Mounika Kandukuri

(a) 12a³bc, 22ab³

The L.C.M. of numerical coefficients = The L.C.M. of 12 and 22.

Since, 12=2x2x3, 22=2×11

Therefore, the L.C.M. of 12 and 22 is 2x2x3x11 = 132

The L.C.M. of literal coefficients = The L.C.M. of a³bc, ab³ = a³b³c

Since, in a³bc, ab³,

The highest power of a is a³.

The highest power of b is b³.

The highest power of c is c.

Thus, the L.C.M. of 12a³bc, 22ab³

= The L.C.M. of numerical coefficients × The L.C.M. of literal coefficients

= 132 × (a³b³c) = 132a³b³c

(b) 11u²v, 121u³v

The L.C.M of numerical coefficients = The lowest common multiple of 11, 121

Since, 11=1×11, 121=11×11

The lowest common multiple of 11, 121 = 11²x1 = 121

The L.C.M. of literal coefficients = The L.C.M. of u²v, u³v = u³v

Since, in u²v, u³v

The highest power of u is u³.

The highest power of v is v.

Thus, L.C.M of 11u²v, 121u³v = L.C.M. of numerical coefficients x L.C.M. of literal coefficients

= 121 x u³v = 121u³v.

(c) 81p⁵qr²s³t, 12pq⁴s⁵

The Least Common Multiple of numerical coefficients = L.C.M of 81, 12

Since, 81=3x3x3x3, 12=3×4

Therefore, the L.C.M. of 81, 12 = 3⁴ x 4 = 3x3x3x3x4 = 324

The lowest common multiple of literal coefficients = L.C.M of p⁵qr²s³t, pq⁴s⁵ = p⁵q⁴r²s⁵t

Since, in p⁵qr²s³t, pq⁴s⁵

The highest power of p is p⁵.

The highest power of q is q⁴.

The highest power of r is r².

The highest power of s is s⁵.

The highest power of t is t.

Thus, L.C.M of 81p⁵qr²s³t, 12pq⁴s⁵ = L.C.M. of numerical coefficients × L.C.M. of literal coefficients

= 324 x p⁵q⁴r²s⁵t = 324p⁵q⁴r²s⁵t.

(d) 117mn, 65m³no

The least common multiple of numerical coefficients = l.c.m of 117, 65

Since, 117=13x3x3, 65=13×5

The l.C.M of 117, 65 = 13×3²x5 = 13 x 3 x 3x 5 = 585

The Lowest common multiple of literal coefficients = L.C.M of mn, m³no = m³no

Since, in mn, m³no

The highest power of m is m³.

The highest power of n is n.

The highest power of o is o.

Thus, the L.C.M of 117mn, 65m³no = L.C.M. of numerical coefficients × L.C.M. of literal coefficients

= 585 x m³no = 585m³no.

(e) x²y⁴z, x³y²z

The L.C.M of literal coefficients = Lowest common multiple of x²y⁴z, x³y²z = x³y⁴z

Since, in x²y⁴z, x³y²z

The highest power of x is x³.

The highest power of y is y⁴.

The highest power of z is z.

The least common multiple of numerical coefficients = 1.

Therefore, L.C.M of x²y⁴z, x³y²z = L.C.M. of numerical coefficients × L.C.M. of literal coefficients

= 1 x x³y⁴z = x³y⁴z.

(a) 14xy²z, 10xy, 2xyz³

The L.C.M of numerical coefficients = L.C.M of 14, 10, 2

Since, 14=2×7, 10=2×5, 2=2×1

Thus, Least common multiple of 14, 10, 2 = 2x7x5x1 = 60

The L.C.M of literal coefficients = L.C.M of xy²z, xy, xyz³ = xy²z³

Since, in xy²z, xy, xyz³

The highest power of x is x.

The highest power of y is y².

The highest power of z is z³.

Therefore, the Lowest common multiple of 14xy²z, 10xy, 2xyz³ = L.C.M. of numerical coefficients × L.C.M. of literal coefficients

= 60 x xy²z³ = 60xy²z³.

(b) 65a²b⁵c, 10ab, 5ab²

The L.C.M of numerical coefficients = L.C.M of 65, 10, 5

Since, 65=5×13, 10=2×5, 5=1×5.

Thus, lowest common multiple of 65, 10, 5 = 5x13x2 = 130

The least common multiple of literal coefficients =L.C.M of a²b⁵c, ab, ab² = a²b⁵c

The highest power of a is a².

The highest power of b is b⁵.

The highest power of c is c.

Therefore, L.C.M of 65a²b⁵c, 10ab, 5ab² = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 130 x a²b⁵c = 130a²b⁵c.

(c) 18pqr, 30pqr, 5pqr

The L.C.M of numerical coefficients = L.C.M of 18, 30, 5

Since, 18=2x3x3, 30=3x2x5, 5=1×5

L.C.M of 18, 30, 5 = 2x3x3x5 = 90

The L.C.M of literal coefficients = L.C.M of pqr, pqr, pqr = pqr.

Since in, pqr, pqr, pqr

The highest power of p is p.

The highest power of q is q.

The highest power of r is r.

Therefore, L.C.M of 18pqr, 30pqr, 5pqr = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 90 x pqr = 90pqr

(d) mn²o, m²n, op³

The L.C.M of literal coefficients = Lowest common multiple of mn²o, m²n, op³ = m²n²op³

Since in, mn²o, m²n, op³

The highest power of m is m².

The highest power of n is n².

The highest power of o is o.

The highest power of p is p³.

The L.C.M of numerical coefficients = 1.

Therefore, L.C.M of mn²o, m²n, op³ = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 1 x m²n²op³ = m²n²op³.

(e) 9u²v, 5uv³, 4u³v

The L.C.M of numerical coefficients = least common multiple of 9, 5, 4

Since, 9=3×3, 5=1×5, 4=2×2

Thus, L.C.M of 9, 5, 4 = 3²x5x2² = 180

The L.C.M of literal coefficients =L.C.M of u²v, uv³, u³v = u³v³

Since in, u²v, uv³, u³v,

The highest power of u is u³.

The highest power of v is v³.

Therefore, L.C.M of 9u²v, 5uv³, 4u³v = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 180 x u³v³ = 180u³v³.

(a) 32y⁴, 40xy², 20x²y²

The L.C.M of numerical coefficients = L.C.M of 32, 40, 20

Since, 32=2x2x2x4, 40=2x4x5, 20=4×5

Thus, L.C.M of 32, 40, 20 = 2x8x2x5x2 = 320

The L.C.M of literal coefficients = least common multiple of y⁴, xy², x²y² = x²y⁴

Since in, y⁴, xy², x²y²

The highest power of y is y⁴.

The highest power of x is x².

Therefore, L.C.M of 32y⁴, 40xy², 20x²y²= L.C.M of numerical coefficients x L.C.M of literal coefficients

= 320 x x²y⁴ = 320x²y⁴.

(b) 27u², 18u

The L.C.M of numerical coefficients = L.C.M of 27, 18

Since, 27=3x3x3, 18=3x3x2

Thus, L.C.M of 27, 18 = 3x3x3x2 = 56

The L.C.M of literal coefficients = lowest common multiple of u², u = u²

Since in, u², u

The highest power of u is u²

Therefore, L.C.M of 27u², 18u= L.C.M of numerical coefficients x L.C.M of literal coefficients

= 56 x u² = 56u²

(c) 3x⁴y⁴z³, 5x²y³z⁴

The L.C.M of numerical coefficients = L.C.M of 3, 5

Since, 3=1×3, 5=1×5

Thus, L.C.M of 3,5 = 3×5 = 15

The least common multiple of literal coefficients = L.C.M of x⁴y⁴z³, x²y³z⁴ = x⁴y⁴z⁴

Since in, x⁴y⁴z³, x²y³z⁴

The highest power of x is x⁴

The highest power of y is y⁴

The highest power of z is z⁴

Therefore, L.C.M of 3x⁴y⁴z³, 5x²y³z⁴= L.C.M of numerical coefficients x L.C.M of literal coefficients

= 15 x x⁴y⁴z⁴ = 15x⁴y⁴z⁴.

(d) 2pq⁵r³s, 21p³qrs, 4pq²rs², 9p²s

The L.C.M of numerical coefficients = L.C.M of 2, 21, 4, 9

Since, 2=2×1, 21=3×7, 4=2×2, 9=3×3

Thus, L.C.M of 2, 21, 4, 9 = 2x3x7x2x3 = 252

The L.C.M of literal coefficients = least common multiple of pq⁵r³s, p³qrs, pq²rs², p²s = p³q⁵r³s²

Since in, pq⁵r³s, p³qrs, pq²rs², p²s

The highest power of p is p³

The highest power of q is q⁵

The highest power of r is r³

The highest power of s is s²

Therefore, L.C.M of 2pq⁵r³s, 21p³qrs, 4pq²rs², 9p²s= L.C.M of numerical coefficients x L.C.M of literal coefficients

= 252 x p³q⁵r³s² = 252p³q⁵r³s².

(e) 36mn², 27m³no², 8mo

The L.C.M of numerical coefficients = L.C.M of 36, 27, 8

Since, 27=3x3x3, 36=2x2x3x3, 8=2x2x2

Thus, L.C.M of 36, 27, 8 = 3x3x3x2x2x2 = 192

The L.C.M of literal coefficients = L.C.M of mn², m³no², mo = m³n²o²

Since in, mn², m³no², mo

The highest power of m is m³.

The highest power of n is n²

The highest power of o is o².

Therefore, L.C.M of 36mn², 27m³no², 8mo= L.C.M of numerical coefficients x L.C.M of literal coefficients

= 192 x m³n²o² = 192m³n²o²

(a) 3a²bc²de, 5a³bc⁴de, 7a²bcde², 13ab³d²

The L.C.M of numerical coefficients = L.C.M of 3, 5, 7, 13

Since, 3=1×3, 5=1×5, 7=1×7, 1=1×13

Thus, Least common multiple of 3, 5, 7, 13 = 13x5x3x7 = 1365

The lowest common multiple of literal coefficients = L.C.M of a²bc²de, a³bc⁴de, a²bcde², ab³d² = a³b³c⁴d²e²

Since, in a²bc²de, a³bc⁴de, a²bcde², ab³d²

The highest power of a is a³

The highest power of b is b³

The highest power of c is c⁴

The highest power of d is d²

The highest power of e is e².

Therefore, least common multiple of 3a²bc²de, 5a³bc⁴de, 7a²bcde², 13ab³d² = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 1365 x a³b³c⁴d²e² = 1365a³b³c⁴d²e².

(b) 12p²q²r, 14qs³t², 16u²tv

The L.C.M of numerical coefficients = L.C.M of 12, 14, 16

Since, 12=2x3x2, 14=2×7, 16=2x2x2x2

Thus, least common multiple of 12, 14, 16 = 2x2x2x2x3x7x4 = 336

The L.C.M of literal coefficients = L.C.M of p²q²r, qs³t², u²tv = p²q²rs³t²v

Since, in p²q²r, qs³t², u²tv

The highest power of p is p²

The highest power of q is q²

The highest power of r is r

The highest power of s is s³

The highest power of t is t²

The highest power of v is v

Therefore, L.C.M of 12p²q²r, 14qs³t², 16u²tv = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 336 x p²q²rs³t²v = 336p²q²rs³t²v

(c) 5x³y⁴, 10xz, 15xy⁴z², 20x²y², 25x³

The L.C.M of numerical coefficients = L.C.M of 5, 10, 15, 20, 25

Since, 5=1×5, 10=2×5, 15=3×5, 20=4×5, 25=5×5

Thus, L.C.M of 5, 10, 15, 20, 25 = 5x2x3x4x5 = 600

The L.C.M of literal coefficients = L.C.M of x³y⁴, xz, xy⁴z², x²y², x³ = x³y⁴z²

Since in x³y⁴, xz, xy⁴z², x²y², x³

The highest power of x is x³

The highest power of y is y⁴

The highest power of z is z²

Therefore, L.C.M of 5x³y⁴, 10xz, 15xy⁴z², 20x²y², 25x³ = L.C.M of numerical coefficients x L.C.M of literal coefficients

= 600 x x³y⁴z² = 600x³y⁴z².

(a) 56a²bc, 24ab²c²

The H.C.F. of numerical coefficients = The H.C.F. of 24 and 56.

Since, 24 = 2 * 2 * 2 * 3, 56 = 2 * 2 * 2 * 7

Therefore, the H.C.F. of 24 and 56 = 2³ = 8

The H.C.F. of literal coefficients = The H.C.F. of a²bc, ab²c² = abc

Since, in a²bc and ab²c², a, b and c are common.

The lowest power of a is a.

The lowest power of b is b.

The lowest power of c is c.

Therefore, the H.C.F. of a²bc, ab²c² is abc.

Thus, the H.C.F. of 56a²bc, 24ab²c²

= The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 8 × (abc)

= 8abc.

(b) 45u²v, 60uv³

The G.C.F of numerical coefficients = The G.C.F of 45, 60.

Since, 45 = 5 * 3 * 3, 60 = 3 * 2 * 5 * 2

Therefore, H.C.F of 45, 60 is 5 * 3 = 15

The H.C.F. of literal coefficients = The H.C.F. of u²v and uv³ = u²v

Since, in u²v, uv³ u, v are common

The lowest power of u is u²

The lowest power of v is v.

Thus, the G.C.F of 45u²v, 60uv³ = The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 15 x u²v

= 15u²v

(c) 99x⁵y²z³, 66xy²z²

The H.C.F of numerical coefficients = The H.C.F. of 99 and 66.

Since, 99 = 3 * 3 * 11, 66 = 3 * 2 * 11

The G.C.F. of 99 and 66 = 3 * 11 = 33

The H.C.F. of literal coefficients = The H.C.F. of x⁵y²z³, xy²z² = xy²z²

The common variables are x, y, z

The lowest power of x is x.

The lowest power of y is y².

The lowest power of z is z³.

Thus, the G.C.F of 99x⁵y²z³, 66xy²z² = 33 * xy²z² = 33xy²z²

(d) 29d⁸, 18c

The H.C.F of numerical coefficients = The H.C.F. of 29, 18

Since, 29 = 3 * 3 * 3, 18 = 2 * 3 * 3

The H.C.F. of 29, 18 = 9

The H.C.F. of d⁸, c = 1.

There are no common variables.

Thus, the G.C.F of 29d⁸, 18c = The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients = 9 x 1

Therefore, G.C.F of 29d⁸, 18c = 9.

(e) 25a², 15ab

The highest common factor of numerical coefficients = The H.C.F. of 25, 15

Since, 25 = 5 * 5, 15 = 3 * 5

The H.C.F. of 25, 15 = 5

The H.C.F. of literal coefficients = The G.C.F of a², b = a.

The only common variable is a.

The lowest power of a is a.

Thus, the highest common factor of 25a², 15ab = The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 5 * a = 5a.

(f) 10s, 2s

The greatest common factor of numerical coefficients = The G.C.F of 10, 2

Since, 10 = 2 * 5, 2 = 2 * 1

The G.C.F of 10, 2 = 2

The H.C.F. of literal coefficients = The highest common factor of s, s is s.

The common variable is s.

The lowest power of s is s.

Thus, the G.C.F of 10s, 2s = The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 2 * s = 2s.

(g) 36p³q⁵r, 6p²q³r⁵

The highest common factor of numerical coefficients = The H.C.F of 36, 6

Since, 36 = 6 * 6, 6 = 1 * 6

The H.C.F of 36, 6 = 6

The H.C.F. of literal coefficients = The greatest common factor of p³q⁵r, p²q³r⁵ = p²q³r

The common variables are p, q, r

The lowest power of p is p².

The lowest power of q is q³.

The lowest power of r is r.

Thus, the H.C.F of 36p³q⁵r, 6p²q³r⁵ = The G.C.F. of numerical coefficients × The G.C.F. of literal coefficients

= 6 * p²q³r = 6p²q³r

(h) 84t²s, 28tu

The G.C.F. of numerical coefficients = The greatest common factor of 84, 28

Since, 84 = 2 * 2 * 3 * 7, 28 = 2 * 2 * 7

The H.C.F of 84, 28 = 14

The H.C.F. of literal coefficients = The highest common factor of t²s, tu = t

The common variable is t.

The lowest power of t is t.

Thus, the G.C.F of 84t²s, 28tu = The H.C.F of numerical coefficients x The G.C.F. of literal coefficients

= 14 * t = 14t

(a) 35p²q³r, 42pq²r³ and 14p³qr²

The H.C.F of numerical coefficients = The H.C.F of 35, 42, 14

Since, 35 = 7 * 5, 42 = 2 * 3 * 7, 14 = 2 * 7

Therefore, the H.C.F. of 35, 42, 14 is 7

The highest common factor of literal coefficients = The G.C.F of p²q³r, pq²r³ and p³qr² = pqr

The common variables are p, q, r

The lowest power of p is p, the lowest power of q is q, the lowest power of r is r.

Thus, the H.C.F of 35p²q³r, 42pq²r³ and 14p³qr² = H.C.F of numerical coefficient x H.C.F of literal coefficients

= 7 x pqr = 7pqr.

(b) 49xy⁴z², 21xy², 7x²z⁴

The H.C.F of numerical coefficients = The H.C.F of 49, 21, 7

Since, 49 = 7 * 7, 21=7*3, 7=1*7

Therefore, the G.C.F of 49, 21, 7 is 7.

The greatest common factor of literal coefficients = The H.C.F of xy⁴z², xy², x²z⁴ = x

Thus, G.C.F of 49xy⁴z², 21xy², 7x²z⁴ = The H.C.F of numerical coefficients * literal coefficients

= 7 * x =7x.

(c) mno², mn, m²n

The H.C.F of literal coefficients = The greatest common factor of mno², mn, m²n = mn

The common variables are m, n.

The lowest power of m is m, n is n.

Thus, H.C.F of mno², mn, m²n = The H.C.F of numerical coefficients x The G.C.F. of literal coefficients

= 1xmn = mn.

(d) 5abc, 25a²b²c², 50ac³

The G.C.F of numerical coefficients = The H.C.F of 5, 25, 50

since , 5=1*5, 25=5*5, 50=5*5*2

Therefore, the H.C.F. of 5, 25, 50 = 5

The G.C.F of literal coefficients = The H.C.F of abc, a²b²c², ac³ = ac

The common variables are a, c.

Thus, the G.C.F of 5abc, 25a²b²c², 50ac³ = H.C.F of numerical coefficients x H.C.F of literal coefficients

= 5 x ac = 5ac.

(e) m³n³a³, man, a²m²n²

The greatest common factor of literal coefficients = The H.C.F of m³n³a³, man, a²m²n² = amn

The common variables are a, m, n.

The H.C.F of m³n³a³, man, a²m²n² = H.C.F of numerical coefficients x H.C.F of literal coefficients

= 1 x amn = amn.

(f) 6xy, 14x²y, 42xy²

The highest common factor of numerical coefficients = The H.C.F of 6, 14, 42

Since, 6=2*3, 14=7*2, 42=2*7*3

Thus, H.C.F of 6, 14, 42 = 2

The H.C.F of literal coefficients = G.C.F of xy, x²y, xy² = xy

Therefore, G.C.F of 6xy, 14x²y, 42xy² = H.C.F of numerical coefficients x H.C.F of literal coefficients

= 2 x xy = 2xy.

(g) x²y³, x⁵, yz

The G.C.F of literal coefficients = H.C.F of x²y³, x⁵, yz = 1.

There are no common variables.

Thus, H.C.f of x²y³, x⁵, yz = H.C.F of numerical coefficients x H.C.F of literal coefficients= 1* 1 = 1.

(h) 8abc, 20abc, 28abc

The G.C.F of numerical coefficients = The H.C.F of 8, 20, and 28 = 2

Since, 8=2*2*2, 20=10*2, 28=2*7*2

H.C.F of literal coefficients = G.C.F of abc, abc, abc = abc

Thus G.C.F of 8abc, 20abc, 28abc = H.C.F of numerical coefficients x H.C.F of literal coefficients = 2*abc = 2abc.

Given data is the Rakesh yearly budget.

Yearly Salary = $ 52007

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

Calculation of central angles:

Steps of constructing a pie chart:

1. Draw a circle of any convenient radius.

2. Draw a horizontal radius of the circle.

3. Draw sectors starting from the horizontal radius with central angles of 83.68°, 30.92°, 63.33°, 92°, and 89.98° respectively.

4. Shade the sectors differently using different colors and label them.

Thus, we obtain the required pie chart, shown in the above figure.

Total Number of workers in the company = 19200

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

Calculation of central angles:

Steps for constructing a pie chart

• Draw a circle of any radius.
• Draw a horizontal radius of the circle.
• draw sectors starting from the horizontal radius with the central angles of 9.37°, 84.37°, 75°, 13.12°, and 178.12°.
• Shade the sectors using different colors and label them.

Thus, we obtain the required pie chart, shown in the above figure.

Given table says the monthly expenditure of a person.

Total expenditure in a month = 500 + 1500 + 850 + 640 + 360 + 300 + 350 + 1700 = 6200

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

Calculation of central angles:

Steps for constructing a pie chart:

1. Draw a circle of any convenient radius.

2. Draw a horizontal radius of the circle.

3. Draw sectors starting from the horizontal radius with central angles of 29°, 87°, 49.35°, 37.16°, 20.9°, 17.41°, 20.32°, and 98.7° respectively.

4. Shade the sectors differently using different colors and label them.

Thus, we obtain the required pie chart, shown in the above figure.

The total number of workers in a factory = 2000

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

Calculation of the central angle of each cadre is as follows:

Construction for creating a pie chart

1. Draw a circle of any convenient radius.

2. Draw a horizontal radius of the circle.

3. Draw sectors starting from the horizontal radius with central angles of 66.24°, 100.8°, 54°, 50.4°, 90°, 27.36°, and 21.6° respectively.

4. Shade the sectors differently using different colors and label them.

Thus, we obtain the required pie chart, shown in the above figure.

The total number of fruits in the hamper = 80

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

Calculation of central angles:

Construction for creating a pie chart

• Draw a circle of any convenient radius
• Draw sectors starting from the horizontal radius with central angles of 54°, 90°, 22.5°, 67.5°, 81°, and 45° respectively.
• Shade the sectors differently using different colors and label them.

Thus, we obtain the required pie chart, shown in the above figure.

Given data is about the hours spent by working women.

Total number of hours = 24

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

Calculation of central angles:

Construction for creating a pie chart

Steps of construction:

1. Draw a circle of any convenient radius.

2. Draw a horizontal radius of the circle.

3. Draw sectors starting from the horizontal radius with central angles of 12°, 30°, 105°, 75°, and 30° respectively.

4. Shade the sectors differently using different colors and label them.

Thus, we obtain the required pie chart, shown in the above figure.

From the pie chart, we can observe that,

(i) Pie chart contains the 500 school students’ favorite activity.

(ii) Spending time with the smartphone is the favorite past activity for most of the school students.

(iii) Out of 500 students, 50 students like reading books.

(iv) 100 students like music. The percentage of students who like music is (100 / 500) * 100 = 20%

(v) 70 students like dance. The percentage of students who like dance is (70 / 500) * 100 = 14%

(vi) 130 students like playing outdoor games. The percentage of students who like playing outdoor games is (130 / 500) * 100 = 26%

The pie chart is given about the sales of different brands of smartphones in an e-commerce site during a festive season.

(i) The different brands mentioned in the chart are Motorola, Coolpad, Apple, HTC, Sony, Nokia, Lenovo, and Samsung.

(ii) A total of 10,000 smartphones are sold during the festive season in an e-commerce season.

(iii) A total of 550 Nokia brand smartphones are sold.

(iv) Lenovo has the highest sales and Coolpad has the lowest sales.

(v) The number of Apple smartphones sold is 600, Motorola smartphones are 664. So, the difference between the sales of Motorola and apple is 664 – 600 = 64.

(vi) The sales of Sony is 1500, Samsung is 2500. The ratio between the sales of sony and Samsung is 1500 : 2500 = 3 : 5.

(i) 125 students come to school by bicycle.

(ii) (115 + 100 + 125) = 340 students do not walk to school.

(iii) 115 students come to school by bus.

(iv) 100 students come to school by car.

(v) 160 students come to school by walk. The central angle for students come to school by walk = (160 / 500) * 360 = 0.32 * 360 = 115.2°

(vi) 100 students come to school by car, 115 students come to school by bus. The ratio is 100 : 115 = 20 : 23

Total quantity = 95

Calculation of central angles:

Construction for creating a pie chart

Steps of construction:

1. Draw a circle of any convenient radius.

2. Draw a horizontal radius of the circle.

3. Draw sectors starting from the horizontal radius with central angles of 113.6°, 75.7°, 132.6°, and 37.8° respectively.

4. Shade the sectors differently using different colors and label them.

Thus, we obtain the required pie chart, shown in the above figure.

Given the central angle of the sector = 152°

(152 / 360) * 100 = 42 %

The percentage of students interested in playing cricket is 42 %.

Given that,

The central angle of the sector for walk= 52°

The total number of students = 1000

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

52° = (Value of the component / 1000) * 360°

52 / 360 = Value of the component / 1000

0.1444 = Value of the component / 1000

Value of the component = 0.144.4 * 1000

Value of the component = 144.4

The number of students who opted walking to reach the school = 144

Given that,

The percentage of the number of people favorite food is biryani = 40%

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

= (40 / 100) * 360

= 144°

The central angle of the sector whose favorite food is biryani is 144°.

Given that,

Person monthly income = $ 5200

Expenditure on food = $ 500

The central angle of food = (Expenditure of food/person monthly income) * 360°

= (500 / 5200) * 360°

= 0.096 * 360°

= 34.61°

The central angle of the sector representing food expenses in the pie chart = 34.61°.

The given table represents the percentage of buyers of six different brands of bathing soaps.

Central angle for a component = (Value of the component / Sum of the values of all components) * 360°

Calculation of central angles:

Construction for creating a pie chart

Steps of construction:

1. Draw a circle of any convenient radius.

2. Draw a horizontal radius of the circle.

3. Draw sectors starting from the horizontal radius with central angles of 36°, 54°, 18°, 144°, 54°, and 54° respectively.

4. Shade the sectors differently using different colors and label them.

Thus, we obtain the required pie chart, shown in the above figure.

Given that,

The marks obtained by Aravind in the annual examination.

On graph paper, draw horizontal and vertical lines.

Take subjects along the x-axis at equal gaps, marks along the y-axis.

Choose 1 small division = 10 units

Then the heights of the bars are:

1 small division = 10 dollars

Then, the heights of the bars are:

French marks = (1/10 * 70) = 7 small divisions

Marks in English = (1/10 * 80) = 8 small divisions

Marks in Maths = (1/10 * 50) = 5 small divisions

Marks in Science = (1/10 * 75) = 7.5 small divisions

Marks in Social Sciences = (1/10 * 60) = 6 small divisions

Bar graph showing Aravind marks in Annual Examination

At the points marked in Step 2, draw bars of equal width and of heights calculated.

Given that,

The population of seven Indian states estimated in 2010.

Draw horizontal, vertical lines on a graph paper.

Take states along the x-axis at equal gaps, the population in lakhs along the y-axis.

Choose 1 small division = 100 lakhs

Then, the heights of the bars are:

Population in Andhra Pradesh = (1/100 * 520) = 5.2 small divisions

Population in Assam = (1/100 * 400) = 4 small divisions

Population in Bihar = (1/100 * 250) = 2.5 small divisions

Population in Telangana = (1/100 * 800) = 8 small divisions

Population in Tamil Nadu = (1/100 * 750) = 7.5 small divisions

Population in Rajasthan = (1/100 * 690) = 6.8 small divisions

Bar graph showing the population in seven Indian states.

The favorite sports of students according to the survey are given.

(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.

Take Sport name along the x-axis, the total number of students along the y-axis.

On a scale, choose 1 small division = 5 units

Number of students whose favorite game cricket = (1/5 * 45) = 9 small divisions

Number of students whose favorite game football = (1/5 * 53) = 10.6 small divisions

Number of students whose favorite game basketball = (1/5 * 70) = 14 small divisions

Number of students whose favorite game volleyball = (1/5 * 44) = 8.8 small divisions

Number of students whose favorite game chess = (1/5 * 66) = 13.2 small divisions

Number of students whose favorite game tabletennis = (1/5 * 22) = 4.4 small divisions

Number of students whose favorite game badminton = (1/5 * 33) = 6.6 small divisions

Bar graph showing students favorite game

From the graph,

(b) The most preferred sport is basketball.

(c) Basketball, Chess, Football, Cricket, Volleyball, Badminton, Table tennis.

(d) Cricket, Volleyball are two sports that are almost equally preferred.

(e) The range of the graph is 70 – 22 = 48

Given that,

The number of cars produced in a factory for five consecutive weeks.

Draw two perpendicular lines, x-axis, y-axis on a graph paper.

Take weeks along the x-axis at equal gaps.

Take the number of cars produced per week along the y-axis.

Scale on the y-axis is 1 small division = 200 cars.

According to the scale, draw rectangular bars for each week.

Fill color in it to show the variation.

Column graph or bar graph representing the number of cars produced by a compy for five consecutive weeks.

Draw two perpendicular lines on a graph paper.

Take fruit on the x-axis, people number on the y -axis.

Choose 1 small division = 5 units

Then, the heights of the bars are:

Number of people who like apple = (1/5 * 35) = 7 small divisions

Number of people who like orange = (1/5 * 30) = 6 small divisions

Number of people who like banana = (1/5 * 10) = 2 small divisions

Number of people who like kiwifruit = (1/5 * 25) = 5 small divisions

Number of people who like blueberry = (1/5 * 40) = 8 small divisions

Number of people who like grapes = (1/5 * 5) = 1 small divisions

The nicest fruit according to the survey is blueberry.

From the given graph we can observe that,

(i) The graph tells about the number of families containing 2, 3, 4, 5 members each.

(ii) 40 families have 3 members.

(iii) No one lives alone.

(iv) Family having 3 members is the most common type.

The given graph represents the life expectancy of various countries.

From the given graph we can observe that,

(i) Graph tells about the life expectancy of five different countries in a particular year.

(ii) Japan has the highest life expectancy which is 70 years. Cambodia has the lowest life expectancy which is 40 years.

(iii) India life expectancy is 63 years.

(iv) Range is 70 – 40 = 30.

Given that,

The market position of different types of lotions.

On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.

Take brands along the x-axis at equal gaps, percentage of buyers along the y-axis.

Scale, choose 1 small division = 5 units

Then, the heights of the bars are:

Brand A position in market = (1/5 * 50) = 10

Brand B position in market = (1/5 * 20) = 4

Brand C position in market = (1/5 * 40) = 8

Brand D position in market = (1/5 * 30) = 6

Brand E position in market = (1/5 * 60) = 12

Other brand position in market = (1/5 * 10) = 2

Bar graph showing the brands position in the market.

given graph shows the number of electric bulbs sold in a shop for a week.

By observing the graph,

(i) On Tuesday, the sale was minimum.

(ii) On Saturday, the sale was maximum.

(iii) The sale during the week is 1040 bulbs.

(iv) The minimum sale during the week = 100 bulbs.

The maximum sale during the week = 275 bulbs.

Therefore, minimum sale : maximum sale = 100 : 275 = 20 : 55.

On a graph paper, take the mode of transport on the x-axis, the number of students on y-axis.

Take 1 small division = 50 students

Then, the heights of the bars are:

Number of students used school bus = (1/ 50 * 640) = 12.8

Number of students used bicycle = (1/ 50 * 500) = 10

Number of students used private bus = (1/ 50 * 220) = 4.4

Number of students used rickshaw = (1/ 50 * 120) = 2.4

Number of students reach school by foot = (1/ 50 * 350) = 7

The given bar graph represents the favorite subjects for the students of a class.

By observing the graph, we can say that

(i) The popularity of the subjects is taken along the x-axis on the bar graph.

(ii) The second most popular subject is social.

(iii) The less popular subject is language.

(iv) Language is a favorite subject for 20 students in the class.

(v) For 40 students maths is the favorite subject.

Given that,

The marks of 40 students of a class in English are 90, 36, 55, 78, 85, 96, 56, 77, 63, 53, 75, 56, 91, 86, 82, 29, 46, 40, 88, 76, 77, 79, 83, 85, 69, 64, 45, 67, 62, 60, 36, 39, 48, 49, 50, 73, 86, 88, 89, 92

The ascending order of the students marks are 29, 36, 36, 39, 40, 45, 46, 48, 49, 50, 53, 55, 56, 56, 60, 62, 63, 64, 67, 69, 73, 75, 76, 77, 77, 78, 79, 82, 83, 85, 85, 86, 86, 88, 88, 89, 90, 91, 92, 96

From the above data analysis, we can say that

The lowest marks obtained is 29.

The highest marks obtained is 96.

The range of the data is 96 – 29 = 67.

Given that,

The weight os 20 students is 56, 68, 43, 55, 58, 59, 52, 50, 46, 49, 50, 44, 45, 68, 67, 64, 60, 62, 55, 49

The ascending order of the students weight is 43, 44, 45, 46, 49, 49, 50, 50, 52, 55, 55, 56, 58, 59, 60, 62, 64, 67, 68, 68

From the above data analysis, we can say that

The lowest weight of the student is 43 Kgs

The highest weight of the students is 68 Kgs

Range is 68 – 43 = 25.

Given that,

The noted outcomes of a dice are 1, 5, 6, 2, 3, 4, 2, 5, 6, 1, 5, 2, 3, 4, 5, 6, 5, 2, 1, 4, 3, 6, 2, 5, 1

Arranging the data in ascending order, we get the observations as 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6

The lowest value is 1, the highest value is 6, the range is 6 – 1 = 5

From the data, we find that

1 was observed 4 times,

2 was observed 5 times,

3 was observed 3 times,

4 was observed 3 times,

5 was observed 6 times,

6 was observed 4 times,

We may represent the above data in a tabular form, showing the frequency of each observation. The number of times data occurs in a data set is known as the frequency of data

This tabular form of representation is called a frequency distribution.

Given that,

The daily earnings of 30 stores in a market is 720, 450, 420, 560, 650, 750, 410, 420, 720, 600, 590, 595, 585, 860, 460, 489, 700, 710, 890, 850, 840, 845, 888, 830, 720, 420, 420, 560, 550, 555

The ascending order of the data is 410, 420, 420, 420, 420, 450, 460, 489, 550, 555, 560, 560, 585, 590, 595, 600, 650, 700, 710, 720, 720, 720, 750, 830, 840, 845, 850, 860, 888, 890

The lowest value is 410 dollars, the highest value is 890 dollars and the range is 890 – 410 = 480.

From the data, we can design a frequency distribution table.

Given that,

The heights of 23 students in a class are 129, 126, 130, 131, 145, 140, 139, 129, 128, 125, 141, 140, 130, 131, 126, 127, 126, 128, 125, 131, 133, 135, 136

The ascending order of the data is

125, 125, 126, 126, 126, 127, 128, 128, 129, 129, 130, 130, 131, 131, 131, 133, 135, 136, 139, 140, 140, 141, 145

The lowest height of a student is 125 cm, the top height of a student is 145 cm, range is 145 – 125 = 20

From the data, we can design a frequency distribution table.

Given that,

The marks scored by 50 students are 15, 20, 25, 12, 18, 45, 40, 30, 20, 10, 22, 33, 44, 46, 34, 43, 32, 23, 22, 25, 16, 17, 18, 12, 21, 29, 28, 34, 33, 30, 32, 41, 49, 48, 44, 46, 41, 47, 14, 40, 12, 11, 18, 30, 31, 35, 37, 38, 39, 45

The ascending order of the data is 10, 11, 12, 12, 12, 14, 15, 16, 17, 18, 18, 18, 20, 20, 21, 22, 22, 23, 25, 25, 28, 29, 30, 30, 30, 31, 32, 32, 33, 33, 34, 34, 35, 37, 38, 39, 40, 40, 41, 41, 43, 44, 44, 45, 45, 46, 46, 47, 48, 49

From the data, we can say that the least marks scored are 10, the highest marks scored is 49.

Range is 49 – 10 = 39

The intervals should separate the scale into equal parts. We could choose intervals of 4. We then begin the scale with 10 and end with 49.

Frequency Distribution Table for Grouped Data is

Given that,

The weekly wages of 26 workers of a factory are 650, 620, 750, 800, 610, 600, 640, 650, 660, 670, 680, 710, 720, 730, 710, 750, 650, 660, 630, 500, 700, 750, 760, 780, 777, 720

The ascending order of the data is

500, 600, 610, 620, 630, 640, 650, 650, 650, 660, 660, 670, 680, 700, 710, 710, 720, 720, 730, 750, 750, 750, 760, 777, 780, 800

The lowest value is 500, the highest value is 800, and the range is 800 – 500 = 300

The intervals should separate the scale into equal parts. We could choose intervals of 50. We then begin the scale with 500 and end with 800.

Frequency Distribution Table for Grouped Data is

Given that,

The weekly pocket expenses of 30 students are 80, 50, 60, 65, 55, 52, 51, 60, 64, 104, 90, 89, 98, 56, 65, 57, 75, 76, 67, 53, 70, 74, 75, 86, 93, 98, 82, 83, 94, 96

The ascending order of the data is 50, 51, 52, 53, 55, 56, 57, 60, 60, 64, 65, 65, 67, 70, 74, 75, 75, 76, 80, 82, 83, 86, 89, 90, 93, 94, 96, 98, 98, 104

The lowest value is 50, the highest value is 104, the range is 104 – 50 = 54

The intervals should separate the scale into equal parts. We could choose intervals of 5. We then begin the scale with 50 and end with 104.

Frequency Distribution Table for Grouped Data is

Given that,

The number of members in 20 families are 5, 8, 2, 4, 3, 6, 7, 9, 10, 2, 3, 4, 5, 5, 6, 7, 5, 8, 2, 4

The ascending order of the data is 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 7, 7, 8, 8, 9, 10

The lowest number is 2, the highest number is 10.

Range is 10 – 2 = 8

We may represent the above data in a tabular form, showing the frequency of each observation. The number of times data occurs in a data set is known as the frequency of data. This tabular form of representation is called a frequency distribution.

Given that,

The numbers of newspapers sold at a local shop are 15, 5, 8, 50, 6, 51, 16, 61, 20, 23, 32, 75, 57, 89, 56, 65, 22, 23, 25, 26, 58, 91, 56, 45, 65, 75, 89, 92, 65, 52

The ascending order of the data is 5, 6, 8, 15, 16, 20, 22, 23, 23, 25, 26, 32, 45, 50, 51, 52, 56, 56, 57, 58, 61, 65, 65, 65, 75, 75, 89, 89, 91, 92

The lowest value is 5, highest value is 92, range is 92 – 5 = 87.

The intervals should separate the scale into equal parts. We could choose intervals of 8. We then begin the scale with 5 and end with 92.

Frequency Distribution Table for Grouped Data is

Given that,

The electricity bills of 25 houses for a month are 800, 750, 300, 250, 261, 320, 650, 590, 450, 270, 850, 450, 541, 200, 220, 260, 590, 240, 350, 461, 420, 700, 605, 630, 400

The ascending order of the data is 200, 220, 240, 250, 260, 261, 270, 300, 320, 350, 400, 420, 450, 450, 461, 541, 590, 590, 605, 630, 650, 700, 750, 800, 850

The lowest value is 200, the highest value is 850, the range is 850 – 200 = 650

The intervals should separate the scale into equal parts. We could choose intervals of 50. We then begin the scale with 200 and end with 650.

Frequency Distribution Table for Grouped Data is

Given that,

The weekly expenses of 30 students are 50, 150, 20, 30, 60, 65, 55, 36, 35, 95, 85, 65, 45, 75, 37, 78, 54, 64, 39, 40, 46, 52, 102, 154, 130, 120, 118, 115, 126, 85

The ascending order of the data is 20, 30, 35, 36, 37, 39, 40, 45, 46, 50, 52, 54, 55, 60, 64, 65, 65, 75, 78, 85, 85, 95, 102, 115, 118, 120, 126, 130, 150, 154

The lowest value is 20, the highest value is 154, the range is 154 – 20 = 134

The intervals should separate the scale into equal parts. We could choose intervals of 15. We then begin the scale with 20 and end with 154.

Frequency Distribution Table for Grouped Data is

 (i) (0, 8)

Here, the x coordinate is 0, y coordinate is 8. The point lies on the y-axis.

(ii) (5, -7)

Here, the abscissa is 5, the ordinate is -7. So, the point lies in the fourth quadrant.

(iii) (-8, 2)

Its x coordinate is -8, y coordinate is 2. So, the point lies in the second quadrant.

(iv) (6, 7)

Its x coordinate is 6, y coordinate is 7. The point lies in the first quadrant.

(v) (-9, -15)

Its x coordinate is -9, y coordinate is -15. So, this ordered pair lies in the third quadrant.

(i) (-16, 25)

As this point abscissa is negative, and the ordinate is positive, the point lies in quadrant II.

(ii) (58, -36)

The x coordinate is positive, y coordinate value is negative. So, the point lies in the fourth quadrant.

(iii) (27, 42)

Both the coordinates of the ordered pair are positive. So, the point lies in quadrant I.

(iv) (-36, 49)

This point x coordinate value is negative, y coordinate value is positive. So, the point lies in the second quadrant.

(v) (-81, -150)

As this point both coordinates are negative values, the point lies in the third quadrant.

If the point has a y coordinate value is 0, then that point lies on the x-axis.
The list of points which lies on the x-axis are (ii) (29, 0), (iv) (-45, 0), (vii) (82, 0), and (viii) (-10, 0).

If the point has an x coordinate or the abscissa value is 0, then that point lies on the y-axis.
The list of points that lies on the y-axis are (i) (0, 52), (ii) (0, 75), (v) (0, -26).

Draw X’OX, Y’OY on a graph paper.

(i) The point (5, 6) lies in the first quadrant. And take 5 units along the x-axis towards the right from the origin, 6 units along the y-axis upwards from the origin to get the point.

(ii) As this point both coordinates are negative, it lies in the third quadrant. Take 8 units along the x-axis towards the right from the origin, 6 units along the y-axis downwards to plot the point (-8, -6).

(iii) This point lies in the fourth quadrant. Take 2 units on the x-axis, -7 units on the y-axis to get the point (2, -7).

(iv) This ordered pair x coordinate is zero. So the point lies on the y-axis. Take 3 units on the y-axis to get the point (0, 3).

(v) Point lies in the second quadrant. Take -1 unit along the x-axis, 5 units along the y-axis to get the point (-1, 5).

(vi) This point y-coordinate is 0. So, the point lies on the x-axis. Take 6 units along the x-axis and mark that point as (6, 0).

(vii) Both coordinates of the point lies in the first quadrant. Take 5 units along the x-axis, 4 units along the y-axis to plot the point (5, 4).

(i) Point A (56, -82) lies in Quadrant IV i.e d. Because x coordinate value is positive, y coordinate is negative.

(ii) Point B (56, 82) lies in quadrant I i.e a. Because both coordinates are positive.

(iii) Point C (-56, -82) lies in quadrant III i.e c. Because both coordinates are negative.

(iv) Point D (-56, 82) lies in quadrant II i.e b. Because x coordinate is negative, y coordinate value is positive.

(v) The abscissa of a point is its distance from the origin.

(vi) The ordinate of a point is its distance from the origin.

(vii) The equation x = 0 represents the x-axis.

(viii) Another name for abscissa is x coordinate.

From the coordinate graph, we can find the coordinates of the points P, Q, R, S, T, U, and V.

P coordinates are (-3, 3). Its x coordinate value is -3, y coordinate value is 3.

Q coordinates are (1, 6). Its abscissa value is 1, the ordinate value is 6.

R coordinates are (6, -8). Its x coordinate value is 6, y coordinate value is -8.

S coordinates are (-6, -3). Its x coordinate is -6, y coordinate is -3.

T coordinates are (6, -6). Its abscissa is 6, ordinate is -6.

U coordinates are (5, 3). Its x coordinate is 5, y coordinate is 3.

V coordinates are (-9, -6). Its x coordinate is -9, y coordinate is -6.

The point P (4, 0) lies on the x-axis. Point Q (8, 3) lies in the first quadrant, point R (2, 6) lies in the first quadrant. The point S (-7, 5) lies in quadrant II, T (3, -1) lies in quadrant IV.
Draw two perpendicular lines represents the x-axis, y-axis. Plot these points on the graph.

Any points whose x coordinate value is 0, then that point lies on the y-axis. If a point y coordinate value is 0, then that point lies on the x-axis.

So, from the above list, (0, 53), (0, 52) lies on the y-axis. The points (86, 0), (71, 0), (289, 0) lies on the x-axis.

The point (0, 0) is called the origin which is the midpoint of both axes. In the point (42, 63) none of the coordinate values is 0, hence it does not lie on the axes.

Given points are A (5, 0), B (5, 5), C (0, 5).

Plot those points on the graph having X’OX, Y’OY as axes. Join those points.

By joining the points, we will get a square having 5 units side length.

The coordinates of P are (5, 6). Its x coordinate is 5, y coordinate is 6.

Q coordinates are (-5, 6). Its abscissa is -5, ordinate is 6.

R coordinate values are (6, -4). Its x coordinate is 6, y coordinate is -4.

S coordinates are (-2, -8). Its abscissa is -2, ordinate is -8.

T coordinate values are (9, -6). Its x coordinate is 9, y coordinate is -6.

U coordinate values are (-3, 3). Its x coordinate is -3, y coordinate is 3

V coordinate values are (-3, -5). Its x coordinate is -3, y coordinate is -5.

(i) The coordinates of G are (3, -4).

(ii) Point identified by coordinates (-1, 5) is C.

(iii) The coordinates of B are (-3, -4).

(iv) The point identified by (3, 3) is A.

(v) The abscissa of point H is -8.

(vi) The ordinate of point E is -3.

(vii) The abscissa of point E is 4.

(viii) The coordinates of point F are (0, -4)

(ix) The point identified by (-9, 7) is D.

(x) The point having coordinates (-3, -4) is B.

Given function is y = 2x

Find the value of y for the corresponding values of the x.

If x = 0, then y = 2 * 0 = 0

If x = 1, then y = 2 * 1 = 2

If x = -1, then y = 2 * -1 = -2

If x = 2, then y = 2 * 2 = 4

Plot the points O (0, 0), A (1, 2), B (-1, -2), C (2, 4) on the graph paper.

Join those points to get a line equation.

Linear equation forms a straight line.

By observing the graph, the value of y when x = 1.5 is 3, -2 is -4, 2.5 is 5.

Given function is A = x⁴

Substitute x =1, 2, 3 in the function to get the A value.

If x = 1, then A = 1⁴ = 1

If x = 2, then A = 2⁴ = 16

If x = 3, then A = 3⁴ = 81

Now plot the points A (1, 1), Q (2, 16), C (3, 81) on a graph paper.

Join those points to get a line equation.

By observing the graph, the value of y when x = 2.5 is 39, 0 is 0, 3.5 is 150.

Given that,

When T = 0.5, D = 3 * 0.50 = 1.5

When T = 1, D = 3 * 1 = 3

When T = 1.5, D = 3 * 1.5 = 4.5

When T = 2, D = 3 * 2 = 6

When T = 2.5, D = 3 * 2.5 = 7.5

Now plot the points A (0.5, 1.5), B (1, 3), C (1.5, 4.5), D (2, 6), E (2.5, 7.5) on the graph.
Join those points to get a graph equation.

Take time along the x-axis, distance along the y-axis.

On the x-axis: 1 unit = 30 seconds.

On the y-axis: 1 unit = 1 unit.

Reading off the values from the graph.

On the x-axis, take the point L at t = 3.

Draw LP ⊥ x-axis, meeting the graph at P.

Clearly, PL = 9 units.

Therefore, t = 3 ⇒ D = 9.

On the x-axis, take the point M at t = 4

Draw MQ ⊥ y-axis, meeting the graph at Q.

Clearly, QM = 12 units.

Therefore, t = 4 ⇒ D = 12.

Add the numbers in the column which is completely filled with numbers.

Calculate the magic constant

magic constant = [n (n² + 1)] / 2

Sum = [4(4² + 1)] / 2

= [4 (16 + 1)] / 2

= [4 * 17] / 2

= 68 / 2 = 34

The sum of all rows, columns, and diagonals is 34.

Sum up the numbers in the column which is completely filled with numbers. Proceed with completing the row or column which has three numbers and only one cell empty.

Sum = 3 + 14 + 13 + 0 + 8 + 5 + 6 + 11 + 4 + 9 + 10 + 7 + 15 + 2 + 1 + 12 = 120

The simple thing is place 0, 1, 2 at the vertices of the triangle. and remaining on other unfilled fields.

First of all cross out all empty cells around the square 0.

The hint is Mine is available at the lower-left corner of the grid.

Based on this hint and data provided on the question, 8 mines are represented using the M letter in the cells.

We can say that each number is equal to the product of two nearest numbers in the row just above it.

So, A = 2 * 4 = 8

B = 8 * 2 = 16

C = 2 * 16 = 32

D = 16 * 64 = 1024

E = 64 * B = 64 * 16 = 1024

The simple trick is to take one row or column which are having more filled numbers and less unfilled numbers.

In this question, the column which is having less unfilled cells is the first column. Continue the process to get the solution easily.

We can figure it out as 5 + 3 = 8, 8 * 2 = 16

In the same way 8 + 12 = 20, 20 * (2 + 2) = 80

34 + 27 = 61 * (4 + 3) = 61 * 7 = 427

Missing number = 16 + 21 = 37 * (7 + 4) = 37 * 11 = 407

∴ Missing number in the image is 407

100 = 177 – 77

= (7 + 7) x (7 + (1 : 7))

So, (7 + 7) x (7 + (1 : 7)) = 14 x (7 + (1 : 7))

= 14 x ((49 + 1) : 7)

= 14 x (50 : 7)

= 2 x 50 = 100

The simple trick is (5 + 7) / 2 = 6

(4 + 4) / 2 = 4

(8 + 2) / 2 = 10 / 2 = 5

∴ Missing number is 5.

We can solve this by using an easy equation 3 (x + 100000) = 10x + 1

3x + 300000 = 10x + 1

10x – 3x = 300000 – 1

7x = 299,999

x = 299,999 / 7

x = 72,857

∴ 5-digit number is 72,857.

If a number is divisible by 36, then it is divisible by both 4 and 9.

Now, for 19x9y to be divisible by 4, we must have y = 2 or x = 6 (since the number formed by the last two digits must be divisible by 4).

Also, for 19x9y to be divisible by 9, the sum of the digits must be divisible by 9.

Therefore, when y = 2, then x = 6 and when y = 6, then x = 2.

Since, x > y, we have x = 6, y = 2.