Discussion
Home ‣ Aptitude ‣ Problems on Numbers See What Others Are Saying!
- Question
Which one of the following is not a prime number?
Options- A. 91
- B. 71
- C. 41
- D. 31
- Correct Answer
- 91
ExplanationPrime Numbers :: Numbers which are divisible by only 1 and itself are Prime Numbers.
It's answer will be 91.
Because 91 can be divisible by 7,13,91,1.
It is quite clear that prime number should be divisible only by itself and by 1.
- 1. What least number must be subtracted from 1294 so that the remainder when divided by 9, 11, 13 will leave in each case the same remainder 6?
Options- A. 0
- B. 1
- C. 2
- D. 3 Discuss
- 2. The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder is?
Options- A. 1677
- B. 1683
- C. 2523
- D. 3363 Discuss
- 3. If the HCF of two positive integers is 26. Then their LCM cannot be?
Options- A. 78
- B. 104
- C. 144
- D. 234 Discuss
- 4. The LCM and HCF of two number are 240 and 16, respectively,If the two numbers are in the ratio 3 : 5, the numbers are?
Options- A. 24, 40
- B. 21, 35
- C. 36, 60
- D. 48, 80 Discuss
- 5. What are the values of c when the HCF of x 3 + c x 2 - x + 2c and x 2 + cx - 2 over the rational is a linear polynomial?
Options- A. +1
- B. +2
- C. +1
- D. +4 Discuss
- 6. What is the greatest number which divides 390, 480 and 620 so as to leave the same remainder in each case?
Options- A. 15
- B. 10
- C. 20
- D. 19 Discuss
- 7. Find the two numbers whose L.C.M. is 1188 and H.C.F. is 9?
Options- A. 27, 396
- B. 9, 27
- C. 36,99
- D. Data inadequate Discuss
- 8. The Sum of two number is 1215 and their H.C.F is 81, how many pairs of such number can be formed? Find them?
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 9. 21 mangoes trees, 42 apples trees and 56 oranges trees have to be planted in rows such that each row contains the same number of trees one variety only. Minimum number of rows in which the above trees may be planted is?
Options- A. 15
- B. 17
- C. 3
- D. 20 Discuss
- 10. Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length, What is the greatest possible length of each plank?
Options- A. 7 m
- B. 14 m
- C. 42 m
- D. 63 m Discuss
More questions
Correct Answer: 1
Explanation:
L. C .M of 9, 11, 13 is 1287
On dividing 1294 by 1287, the remainder is 7 .
? 1 must be subtracted from 1294, so that 1293 when divided by 9, 11, 13 leaves in each case the same remainder 6 .
Correct Answer: 1683
Explanation:
L.C.M of 5, 6 , 7, 8 is 840
So, the number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2
? Required number = (840 x 2 + 3 ) = 1683
Correct Answer: 144
Explanation:
LCM of the number is always divisible by the HCF of the same numbers.
So, 78, 104 and 234 are all divisible by 26. Whereas, 144 is not divisible by 26.
Thus, 144 cannot be the LCM of the number whose HCF is 26.
Correct Answer: 48, 80
Explanation:
Suppose two numbers are 3N and 5N.
Then, 3N x 5N = HCF x LCM
? 15N2 = 16 x 240
? N2 = 256
? N = 16
So, the number are 3N = 3 x 16 = 48 and 5N = 5 x 16 = 80.
Correct Answer: +1
Explanation:
Find HCF of x3 +c x2 -x + 2c and x2 + cx - 2 by long division method and get remainder
? Remainder = 2c - 2/c
Since,remainder should be zero
? 2c2 - 2 = 0
? 2(c2 -1) = 0 ? c= ± 1
Correct Answer: 10
Explanation:
The HCF of [(480 - 390), (620 - 480), (620 - 390)]
= HCF of 90, 140, 230 = 10
Correct Answer: 27, 396
Explanation:
Let the two number be 9a and 9b where a and b are two numbers prime to each other. The L.C.M. of 9a and 9b is 9ab.
? 9ab = 1188
? ab = 132
Now, the possible pairs of factors of 132 are 1 x 132, 2 x 66, 3 x 44, 6 x 22, 11 x 12 of these pairs (2, 66) and (6, 22) are not prime to each other and, therefore, not admissible.
Hence the admissible pairs are
(1, 132), (3, 44), (4, 33), (11, 12)
? a = 1, b = 132; or
a=3, b=44 or
a=4, b=33 or
a=11, b = 12
Hence, the possible number are 9, 9 x 13, 9 x 3, 9 x 44, 9 x 4, 9 x 33, 9 x 11, 9 x 12
So option A is correct 27, 396.
Correct Answer: 4
Explanation:
Let the number be 81a and 81b where a and b are two numbers prime to each other.
? 81a + 81b = 1215
? a + b = 1215/81 = 15
Now, find two numbers, whose sum is 15, the possible pairs are (14, 1) , (13, 2), (12, 3), (11, 4), (10, 5), (9, 6 ), (8, 7) of these the only pairs of numbers that are prime to each other are (14, 1) , (13, 2), (11, 4), and (8, 7).
Hence, the required number are
(14 x 81, 1 x 81), (13 x 81, 2 x 81), (11 x 81, 4 x 81), (8 x 81, 7 x 81)
or (1134, 81), (1053, 162), (891, 324), (648, 567)
So, there are four such pairs .
Correct Answer: 17
Explanation:
H.C.F. of 21, 42, 56 = 7
Number of rows of mango trees, apple trees and oranges trees are 21/7 = 3, 42/7 = 6 and 56/7 = 8
? Required number of rows = (3 + 6 + 8 ) = 17
Correct Answer: 7 m
Explanation:
Greatest possible length of each plank = H.C.F of 42, 49, 63 = 7 m
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.