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Question 1: Number of students who have opted for subjects A, B and C are 60, 84 and 108 respectively. The examination is to be conducted for these students such that only the students of the same subject are allowed in one room. Also, the number of students in each room must be the same. What is the minimum number of rooms that should be arranged to meet all these conditions?
(a)
28
(b) 
60
(c) 
12
(d) 
21

Correct Answer is Option (d). 

  • As we can see here that the total number of students are = 60+84+108 = 252
  • Now given condition is that in one room only the students of the same subject can be there and the number of rooms should be minimum that means the number of students in a particular room will be maximum.
  • This Maximum number of students will be HCF (Highest common factor) of 60, 84 and 108 and that will be 12.
  • Hence, number of rooms will be = 252/12 = 21 


Question 2: A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?
(a) 
30
(b) 
24
(c) 
20
(d) 
60

Correct Answer is Option (a).

  • A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. 
  • So red light fashes after every 1/3 min and green light flashes every 2/5 min. LCM of both the fractions is 2 min.
  • Hence, they flash together after every 2 min. So in an hour, they flash together 30 times. 


Question 3: A is the set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5, respectively. How many integers between 0 and 100 belong to set A?
(a) 0
(b) 1
(c) 2
(d) None of these

Correct Answer is Option (b).

  • Let the number ‘n’ belong to set A.
  • Hence, the remainder when n is divided by 2 is 1
  • The remainder when n is divided by 3 is 2
  • The remainder when n is divided by 4 is 3
  • The remainder when n is divided by 5 is 4 and
  • The remainder when n is divided by 6 is 5
  • So, when (n+1) is divisible by 2,3,4,5 and 6.
  • Hence, (n+1) is of form 60k for some natural number k and n is of the form 60k-1
  • Between numbers 0 and 100, only 59 is of the form above and hence the correct answer is 1 


Question 4: In Sivakasi, each boy’s quota of match sticks to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25, he can fill 3 more boxes with the total number of sticks assigned to him. Which of the following is the possible number of sticks assigned to each boy?
(a) 
200
(b) 
150
(c) 
125
(d) 
175

Correct Answer is Option (b).

  • Let the number of sticks assigned to each boy be N.
  • Let the number of boxes be M.
  • So, the number of sticks per box = N/M
  • Now, if he reduces the number of sticks in each box, the equation becomes N/(M+3) = N/M – 25
  • So, 25 = N/M – N/(M+3)
  • From the options, if N = 150, then, we get 25 = 150 [ 1/M – 1/(M+3) ]
     1/6 = 1/M – 1/(M+3) ⇒ M = 3
  • So, the number of sticks assigned to each boy = 150 


Question 5: A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is
(a) 
12 × 81
(b) 
16 × 192
(c) 
20 × 125
(d) 
24 × 216

Correct Answer is Option (a).

  • Any of the 4 colours can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe.
  • The third stripe can again be coloured in 3 ways (we can repeat the colour of the first stripe but not use the colours of the second stripe).
  • Similarly, there are 3 ways to colour each of the remaining stripes.
  • ∴ The number of ways the flag can be coloured is 4(3)5 = (12) (3)4 = 12 x 81

Question for Practice Questions: HCF & LCM
Try yourself:The traffic lights at three different road crossings change after every 48 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 8:20:00 hrs, when will they again change simultaneously?
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Question for Practice Questions: HCF & LCM
Try yourself:A wine seller had three types of wine. 403 liters of 1st kind, 434 liters of 2nd kind and 465 liters of 3rd kind. Find the least possible number of casks of equal size in which different types of wine can be filled without mixing.
View Solution

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FAQs on Practice Questions: HCF & LCM - CSAT Preparation - UPSC

1. What is the meaning of HCF and LCM?
Ans. HCF stands for Highest Common Factor and LCM stands for Lowest Common Multiple. HCF is the largest number that divides two or more numbers without leaving any remainder, while LCM is the smallest number that is divisible by two or more numbers.
2. How do you find the HCF of two numbers?
Ans. To find the HCF of two numbers, you can use the method of prime factorization. Prime factorize both numbers and find the common prime factors. Multiply these common prime factors to get the HCF of the two numbers.
3. Can the HCF of two numbers be greater than the numbers themselves?
Ans. No, the HCF of two numbers cannot be greater than the numbers themselves. The HCF is always a factor of the given numbers, so it will always be less than or equal to the numbers.
4. Is it possible for two numbers to have more than one LCM?
Ans. No, two numbers cannot have more than one LCM. The LCM is unique for any two numbers. It is the smallest number that is divisible by both numbers, so there can only be one LCM.
5. Can you find the LCM of three or more numbers using prime factorization?
Ans. Yes, you can find the LCM of three or more numbers using prime factorization. Prime factorize all the numbers and then multiply the highest powers of all the common prime factors to get the LCM.
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