AP Police SI Mains 2019 Arithmetic and Reasoning

Let say,

number of boys who can swim is a and who cant is b.

And number of girls who can swim is c and who cant is d.

The number of boys who can swim is one more than the number of girls who can swim a = c+1

The number of girls who cannot swim is one more than the number of boys that cannot swim => d = b+1

The difference between number of boys who can swim and number of girls who cannot swim is two => a - d = 2

Let say,he bought x number of pens and y number of books.

each pen cost 20 rs.and each book cost 35 rs.

So,it will cost a total of $$(20x+35y)$$rs.

So, According to the question,

$$350\ge(20x+35y). $$

by using trial and error method,

if we put x=10 and y=4 then,

$$20x+35y=200+140=340$$

which is less than 350.

But if we put x=14 and y=2 then,

$$20x+35y=280+70=350.$$

So, we can utilize the 350 Rs. full only when he purchase 14 number of pen and 2 number of books.

So, no rupees will left with him.

A is correct choice.

if we put $$\left(-\ \frac{\ 1}{10},+\ \frac{\ 1}{10}\right)$$  in $$2x\ -\ 4y>1$$,it does not satisfy the equation.

So, C is correct choice.

Let say, initially A had a Rs and B had b Rs.

from condition 1 :

$$B+30=2\times\ \left(A-30\right).$$

or,$$2A-B=90.$$

from condition 2:

$$A+10=3\times\ \left(B-10\right).$$

or,$$3B-A=40.$$

By solving these 2 equations we get,

$$A=62$$ and $$B=34$$.

So, C is correct choice.

Here first term ,a$$=2020$$ and common difference, d$$=4$$.

and last term$$=2124.$$

So,$$a+(n-1)d=l$$

or,$$2020+(n-1)4=2124$$

or,$$(n-1)4=104$$

or,$$n=27.$$

So,B is correct choice.

If we consider a 7 day cycle then next day should be a Monday.

Here 15th of January is Monday.

But 1952 is a leap year .

So there would be one extra day 

Which should be next day of the week.

So Tuesday is the answer.

C is correct choice.

Formula to calculate day :

(year code+Month code+Century code+Date Number-Leap year code) mod 7.

Year code:

$$(YY+(YY\div4)) mod 7$$

Here,for 1600th year,

$$(00+(00/4))mod7$$=00.

year code is 0.

Month code:

from January to December the month codes are as : 033614625035(we know it).

here month code is 5.

Century Code:

Century code for 1600 years is 6.

So, required date would be at=(0+5+6+31-0)mod 7=0(as 1600 is a leap year and month is not in January or February so we will subtract 0).

So, it will be Sunday not Saturday.

D is correct choice.

Hour hand covers 360° in 12 hour or 720 minutes.

Also, Minute hand covers 360° in 60 minutes.

So, hour hand covers (1/2)° in 1 minute.

and minute hand covers 6° in 1 minute.

Let say, time in the clock when they coincide : hour hand moved H hour and minute hand moved M minutes.

So,

Angle between hands of a clock

When the minute hand is behind the hour hand, the angle between the two hands at M minutes past HH 'o clock

$$=(30H-6M)+(M/2)°$$

$$=(30H-(11M/2))°$$

When the minute hand is ahead of the hour hand, the angle between the two hands at M minutes past HH 'o clock
$$=((6M−30H)−M/2)°$$

$$=((11M/2)-30H)°$$

So, taking mod:

$$|60H-11M|=2\theta$$..........................(1)

So, hour hand and minute hand will coincide 11 times :

between 1 and 2 , 2 and 3 , 3 and 4, 4 and 5 , 5 and 6, 6 and 7 , 7 and 8 , 8 and 9, 9 and 10, 10 and 11, 11 and 12.

1 and 2 :

between 1 and 2 hour hand will pass 1 O'clock,

So,H=1 and angle should be 0.

using equation (1):$$60\times\ 1-11M=0\ or\ M=60\div11=5\frac{5\ }{11}.$$

So,between 1 and 2 , hour hand and minute hand will coincide at $$5\frac{5\ }{11}$$ minutes past 1.

2 and 3:

$$60\times2-11M=0\ or\ M=120\div11=10\frac{10\ }{11}.$$

They will coincide $$10\frac{10\ }{11}$$ minutes past 2.

3 and 4:

$$60\times3-11M=0\ or\ M=\frac{180}{11}=16\frac{4\ }{11}.$$

They will coincide $$16\frac{4\ }{11}$$ minutes past 3.

4 and 5:

$$60\times4-11M=0\ or\ M=\frac{240}{11}=21\frac{9\ }{11}.$$

They will coincide $$21\frac{9\ }{11}$$ minutes past 4.

5 and 6:

$$60\times5-11M=0\ or\ M=\frac{300}{11}=27\frac{3\ }{11}.$$

They will coincide $$27\frac{3\ }{11}$$ minutes past 5.

6 and 7:

$$60\times6-11M=0\ or\ M=\frac{360}{11}=32\ \frac{\ 8}{11}.$$

They will coincide $$32\ \frac{\ 8}{11}$$ minutes past 6.

7 and 8:

$$60\times7-11M=0\ or\ M=\frac{420}{11}=38\frac{2\ }{11}.$$

They will coincide $$38\frac{2\ }{11}$$ minutes past 7.

8 and 9:

$$60\times8-11M=0\ or\ M=\frac{480}{11}=43\frac{7\ }{11}.$$

They will coincide $$43\frac{7\ }{11}$$ minutes past 8.

9 and 10:

$$9\times60-11M=0\ or\ M=\frac{540}{11}=49\frac{1\ }{11}.$$

They will coincide $$49\frac{1\ }{11}$$ minutes past 9.

10 and 11:

$$60\times10-11M=0\ or\ M=\frac{600}{11}=54\frac{6\ }{11}.$$

They will coincide $$54\frac{6\ }{11}$$ minutes past 10.

11 and 12:

$$60\times11-11M=0\ or\ M=\frac{660}{11}=60.$$

So, last time they will coincide 60 minutes past 11O'clock which means at 12 noon.

So, Total 11 times they will coincide.

B is correct choice.

At the time 3:00 O'clock hour hand is 15 minutes away from centre.

But , minutes hand gain 5 minutes in every 60 minutes.So,to be in staright line it should cover $$(15-5)=10 minutes$$ away from 4 O'clock hand.

and in 60 seconds it will gain 55 seconds.

So, They will be in a straight line at 

$$4 O'clock- 10 minutes-55seconds$$

at 3:49:05.

A is correct choice.

It gain 15 minutes in 24 hour or 1440 minutes.

or,in 60 minutes it gain 0.625 minutes.

so,From 8AM of 1.1.2018 to 8AM 4.1.2018, the 3 days have been covered .

So,$$(15×3)=45minutes$$ shall count.

Now from 8AM to 10AM ,it is 2 hour.

So,the clock will gain $$(0.625×2)=1.25$$ minutes.

So,it will show 10:46 AM in the clock.

C is correct choice.