Ratio - Aptitude MCQ Questions and Solutions with Explanations

10 : 12 :: 15 : a
12 × 15 = 10 × a
a = 18

\[\begin{array}{*{20}{c}} {{\text{Copper}}}&{}&{{\text{Zinc}}}&{} \\ 4&:&3&{ \Rightarrow {7_{ \times 5}}} \\ 5&:&2&{ \Rightarrow {7_{ \times 6}}} \end{array}\]
\[\begin{array}{*{20}{c}} {{\text{Copper}}\,\,\,\,\,{\text{Zinc}}} \\ {20\,\,\,\,\,:\,\,\,\,\,15} \\ {30\,\,\,\,\,:\,\,\,\,\,12} \\ {\overline {\,50\,\,\,\,\,:\,\,\,\,\,27\,} } \end{array}\]
$${\text{Zinc }}\% = \frac{{27}}{{77}} \times 100 = 35\% $$

\[\begin{array}{*{20}{c}} {}&{{\text{Juice}}}&:&{{\text{Water}}}&{{\text{Total}}} \\ {{{\text{I}}^{{\text{st}}}}}&{{5_{ \times 11}}}&:&{{2_{ \times 11}}}&{{7_{ \times 11}}} \\ {{\text{I}}{{\text{I}}^{{\text{nd}}}}}&{{7_{ \times 7}}}&:&{{4_{ \times 7}}}&{{{11}_{ \times 7}}} \end{array}\]
Capacity of both cup is same
\[\begin{array}{*{20}{c}} {}&{{\text{Juice :}}\,{\text{Water}}\,\,\,{\text{Total}}} \\ {{{\text{I}}^{{\text{st}}}}}&{55\,\,\,:\,\,\,22\,\,\,\,\,\,\,\,\,\,77} \\ {{\text{I}}{{\text{I}}^{{\text{nd}}}}}&{49\,\,\,:\,\,\,28\,\,\,\,\,\,\,\,\,\,77} \\ {{\text{Total}}}&{\overline {\underline {\,104\,\,:\,\,50\,\,\,\,\,\,\,\,\,154\,} } } \end{array}\]
Final ratio of water to juice in cup = 50 : 104 = 25 : 52

\[\begin{array}{*{20}{c}} {}&{{\text{Rs}}{\text{. 1}}}&:&{50\,{\text{Paise}}}&:&{25{\text{ Paise}}} \\ {{\text{Value of coins}}}&{8x}&:&{4x}&:&{3x} \\ {{\text{Number of coins}}}&{8x \times 1}&:&{4x \times 2}&:&{3x \times 4} \\ {}&{8x}&:&{8x}&:&{12x} \end{array}\]
∴ Total coins ⇒ 8x + 8x + 12x = 28x
28x = 280   (Given)
x = \[\frac{{280}}{{28}}\] = 10
∴ Number of 50 paise coins are = 8x = 8 × 10 = 80

Let total students = 72 units

39 units → 260
1 unit → $$\frac{{20}}{3}$$
45 units → $$45 \times \frac{{20}}{3}$$
Boys = 300

\[\begin{array}{*{20}{c}} {}&{{\text{A}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{B}}\,} \\ {{\text{Student}}}&{x - 5 = y + 5} \end{array}\]
$$\eqalign{ & x - y = 10\,.\,.\,.\,.\left( {\text{i}} \right) \cr & x + 25 = 2\left( {y - 25} \right) \cr & x - 2y = - 75\,.\,.\,.\,.\left( {{\text{ii}}} \right) \cr & {\text{Subtract both equation}} \cr & x - y = 10 \cr & x - 2y = - 75 \cr & \underline {\, - \,\,\, + \,\,\,\,\,\,\, + \,\,\,\,\,} \cr & \,\,\,\,\,\,\,\,\,\,y = 85 \cr & x = 95 \cr & x:y = 95:85 = 19:17 \cr} $$

In first case
→ 3 : 4 : 5 ⇒ 12 × 47
Second case
→ $$\frac{1}{3}:\frac{1}{4}:\frac{1}{5}$$
⇒ 20 : 15 : 12
⇒ 47 × 12
First case → 141 : 188 : 235
Second case → 240 : 180 : 144
(240 - 141) units → Rs. 1188
99 units → Rs. 1188
1 unit → Rs. 12
Total income = 564 units
= 564 × 12 = 6768

If salary of D = 4m
4m + 0.4(4m) = 560
4m + 1.6m = 560
m = $$\frac{{560}}{{5.6}}$$
m = 100
Original salary of B, C, D, E = 2m + 3m + 4m + 5m
= 14m
= 14 × 100
= 1400
Hence, the sum of the original salaries of B, C, D and E is Rs. 1400

$$\eqalign{ & A + B + C + D = 46800 \cr & \frac{{A + D}}{{B + C}} = \frac{8}{5} \cr & \frac{B}{C} = \frac{{5P}}{{4P}},\,\,B + C = 9P \cr & \frac{{A + D}}{{9P}} = \frac{8}{5} \cr & A + D = \frac{{72}}{5}P \cr & \frac{{72}}{5}P + 9P = 46800 \cr & 117P = 46800 \times 5 \cr & P = 2000 \cr & B = 5P = 5 \times 2000 = 10000 \cr & C = 4P = 4 \times 2000 = 8000 \cr & A = 18400 \cr & D = 46800 - 10000 - 8000 - 18400 = 10400 \cr & A - B = x = 18400 - 10000 = 8400 \cr & D - C = y = 10400 - 8000 = 2400 \cr & x - y = 8400 - 2400 = 6000 \cr} $$

(a² + b²) : (b² + c²) : (c² + a²) = 34 : 61 : 45
Add all
2(a² + b² + c²) = 34 + 61 + 45
2(a² + b² + c²) = 140
(a² + b² + c²) = 70
c² = 70 - 34
c² = 36
c = 6
b² = 70 - 45
b² = 25
b = 5
a² = 70 - 61
a² = 9
a = 3
(b - a) : (c - b) : (c - a)
= (5 - 3) : (6 - 5) : (6 - 3)
= 2 : 1 : 3