It is given that By cross multiplication \[\begin{array}{*{35}{l}} {{x}^{3}}~+\text{ }3x\text{ }=\text{ }3a{{x}^{2}}~+\text{ }a  \\ {{x}^{3}}~\text{ }3a{{x}^{2}}~+\text{ }3x\text{ }\text{ }a\text{...

It is given that

It is given that = RHS

It is given that

It is given that \[(\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{2}{{\mathbf{y}}^{\mathbf{2}}}):\text{ }(\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{...

It is given that a: b = \[9:10\] So we get a/b = \[9/10\] = \[5\]

Consider So we get x = k (b + c – a) y = k (c + a – b) z = k (a + b – a) Here   = k Therefore, it is proved.

We know that x/a = y/b = k So we get x = ak, y = bk Here Here LHS = RHS Therefore, it is proved.

It is given that x/a = y/b = z/c = k So we get x = ak, y = bk, z = ck Here = \[{{k}^{3}}\] Hence, LHS = RHS.

It is given that a/b = c/d = e/f = k So we get a = k, c = dk, e = fk Therefore, it is proved. = k Therefore, it is proved.

It is given that q is the mean proportional between p and r q2 = pr Here LHS = \[{{p}^{2}}~\text{ }3{{q}^{2}}~+\text{ }{{r}^{2}}\] We can write it as \[=\text{ }{{p}^{2}}~\text{ }3pr\text{ }+\text{...

Consider x and y as the two numbers Mean proportion = \[16\] Third proportion = \[128\] \[\begin{array}{*{35}{l}} \surd xy\text{ }=\text{ }16  \\ xy\text{ }=\text{ }256  \\ \end{array}\] Here...

It is given that a, b, c, d, e are in continued proportion We can write it as a/b = b/c = c/d = d/e = k \[d\text{ }=\text{ }ek,\text{ }c\text{ }=\text{ }e{{k}^{2}},\text{ }b\text{ }=\text{...

It is given that \[\mathbf{2},\text{ }\mathbf{6},\text{ }\mathbf{p},\text{ }\mathbf{54}\] and q are in continued proportion We can write it as \[2/6\text{ }=\text{ }6/p\text{ }=\text{ }p/54\text{...

It is given that \[\left( \mathbf{a}\text{ }+\text{ }\mathbf{2b}\text{ }+\text{ }\mathbf{c} \right),\text{ }\left( \mathbf{a}\text{ }\text{ }\mathbf{c} \right)\text{ }\mathbf{and}\text{ }\left(...

Consider x be added to each number So the numbers will be \[15\text{ }+\text{ }x,\text{ }17\text{ }+\text{ }x,\text{ }34\text{ }+\text{ }x\text{ }and\text{ }38\text{ }+\text{ }x\] Based on the...

Consider the number of passed = \[3x\] Number of failed = x So the total candidates appeared = \[3x\text{ }+\text{ }x\text{ }=\text{ }4x\] In the second case Number of candidates appeared =...

Consider \[5x\] and \[6x\] as the savings of Lokesh and his sister. Lokesh should save Rs y more Based on the problem \[\left( 5x\text{ }+\text{ }y \right)/\text{ }\left( 6x\text{ }+\text{ }30...

Consider the two shorter sides of a right-angled triangle as \[5x\] and \[12x\] So the third longest side = \[13x\] It is given that \[5x\text{ }+\text{ }12x\text{ }+\text{ }13x\text{ }=\text{...

It is given that a: b = \[3:5\] We can write it as a/b = \[3/5\] Here \[\left( 3a\text{ }+\text{ }5b \right):\text{ }\left( 7a\text{ }\text{ }2b \right)\] Now dividing the terms by b Here \[\left(...

It is given that \[\left( 7p\text{ }+\text{ }3q \right):\text{ }\left( 3p\text{ }\text{ }2q \right)\text{ }=\text{ }43:\text{ }2\] We can write it as \[\left( 7p\text{ }+\text{ }3q \right)/\text{...

\[\begin{array}{*{35}{l}} {{\left( a\text{ }+\text{ }b \right)}^{2}}:\text{ }{{\left( a\text{ }\text{ }b \right)}^{2}}  \\ ({{a}^{2}}~\text{ }{{b}^{2}}):\text{ }({{a}^{2}}~+\text{ }{{b}^{2}})  \\...

It is given that If \[x\text{ }+\text{ }y\text{ }+\text{ }z\text{ }\ne \text{ }0\] Therefore, it is proved.

It is given that By cross multiplication \[\begin{array}{*{35}{l}}    6x\text{ }\text{ }6\text{ }=\text{ }5x\text{ }+\text{ }5  \\    6x\text{ }\text{ }5x\text{ }=\text{ }5\text{ }+\text{ }6  \\   ...

It is given that By further calculation \[\begin{array}{*{35}{l}} 2x/4\text{ }=\text{ }2y/3  \\ x/2\text{ }=\text{ }y/3  \\ \end{array}\] By cross multiplication \[x/y\text{ }=\text{ }2/3\] Hence,...

It is given that By cross multiplication \[a\text{ }+\text{ }b\text{ }=\text{ }5a\text{ }\text{ }5b\] We can write it as \[\begin{array}{*{35}{l}} 5a\text{ }\text{ }a\text{ }\text{ }5b\text{ }\text{...

It is given that We get \[\begin{array}{*{35}{l}} 2ax\text{ }=\text{ }{{x}^{2}}~+\text{ }1  \\ {{x}^{2}}~\text{ }2ax\text{ }+\text{ }1\text{ }=\text{ }0  \\ \end{array}\] Therefore, it is...

So we get \[\begin{array}{*{35}{l}} 3x\text{ }=\text{ }a  \\ x\text{ }=\text{ }a/3  \\ \end{array}\] So we get x = \[3a\] Therefore, x= \[a/3,3a\].

x = \[1/5\]

By cross multiplication \[\begin{array}{*{35}{l}} 81{{x}^{2}}~\text{ }45\text{ }=\text{ }36{{x}^{2}}  \\ 81{{x}^{2}}~\text{ }36{{x}^{2}}~=\text{ }45  \\ \end{array}\] So we get...

By cross multiplication \[\begin{array}{*{35}{l}} 81{{x}^{2}}~\text{ }45\text{ }=\text{ }36{{x}^{2}}  \\ 81{{x}^{2}}~\text{ }36{{x}^{2}}~=\text{ }45  \\ \end{array}\] So we get 45x2 = 45...

By cross multiplication \[\begin{array}{*{35}{l}} 50x\text{ }\text{ }75\text{ }=\text{ }12x\text{ }+\text{ }1  \\ 50x\text{ }\text{ }12x\text{ }=\text{ }1\text{ }+\text{ }75  \\ \end{array}\] So we...

By cross multiplication \[8\text{ }+\text{ }4x\text{ }=\text{ }2\text{ }\text{ }x\] So we get \[\begin{array}{*{35}{l}} 4x\text{ }+\text{ }x\text{ }=\text{ }2\text{ }\text{ }8  \\ 5x\text{ }=\text{...

\[\begin{array}{*{35}{l}}    =\text{ }2\left( a\text{ }\text{ }b \right)/\text{ }\left( a\text{ }\text{ }b \right)  \\    =\text{ }2  \\ \end{array}\]

\[\begin{array}{*{35}{l}}    =\text{ }2\left( a\text{ }\text{ }b \right)/\text{ }\left( a\text{ }\text{ }b \right)  \\    =\text{ }2  \\ \end{array}\]

It is given that \[(\mathbf{11}{{\mathbf{a}}^{\mathbf{2}}}~+\text{ }\mathbf{13}{{\mathbf{b}}^{\mathbf{2}}})\text{ }(\mathbf{11}{{\mathbf{c}}^{\mathbf{2}}}~\text{...

It is given that (ma + nb): b :: (mc + nd): d We can write it as (ma + nb)/ b = (mc + nd)/ d By cross multiplication mad + nbd = mbc + nbd Here mad = mbc ad = bc By further calculation a/b = c/d...

It is given that (pa + qb): (pc + qd) :: (pa – qb): (pc – qd) We can write it as Therefore, it is proved that a: b :: c: d.

It is given that \[~\left( \mathbf{4a}\text{ }+\text{ }\mathbf{5b} \right)\text{ }\left( \mathbf{4c}\text{ }\text{ }\mathbf{5d} \right)\text{ }=\text{ }\left( \mathbf{4a}\text{ }\text{ }\mathbf{5d}...

Therefore, it is proved. Therefore, it is proved.

(iii) We know that If a: b :: c: d we get a/b = c/d By multiplying \[2/3\] \[2a/3b\text{ }=\text{ }2c/3d\] By applying componendo and dividendo \[\left( 2a\text{ }+\text{ }3b \right)/\text{ }\left(...

(i) We know that If a: b :: c: d we get a/b = c/d By multiplying \[2/5\] \[2a/5b\text{ }=\text{ }2c/5d\] By applying componendo and dividendo \[\left( 2a\text{ }+\text{ }5b \right)/\text{ }\left(...

It is given that a, b, c, d are in continued proportion Here we get a/b = b/c = c/d = k \[c\text{ }=\text{ }dk,\text{ }b\text{ }=\text{ }ck\text{ }=\text{ }dk\text{ }.\text{ }k\text{ }=\text{...

It is given that a, b, c, d are in continued proportion Here we get a/b = b/c = c/d = k \[c\text{ }=\text{ }dk,\text{ }b\text{ }=\text{ }ck\text{ }=\text{ }dk\text{ }.\text{ }k\text{ }=\text{...

It is given that a, b, c, d are in continued proportion Here we get a/b = b/c = c/d = k \[c\text{ }=\text{ }dk,\text{ }b\text{ }=\text{ }ck\text{ }=\text{ }dk\text{ }.\text{ }k\text{ }=\text{...

It is given that a, b, c are in continued proportion So we get a/b = b/c = k (v) LHS = \[abc\text{ }{{\left( a\text{ }+\text{ }b\text{ }+\text{ }c \right)}^{3}}\] We can write it as \[=\text{...

It is given that a, b, c are in continued proportion So we get a/b = b/c = k   (iii) \[~a:\text{ }c\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}):\text{ }({{b}^{2}}~+\text{ }{{c}^{2}})\] We...

It is given that a, b, c are in continued proportion So we get a/b = b/c = k Therefore, LHS = RHS. Therefore, LHS = RHS.

It is given that a, b, c are in continued proportion \[\frac{p{{a}^{2}}+qab+r{{b}^{2}}}{p{{b}^{2}}+qbc+r{{c}^{2}}}=\frac{a}{c}\] Consider a/b = b/c = k So we get a = bk and b = ck ….. (1) From...

It is given that x, y, z are in continued proportion Consider x/y = y/z = k So we get y = kz \[x\text{ }=\text{ }yk\text{ }=\text{ }kz\text{ }\times \text{ }k\text{ }=\text{ }{{k}^{2}}z\] Therefore,...

It is given that a, b, c, d are in proportion Consider a/b = c/d = k a = b, c = dk Therefore, LHS = RHS. So we get = d2 (1 + k2) + b2 (1 + k2) = (1 + k2) (b2 + d2) RHS = a2 + b2 + c2 + d2 We can...

It is given that a, b, c, d are in proportion Consider a/b = c/d = k a = b, c = dk Therefore, LHS = RHS. Therefore, LHS = RHS.

It is given that a, b, c, d are in proportion Consider a/b = c/d = k a = b, c = dk (i) LHS = \[\left( 5a\text{ }+\text{ }7b \right)\text{ }\left( 2c\text{ }\text{ }3d \right)\] Substituting the...

Consider ax = by = cz = k It can be written as x = k/a, y = k/b, z = k/c

Consider a/b = c/d = e/f = k So we get a = bk, c = dk, e = fk Therefore, LHS = RHS. So we get \[=\text{ }bdf\text{ }{{\left( k\text{ }+\text{ }1\text{ }+\text{ }k\text{ }+\text{ }1\text{ }+\text{...

Consider a/b = c/d = e/f = k So we get a = bk, c = dk, e = fk (i) LHS = \[({{b}^{2}}~+\text{ }{{d}^{2}}~+\text{ }{{f}^{2}})\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~+\text{ }{{e}^{2}})\] We can write it...

Therefore, LHS = RHS.

It is given that x/a = y/b = z/c We can write it as x = ak, y = bk and z = ck Therefore, LHS = RHS. Therefore, LHS = RHS.

It is given that a + c = mb and \[\mathbf{1}/\mathbf{b}\text{ }+\text{ }\mathbf{1}/\mathbf{d}\text{ }=\text{ }\mathbf{m}/\mathbf{c}\] a + c = mb Dividing the equation by b a/b + c/d = m ……. (1)...

It is given that y is mean proportional between x and z We can write it as \[{{y}^{2}}~=\text{ }xz\]…… (1) Consider LHS = \[xyz\text{ }{{\left( x\text{ }+\text{ }y\text{ }+\text{ }z \right)}^{3}}\]...

It is given that b is the mean proportional between a and c \[{{b}^{2}}~=\text{ }ac\]…. (1) Here (ab + bc) is the mean proportional between \[({{\mathbf{a}}^{\mathbf{2}}}~+\text{...

Solution: It is given that b is the mean proportional between a and c We can write it as b2 = a × c b2 = ac ….. (1) We know that a, c, \[{{\mathbf{a}}^{\mathbf{2}}}~+\text{...

  Consider a and b as the two numbers It is given that \[28\] is the mean proportional \[a:\text{ }28\text{ }::\text{ }28:\text{ }b\] We get \[ab\text{ }=\text{ }{{28}^{2}}~=\text{ }784\] Here...

Consider x be added to each number \[16\text{ }+\text{ }x\text{ },\text{ }26\text{ }+\text{ }x\text{ }and\text{ }40\text{ }+\text{ }x\] are in continued proportion It can be written as \[\left(...

It is given that \[\mathbf{x}\text{ }+\text{ }\mathbf{5}\] is the mean proportion between \[\mathbf{x}\text{ }+\text{ }\mathbf{2}\text{ }\mathbf{and}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{9}\]...

It is given that \[\mathbf{k}\text{ }+\text{ }\mathbf{3},\text{ }\mathbf{k}\text{ }+\text{ }\mathbf{2},\text{ }\mathbf{3k}\text{ }\text{ }\mathbf{7}\text{ }\mathbf{and}\text{ }\mathbf{2k}\text{...

Consider x be subtracted from each term \[23\text{ }\text{ }x,\text{ }30\text{ }\text{ }x,\text{ }57\text{ }\text{ }x\text{ }and\text{ }78\text{ }\text{ }x\] are proportional It can be written as...

Consider x to be added to \[\mathbf{5},\text{ }\mathbf{11},\text{ }\mathbf{19}\text{ }\mathbf{and}\text{ }\mathbf{37}\] to make them in proportion \[5\text{ }+\text{ }x:\text{ }11\text{ }+\text{...

It is given that a, \[12,16\] and b are in continued proportion \[a/12\text{ }=\text{ }12/16\text{ }=\text{ }16/b\] We know that \[a/12\text{ }=\text{ }12/16\] By cross multiplication \[a/12\text{...

(iii) Consider x as the mean proportion of \[8.1\text{ }and\text{ }2.5\] \[8.1:\text{ }x\text{ }::\text{ }x:\text{ }2.5\] It can be written as \[\begin{array}{*{35}{l}} {{x}^{2}}~=\text{ }8.1\text{...

(i) Consider x as the mean proportion of 5 and 80 \[5:\text{ }x\text{ }::\text{ }x:\text{ }80\] It can be written as \[\begin{array}{*{35}{l}} {{x}^{2}}~=\text{ }5\text{ }\times \text{ }80\text{...

(iii) Consider x as the third proportional to \[~Rs.\text{ }3\text{ }and\text{ }Rs.\text{ }12\] \[3:\text{ }12\text{ }::\text{ }12:\text{ }x\] It can be written as \[\begin{array}{*{35}{l}} 3\text{...

(i) Consider x as the third proportional to \[5,10\] \[5:\text{ }10\text{ }::\text{ }10:\text{ }x\] It can be written as \[\begin{array}{*{35}{l}} 5\text{ }\times \text{ }x\text{ }=\text{ }10\text{...

(iii) \[1.5,\text{ }2.5,\text{ }4.5\] Consider x as the fourth proportional to \[1.5,\text{ }2.5,\text{ }4.5\] \[1.5:\text{ }2.5\text{ }::\text{ }4.5:\text{ }x\] We can write it as \[1.5\text{...

(i) \[\mathbf{3},\text{ }\mathbf{12},\text{ }\mathbf{15}\] Consider x as the fourth proportional to \[\mathbf{3},\text{ }\mathbf{12},\text{ }\mathbf{15}\] \[3:\text{ }12\text{ }::\text{ }15:\text{...

(iii)\[~2.5:\text{ }1.5\text{ }=\text{ }x:\text{ }3\] We can write it as \[1.5\text{ }\times \text{ }x\text{ }=\text{ }2.5\text{ }\times \text{ }3\] So we get \[\begin{array}{*{35}{l}} x\text{...

(i)\[\mathbf{10}:\text{ }\mathbf{35}\text{ }=\text{ }\mathbf{x}:\text{ }\mathbf{42}\] We can write it as \[35\text{ }\times \text{ }x\text{ }=\text{ }10\text{ }\times \text{ }42\] So we get...

Consider number of passes = \[4x\] Number of failures = x Total number of students appeared = \[4x\text{ }+\text{ }x\text{ }=\text{ }5x\] In case \[2\] Number of students appeared = \[5x\text{...

(i) Consider the monthly pocket money of Ravi and Sanjeev as \[5x\] and \[7x\] Their expenditure is \[3y\] and \[5y\] respectively. \[5x\text{ }\text{ }3y\text{ }=\text{ }80\] …… (1) \[7x\text{...

(i) It is given that Mixture of milk to water = \[45\] litres Ratio of milk to water = \[13:2\] Sum of ratio = \[13\text{ }+\text{ }2\text{ }=\text{ }15\] Here the quantity of milk = \[\left(...

(i) It is given that Ratio between A, B and C = \[7:\text{ }5:\text{ }4\] Consider A share = \[7x\] B share = \[5x\] C share = \[4x\] So the total sum =\[~7x\text{ }+\text{ }5x\text{ }+\text{...

It is given that Ratio of three numbers \[=\text{ }1/2:\text{ }1/3:\text{ }1/4\] \[\begin{array}{*{35}{l}} =\text{ }\left( 6:\text{ }4:\text{ }3 \right)/\text{ }12  \\ =\text{ }6:\text{ }4:\text{...

(i) It is given that Perimeter of triangle = \[30\] cm Ratio among sides = \[7:5:3\] Here the sum of ratios = \[7\text{ }+\text{ }5\text{ }+\text{ }3\text{ }=\text{ }15\] We know that Length of...

(i) Ratio of original and reduced weight of woman = \[7:5\] Consider original weight = \[7x\] Reduced weight = \[5x\] Here original weight = \[91\]  kg So the reduced weight = \[~\left( 91\text{...

(i) Ratio = \[\mathbf{8}:\text{ }\mathbf{7}\] Consider the numbers as \[8x\] and \[7x\] Using the condition \[\left[ 8x\text{ }\text{ }25/2 \right]/\text{ }\left[ 7x\text{ }\text{ }25/2...

(iii) \[\left( x\text{ }+\text{ }2y \right)/\text{ }\left( 2x\text{ }\text{ }y \right)\text{ }=\text{ }{{3}^{2}}/\text{ }{{2}^{2}}\] So we get \[\left( x\text{ }+\text{ }2y \right)/\text{ }\left(...

(i) \[\left( x\text{ }\text{ }9 \right)/\text{ }\left( 3x\text{ }+\text{ }6 \right)\text{ }=\text{ }{{\left( 4/9 \right)}^{2}}\] So we get \[\left( x\text{ }\text{ }9 \right)/\text{ }\left( 3x\text{...

(i) \[(\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{xy}):\text{ }(\mathbf{3xy}\text{ }\text{ }{{\mathbf{y}}^{\mathbf{2}}})\text{ }=\text{ }\mathbf{12}:\text{ }\mathbf{5}\] We can write it...

(i) \[\mathbf{3x}\text{ }+\text{ }\mathbf{5y}/\text{ }\mathbf{3x}\text{ }\text{ }\mathbf{5y}\text{ }=\text{ }\mathbf{7}/\mathbf{3}\] By cross multiplication \[9x\text{ }+\text{ }15y\text{ }=\text{...

(i) We know that \[\begin{array}{*{35}{l}} A:\text{ }B\text{ }=\text{ }1/4\text{ }\times \text{ }5/1\text{ }=\text{ }5/4  \\ B:\text{ }C\text{ }=\text{ }1/7\text{ }\times \text{ }6/1\text{ }=\text{...

(i) It is given that \[\mathbf{A}:\text{ }\mathbf{B}\text{ }=\text{ }\mathbf{2}:\text{ }\mathbf{3},\text{ }\mathbf{B}:\text{ }\mathbf{C}\text{ }=\text{ }\mathbf{4}:\text{ }\mathbf{5}\text{...

It is given that \[\mathbf{2}:\text{ }\mathbf{3},\text{ }\mathbf{17}:\text{ }\mathbf{21},\text{ }\mathbf{11}:\text{ }\mathbf{14}\text{ }\mathbf{and}\text{ }\mathbf{5}:\text{ }\mathbf{7}\] We can...

(iii) \[\mathbf{1}/\mathbf{9}:\text{ }\mathbf{2}\] We know that Reciprocal ratio of \[1/9:\text{ }2\text{ }=\text{ }2:\text{ }1/9\text{ }=\text{ }18:\text{ }1\]

(i) \[\mathbf{4}:\text{ }\mathbf{7}\] We know that Reciprocal ratio of \[4:\text{ }7\text{ }=\text{ }7:\text{ }4\] (ii) \[{{\mathbf{3}}^{\mathbf{2}}}:\text{ }{{\mathbf{4}}^{\mathbf{2}}}\] We know...

(iii) \[27{{a}^{3}}:\text{ }64{{b}^{3}}\] We know that Sub-triplicate ratio of \[27{{a}^{3}}:\text{ }64{{b}^{3}}~=\text{ }{{[{{\left( 3a \right)}^{3}}]}^{1/3}}:\text{ }{{[{{\left( 4b...

(i) \[\mathbf{1}:\text{ }\mathbf{216}\] We know that Sub-triplicate ratio of \[1:\text{ }216\text{ }=\sqrt[3]{1}:\sqrt[3]{216}\] By further calculation \[\begin{array}{*{35}{l}} =\text{...

(iii) \[9{{a}^{2}}:\text{ }49{{b}^{2}}\] We know that Sub-duplicate ratio of \[9{{a}^{2}}:\text{ }49{{b}^{2}}~=\text{ }\surd 9{{a}^{2}}:\text{ }\surd 49{{b}^{2}}~=\text{ }3a:\text{ }7b\]

(i) \[\mathbf{9}:\text{ }\mathbf{16}\] We know that Sub-duplicate ratio of \[9:\text{ }16\text{ }=\text{ }\surd 9:\text{ }\surd 16\text{ }=\text{ }3:\text{ }4\] (ii) \[{\scriptscriptstyle 1\!/\!{...

(iii)\[~{{1}^{3}}:\text{ }{{2}^{3}}\] We know that Triplicate ratio of \[{{1}^{3}}:\text{ }{{2}^{3}}~=\text{ }{{({{1}^{3}})}^{3}}:\text{ }{{({{2}^{3}})}^{3}}~=\text{ }{{1}^{3}}:\text{...

(i) \[\mathbf{3}:\text{ }\mathbf{4}\] We know that Triplicate ratio of \[3:\text{ }4\text{ }=\text{ }{{3}^{3}}:\text{ }{{4}^{3}}~=\text{ }27:\text{ }64\] (ii) \[{\scriptscriptstyle 1\!/\!{...

iii) \[5a:\text{ }6b\] We know that Duplicate ratio of \[5a:\text{ }6b\text{ }=\text{ }{{\left( 5a \right)}^{2}}:\text{ }{{\left( 6b \right)}^{2}}~=\text{ }25{{a}^{2}}:\text{ }36{{b}^{2}}\]

(i) \[\mathbf{2}:\text{ }\mathbf{3}\] We know that Duplicate ratio of \[2:\text{ }3\text{ }=\text{ }{{2}^{2}}:\text{ }{{3}^{2}}~=\text{ }4:\text{ }9\] (ii) \[\surd \mathbf{5}:\text{ }\mathbf{7}\] We...

(iii) \[\left( \mathbf{a}\text{ }\text{ }\mathbf{b} \right):\text{ }\left( \mathbf{a}\text{ }+\text{ }\mathbf{b} \right),\text{ }{{\left( \mathbf{a}\text{ }+\text{ }\mathbf{b}...

(i) \[\mathbf{2}:\text{ }\mathbf{3}\text{ }\mathbf{and}\text{ }\mathbf{4}:\text{ }\mathbf{9}\] We know that Compound ratio \[\begin{array}{*{35}{l}} ~=\text{ }2/3\text{ }\times \text{ }4/9  \\...

It is given that Copper = \[\mathbf{27}\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\] kg = \[55/2\]kg Tin = \[\mathbf{2}\text{ }{\scriptscriptstyle 3\!/\!{ }_4}\] kg = \[11/4\] kg We know that Total...