Problems on LCM and HCF Part 2

Problems on LCM and HCF Part 2


21. 
The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:
A.15 cm
B.25 cm
C.35 cm
D.42 cm
Answer: Option C
Explanation:
Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.
22. 
Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:
A.75
B.81
C.85
D.89
Answer: Option C
Explanation:
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
First number =551= 19;    Third number =1073= 37.
2929
 Required sum = (19 + 29 + 37) = 85.
23. 
Find the highest common factor of 36 and 84.
A.4
B.6
C.12
D.18
Answer: Option C
Explanation:
36 = 22 x 32
84 = 22 x 3 x 7
 H.C.F. = 22 x 3 = 12.
24. 
Which of the following fraction is the largest ?
A.
7
8
B.
13
16
C.
31
40
D.
63
80
Answer: Option A
Explanation:
L.C.M. of 8, 16, 40 and 80 = 80.
7=70;  13=65;  31=62
88016804080
Since,70>65>63>62, so7>13>63>31
808080808168040
So,7is the largest.
8
25. 
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
A.504
B.536
C.544
D.548
Answer: Option D
Explanation:
Required number = (L.C.M. of 12, 15, 20, 54) + 8
   = 540 + 8
   = 548.

26. 
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
A.123
B.127
C.235
D.305
Answer: Option B
Explanation:
Required number = H.C.F. of (1657 - 6) and (2037 - 5)
  = H.C.F. of 1651 and 2032 = 127.
27. 
Which of the following has the most number of divisors?
A.99
B.101
C.176
D.182
Answer: Option C
Explanation:
99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
28. 
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
A.28
B.32
C.40
D.64
Answer: Option C
Explanation:
Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8.
 The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.
29. 
The H.C.F. of9,12,18and21is:
10253540
A.
3
5
B.
252
5
C.
3
1400
D.
63
700
Answer: Option C
Explanation:
Required H.C.F. =H.C.F. of 9, 12, 18, 21=3
L.C.M. of 10, 25, 35, 401400
30. 
If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
A.
55
601
B.
601
55
C.
11
120
D.
120
11
Answer: Option C
Explanation:
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
 The required sum =1+1=a + b=55=11
abab600120

No comments:

Post a Comment