H.C.F and L.C.M
Q1. 128352/238368 when reduced to its lowest terms is
(a) 5/11
(b) 7/13
(c) 9/17
(d) 11/13
Answer: (b) 7/13
128352 ) 238368 ( 1
128352
110016) 128352 ( 1
110016
18336) 110016 ( 6
110016
∴ H.C.F = 18336
∴ | 128352 ÷ 18336 | = | 7 |
238368 ÷ 18336 | 13 |
Q2. The L.C.M of 3/4, 6/7, 9/8 is
(a) 12
(b) 15
(c) 18
(d) 21
Answer: (c) 18
L.C.M = | L.C.M of numerators |
H.C.F of denominator |
= | L.C.M of 3, 6, 9 |
H.C.F of 4, 7, 8 |
= | 18 |
1 |
= 18
Q3. The ratio between two numbers is 15:11. If their H.C.F is 13, then the numbers are
(a) 141, 135
(b) 174, 145
(c) 195, 143
(d) 201, 157
Answer: (c) 195, 143
Let, the numbers be 15x, 11x
∴ H.C.F of 15x, 11x = x
A/Q x = 13
∴ The numbers are 15 X 13, 11 X 13 = 195, 143
Q4. The sum of two numbers is 495 and their H.C.F is 33. The numbers of such pairs is
(a) 3
(b) 4
(c) 6
(d) 7
Answer: (b) 4
Let the numbers be 33x and 33y
Now, 33x + 33y = 495
⇒ 33(x+y) = 495
⇒ x+y = 495/33 = 15
∴ (x = 1, y = 14), (x = 2, y = 13), (x = 4, y = 11), (x = 7, y =8)
∴ Possible numbers of pairs = 4
Q5. The H.C.F of two numbers is 25. The number which can be their L.C.M is
(a) 80
(b) 95
(c) 116
(d) 125
Answer: (d) 125
H.C.F is always divides L.C.M completely
∴ L.C.M = 125
Q6. A, B and C go walking round a circle 2 km in circumference at the rates of 20 m/min, 40m/min and 50 m/min respectively. If they all start together and walk in the same direction, when will they be together at the same place ?
(a) 110 min
(b) 145 min
(c) 170 min
(d) 200 min
Answer: (d) 200 min
Time taken by A, B, C to make a round
= | 2000 | min, | 2000 | min, | 2000 | min, |
20 | 40 | 50 |
= 100 min, 50 min, 40min
∴ Required time = L.C.M of 100, 50, 40 = 2X5X5X2X2 = 200 min
Q7. Which greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively ?
(a) 45
(b) 51
(c) 57
(d) 63
Answer: (a) 45
Required number = H.C.F of (3026 - 11) and (5053 - 13)
= H.C.F of 3015 and 5040
3015 ) 5040 ( 1
3015
2025) 3015 ( 1
2025
990) 2025 ( 2
1980
45) 990 ( 22
90
∴ H.C.F of 3015 and 5040 = 45
Q8. The greatest number of 5 digits, which is divisible by 15, 20, 25 is
(a) 99000
(b) 99900
(c) 99990
(d) 99999
Answer: (b) 99900
L.C.M of 15, 20, 25 = 5X3X4X5 = 300
Greatest number of 5 digits = 99999
Now 99999 ÷ 300 will give remainder 99
∴ Required number = 99999 - 99 = 99900
Q9. What is the greatest number of 3 digits which when divided by 6, 9, 12 leaves a remainder 3 in each case ?
(a) 850
(b) 925
(c) 975
(d) 1050
Answer: (c) 975
L.C.M of 6, 9, 12 is
= 3X2X3X2
= 36
Greatest number of 3 digit = 999
Now 999 ÷ 36 will give remainder 27
∴ Required number = 999 - 27 + 3 = 975
Q10. If two numbers are in the ratio of 3:2 and the product of their H.C.F and L.C.M is 3750, then the sum of the numbers is
(a) 125
(b) 133
(c) 147
(d) 155
Answer: (a) 125
Let, the numbers be 3x and 2x
∴ H.C.F = x
L.C.M = 6x
A/Q x . 6x = 3750
⇒ 6x2 = 3750
⇒ x2 = 625
⇒ x = √625 = 25
∴ The numbers are 75 and 50
∴ Sum = 75 + 50 = 125
Practice Test Exam