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46.
6897 is divisible by :
- A.11 only
- C.Both 11 and 19
- B.19 only
- D.neither 11 nor 19
- Answer & Explanation
- Report
Answer : [C]
Explanation :
Explanation :
| Clearly, 6897 is divisible by both 11 and 19. |
47.
If the number 357*25* is divisible by both 3 and 5, then the missing digits in the unit’s place and the thousandth place respectively are:
- A.0, 6
- C.5, 4
- B.5, 6
- D.None of these
- Answer & Explanation
- Report
Answer : [B]
Explanation :
Explanation :
| Let the required number be 357y25x. | |||||
| Then, for divisibility by 5, we must have x = 0 or x = 5. | |||||
| Case I. When x = 0 | |||||
| Then, sum of digits = (22 + y). For divisibility by 3, (22 + y) must be divisible by 3. | |||||
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| Case II. When x = 5. | |||||
| Then, Sum of digits = (27 + y). For divisibility by 3, we must have y = 0 or 3 or 6 or 9. | |||||
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| So, correct answer is (b). | |||||
48.
If -1 ≤ x ≤ 2 and 1 ≤ y ≤ 3, then least possible value of (2y - 3x) is:
- A.0
- C.-4
- B.-3
- D.-5
- Answer & Explanation
- Report
Answer : [C]
Explanation :
Explanation :
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49.
3578 + 5729 - ?486 = 5821
- A.1
- C.3
- B.2
- D.None of these
- Answer & Explanation
- Report
Answer : [C]
Explanation :
Explanation :
| Let 3578 + 5729 - x486 = 5821 |
| Then, 9307 - x486 = 5821 |
| x486 = (9307 - 5821) |
| x486 = 3486 |
| x = 3 |
50.
A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as the remainder. The number is :
- A.1220
- C.22030
- B.1250
- D.220030
- Answer & Explanation
- Report
Answer : [D]
Explanation :
Explanation :
| Required number = (555 + 445) x 2 x 110 + 30 = 220000 + 30 = 220030. |
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