Explanation:
Let x be the number and y be the quotient. Then,
x = 357 x y + 39
= (17 x 21 x y) + (17 x 2) + 5
= 17 x (21y + 2) + 5)
Required remainder = 5.
[B].
[C].
[D].
Answer: Option B
Explanation:
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.
[B].
[C].
[D].
Answer: Option B
Explanation:
Number = 269 x 68 + 0 = 18292
67) 18292 (273
134
—-
489
469
—-
202
201
—
1
—
Therefore, Required remainder = 1