AP Police SI Mains 2019 Arithmetic and Reasoning

According to 1st condition,

$$\left(\frac{\ 1}{y}-\frac{1\ }{2}\right)=\left(\frac{1\ }{3}-\frac{1\ }{y}\right).$$

or,$$\frac{\ 2}{y}=\frac{1\ }{3}+\frac{1\ }{2}=\frac{5\ }{6}.$$

or,$$y=\frac{12\ }{5}.$$

According to second condition,

$$\frac{1\ }{y}-\frac{1\ }{2-x}=\frac{1\ }{3-x}-\frac{1\ }{y}.$$

or,$$\frac{2\ }{y}=\frac{1\ }{3-x}+\frac{1\ }{2-x}.$$

or,$$\frac{5}{6}=\frac{3-x+2-x}{\left(3-x\right)\left(2-x\right)}=\frac{\left(5-2x\right)}{\left(x^2-5x+6\right)}.$$

or,$$\left(5x^2-25x+30\right)=30-12x.$$

or,$$5x^2=13x.$$

or,$$x=2.6.$$

A is correct choice.

Let say, A,B,C require a,b and c hour to fill the pool.

According to question,

b=20+a  and  b=c+14.

So,$$\ 20+a=c+14.$$

or,$$c-a=6.$$..........................(1)

Again, A and B together can fill the pool in half the time that is required for C to fill the pool.

So,$$\ \frac{ab\ }{a+b}=\frac{c\ }{2}.$$

Putting value from (1) and b=20+a :

$$\ \frac{a\left(20+a\right)\ }{a+20+a}=\frac{a+6\ }{2}.$$

or,$$\ \frac{a\left(20+a\ \right)}{a+10}=\frac{a+6\ }{1}.$$

or,$$\ a^2+20a=\left(a+6\right)\left(a+10\right).$$

or,$$\ a^2+20a=a^2+6a+10a+60.$$

or,$$20a=16a+60.$$

or,$$4a=60.$$

or,$$a=15.$$

So,b=35 and c=21.

So,together they will do in 1 hour$$\left(\ \frac{\ 1}{15}+\frac{1\ }{35}+\frac{\ 1}{21}\right)$$ part

or $$\frac{\ 1}{15}+\frac{1\ }{35}+\frac{\ 1}{21}=\frac{\ 7+3+5}{105}=\frac{15\ }{105}=\frac{1}{7}$$part.

So,they together will complete in 7 hour.

Let,Srinu takes x hour to complete work.

then Mohan takes x/3 hour to complete the work.

According to question,

$$x\ -40=\frac{x\ }{3}.$$

or,$$x\ -\frac{x\ }{3}=40.$$

or,$$\frac{2x\ }{3}=40.$$

or,$$x\ =60.$$

Then,they will do in 1 hour $$\ \frac{\ 1}{20}+\frac{1\ }{60}=\frac{3+1\ }{60}=\frac{4\ }{60}=\frac{1`\ }{15}$$ part.

So, they together will complete in 15 hour.

C is correct choice.

A can do full picture in 30 min.

B can erase it in 40 min.

A's rate =$$\frac{1\ }{30}$$ per min.

B's rate=$$\frac{1\ }{40}$$ per min.

So, 1st min A will do=$$\frac{1\ }{30}$$.

2nd min work done after B erase=$$\frac{1\ }{30}-\frac{1\ }{40}=\frac{40-30\ }{120}=\frac{10\ }{120}.$$

3rd min,work done=$$\frac{10\ }{120}+\frac{1\ }{30}\frac{10+4}{120}=\frac{14\ }{120}.$$

4th min=$$\frac{14\ }{120}-\frac{1}{40}=\frac{14-3}{120}=\frac{11}{120}.$$

5th min=$$\frac{11}{120}+\frac{1}{30}=\frac{11+4}{120}=\frac{15}{120}.$$

6thmin=$$\frac{15}{120}-\frac{1}{40}=\frac{15-3}{120}=\frac{12}{120}.$$

So, every even number minutes it gain by 1 unit of work.

So, 2,4,6,......,232nd min it will complete ==> 10,11,12,.......,116th part of work.

But in 233rd min it will complete$$\frac{116\ }{120}+\frac{1\ }{30}=\frac{\ 116+4}{120}=\frac{120\ }{120}=1.$$

So, full job will done in 233 min.

D is correct choice.

18m+12w=>18days.

2w=1m.

18m+6m=>18days.

So,24 man can do the task in 18 days.

So,8men can do it in=(24×18)/8=54 days.

B is correct choice.

A boy can do in 1 day (1/72) part of work.

A man can do in 1 day (1/12) part of work.

A women can do in 1 day (1/48)part of work.

6 boys can do in 1 day (6/72)=1/12 part of work.

let say ,x number of women needed.

So,6 boys,1 man and x women can do work in 1 day=(1/12+1/12+x/48) unit.

so, in 2 days they will do=(1/6+1/6+x/24).

So,(1/3+x/24)=1.

or,x/24=1-1/3=2/3.

or, x=16.

B is correct choice.

We know ,unit of work=M×D×H

M=number of men

H=number of hour

D=number of days.

if 64 men complete 40% of work in 5 days,

then work done=64×5×8=2560 unit.

so,40% work=2560 units.

so,60% work=2560(60/40)=3840 units.

let say, to complete the  work in 4 days they need to work for x hour per day .

So,64×4×x=3840

or, x=15.

D is correct choice.

A and B together can do the work in 16 days.

So, in 1 day they together can do (1/16) part of the work.

A can do (1/32) part of work in 1 day.

So,B can do $$(1/16-1/32)=1/32$$ part of work in 1 day.

So,if B do work in alternative days then they together have to work for 32 days i.e.

$$32×(1/32)=1$$.

D is correct choice.

Factors of the number are

$$576 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^6 \times 3^2$$

$$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^5 \times 3$$

$$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$$

LCM of 96,144 and n is 576.

i.e.

$$n = 2^6 \times k = 64k$$
....(1)

HCF is 48.

$$HCF(96, 144, 64k) = 2^4 \times 3$$

$$48 = 2^4 \times 3$$

This only happen if k=3

So, substituting k=3 in (1),

$$n=64×3=192.$$

D is correct choice.

To produce a 0 ,we need at least one 5 and at least one 2 .

But in the given question, all the numbers are odd number ,so there is no 2 exist in the product.

So, the above product can not produce any number of 0's.

C is correct choice.