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Maths : LCM & HCF (Solved in Steps)

1. Two numbers are in the ratio 2 : 3. If their L.C.M. is 48. what is the sum of the numbers?

A. 28

B. 40

C. 64

D. 42

Answer : Option B

Explanation :

Let the numbers be 2x and 3x
LCM of 2x and 3x = 6x (? LCM of 2 and 3is 6. Hence LCM of 2x and 3x is 6x)
Given that LCM of 2x and 3x is 48
=> 6x = 48
=> x =48#6=8
Sum of the numbers = 2x + 3x = 5x = 5 × 8 = 40


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

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. 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

Answer : Option C

Explanation :

Let the numbers be 2x, 3x and 4x
LCM of 2x, 3x and 4x = 12x
=> 12x = 240
=> x = 240÷12=20
H.C.F of 2x, 3x and 4x = x = 20


4. What is the lowest common multiple of 12, 36 and 20?

A. 160

B. 220

C. 120

D. 180

Answer: Option D

Explanation :

LCM OF 12,36,20= 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. 336

Answer: Option B

Explanation :

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


6. The H.C.F. of two numbers is 5 and their L.C.M. is 150. If one of the numbers is25, then the other is:

A. 30

B. 28

C. 24

D. 20

Answer: Option A

Explanation :

The product of two numbers = Product of their HCF and LCM.
Let one number = x
25×x =5×150
x= (5×150)÷25
X=30


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

Answer: Option D

Explanation :

It is clear that 504 = 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

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. 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

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

A. 1286

B. 1436

C. 1216

D. 1386

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.


 

11. What is the greatest possible length which can be used to measure exactly
the lengths 8 m, 4 m 20 cm and 12 m 20 cm?

A. 10 cm

B. 30 cm

C. 25 cm

D. 20 cm

Answer: Option D

Explanation :

Required length = HCF of 800 cm, 420
cm, 1220 cm = 20 cm


12. The product of two 2 digit numbers is 2028 and their HCF is 13. What are the numbers?

A. 26, 78

B. 39, 52

C. 13, 156

D. 36, 68

Answer: Option B

Explanation :

Let the numbers be 13x and 13y (? HCF of the numbers = 13)
13x × 13y = 2028
=> xy = 12
co-primes with product 12 are (1, 12) and (3, 4) (? we need to take only co-primes with product 12. If we take two numbers with product 12, but not coprime, the HCF will not remain as 13)
Hence the numbers with HCF 13 and product 2028
= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
= (13, 156) and (39, 52)
Given that the numbers are 2 digit numbers
Hence numbers are 39 and 52


13. N is the greatest number which divides 1305, 4665 and 6905 and gives the same remainder in each case. What is the sum of the digits in N?

A. 4

B. 3

C. 6

D. 5

Answer: Option A

Explanation :

If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required greatest number
6905 – 1305 = 5600
6905 – 4665 = 2240
4665 – 1305 = 3360
Hence, the greatest number which divides
1305, 4665 and 6905 and gives the same remainder, N
= HCF of 5600, 2240, 3360
= 1120
Sum of digits in N
= Sum of digits in 1120
= 1 + 1 + 2 + 0
= 4


14. A boy divided the numbers 7654, 8506 and 9997 by a certain largest number and he gets the same remainder in each case. What is the common remainder?

A. 156

B. 199

C. 211

D. 231

Answer: Option B

Explanation:

If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required largest number
9997 – 7654 = 2343
9997 – 8506 = 1491
8506 – 7654 = 852
Hence, the greatest number which divides
7654, 8506 and 9997 and leaves the same remainder
= HCF of 2343, 1491, 852
= 213
Now we need to find out the common remainder.
Take any of the given numbers from 7654,
8506 and 9997, say 7654
7654 ÷ 213 = 35, remainder = 199


15. The ratio of two numbers is 4 : 5. If the HCF of these numbers is 6, what is their LCM?

A. 30

B. 60

C. 90

D. 120

Answer: Option D

Explanation :

Let the numbers be 4k and 5k
HCF of 4 and 5 = 1
Hence HCF of 4k and 5k = k
Given that HCF of 4k and 5k = 6
=> k = 6
Hence the numbers are (4 × 6) and (5 × 6) = 24 and 30
LCM of 24 and 30 = 120


16. 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?

A. 36 minutes 22 seconds

B. 46 minutes 22 seconds

C. 36 minutes 12 seconds

D. 46 minutes 12 seconds

Answer: Option D

Explanation :

LCM of 252, 308 and 198 = 2772
Hence they all will be again at the starting point after 2772 seconds or 46 minutes 12 seconds


17. What is the HCF of 2.04, 0.24 and 0.8?

A. 1

B. 2

C. 0.02

D. 0.04

Answer: Option D

Explanation :

Step 1 : Make the same number of decimal places in all the given numbers
by suffixing zero(s) in required numbers as needed.=> 2.04, 0.24 and 0.80

Step 2 : Now find the HCF of these numbers without decimal.
=>HCF of 204, 24 and 80 = 4

Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers. i.e., here, we need to put the decimal point in the result obtained in step 2 leaving two digits on its right.
=> HCF of 2.04, 0.24 and 0.8 = 0.04


18. If HCF of two numbers is 11 and the product of these numbers is 363, what is the greater number?

A. 9

B. 22

C. 33

D. 11

Answer: Option C

Explanation :

Let the numbers be 11a and 11b
11a × 11b = 363
=> ab = 3
co-primes with product 3 are (1, 3)
Hence the numbers with HCF 11 and product 363
= (11 × 1, 11 × 3)
= (11, 33)
Hence numbers are 11 and 33
The greater number = 33


19. What is the greatest number which on dividing 1223 and 2351 leaves remainders 90 and 85 respectively?

A. 1133

B. 127

C. 42

D. 1100

Answer: Option A

Explanation :

Required number
= HCF of (1223 – 90) and (2351 – 85)
= HCF of 1133 and 2266
= 1133


20. What is the least multiple of 7 which leaves a remainder of 4 when divided by
6, 9, 15 and 18?

A. 364

B. 350

C. 343

D. 371

Answer: Option A

Explanation :

LCM of 6, 9, 15 and 18 = 90
Required Number = (90k + 4) which is a multiple of 7
Put k = 1. We get number as (90 × 1) +
4 = 94. But this is not a multiple of 7
Put k = 2. We get number as (90 × 2) +
4 = 184. But this is not a multiple of 7
Put k = 3. We get number as (90 × 3) +
4 = 274. But this is not a multiple of 7
Put k = 4. We get number as (90 × 4) +
4 = 364. This is a multiple of 7
Hence 364 is the answer.


21. What is the least number which when divided by 8, 12, 15 and 20 leaves in
each case a remainder of 5?

A. 125

B. 117

C. 132

D. 112

Answer: Option A

Explanation :

LCM of 8, 12, 15 and 20 = 120
Required Number = 120 + 5 = 125


22. Three numbers which are co-prime to each other are such that the product of the first two is 119 and that of the last two is 391. What is the sum of the three numbers?

A. 47

B. 43

C. 53

D. 51

Answer: Option A

Explanation :

Since the numbers are co-prime, their HCF = 1
Product of first two numbers = 119
Product of last two numbers = 391
The middle number is common in both of these products.
Hence if we take HCF of 119 and 391, we get the common middle number
HCF of 119 and 391 = 17
=> Middle Number = 17
First Number = 119?17 = 7
Last Number = 391?17 = 23
Sum of the three numbers = 7+17+23 = 47


23. What is the greatest number which divides 24, 28 and 34 and leaves the same remainder in each case?

A. 1

B. 2

C. 3

D. 4

Answer: Option B

Explanation :

If the remainder is same in each case and remainder is not given, HCF of the
differences of the numbers is the required greatest number
34 – 24 = 10
34 – 28 = 6
28 – 24 = 4
Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder
= HCF of 10, 6, 4
= 2


24. Six bells start ringing together and ring at intervals of 4, 8, 10, 12, 15 and 20 seconds respectively. how many times will they ring together in 60 minutes?

A. 31

B. 15

C. 16

D. 30

Answer: Option A

Explanation :

LCM of 4, 8, 10, 12, 15 and 20 = 120
120 seconds = 2 minutes
Hence all the six bells will ring together in every 2 minutes
Hence, number of times they will ring together in 60 minutes
=1+(60÷2)=31


25. The HCF of two numbers is 23 and the other two factors of their LCM are 13 and 14. What is the largest number?

A. 312

B. 282

C. 299

D. 322

Answer: Option D

Explanation :

The HCF of a group of numbers will be always a factor of their LCM
HCF is the product of all common prime factors using the least power of each
common prime factor.
LCM is the product of highest powers of all prime factors
HCF of the two numbers = 23
=> Highest Common Factor in the numbers = 23
Since HCF will be always a factor of LCM,
23 is a factor of the LCM.
Other two factors in the LCM are 13 and 14.
Hence factors of the LCM are 23, 13, 14
So, numbers can be taken as (23 × 13) and (23 × 14) = 299 and 322
Hence, largest number = 322


26. What is the smallest number which when diminished by 12, is divisible 8, 12, 22 and 24?

A. 276

B. 264

C. 272

D. 268

Answer : Option A

Explanation :

Required Number = (LCM of 8, 12, 22 and 24) + 12 = 264 + 12 = 276