1) Find the H.C.F. of 26 * 35 * 52, 28 * 32 * 54 * 72 and 27 * 33* 53 * 74.
a) 26 *35 * 52 * 72.b) 27 *32 * 52 * 74.
c) 27 *35 * 54.
d) 26 *32 * 52.
ANSWER : 26 *32 * 52
SOLUTION:
The H.C.F. of the given numbers is the product of the common factors with least power.
Hence, H.C.F. = 26 *32 * 52.
SOLUTION:
The H.C.F. of the given numbers is the product of the common factors with least power.
Hence, H.C.F. = 26 *32 * 52.
2) Find the highest common factor of 4266 and 7848.
a) 24b) 18
c) 12
d) 6
ANSWER : 18
SOLUTION :
4266 = 2 * 33 * 79
7848 = 23 * 32 * 109
Hence, H.C.F. = 2 * 32 = 18
SOLUTION :
4266 = 2 * 33 * 79
7848 = 23 * 32 * 109
Hence, H.C.F. = 2 * 32 = 18
3) The L.C.M. of 24, 38, and 74 is:
a) 2818b) 912
c) 16872
d) None of these
ANSWER : 16872
SOLUTION :
Factors of the numbers are:
24: - 2*2*2*3
38: - 2*19
74: - 2*37
Thus, L.C.M. is 2*2*2*3*19*37 = 16872
SOLUTION :
Factors of the numbers are:
24: - 2*2*2*3
38: - 2*19
74: - 2*37
Thus, L.C.M. is 2*2*2*3*19*37 = 16872
4) Four numbers are in the ratio 2:3:5:6 and their H.C.F. is 18. The numbers are:
a) 36, 54, 90, 108b) 18, 27, 45, 54
c) 20, 30, 50, 60
d) None of these
ANSWER : 36, 54, 90, 108
SOLUTION :
Let the numbers be 2x, 3x, 5x, 6x. H.C.F. = x = 18
Therefore, the numbers are 36, 54, 90, 108.
SOLUTION :
Let the numbers be 2x, 3x, 5x, 6x. H.C.F. = x = 18
Therefore, the numbers are 36, 54, 90, 108.
5) The H.C.F. of two numbers is 43 and their product is 33282. How many combinations are possible:
a) 1b) 2
c) 3
d) 4
ANSWER : 2
SOLUTION :
Let the numbers be 43x and 43y. Hence, 43x*43y = 33282
xy = 18
Co-primes with the product 18 are (1,18) and (2,9)
There are 2 such combinations possible.
SOLUTION :
Let the numbers be 43x and 43y. Hence, 43x*43y = 33282
xy = 18
Co-primes with the product 18 are (1,18) and (2,9)
There are 2 such combinations possible.
6) If the product of the L.C.M. and H.C.F. of two numbers is 60 and the difference between the two numbers is 4, find the numbers.
a) 6 and 10b) 8 and 12
c) 4 and 8
d) Cannot be determined
ANSWER : 6 and 10
SOLUTION :
H.C.F. * L.C.M = product of the two numbers
H.C.F.*L.C.M = 60 = product of the two numbers
Let one of the numbers be x. The second number is (x+4)
60 = product of the two numbers = x*(x+4)
x2 + 4x – 60 = 0 = (x + 10)*(x – 6)
x = 6
Therefore, the numbers are 6 and 10.
SOLUTION :
H.C.F. * L.C.M = product of the two numbers
H.C.F.*L.C.M = 60 = product of the two numbers
Let one of the numbers be x. The second number is (x+4)
60 = product of the two numbers = x*(x+4)
x2 + 4x – 60 = 0 = (x + 10)*(x – 6)
x = 6
Therefore, the numbers are 6 and 10.
7) The least number which is a multiple of 11 and when divided by 3, 5 and 9 leaves 2 as remainder:
a) 297b) 319
c) 407
d) None of these
ANSWER : 407
SOLUTION :
The L.C.M. of 3, 5 and 9 is 45.
Hence, the number is of the form (45a + 2).
The least value of ‘a’ for which (45a + 2) is multiple of 11 is 9.
Therefore, the number is 407.
SOLUTION :
The L.C.M. of 3, 5 and 9 is 45.
Hence, the number is of the form (45a + 2).
The least value of ‘a’ for which (45a + 2) is multiple of 11 is 9.
Therefore, the number is 407.
8) 3 bells beep at an interval of 12, 20, and 35 minutes. If they beep together at 10 a.m., then they will again beep together at:
a) 12 p.m.b) 1 p.m.
c) 4 p.m.
d) 5 p.m.
ANSWER : 5 p.m.
SOLUTION :
The L.C.M. of 12, 20 and 35 is 420. Hence, all 3 bells beep together after 420 minutes = 7 hours
Hence, the 3 bells will beep together 7 hours after 10 a.m. i.e. at 5 p.m.
SOLUTION :
The L.C.M. of 12, 20 and 35 is 420. Hence, all 3 bells beep together after 420 minutes = 7 hours
Hence, the 3 bells will beep together 7 hours after 10 a.m. i.e. at 5 p.m.
9) A man decides to pave with square tiles his hall which is 4.8 metres long and 5.38 metres wide. Find the largest size of the tile that he could use.
a) 21 cmb) 2 cm
c) 269 cm
d) None of these
ANSWER : 2 cm
SOLUTION :
The largest size of the tile is H.C.F. of 480 cm and 538 cm is 2 cm.
SOLUTION :
The largest size of the tile is H.C.F. of 480 cm and 538 cm is 2 cm.
10) Find the least number that is a perfect cube and can be divisible by 8, 14, and 18.
a) 5832b) 74088
c) 504
d) Cannot be determined
ANSWER : 74088
SOLUTION :
A number that is divisible by 8, 14 and 18 = L.C.M. of 8, 14 and 18 = 504
504 = 2*2*2*3*3*7
Hence, to get a perfect cube 504 should be multiplied by 3*7*7 = 147
The number is 74088.
SOLUTION :
A number that is divisible by 8, 14 and 18 = L.C.M. of 8, 14 and 18 = 504
504 = 2*2*2*3*3*7
Hence, to get a perfect cube 504 should be multiplied by 3*7*7 = 147
The number is 74088.
Sample questions on Problems on H.C.F and L.C.M
1. Find the L.C.M. of 72, 108 and 2100.
3880037800
38880
37870
2. Two numbers are in ratio of 17:13. If their H.C.F. is 19 then find the numbers.
323, 247343,257
354,275
365,278
3. Find the greatest possible length which can be used to measure exactly the length 3 m 84 cm, 5 m 64 cm and 7 m 32 cm.
2432
12
18
4. Find the largest number which divides 62, 132 and 237 to leave the remainder in each case.
3538
42
45
5. The LCM of two numbers is 495 an their HCF is 5. If the sum of the numbers is 10, then their difference is :
1014
20
32
6. If the sum of two numbers is 55 and the HCF and LCM of these numbers are 5 and 120 respectively, then the reciprocals of the numbers is equal to:
45/123907/34
11/120
56/117
7. Amar, Akbar, Anthony start running to save their mother from Robert at the same time in same direction around a circular lake. Amar completes the whole round in 27 seconds, Akbar in 23 seconds and Anthony in 50 seconds, all starting at the same point. After what time will they meet again at starting point?
8 hours 37 minutes 50 seconds7 hours 37 minutes 50 seconds
9 hours 37 minutes 50 seconds
6 hours 37 minutes 50 seconds
8. Traffic Signal turns Green at regular interval of 36 seconds and 45 seconds. If first turn green together at 10 am in morning, at what time will they turn green again for the first time?
10:05 am10:02 am
10:03 am
10:01 am
9. A, B, C start at the same time in same direction to run around a circular lake. A completes the whole round in 12 seconds, B in 23 seconds and C in 45 seconds, all starting at the same point. After what time will they meet again at starting point?
59 minutes60 minutes
69 minutes
70 minutes
0 comments:
Post a Comment