TISSNET 2022

Let the cost price of one article be C.P and selling price of one article be S.P

It is given that,

20C.P = 25S.P

Loss% = $$\ \frac{\ C.P-S.P}{C.P}\times100$$ = $$\ \ \frac{\ \frac{5}{4}S.P-S.P}{\frac{5}{4}S.P}\times\ 100\ $$ = 20%

Answer is option D.

Let the number of benches be 'b' and number of students be 's'
From first statement,
b(1) + 10 = s 
b + 10 = s ....... (i)    
From second statement,
Let us consider x benches are occupied with two students, i.e.
x(2) + (b-x)0 = s
2x = s ......... (ii)
It is given,
b - x = 10 ......... (iii)
Solving (i), (ii) and (iii), we get
x = 20, b = 30 and s = 40
students:benches = 40:30 = 4:3

Answer is option D.

 5 x 2 - 3 = 7
 7 x 2 - 3 = 11
11 x 2 - 3 = 19
19 x 2 - 3 = 35
35 x 2 - 3 = 67
67 x 2 - 3 = 131
131 x 2 - 3 = 259

Answer is option D.

The percentage of tagged tigers in second sample = $$\ \ \frac{\ 5}{50}\times\ 100\ $$ = 10%

It is mentioned that there are 150 tagged tigers and approximately 10% of all the tigers in national park are tagged.

Therefore, approximate number of tigers in national park = 10*150 = 1500

Answer is option D.

Let the incomes of two persons be 9x and 7x.
Let the expenditures of two persons be 4y and 3y
It is given that,
9x - 4y = 7x - 3y = 3000
9x - 4y = 7x - 3y
y = 2x .......... (i)
7x - 3y = 3000 ........... (ii)
Solving (i) and (ii), we get
x = 3000 and y = 6000
Difference in income = 2x = 2*3000 = Rs 6000

Answer is option C.

Let the present ages of Shibu and Sohail be 4x and 5x

It is given that,

$$\ \frac{\ 4x\ +\ 8}{5x\ +\ 8}=\frac{5}{6}$$

24x + 48 = 25x + 40

x = 8

Age difference = x = 8

Answer is option D.

Number of days in leap year = 366
There are 2 odd days in a leap year. For a year to have 53 sundays and 53 mondays, 2 odd days should be sunday and monday.
This is only possible when the first day of the year is sunday.
Therefore, the probability of the first day is sunday = $$\frac{1}{7}$$

Answer is option D.

It is mentioned that four clocks beep for every 25 minutes, 50 minutes, 100 minutes and 125 minutes.

All four clocks beep together for every LCM(25,50,100,125) minutes, i.e. 500 minutes.

500 minutes = 8 hours 20 minutes

All clocks next beep together at 5 am + 8 hours 20 minutes, i.e. 1:20 pm.

Answer is option C.

BC is ground and it is given that $$\angle\ ACB\ =30^{\circ\ }$$
Actual length of the tree = AB + AC
$$\sin30^{\circ\ }=\frac{AB}{AC}$$
AB = 20/2 = 10 cm
Length of the tree = AB + AC = 10 + 20 = 30 cm

Answer is option D.

There are even number of observations in the given data. Therefore, median is the average of X and X+2.
$$\ \frac{\ X+X+2}{2}=65$$
X + 1 = 65
X = 64

Answer is option C.