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