AP Police SI Mains 2019 Arithmetic and Reasoning

Let say,

$$A-x=k\ ,\ A-y=7k\ and\ A-z=4k.$$

So,$$2A=2\left(k+x\right)\ ,\ 2A=2\left(7k+y\right)\ and\ 2A=2\left(4k+z\right).$$

So,$$x+y+z=2k+2x\ ,\ x+y+z=14k+2y\ and\ \ x+y+z=8k+2z.$$.........................................(1)

or,$$3\left(x+y+z\right)=2\left(x+y+z\right)+2k+14k+8k.\ \ \left(by\ adding\ these\ 3\ equations\right)$$

or,$$x+y+z=24k.$$

put this value in (1) :

$$2x=22k\ or\ x=11k.$$

$$2y=10k\ or\ y=5k.$$

and $$2z=16k\ or\ z=8k.$$

So,$$x:y:z=11:5:8.$$

C is correct choice.

Let say,

$$a^x=b^y=c^z=k.$$

or,$$a=k^{\frac{1}{x}},\ b=k^{\frac{1}{y}}\ and\ c=k^{\frac{1}{z}}.$$

and $$\frac{a}{b}=\frac{b}{c}.$$

or,$$b^2=ac.$$

So,$$k^{\frac{2}{y}}=k^{\frac{1}{x}}k^{\frac{1}{y}}.$$

or,$$k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}}.$$

or,$$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}.$$

or,$$2xz=yz+xy.\ \left(multiply\ both\ side\ by\ xyz\right)$$

or,$$yz-xz=xz-xy.$$

or,$$z\left(y-x\right)=x\left(z-y\right).$$

or,$$\frac{\left(y-x\right)}{\left(z-y\right)}=\frac{x}{z}.$$

So, A is correct choice.

Let say, cost price=c and marked price=m.

given that , x-y=20.

No, SP_1=m(1-x) and SP_2=m(1-y) .

So,P1 %= (SP_1/c)-1=((m/c)(1-x))-1.

and P2 %=(SP_2/c)-1=((m/c)(1-y))-1.

so,( P2-P1)

=((m/c)(1-y))-1-((m/c)(1-x))+1

=(m/c)(1-y-1+x)

=(m/c)(x-y).

=20m/c.

So,

20m/c=32.(given in question)

or,c/m=20/32

or,c/m=5/8.

So,c:m=5:8.

C is correct choice.

Let assume any value for a,b and x that satisfy the above conditions.

Let say, a=1 and b=2 and x=5.

Then (a/b)=(1/2)=0.5.

and (a+x)/(b+x)=6/7=0.85.

So,(a+x)/(b+x) >(a/b).

but it is not greater than x or (b/a).

now,(1/x)=1/5=0.2

So,in this case (a+x)/(b+x)>(1/x).

But if we consider x=0.5 :

(a+x)/(b+x)=1.5/2.5=0.6.

but (1/x)=10/5=2.

So, in this case (a+x)/(b+x) is lesser than (1/x).

So, only option C is correct.

Let say, distance cover by them in the playground is d.

So, Speed of A,B,C and D are

(d/1.5),(d/25),(d/35) and (d/20).

So,they will receive money in following ratio:

(d/1.5):(d/25):(d/35):(d/20)

=(1/1.5):(1/25):(1/35):(1/20)

=(2100/1.5):(2100/25):(2100/35):(2100/20)

=1400:84:60:105.

So, 

A will get=1400K/1649.

B will get=84K/1649.

C will get=60K/1649.

D will get=105K/1649.

So, A will get maximum amount.

A is correct choice.

According to given condition:

(15/17)-(5/7)=0.1680.

(12/13)-(2/3)=0.2564.

(14/17)-(4/7)=0.2521.

(13/14)-(3/4)=0.1785.

So, B is correct choice.

Number of men ,women and children are distinct number (given).

And number of women is 7.

Again,

number of men/number of women =

number of women/number of children.

So,number of men×number of children

=$$(number of women)^2$$

=$$7^2$$

=49.

But, 49 has three different factors 1,7 and 49.

as number of men, women and children are distinct, number of children or men can have either 49 or 1 as their value.

So,Total number of visitors were 

=49+7+1=57.

C is correct choice.

Let say ,speed of walking, cycling and running are k,9k and 3k respectively.

Time and speed are inversely proportional when distance is constant.

So,their time taken would be as following:

(1/k),(1/9k) and (1/3k).

So,ratio of time and speed would be :

(1/k)/k ,(1/9k)/9k and (1/3k)/3k

or,$$(1/k^2), (1/81k^2) and (1/9k^2)$$.

So , required ratio would be ==>

$$(1/k^2):(1/81k^2):(1/9k^2)$$

=$$1:(1/81):(1/9)$$

=81:1:9.

C is correct choice.

$$16\frac{2}{3}$$%=50/300.

$$33\frac{1}{3}$$%=100/300.

5%=5/100=15/300.

25%=25/100=75/300.

So, according to question:

(15/100)x=(50/300)+(100/300)+(15/300)+ (75/300).

(15/100)x=(240/300).

or,x=(240×100)/(15×300)=80/15=16/3.

A is correct choice.

Originally expected 5K.

but he wrote (1/5)K=0.2K.

first error,x%=(5K-0.2K)/5K=0.96 or 96%.

next time he wrote (1/25)K=0.04K.

second error,y%=(5K0.04K)/5K=0.992 or 99.2%.

so increase of y over x is =

(0.992-0.96)/0.96=0.033333 or 3.33%.

A is correct choice.