model 4 addition, subtraction, multiplication and division with lcm & hcf Practice Questions Answers Test with Solutions & More Shortcuts

Question : 21

The sum of two numbers is 36 and their H.C.F. is 4. How many pairs of such numbers are possible ?

a) 1

b) 3

c) 2

d) 4

Answer: (b)

HCF of two numbers = 4.

Hence, the numbers can be given by 4x and 4y

where x and y are co-prime.

Then, 4x + 4y = 36

4 (x + y) = 36

x + y = 9

Possible pairs satisfying this condition are : (1, 8), (4, 5), (2, 7)

Question : 22 [SSC CPO S.I.2010]

The LCM of two numbers is 44 times of their HCF. The sum of the LCM and HCF is 1125. If one number is 25, then the other number is

a) 1100

b) 900

c) 975

d) 800

Answer: (a)

Using Rule 1,

If the HCF = H,

then LCM = 44 H

44 H + H = 1125

→ 45 H = 1125

∴ H = $1125/45$ = 25

LCM = 44 × 25 = 1100

Now, First number × Second number = LCM × HCF

→ 25 × Second number = 1100 × 25

∴ Second number = ${1100 × 25}/25$ = 1100

Question : 23

A number x is divisible by 7. When this number is divided by 8, 12 and 16. It leaves a remainder 3 in each case. The least value of x is:

a) 148

b) 150

c) 149

d) 147

Answer: (d)

LCM of 8, 12 and 16 = 48

∴ Required number

= 48a + 3 which is divisible by 7.

x = 48a + 3 = (7 × 6a) + (6a + 3) which is divisible by 7.

i.e. 6a + 3 is divisible by 7.

When a = 3, 6a + 3 = 18 + 3 = 21

which is divisible by 7.

∴ x = 48 × 3 + 3 = 144 + 3 = 147

Question : 24 [SSC CGL Tier-II 2015]

Let x be the smallest number, which when added to 2000 makes the resulting number divisible by 12, 16, 18 and 21. The sum of the digits of x is

a) 7

b) 6

c) 5

d) 4

Answer: (a)

212, 16, 18, 21
26, 8, 9, 21
33, 4, 9, 21
1, 4, 3, 7

LCM = 2 × 2 × 3 × 4 × 3 × 7 = 1008

Multiple of 1008 = 2016

∴ Required number = 2016 – 2000 = 16 = x

∴ Sum of digits of x = 1 + 6 = 7

Question : 25 [SSC CGL 2012]

The LCM of two positive integers is twice the larger number. The difference of the smaller number and the GCD of the two numbers is 4. The smaller number is :

a) 12

b) 8

c) 6

d) 10

Answer: (b)

Let the numbers be x H and yH

where H is the HCF and yH > x H.

LCM = xy H

xyH = 2yH

→ x = 2

Again, x H – H = 4

→ 2H – H = 4 ⇒ H = 4

∴ Smaller number = x H = 8

IMPORTANT quantitative aptitude EXERCISES

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