Thursday, 19 February 2015

Mathematics for Competitive Exam

Problem on HCF.& LCM

1. Two numbers are in the ratio 2 : 3. If their L.C.M. is 48. what is sum of the numbers?
A. 28
B. 40
C. 64
D. 42

2. What is the greatest number of four digits which is divisible by 15, 25, 40 and 75 ?
A. 9800
B. 9600
C. 9400
D. 9200


3. Three numbers are in the ratio of 2 : 3 : 4 and their L.C.M. is 240. Their H.C.F. is:
A. 40
B. 30
C. 20
D. 10


4. What is the lowest common multiple of 12, 36 and 20?
A. 160
B. 220
C. 120
D. 180


5. What is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder?
A. 1108
B. 1683
C. 2007
D. 3363


6. The H.C.F. of two numbers is 5 and their L.C.M. is 150. If one of the numbers is 25, then the other is:
A. 30
B. 28
C. 24
D. 20


7. 504 can be expressed as a product of primes as
A. 2 × 2 × 3 × 3 × 7 × 7
B. 2 × 3 × 3 × 3 × 7 × 7
C. 2 × 3 × 3 × 3 × 3 × 7
D. 2 × 2 × 2 × 3 × 3 × 7



8. Which of the following integers has the most number of divisors?
A. 101
B. 99
C. 182
D. 176

9. The least number which should be added to 28523 so that the sum is exactly divisible by 3, 5, 7 and 8 is
A. 41
B. 42
C. 32
D. 37

10. What is the least number which when doubled will be exactly divisible by 12, 14, 18 and 22 ?
A. 1286
B. 1436
C. 1216



1. Answer : Option B
Explanation :
Let the numbers be 2x and 3x

LCM of 2x and 3x = 6x (∵ LCM of 2 and 3 is 6. Hence LCM

of 2x and 3x is 6x)


Given that LCM of 2x and 3x is 48

=> 6x = 48

=> x = 486 = 8

Sum of the numbers = 2x + 3x = 5x = 5 × 8 = 40

2. Answer : Option B
Explanation :
Greatest number of four digits = 9999

LCM of 15, 25, 40 and 75 = 600

9999 ÷ 600 = 16, remainder = 399

Hence, greatest number of four digits which is divisible by 15, 25, 40 and 75

= 9999 - 399 = 9600

3. Answer : Option C
Explanation :
Let the numbers be 2x, 3x and 4x

LCM of 2x, 3x and 4x = 12x

=> 12x = 240

=> x = 24012 = 20

H.C.F of 2x, 3x and 4x = x = 20

4. Answer : Option D
Explanation :
  2 
12, 36, 20
2
6, 18, 10
3
3, 9, 5
1, 3, 5
LCM = 2 × 2 × 3 × 1 × 3 × 5 = 180

4. Answer : Option B
Explanation :
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Solution 1
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LCM of 5, 6, 7 and 8 = 840

Hence the number can be written in the form (840k + 3) which is divisible by 9

If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9

If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9

Hence 1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3,

but when divided by 9 leaves no remainder

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Solution 2 - Hit and Trial Method
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Just see which of the given choices satisfy the given condtions

Take 3363. This is not even divisible by 9. Hence this is not the answer

Take 1108. This is not even divisible by 9. Hence this is not the answer

Take 2007. This is divisible by 9.

2007 ÷ 5 = 401, remainder = 2 . Hence this is not the answer

Take 1683. This is divisible by 9.

1683 ÷ 5 = 336, remainder = 3

1683 ÷ 6 = 280, remainder = 3

1683 ÷ 7 = 240, remainder = 3

1683 ÷ 8 = 210, remainder = 3

Hence 1683 is the answer

6. Answer : Option A
Explanation :
Product of two numbers = Product of their HCF and LCM.

Let one number = x25×x=5×150x=5×15025=30
7. Answer : Option D
Explanation :
It is clear that 504 = 2 × 2 × 2 × 3 × 3 × 7

8. Answer : Option D
Explanation :
99 = 1 × 3 × 3 × 11

=> Divisors of 99 are 1, 3, 11, 9, 33 and 99

101 = 1 × 101

=> Divisors of 101 are 1 and 101

182 = 1 × 2 × 7 × 13

=> Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182

176 = 1 × 2 × 2 × 2 × 2 × 11

=> Divisors of 176 are 1, 2, 11, 4, 22, 8, 44, 16, 88, 176

Hence 176 has most number of divisors

9. Answer : Option D
Explanation :
LCM of 3, 5, 7 and 8 = 840

28523 ÷ 840 = 33 remainder = 803

Hence the least number which should be added = 840 - 803 = 37

10. Answer : Option D
Explanation :
LCM of 12, 14, 18 and 22 = 2772

Hence the least number which will be exactly divisible by 12, 14, 18 and 22 = 2772

2772 ÷ 2 = 1386

=> 1386 is the number which when doubled, we get 2772

Hence, 1386 is the least number which when doubled will be exactly divisible by 12, 14, 18 and 22 ?


D. 1386


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