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Ratio
Question 66 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER] |
${\text{If }}\frac{x}{2} = \frac{y}{3} = \frac{z}{4} = \frac{{2x - 3y + 5z}}{k}{\text{, }}$ then the value of k is -
12 | |
16 | |
15 | |
18 |
Question 66 Explanation:
\begin{align} & {\text{Let }}\frac{x}{2} = \frac{y}{3} = \frac{z}{4} = l. \cr & {\text{Then, }} \cr & = x = 2l, y = 3l, z = 4l. \cr & \therefore \frac{x}{2} = \frac{{2x - 3y + 5z}}{k} \cr & \Rightarrow \frac{{2l}}{2} = \frac{{2 \times 2l - 3 \times 3l + 5 \times 4l}}{k} \cr & \Rightarrow k = 4 - 9 + 20 = 15. \cr\end{align}
Question 67 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER] |
x, y, z, u are real numbers such that x : y = y : z = z : u and x : u = 64 : 27. the value of x : z is -
3 : 4 | |
16 : 9 | |
64 : 27 | |
4 : 3 |
Question 67 Explanation:
\begin{align} & {\text{Let}}\, \frac{x}{y} = \frac{y}{z} = \frac{z}{u} = k. \cr & {\text{Now, }}\frac{x}{u} = \frac{{64}}{{27}} \cr & \Rightarrow \frac{x}{y} \times \frac{y}{z} \times \frac{z}{u} = \frac{{64}}{{27}} \cr & \Rightarrow {k^3} = {\left( {\frac{4}{3}} \right)^3} \cr & \Rightarrow k = \frac{4}{3}. \cr & {\text{So}}, \, x:y = y:z = z:u = 4:3. \cr & \therefore \frac{x}{z} = \frac{x}{y} \times \frac{y}{z} = \frac{4}{3} \times \frac{4}{3} = \frac{{16}}{{9.}} \cr\end{align}
Question 68 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER] |
If an amount of Rs. 1, 50, 000 is shared among A, B, and C in the ratio 2 : 3 : 5, then A receives the same amount as he would receive if another sum of money is shared between A, B and C in the ratio of 5 : 3 : 2. The ratio of Rs. 1, 50, 000 to the second amount of money is -
2 : 3 | |
5 : 2 | |
3 : 2 | |
5 : 3 |
Question 68 Explanation:
Let the 2nd amount be Rs. x \begin{align} & {\text{Then, }} \cr & {\text{A's share from }}{{\text{1}}^{{\text{st}}}}{\text{ amount}} \cr & = {\text{Rs}}{\text{.}}\left( {150000 \times \frac{2}{{10}}} \right) \cr & = {\text{Rs}}{\text{. }}30000. \cr & {\text{A's share from }}{{\text{2}}^{{\text{nd}}}}{\text{ amount}} \cr & = {\text{Rs}}{\text{.}}\left( {x \times \frac{5}{{10}}} \right) \cr & = {\text{Rs}}{\text{. }}\frac{x}{2}. \cr & \therefore \frac{x}{2} = 30000\, {\text{or}}\, x = 60000. \cr & {\text{Required ratio}} \cr & = 150000:60000 \cr & = 5:2 \cr\end{align}
Question 69 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER] |
At a casino in Mumbai, there are 3 tables A, B and C. The payoffs at A is 10 : 1, at B is 20 : 1 and C is 30 :1. If a man bets Rs. 200 at each table and win at two of the tables, what is the maximum and minimum difference between his earnings can be?
Rs. 2000 | |
Rs. 4500 | |
Rs. 2500 | |
Rs. 4000 |
Question 69 Explanation:
Maximum earning will be only possible when he will won on the maximum yielding table.
A → 10 :1
B → 20 :1
C → 30 :1
i.e., he won B and C but lost on A,
20 *200 +30 *200 - 1 *200 = 9800
Minimum earnings will be when he won on table A and B and lose on table 3.
10 *200 + 20 *200 - 1 *200 = 5800
Therefore, difference = 9800-5800 = Rs. 4000.
Alternatively,
The difference,
= [(30 +20 -1) - (10 +20 -1)] * 200
= Rs. 4000.
Question 70 [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER] |
Income of A and B are in the ratio 4 : 3 and their annual expenses are in the ratio 3 : 2. If each save Rs. 60000 at the end of the year, the annual income of A is =?
Rs. 150000 | |
Rs. 360000 | |
Rs. 240000 | |
Rs. 120000 |
Question 70 Explanation:
| A | : | B | |
| Income | 4x | : | 3x |
| Expenses | 3y | : | 2y |
| Saving | 6000 | : | 6000 |
⇒ x = 60000
∴ Income of A = 4 $\times$ 60000 = 240000
There are 70 questions to complete.