# Worksheet on H.C.F. and L.C.M. of Monomials | GCF and LCM of Monomials Worksheet

Students who are seeking help on finding the highest common factor and lowest common multiple can refer to Worksheet on H.C.F. and L.C.M. of Monomials. The questions in this article are based on the two or more monomials G.C.F and L.C.M. Practice all the questions provided in H.C.F and L.C.M of Monomials Worksheet and get a better idea of the concepts.

For the sake of your comfort & convenience, we have given the step by step solutions to all sample problems. Solve the questions on the greatest common factor and least common multiple of monomials on your own, verify with the solutions provided, and test your knowledge and preparation level.

1. Find the greatest common factor (G.C.F) and lowest common multiple (L.C.M) of two monomials:

(a) 15x²y, 3x³y

(b) 45pq³r⁴, 18p³q²r

(c) 9u²v, 90uv³

Solution:

(a) 15x²y, 3x³y

Numerical coefficients = 15, 3

Since, 15=3×5, 3=3×1.

L.C.M of 15, 3 = 3×5 = 15

G.C.F of 15, 3 = 3

Literal coefficients = x²y, x³y

Since in, x²y, x³y

The highest power and lowest powers of x is x³ & x².

The highest power and lowest powers of y is y & y.

Thus, L.C.M of x²y, x³y = x³y

H.C.F of x²y, x³y = x²y

Therefore, L.C.M of 15x²y, 3x³y = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 15 x x³y = 15x³y.

G.C.F of 15x²y, 3x³y = H.C.F of numerical coefficients x H.C.F. of literal coefficients

= 3 x x²y = 3x²y.

(b) 45pq³r⁴, 18p³q²r

Numerical coefficients = 45, 18

Since, 45=9*5, 18=9*2

The least common multiple of 45, 18 = 9*5*2 = 90

The G.C.F of 45, 18 = 9

Literal coefficients = pq³r⁴, p³q²r

Since in, pq³r⁴, p³q²r

The highest power and lowest powers of p is p³ & p.

The highest power and lowest powers of q is q³ & q².

The highest power and lowest powers of r is r⁴ & r.

Thus, L.C.M of pq³r⁴, p³q²r = p³q³r⁴

G.C.F of pq³r⁴, p³q²r = pq²r

Therefore, L.C.M of 45pq³r⁴, 18p³q²r = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 90 x p³q³r⁴ = 90p³q³r⁴

The H.C.F of 45pq³r⁴, 18p³q²r = H.C.F of numerical coefficients x H.C.F. of literal coefficients

= 9 x pq²r = 9pq²r.

(c) 9u²v, 90uv³

Numerical coefficients = 9, 90

Since, 9=9*1, 90=9*10

The L.C.M of 9, 90 = 9*10 = 90

The H.C.F of 9, 90 = 9

Literal coefficients = u²v, uv³

Since in, u²v, uv³

The highest power and lowest powers of u is u² & u.

The highest power and lowest powers of v is v³ & v.

Thus, L.C.M of u²v, uv³ = u²v³

H.C.F of u²v, uv³ = uv

Therefore, the lowest common multiple of 9u²v, 90uv³ = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 90 x u²v³ = 90u²v³

H.C.F of 9u²v, 90uv³ = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 9 x uv = 9uv.

2. Find the H.C.F and L.C.M of three monomials:

(a) 14c²d³e, 38fe²g, 28b²c³d

(b) 32x⁵yz², 40x⁴yz², 24x²yz³

(c) 3pq³, 2p²r and 5p³q²r²

Solution:

(a) 14c²d³e, 38fe²g, 28b²c³d

Numerical coefficients = 14, 38, 28

Since, 14=2×7, 38=2×19, 28=2x2x7

Thus, greatest common factor of 14, 38, 28 = 2

The least common multiple of 14, 38, 28 = 2x7x19x2 = 532

Literal coefficients = c²d³e, fe²g, b²c³d

since in, c²d³e, fe²g, b²c³d

The highest power and lowest powers of b is b² & b².

The highest power and lowest powers of c is c³ & c².

The highest power and lowest powers of d is d³ & d.

The highest power and lowest powers of e is e² & e.

The highest power and lowest powers of f is f & f.

The highest power and lowest powers of g is g & g.

Thus, L.C.M of c²d³e, fe²g, b²c³d = b²c³d³e²fg

H.C.F of c²d³e, fe²g, b²c³d = c²

Therefore, Lowest common multiple of 14c²d³e, 38fe²g, 28b²c³d = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 532 x b²c³d³e²fg = 532b²c³d³e²fg

H.C.F of 14c²d³e, 38fe²g, 28b²c³d = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 2 x c² = 2c².

(b) 32x⁵yz², 40x⁴yz², 24x²yz³

Numerical coefficients = 32, 40, and 24

Since, 32=2×8, 40=8×5, 24=3×8

The G.C.F of 32, 40, and 24 = 8

The L.C.M of 32, 40, and 24 = 8x3x2x5 = 240

Literal coefficients = x⁵yz², x⁴yz², x²yz³

The highest power and lowest powers of x is x⁵ & x².

The highest power and lowest powers of y is y & y.

The highest power and lowest powers of z is z³ & z².

Thus, L.C.M of x⁵yz², x⁴yz², x²yz³ = x⁵yz³

H.C.F of x⁵yz², x⁴yz², x²yz³ = x²yz²

L.C.M of 32x⁵yz², 40x⁴yz², 24x²yz³ = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 240 x x⁵yz³ = 240x⁵yz³

H.C.f of 32x⁵yz², 40x⁴yz², 24x²yz³ = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 8 x x²yz² = 8x²yz².

(c) 3pq³, 2p²r and 5p³q²r²

Numerical coefficients = 3, 2, 5

H.C.F of 3, 2, 5 = 1

Least common multiple of 3, 2, 5 = 30

Literal coefficients = pq³, p²r and p³q²r²

Since in, pq³, p²r and p³q²r²

The highest power and lowest powers of p is p³ & p.

The highest power and lowest powers of q is q³ & q².

The highest power and lowest powers of r is r² & r.

Thus, L.C.M of pq³, p²r and p³q²r² = p³q³r²

H.C.F of pq³, p²r and p³q²r² = pr

Therfore, L.C.M of 3pq³, 2p²r and 5p³q²r² = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 30 x p³q³r² = 30p³q³r²

H.C.f of 3pq³, 2p²r and 5p³q²r² = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 1 x pq²r = pr.

3. Find the least common multiple, highest common factor of each:

(a) 30x³y², 24x²y, 6x²y³

(b) 75a², 24a, 30a⁴b

(c) 4mn, 10n

Solution:

(a) 30x³y², 24x²y, 6x²y³

Numerical coefficients = 30, 24, 6

Since 30=2x5x3, 24=2x2x2x3, 6=2×3

H.C.F of 30, 24, 6 = 2×3 = 6

Least common multiple of 30, 24, 6 = 2x3x5x2x2 = 120

Literal coefficients = x³y², x²y, x²y³

Since in, x³y², x²y, x²y³

The highest power and lowest powers of x is x³ & x².

The highest power and lowest powers of y is y³ & y.

Thus, L.C.M of x³y², x²y, x²y³ = x³y³

H.C.F of x³y², x²y, x²y³ = x²y

Therfore, L.C.M of 30x³y², 24x²y, 6x²y³ = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 120 x x³y³ = 120x³y³

H.C.F of 30x³y², 24x²y, 6x²y³ = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 6 x x²y = 6x²y.

(b) 75a², 24a, 30a⁴b

Numerical coefficients = 75, 24, 30

Since, 75=5x5x3, 24=4x3x2, 30=5x3x2

The highest common factor of 75, 24, 30 = 3

The lowest common multiple of 75, 24, 30 = 5x5x4x2x5x2 = 360

Literal coefficients = a², a, a⁴b

Since in, a², a, a⁴b

The highest power and lowest powers of a is a⁴ & a.

The highest power and lowest powers of b is b & b.

Thus, H.C.F of a², a, a⁴b = a.

L.C.M of a², a, a⁴b = a⁴b

Therfore, L.C.M of 75a², 24a, 30a⁴b = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 360 x a⁴b = 360a⁴b

H.C.F of 75a², 24a, 30a⁴b = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 3 x a = 3a.

(c) 4mn, 10n

Numerical coefficients = 4, 10

Since, 4=2×2, =5×2

The greatest common factor of 4, 10 is 2.

The least common multiple of 4, 10 = 2x5x2 = 20

Literal coefficients = mn, n

Since in, mn, n

The highest power and lowest powers of m is m & m.

The highest power and lowest powers of n is n & n.

Thus, H.C.F of mn, n = n.

L.C.M of mn, n = mn

Therefore, H.CF of 4mn, 10n = H.C.F of numerical coefficients x H.C.F of literal coefficients

= 2 x n = 2n.

L.C.M of 4mn, 10n = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 20 x mn = 20mn.

4. Find L.C.M and G.C.F of the following monomials:

(a) 7x⁴y³, 28xy², 7x²y⁴, 28x³y²

(b) 14p²q³, 22pq², 9p⁴q, 5p³qr²

(c) 12a⁴b³c, 24a²b⁵c, 16a⁴bc², 14ab³c, 50a²b²c²

Solution:

(a) 7x⁴y³, 28xy², 7x²y⁴, 28x³y²

Numerical Coefficients = 7, 28, 7, and 28

Since, 7=1×7, 28=7×4, 7=1×7, and 28=7×4

The H.C.F of 7, 18, 7, and 28 = 7

The L.C.M of 7, 18, 7, and 28 = 7×4 = 28

Literal Coefficients = x⁴y³, xy², x²y⁴, x³y²

Since in, x⁴y³, xy², x²y⁴, x³y²

The lowest and highest power of x is x & x⁴.

The lowest and highest power of y is y² & y⁴.

Thus, least common multiple of x⁴y³, xy², x²y⁴, x³y² = x⁴y⁴

Highest common factor of x⁴y³, xy², x²y⁴, x³y² = xy²

Therefore, G.C.F of 7x⁴y³, 28xy², 7x²y⁴, 28x³y² = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 7 x xy² = 7xy²

L.C.M of 7x⁴y³, 28xy², 7x²y⁴, 28x³y² = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 28 x x⁴y⁴ = 28x⁴y⁴.

(b) 14p²q³, 22pq², 9p⁴q, 5p³qr²

Numerical Coefficients = 14, 22, 9, 5

Since, 14=2×7, 22=2×11, 9=3×3, 5=1×5

Thus, L.C.M of 14, 22, 9, 5 = 2x7x11x9x5 = 6930

H.C.F of G.C.F of 14, 22, 9, 5 = 1

Literal Coefficients = p²q³, pq², p⁴q, p³qr²

Since in, p²q³, pq², p⁴q, p³qr²

The lowest and highest power of p is p & p⁴.

The lowest and highest power of q is q & q³.

The lowest and highest power of r is r² & r².

The L.C.M of p²q³, pq², p⁴q, p³qr² = p⁴q³r²

H.C.F of p²q³, pq², p⁴q, p³qr²=pq

Therfore, G.C.F of 14p²q³, 22pq², 9p⁴q, 5p³qr² = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 1 x pq = pq

L.C.M of 14p²q³, 22pq², 9p⁴q, 5p³qr² = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 6930 x p⁴q³r² = 6930p⁴q³r².

(c) 12a⁴b³c, 24a²b⁵c, 16a⁴bc², 14ab³c, 50a²b²c²

Numerical Coefficients = 12, 25, 16, 14, 50

Since, 12=2x2x3, 24=2x2x2x3, 16=2x2x2x2, 14=2×7, 50=5x5x2

L.C.M of 12, 24, 16, 14, 50 = 4x6x14x25 = 8400

H.C.F of 12, 24, 16, 14, 50 = 2

Literal Coefficients = a⁴b³c, a²b⁵c, a⁴bc², ab³c, a²b²c²

Since in, a⁴b³c, a²b⁵c, a⁴bc², ab³c, a²b²c²

The lowest and highest power of a is a & a⁴.

The lowest and highest power of b is b & b⁵.

The lowest and highest power of c is c & c².

H.C.F of a⁴b³c, a²b⁵c, a⁴bc², ab³c, a²b²c² = abc

L.C.M of a⁴b³c, a²b⁵c, a⁴bc², ab³c, a²b²c² = a⁴b⁵c²

Therfore, H.C.F of 12a⁴b³c, 24a²b⁵c, 16a⁴bc², 14ab³c, 50a²b²c² = G.C.F of numerical coefficients x G.C.F of literal coefficients

= 2 x abc = 2abc

L.C.M of 12a⁴b³c, 24a²b⁵c, 16a⁴bc², 14ab³c, 50a²b²c² = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 8400 x a⁴b⁵c² = 8400a⁴b⁵c².