Discussion
Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
- The HCF of (x 4 - y 4) and (x 6 - y 6) is
Options- A. x2 - y2
- B. x - y
- C. x3 - y3
- D. x4 - y4
- Correct Answer
- x2 - y2
ExplanationLet f (x) = (x4 - y4)
= (x2 - y2) (x2 + y2)
= (x - y) (x + y) (x2 + y2)
and g(x) = (x6 - y6)
= (x3)2 - (y3)2
= (x3 + y3) (x3 - y3)
= (x + y) (x2 - xy + y2) (x - y)(x2 + xy + y2)
= (x - y) (x + y) (x2 - xy + y2)
(x2 + xy + y2)
? HCF of [f(x), g(x)] = (x - y) (x + y)
= x2 - y2 - 1. What is the HCF of 8 (N 5 - N 3 + N ) and 28 (N 6 + 1)?
Options- A. 4 (N4 - N2 + 1)
- B. N3 - N + 4N2
- C. N3 - N + 3N2
- D. None of these Discuss
- 2. If a and b are positive integers, then what is the value of (a / HCF (a,b) , b/HCF (a,b) )?
Options- A. a
- B. b
- C. 1
- D. a/HCF (a,b) Discuss
- 3. If (x - 6) is the HCF of x 2 - 2x - 24 and x 2- kx - 6, then what is the value of k?
Options- A. 3
- B. 5
- C. 6
- D. 8 Discuss
- 4. The HCF and LCM of two number are 21 and 4641, respectively. If one of the numbers lies between 200 and 300, then find the two numbers.
Options- A. 273, 363
- B. 273, 359
- C. 273, 361
- D. 273, 357 Discuss
- 5. 6 bells commence tolling together and toll of intervals are 2, 4, 6, 8, 10 and 12 s, respectively. In 1 h, how many times, do they toll together?
Options- A. 16
- B. 32
- C. 21
- D. 35 Discuss
- 6. The HCF of (x 3 - y 2 - 2x) and (x 3 + x 2) is
Options- A. x3 - x2 - 2x
- B. x2 + x
- C. x4 - x3 - 2x2
- D. x - 2 Discuss
- 7. What is the HCF of a 2 b 4 + 2 a 2 b 2 and (ab) 7 - 4a 2 b 2?
Options- A. ab
- B. a2 b3
- C. a2 b2
- D. a3 b2 Discuss
- 8. Find the largest number which divides 1305, 4665 and 6905 leaving same remainder in each case. Also, find the common Remainder.
Options- A. 1210, 158
- B. 1120, 158
- C. 1120, 185
- D. 1210, 185 Discuss
- 9. The average of 8, 9, 12, 13, 15, 9 is
Options- A. 11
- B. 6
- C. 16
- D. 18 Discuss
- 10. A has 8 pencils, B has 10 pencils and C has 15 pencils, then the average number of pencils with them are
Options- A. 8
- B. 10
- C. 15
- D. lies between 9 and 15 Discuss
Problems on H.C.F and L.C.M
Search Results
Correct Answer: 4 (N4 - N2 + 1)
Explanation:
Let p(N) = 8 (N5 - N3 + N)
= 4 x 2 x N (N4 - N2 + 1)
and q(N) = 28 (N6 + 1)
= 7 x 4 [( N2)3 + (1)3]
= 4 x 7 (N2 + 1) ( N4 - N2 + 1)
? HCF of p(N) and q (N) = 4 (N4 - N2 + 1)
Correct Answer: 1
Explanation:
HCF (a /HCF (a, b) , b/HCF (a,b) it value is always equal to 1;
IIIustration 1 Let the two positive integers be a = 24 and b = 36.
? HCF (24/HCF (24,36) , 36/HCF (24,36)
? HCF (24/12 , 36/12)
? HCF (2, 3) = 1
IIIustration 2
Let the two positive integers be a = 13 and b = 17
? HCF (13 / HCF (13,17) , 17/HCF (13,17)
? HCF (13/1 , 17/1) = 1
Correct Answer: 5
Explanation:
Given that, (x - 6) is the HCF of x2 - 2x - 24 and x2 - kx - 6 i.e., (x - 6) is a factor of both expressions.
Let f (x1) = x12 - 2x1 - 24
and f (x2) = x22 - kx2 - 6
Now f(x1) = f (x2) at (x1 = x2 = 6)
? (6)2 - 2(6) - 24 = (6)2 - k(6) - 6 [by condition]
? 0 = 30 - 6k ? 6 k = 30
? k = 5
Correct Answer: 273, 357
Explanation:
Let the numbers be 21a and 21b, where a and b are co-primes.
Then, 21a x 21b = (21 x 4641)
? ab = 221
Two co-primes with product 221 are 13 and 17.
? Required number = (21 x 13, 21 x 17)
= (273, 357)
Correct Answer:
Explanation:
LCM of 2, 4, 6, 8, 10, 12
=(2 x 2 x 3 x 2 x 5) = 120
? After every 2 min, they toll together.
? Number of times they toll together in one hour = (60/2) + 1 times
= 31 times
Correct Answer: x4 - x3 - 2x2
Explanation:
Let f(x) = x3 - y2 - 2x = x( x2 - x- 2)
= x( x2 - 2x + x - 2)
= x {x (x - 2) + 1 (x -2)}
= x (x +1) (x - 2)
and g(x) = x3 + x2
= x2 (x +1) = x. x (x +1)
? LCM of [ f(x), g(x)] = x (x +1) . x . (x - 2)
= x2(x + 1)(x - 2)
= x2(x2 - x - 2)
= x4 - x3 - 2x2
Correct Answer: a2 b2
Explanation:
a2 b4 + 2 a2 b2 = a2 b2 ( b2 +2).......................(i)
and (ab)7 - 4 a2 b9 = a7 b7 - 4 a2 b9 = a2 b2(a5 b5 - 4b7)................(ii)
from Eqs. (i) and (ii)
HCF = a2 b2
Correct Answer: 1120, 185
Explanation:
Given that
x = 1305, y = 4665, z = 6905
Then ,
|x - y| = |1305 - 4665| = 3360
|y - z] = |4665 - 6905| = 2240
|z - x| =|6905 -1305| = 5600
? Required number = HCF of 3360, 2240 and 5600 = 1120
On dividing 1305 by 1120, remainder is 185.
On dividing 4665 by 1120, remainder is 185.
On dividing 6905 by 1120, remainder is 185.
? Common remainder = 185
Correct Answer: 11
Explanation:
Average = (8 + 9 + 12 + 13 + 15 + 9)/6
= 66 / 6
= 11
Correct Answer: lies between 9 and 15
Explanation:
Average number of pencils
= ( 8 + 10 + 15 ) / 3
=11
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.