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

Question : 1 [SSC CGL Prelim 2005]

The sum of the H.C.F. and L.C.M of two numbers is 680 and the L.C.M. is 84 times the H.C.F. If one of the number is 56, the other is :

a) 84

b) 8

c) 12

d) 96

Answer: (d)

Let HCF be h and LCM be l.

Then, l = 84h and l + h = 680

→ 84h + h = 680

→ h = $680/85$ = 8

∴ l = 680 – 8 = 672

∴ Other number = ${672× 8}/56$ = 96

Question : 2 [SSC CHSL 2015]

The smallest five digit number which is divisible by 12,18 and 21 is :

a) 10224

b) 10080

c) 30256

d) 50321

Answer: (b)

212, 18, 21
36, 9, 21
2, 3, 7

LCM of 12, 18 and 21

= 2 × 3 × 2 × 3 × 7 = 252

Of the options, 10080 ÷ 252 = 40

Question : 3 [SSC FCI 2013]

If A and B are the H.C.F. and L.C.M. respectively of two algebraic expressions x and y, and A + B = x + y, then the value of A3 + B3 is

a) x3 – y3

b) y3

c) x3

d) x3 + y3

Answer: (d)

Let no. are x and y and HCF = A, LCM = B

Using Rule, we have

xy = AB

⇒ x + y = A + B (given) ...(i)

$(x–y)^2$ = $(x+y)^2$ – 4xy

or, $(x–y)^2$ = $(A+B)^2$ – 4 AB

$(x–y)^2$ = $(A–B)^2$

(x–y) = A – B ...(ii)

Using (i) and (ii), we get

x = A and y = B

∴ $A^3 + B^3 = x^3 + y^3$

Question : 4

A number between 1000 and 2000 which when divided by 30, 36 and 80 gives a remainder 11 in each case is

a) 1451

b) 1712

c) 1641

d) 1523

Answer: (a)

We find LCM of 30, 36 and 80.

230, 36, 80
215, 18, 40
315, 9, 20
55, 3, 20
1, 3, 4

LCM = 2 × 2 × 3 × 3 × 4 × 5 = 720

∴ Required number = 2 × 720 + 11

= 1440 + 11 = 1451

Question : 5 [SSC CGL 2011]

The sum of a pair of positive integer is 336 and their H.C.F. is 21. The number of such possible pairs is

a) 2

b) 4

c) 3

d) 5

Answer: (b)

Let the numbers be 21x and 21y

where x and y are prime to each other.

21x + 21y = 336

21 (x + y) = 336

x + y = $336/21$ = 16

∴ Possible pairs = (1, 15), (5, 11), (7, 9), (3, 13)

IMPORTANT quantitative aptitude EXERCISES

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