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Ratio
Question 156 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER] |
If 8a = 9b then the ratio of a/9 to b/8 is
64 : 81 | |
2 : 1 | |
1 : 2 | |
1 : 1 |
Question 156 Explanation:
\begin{align} & = {\text{8a}} = {\text{9b}} \Rightarrow a = \frac{9}{8}b. \cr & \therefore \frac{{\text{a}}}{{\text{9}}}:\frac{{\text{b}}}{{\text{8}}} = \frac{{\left( {\frac{9}{8}b} \right)}}{9}:\frac{{\text{b}}}{{\text{8}}} \cr & = \frac{{\text{b}}}{{\text{8}}}:\frac{{\text{b}}}{{\text{8}}} = 1:1 \cr\end{align}
Question 157 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER] |
If a : (b + c) = 1 : 3 and c : (a + b) = 5 : 7, then b : (a + c) is equal to.
2 : 1 | |
1 : 3 | |
2 : 3 | |
1 : 2 |
Question 157 Explanation:
\begin{align} & = \frac{a}{{b + c}} = \frac{1}{3} \cr & \Rightarrow a = \frac{{b + c}}{3} \cr & \Rightarrow \frac{c}{{a + b}} = \frac{5}{7} \cr & \Rightarrow 7c = 5a + 5b \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{{5\left( {b + c} \right)}}{3} + 5b \cr & \Rightarrow 7c - \frac{5}{3}c = 5b + \frac{5}{3}b \cr & \Rightarrow \frac{{16c}}{3} = \frac{{20b}}{3} \cr & \Rightarrow 16c = 20b \cr & \Rightarrow b = \frac{4}{5}c. \cr & a = \frac{{b + c}}{3} = \frac{{\frac{4}{5}c + c}}{3} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{{9c}}{5} \times \frac{1}{3} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{3}{5}c. \cr & \therefore \frac{b}{{a + c}} = \frac{{\left( {\frac{4}{5}c} \right)}}{{\left( {\frac{3}{5}c + c} \right)}} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{{4c}}{5} \times \frac{5}{{8c}} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{1}{2} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 1:2 \cr\end{align}
Question 158 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER] |
If two times A is equal to three times of B and also equal to four times of C, then A:B:C is
2:3:4 | |
3:4:2 | |
6:4:3 | |
4:6:3 |
Question 158 Explanation:
\begin{align} & 2A = 3B \cr & Or, \, B = \left( {\frac{2}{3}} \right)A;\, {\text{and}} \cr & 2A = 4C \cr & Or, \, C = \left( {\frac{1}{2}} \right)A; \cr & {\text{Hence}}, \cr & A:B:C = A:\frac{{2A}}{3}:\frac{A}{2} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 1:\frac{2}{2}:\frac{1}{2} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 6:4:3 \cr\end{align}
Question 159 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER] |
If x : y = 3 : 1, then x3 - y3 : x3 + y3 =?
14 : 13 | |
13 : 14 | |
10 : 11 | |
11 : 10 |
Question 159 Explanation:
\begin{align} & = \frac{x}{y} = \frac{3}{1} \cr & \Rightarrow \frac{{{x^3}}}{{{y^3}}} = \frac{{27}}{1} \cr & \Rightarrow \frac{{{x^3} - {y^3}}}{{{x^3} + {y^3}}} = \frac{{\left( {\frac{{{x^3}}}{{{y^3}}}} \right) - 1}}{{\left( {\frac{{{x^3}}}{{{y^3}}}} \right) + 1}} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{{27 - 1}}{{27 + 1}} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{{26}}{{28}} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{{13}}{{14}} \cr & = \left( {{x^3} - {y^3}} \right):\left( {{x^3} + {y^3}} \right) \cr & = 13:14. \cr\end{align}
Question 160 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER] |
Two numbers are such that the ratio between them is 4 : 7. If each is increase by 4, the ratio become 3 : 5. The larger number is =?
56 | |
64 | |
36 | |
48 |
Question 160 Explanation:
A : B
4x : 7x
Now 4 is added to each number \begin{align} & \frac{{4x + 4}}{{7x + 4}} = \frac{3}{5} \cr & \Rightarrow 20x + 20 = 21x + 12 \cr & \Rightarrow x = 8 \cr\end{align} ∴ Smaller number is 4 $\times$ 8 = 32
Larger number is 7 $\times$ 8 = 56
There are 160 questions to complete.