17. Rs.5625 is divided among A, B and C so that A may receive $\displaystyle\frac{1}{2}$ as much as B and C together receive, B receives $\displaystyle\frac{1}{4}$ of what A and C together receive. The share of A is more than that of B by :
a. Rs.750
b. Rs.775
c. Rs.1500
d. Rs.1600

Answer: A
Explanation:
$A = \dfrac{1}{2}\left( {B + C} \right) \Rightarrow B + C = 2A$
$ \Rightarrow $ $A + B + C = 3A$
Given, $A + B + C = 5625$
$ \Rightarrow $ $3a = 5625$
$\therefore$ $A =1875$
Again, $B = \dfrac{1}{4}(A + C)$ $\Rightarrow A+C=4B$
$ \Rightarrow $$A + B+ C = 5B$
$5B=5625$ $ \Rightarrow$ $B = 1125$
Then, $A$'s share is more than that of B by $\text{Rs}.(1875 - 1125)= \text{Rs}.750$

18. 729 ml.of a mixture contains milk and water in the ratio of 7:2. How much more water is to be added to get a new mixture containing milk and water in the ratio of 7:3 ?
a. 60 ml
b. 70 ml
c. 81 ml
d. 90 ml

Answer: C
Explanation:
Milk = $\left( {729 \times \displaystyle\frac{7}{9}} \right) = 567ml$
Water = $\left( {729 \times \displaystyle\frac{2}{9}} \right) = 162ml$
Let the amount of water to be added $=x$
$\displaystyle\frac{{567}}{{162 + x}} = \displaystyle\frac{7}{3} \Rightarrow 7(162 + x) = 3 \times 567$
$ \Rightarrow$ $7x = 1701 - 1134$
$ \Rightarrow$ $x = \displaystyle\frac{{567}}{7} = 81$ml

19. A and B are two alloys of gold and copper prepared by mixing metals in proportions 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the proportion of gold and copper in C will be :
a. 5 : 9
b. 5 : 7
c. 7 : 5
d. 9 : 5

Answer: C
Explanation:
Weighted average rule can be used here. And only one component should be considered.
$A = \dfrac{{mx + ny}}{{m + n}}$
Here, $m, n$ are quantitites. $x, y$ are averages or proportions.
Gold proportion in the first alloy = $\dfrac{7}{{7 + 2}} = \dfrac{7}{9}$
Gold proportion in the second alloy = $\dfrac{7}{{7 + 11}} = \dfrac{7}{18}$
Proportion of Gold = $\dfrac{{1 \times \dfrac{7}{9} + 1 \times \dfrac{7}{{18}}}}{{1 + 1}}$ $ = \dfrac{{\dfrac{{14 + 7}}{{18}}}}{2}$ $ = \dfrac{{21}}{{36}} = \dfrac{7}{{12}}$
Gold : Copper = $7 : (12 - 7) = 7 : 5$

Alternative method:
Gold in C = $\left( {\displaystyle\frac{7}{9} + \displaystyle\frac{7}{{18}}} \right) = \displaystyle\frac{{21}}{{18}} = \displaystyle\frac{7}{6}$
Copper in C = $\left( {\displaystyle\frac{2}{9} + \displaystyle\frac{{11}}{{18}}} \right) = \displaystyle\frac{{15}}{{18}} = \displaystyle\frac{5}{6}$
Gold : Copper = $\displaystyle\frac{7}{6}:\displaystyle\frac{5}{6} = 7:5$

20. The students in three classes are in the ratio 2:3:5. If 20 students are increased in each class, the ratio changes to 4:5:7. The total number of students before the increase were :
a. 10
b. 90
c. 100
d. None of these

Answer: C
Explanation:
Let the number of students be $2x$, $3x$ and $5x$.
Then $(2x + 20) : (3x + 20):(5x + 20) = 4 : 5 : 7$
So, $\displaystyle\frac{{2x + 20}}{4} = \displaystyle\frac{{3x + 20}}{5} = \displaystyle\frac{{5x + 20}}{7}$
$ \Rightarrow$ $5(2x + 20)=4(3x + 20)$ ($\because $ equating any two fractions)
$ \Rightarrow $ $x = 10$
$\therefore $ Total number of students before increase $=10x = 100$

21. A right cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is :
a. 3:5
b. 2:5
c. 3:1
d. 1:3

Answer: D
Explanation:
Let the heights of the cylinder and cone be h and H respectively. Then,
$\pi {r^2}h = \displaystyle\frac{1}{3}\pi {r^2}H$
$\displaystyle\frac{h}{H} = \displaystyle\frac{1}{3}$
So, their heights are in the ratio $1 : 3$

22. A circle and square have same area. Therefore, the ratio of the side of the square and the radius of the circle is :
a. $\sqrt \pi :1$
b. $1:\sqrt \pi $
c. $1:\pi $
d. $\pi :1$

Answer: A
Explanation:
Let the side of the square be $x$ and let the radius of the circle be $y$.
Then, ${x^2} = \pi {y^2}$
$ \Rightarrow \dfrac{{{x^2}}}{{{y^2}}} = \dfrac{\pi }{1}$
$ \Rightarrow \dfrac{x}{y} = \sqrt {\dfrac{\pi }{1}} = \dfrac{{\sqrt \pi }}{1}$
$\Rightarrow x : y = \sqrt \pi :1$

23. In a class, the number of boys is more than the number of girls by 12% of the total strength. The ratio of boys to girls is :
a. 11:4
b. 14:11
c. 25:28
d. 28:25

Answer: B
Explanation:
Let the number of boys and girls be $x$ and $y$ respectively.
$ \Rightarrow $ $(x-y) = 12\% (x+y)$
$ \Rightarrow $ $x - y = \dfrac{12}{{100}}\left( {x + y} \right)$ $ = \dfrac{3}{{25}}\left( {x + y} \right)$
$ \Rightarrow $ $25x - 25y=3x + 3y$
$ \Rightarrow $ $22x = 28y$
$ \Rightarrow $ $\displaystyle\frac{x}{y} = \displaystyle\frac{{28}}{{22}}$ $= \displaystyle\frac{{14}}{{11}} = 14:11$

24. A, B and C do a work in 20, 25 and 30 days respectively. They undertook to finish the work together for Rs.2220, then the share of A exceeds that of B by :
a. Rs.120
b. Rs.180
c. Rs.300
d. Rs.600

Answer: B
Explanation: NoteRemuneration is inversely proportional to the days taken to complete the work.
Ratio of shares of A,B & C = $\dfrac{1}{{20}}:\dfrac{1}{{25}}:\dfrac{1}{{30}} = \dfrac{{15:12:10}}{{300}}$ ($\because $ by taking LCM of 20, 25, 30 and multiplying the given ratios)
Sum of the ratios $= 15 + 12 + 10 = 37$.
A's share = Rs.$\left( {2220 \times \displaystyle\frac{{15}}{{37}}} \right) = Rs.900$
B's share = Rs.$\left( {2220 \times \displaystyle\frac{{12}}{{37}}} \right) = Rs.720$
Thus, the share of A exceeds that of B by $\text{Rs}.(900 - 720) = \text{Rs}.180$