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The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly. There are two methods of finding the H.C.F. of a given set of numbers:
- Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.
- Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.
Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number. Similarly, the H.C.F. of more than three numbers may be obtained.
Least Common Multiple (L.C.M.):
The least number which is exactly divisible by each one of the given numbers is called their L.C.M. There are two methods of finding the L.C.M. of a given set of numbers:
- Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.
- Division Method (short-cut): Arrange the given numbers in a rwo in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.
Product of two numbers = Product of their H.C.F. and L.C.M.
Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.
H.C.F. and L.C.M. of Fractions:
- H.C.F = H.C.F of Numerators / L.C.M of Denominators
- L.C.M = L.C.M of Numerators / H.C.F of Denominators
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Hcf and Lcm – Question and Answers
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Question 1 |
A teacher can divide her class into groups of 5,13, and 17. What is the smallest possible strength of the class?
835 | |
940 | |
1105 | |
1220 |
Question 1 Explanation:
Smallest possible strength of the class will be L.C.M of 5,13,17 ans:1105
Question 2 |
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
544 | |
548 | |
504
| |
536 |
Question 2 Explanation:
Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.
= 540 + 8
= 548.
Question 3 |
Find the highest common factor of 36 and 84.
4 | |
6 | |
12 | |
18 |
Question 3 Explanation:
36 = 22 x 32
84 = 22 x 3 x 7
H.C.F. = 22 x 3 = 12.
84 = 22 x 3 x 7
H.C.F. = 22 x 3 = 12.
Question 4 |
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:
85 | |
75 | |
65 | |
60 |
Question 4 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 = 551/29 = 19; Third number = 1073/29 = 37.
Required sum = (19 + 29 + 37) = 85.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
First number = 551/29 = 19; Third number = 1073/29 = 37.
Required sum = (19 + 29 + 37) = 85.
Question 5 |
The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:
15 cm | |
35 cm | |
25 cm | |
52 cm |
Question 5 Explanation:
Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.
Question 6 |
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively is:
127 | |
305 | |
235 | |
123 |
Question 6 Explanation:
Required number = H.C.F. of (1657 – 6) and (2037 – 5)
= H.C.F. of 1651 and 2032 = 127.
= H.C.F. of 1651 and 2032 = 127.
Question 7 |
Which of the following has the most number of divisors?
176 | |
182 | |
99 | |
101 |
Question 7 Explanation:
Option A, 176 = 1 x 2 x 2 x 2 x 2 x 11.
Option B, 182 = 1 x 2 x 7 x 13.
Option C, 99 = 1 x 3 x 3 x 11.
Option D, 101 = 1 x 101.
Divisors of 99 are 1, 3, 9, 11, 33, 99.
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 have the most number of divisors
Option B, 182 = 1 x 2 x 7 x 13.
Option C, 99 = 1 x 3 x 3 x 11.
Option D, 101 = 1 x 101.
Divisors of 99 are 1, 3, 9, 11, 33, 99.
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 have the most number of divisors
Question 8 |
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
30 | |
40 | |
38 | |
36 |
Question 8 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.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.
Question 9 |
The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
267 | |
318 | |
308 | |
279 |
Question 9 Explanation:
Other number = (11 x 7700/275) = 308.
Question 10 |
What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?
1260 | |
630 | |
2524
| |
None of these |
Question 10 Explanation:
L.C.M. of 12, 18, 21 30 2 | 12 – 18 – 21 – 30
—————————-
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 – 9 – 21 – 15
—————————- Required number = (1260 ÷ 2) | 2 – 3 – 7 – 5
= 630.
—————————-
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 – 9 – 21 – 15
—————————- Required number = (1260 ÷ 2) | 2 – 3 – 7 – 5
= 630.
Question 11 |
The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
12 | |
68 | |
48 | |
78 |
Question 11 Explanation:
Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
Question 12 |
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
1008 | |
1032 | |
1015 | |
1022 |
Question 12 Explanation:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
= 1008 + 7
= 1015
Question 13 |
252 can be expressed as a product of primes as:
2 x 3 x 3 x 3 x 7 | |
3 x 3 x 3 x 3 x 7 | |
2 x 2 x 2 x 3 x 7 | |
2 x 2 x 3 x 3 x 7 |
Question 13 Explanation:
Clearly, 252 = 2 x 2 x 3 x 3 x 7.
Question 14 |
A, B and C start at the same time in the same direction to run around a circular stadium. A completes a
round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time
will they again at the starting point ?
26 minutes and 18 seconds | |
45 minutes | |
46 minutes and 12 seconds | |
42 minutes and 36 second |
Question 14 Explanation:
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
Question 15 |
The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leave no remainder, is:
1677 | |
2523 | |
1683 | |
3363 |
Question 15 Explanation:
L.C.M. of 5, 6, 7, 8 = 840.
Required 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.
Required 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.
Question 16 |
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
13 | |
33 | |
23 | |
43 |
Question 16 Explanation:
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 – 37) = 23.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 – 37) = 23.
Question 17 |
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
114 | |
294 | |
364 | |
364 |
Question 17 Explanation:
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
Question 18 |
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
123 | |
127 | |
235 | |
305 |
Question 18 Explanation:
Required number = H.C.F. of (1657 – 6) and (2037 – 5)
= H.C.F. of 1651 and 2032 = 127.
= H.C.F. of 1651 and 2032 = 127.
Question 19 |
Which of the following has the most number of divisors?
99 | |
101 | |
176 | |
182 |
Question 19 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.
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.
Question 20 |
The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
28 | |
32 | |
40 | |
64 |
Question 20 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.
Then, their L.C.M. = 6x. .
So, 6x = 48 or x = 8..
The numbers are 16 and 24..
Hence, required sum = (16 + 24) = 40.
Question 21 |
The H.C.F.
| 9 | , | 12 | , | 18 | and | 21 | is: |
| 10 | 25 | 35 | 40 |
3/5 | |
252/5 | |
3/1400 | |
63/700 |
Question 21 Explanation:
Required H.C.F. = H.C.F. of 9, 12, 18, 21 = 3
L.C.M. of 10, 25, 35, 40 1400
L.C.M. of 10, 25, 35, 40 1400
Question 22 |
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:
55
601
| |
601
55
| |
11
120
| |
120
11
|
Question 22 Explanation:
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum = 1/a+1/b = a+b/ab == 55/600 = 11/200
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum = 1/a+1/b = a+b/ab == 55/600 = 11/200
Question 23 |
The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:
15 cm | |
25 cm | |
35 cm | |
42 cm |
Question 23 Explanation:
Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.
Question 24 |
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:
75 | |
81 | |
85 | |
89 |
Question 24 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 = (551/29)=19
Third number = (1073/29)=37
Required sum = (19 + 29 + 37) = 85.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
First number = (551/29)=19
Third number = (1073/29)=37
Required sum = (19 + 29 + 37) = 85.
Question 25 |
Find the highest common factor of 36 and 84.
4 | |
6 | |
12 | |
18 |
Question 25 Explanation:
36 = 22 x 32
84 = 22 x 3 x 7
H.C.F. = 22 x 3 = 12.
84 = 22 x 3 x 7
H.C.F. = 22 x 3 = 12.
Question 26 |
Which of the following fraction is the largest ?
7/8 | |
13/16 | |
31/40 | |
63/80 |
Question 26 Explanation:
L.C.M. of 8, 16, 40 and 80 = 80.
7/8 = 70/80 ; 13/16 = 65/80 ; 31/40 = 62/80
Since,70/80>65/80>63/80>62/80>, so 7/8,>13/16>63/80>31/40
So, 7/8 is the largest.
Since,70/80>65/80>63/80>62/80>, so 7/8,>13/16>63/80>31/40
So, 7/8 is the largest.
Question 27 |
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
504 | |
536 | |
544 | |
548 |
Question 27 Explanation:
Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.
= 540 + 8
= 548.
Question 28 |
The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
279 | |
283 | |
308 | |
318 |
Question 28 Explanation:
Other number = (11 x 7700/275) = 308.
Question 29 |
What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?
196 | |
630 | |
1260 | |
2520 |
Question 29 Explanation:
L.C.M. of 12, 18, 21 30 2 | 12 – 18 – 21 – 30
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 – 9 – 21 – 15
—————————- Required number = (1260 ÷ 2) | 2 – 3 – 7 – 5
= 630.
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 – 9 – 21 – 15
—————————- Required number = (1260 ÷ 2) | 2 – 3 – 7 – 5
= 630.
Question 30 |
The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
12 | |
16 | |
24 | |
48 |
Question 30 Explanation:
Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
Question 31 |
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
1008 | |
1015 | |
1022 | |
1032 |
Question 31 Explanation:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
= 1008 + 7
= 1015
Question 32 |
252 can be expressed as a product of primes as:
2 x 2 x 3 x 3 x 7 | |
2 x 2 x 2 x 3 x 7 | |
3 x 3 x 3 x 3 x 7 | |
2 x 3 x 3 x 3 x 7 |
Question 32 Explanation:
Clearly, 252 = 2 x 2 x 3 x 3 x 7.
Question 33 |
Find the lowest common multiple of 24, 36 and 40.
120 | |
240 | |
360 | |
480 |
Question 33 Explanation:
2 | 24 – 36 – 40
——————–
2 | 12 – 18 – 20
——————–
2 | 6 – 9 – 10
——————-
3 | 3 – 9 – 5
——————-
| 1 – 3 – 5
L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
——————–
2 | 12 – 18 – 20
——————–
2 | 6 – 9 – 10
——————-
3 | 3 – 9 – 5
——————-
| 1 – 3 – 5
L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
Question 34 |
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
3 | |
13 | |
23 | |
33 |
Question 34 Explanation:
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 – 37) = 23.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 – 37) = 23.
Question 35 |
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:
1677 | |
1683 | |
2523 | |
3363 |
Question 35 Explanation:
L.C.M. of 5, 6, 7, 8 = 840.
Required 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.
Required 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.
Question 36 |
A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?
26 minutes and 18 seconds | |
42 minutes and 36 seconds | |
45 minutes | |
46 minutes and 12 seconds |
Question 36 Explanation:
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
Question 37 |
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
101 | |
107 | |
111 | |
185 |
Question 37 Explanation:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
Question 38 |
Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
40 | |
80 | |
120 | |
200 |
Question 38 Explanation:
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
Question 39 |
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
74 | |
94 | |
184 | |
364 |
Question 39 Explanation:
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
Question 40 |
The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
276 | |
299 | |
322 | |
345 |
Question 40 Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.
Larger number = (23 x 14) = 322.
Question 41 |
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
4 | |
10 | |
15 | |
16 |
Question 41 Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together 30/2 + 1 = 16 times.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together 30/2 + 1 = 16 times.
Question 42 |
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
4 | |
5 | |
6 | |
8 |
Question 42 Explanation:
N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
Question 43 |
The least number which when divided by 16, 18 and 21, leave the remainder 3, 5 and 8 respectively is:
982 | |
893 | |
1024 | |
995 |
Question 43 Explanation:
16———-3
18———-5 CD = 13
21———-8
LCM od 16,18,21 = 1008
1008-13 = 995
18———-5 CD = 13
21———-8
LCM od 16,18,21 = 1008
1008-13 = 995
Question 44 |
Four bells begin to toll together respectively at the intervals of 8, 10, 12 and 16 seconds. After how many seconds will they toll together again?
246 seconds | |
242 seconds | |
240 seconds | |
243 seconds |
Question 44 Explanation:
LCM = 240
Question 45 |
A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is?
31 | |
41 | |
51 | |
61 |
Question 45 Explanation:
LCM = 30 => 30 + 1 = 31
Question 46 |
Five bells first begin to toll together and then at intervals of 5, 10, 15, 20 and 25 seconds respectively. After what interval of time will they toll again together? s?
5 min | |
5.5 min | |
5.2 min | |
61 |
Question 46 Explanation:
LCM = 300/60 = 5 min
Question 47 |
HCF and LCM two numbers are 12 and 396 respectively. If one of the numbers is 36, then the other number is?
36 | |
66 | |
132 | |
264 |
Question 47 Explanation:
12 * 396 = 36 * x
x = 132
x = 132
Question 48 |
The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 10, then their difference is:
10 | |
46 | |
70 | |
90 |
Question 48 Explanation:
Let the numbers be x and (100 – x).
Then, x(100 – x) = 5 * 495
x2 – 100x + 2475 = 0
(x – 55)(x – 45) = 0
x = 55 or 45
The numbers are 45 and 55.
Required difference = 55 – 45 = 10.
Then, x(100 – x) = 5 * 495
x2 – 100x + 2475 = 0
(x – 55)(x – 45) = 0
x = 55 or 45
The numbers are 45 and 55.
Required difference = 55 – 45 = 10.
Question 49 |
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 reciprocal of the numbers is equal to:
55/601 | |
601/55 | |
11/120 | |
120/11 |
Question 49 Explanation:
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 * 120 = 600.
Required sum = 1/a + 1/b = (a + b)/ab = 55/600 = 11/120.
Then, a + b = 55 and ab = 5 * 120 = 600.
Required sum = 1/a + 1/b = (a + b)/ab = 55/600 = 11/120.
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