TISSNET 2023

Let the cost price of 1000 grams be Rs. 1000,

The weighing scale shows 1200 grams for 1000 grams which means he sells 1000 gms at a cost price of 1200 gms.

He will get Rs. 1200 for selling 1000 grams.

It is given that he further marks up his cost by 10%, he is selling 1000 gms at 1200*(1.1), i.e. Rs 1320

Therefore, the net profit percentage = $$\frac{1320-1000\ }{1000}\times100$$ = 32%

The answer is option B.

The value of  $$3^5+\dfrac{1}{3^5}$$

=>  $$\dfrac{3^5}{1}+\dfrac{1}{3^5}$$

LCM is $$3^5$$ 

=> $$\dfrac{\left(3^5\right)^2+1}{3^5}$$

=> $$\ \dfrac{3^{10}+1}{3^5}$$ 

The answer is option B.

We can write any fraction where the denominator is a hundred as given below

$$\ \dfrac{1}{2}$$ = $$\ \dfrac{50}{100}$$

$$\ \dfrac{1}{4}$$ = $$\ \dfrac{25}{100}$$

To find $$\left(\dfrac{1}{4}\right)^2$$

=> $$\left(\dfrac{25}{100}\right)^2$$

=> $$\ \dfrac{625}{10000}$$ 

=> 0.0625

The answer is option C.

There are 31 possible combinations where all three numbers are identical(e.g. 0-0-0, 1-1-1, 2-2-2,.......,30-30-30).

Total combinations possible for the lock when numbers are from 0 to 30 = 31*31*31

Hence, the probability of opening the lock by trying all the combinations with identical numbers is

=> possible outcomes / total number of outcomes = $$\dfrac{31}{31\times\ 31\times\ 31}$$ = $$\dfrac{1}{31\times\ 31}$$

=> $$\dfrac{1}{961}$$ 

The answer is option A.

Let the speed of the group when going up be A and therefore, the speed of the group when going down = $$\dfrac{3A}{2}$$

=> Time taken by both routes is the same, and it is given that they go up in 2 days.

=> Speed of group when going up = A = 4km/day

Therefore, speed of the group when going down = $$\dfrac{3A}{2}$$ = 6km/day

Since, the time taken is the same, time taken going down = 2 days

Hence, the length of route down the mountain = 6*2 = 12 km 

The answer is option A.

Since both tasks are independent, therefore the sum of the probability of completion of task 1 + probability of not completing of task 1 = 1

The probability that task 1 will be completed on time is 7/8, and the probability that it will not be completed on time is $$1-\dfrac{7}{8}$$, i.e. $$\dfrac{1}{8}$$

Also, the probability that task 2 will be completed on time is 3/5, and the probability that task 2 will not be completed on time is $$1-\dfrac{3}{5}$$, i.e. $$\dfrac{2}{5}$$

To find the probability that task 1 will be completed on time but task 2 will not, we can multiply these probabilities together:

Probability = (Probability of task 1 being completed on time) * (Probability of task 2 not being completed on time)

Probability = $$\left(\dfrac{7}{8}\right)\left(\dfrac{2}{5}\right)=\dfrac{7}{20}$$

Therefore, the probability that task 1 will be completed on time, but task 2 will not, is $$\dfrac{7}{20}$$

The answer is option D.

Take a closer look at the data provided in the table:

The table shows that the customer had Rs. 100 in his account initially and made a total of 6 withdrawals amounting to Rs. 100. The table did not show that the customer made any deposits or added any interest to the account during this time.

The "Debit" column shows the amount of each withdrawal, while the "Credit" column is not present, suggesting that no money was added to the account. The "Balance" column shows the remaining balance in the account after each withdrawal.

When we add up the "Debit" column, we get a total of Rs. 100, which matches the total amount of the withdrawals. However, there is no  "Credit" column, so the total credit is zero. When we subtract the total debit from the total credit, we get a balance of -Rs. 100. This means that the customer owes Rs. 100 to the bank.

However, the customer inferred that he owes only Re. 1 to the bank. This inference is incorrect, as the total balance owed to the bank is -Rs. 100, not -Rs. 1. Therefore, the customer was not right in making this inference from the data.

In this case, the radius of the circle traced by the minute hand is 1.5 cm.

Circumference = 2πr = 2π(1.5) = 3π cm

The minute hand covers this distance in 60 minutes.

Let x be the distance covered by the tip of the minute hand in 40 minutes.

$$\dfrac{x}{3\pi\ }=\dfrac{40}{60}$$

x = 2π = 6.28 cm

Therefore, the tip of the minute hand moves a distance of 6.28 cm in 40 minutes.

The answer is option A.

Closing prices for stocks on each day of the week are given
Monday - 17.94
Tuesday - 18.19
Wednesday - 18.06
Thursday - 17.54
Friday - 17.38
Given average is 18.43 and the standard deviation is 0.27.
Two standard deviations = (2)(0.27) = 0.54
Two standard deviations below the mean = 18.43 - 0.54 = 17.89
Two standard deviations above the mean = 18.43 + 0.54 = 18.97
Values should lie between 17.89 and 18.97.
Days during this week, the closing price falls within two standard deviations from the average: Monday, Tuesday, and Wednesday values are in the given range.
Therefore, the answer is 3.

It is given, Raman bought A at a price inclusive of 12% GST = Rs 1680

Let us assume the selling price of A be P then $$P\left(1+\ \frac{GST\ \%}{100}\right)$$ = 1680

=> $$P\left(1+\ \frac{12\ }{100}\right)$$ = 1680

=> $$P\ =\ 1680\times\frac{100\ }{112}$$

Therefore, P = Rs. 1500 and GST on item A = Rs. 180

Similarly, for item B

Let us assume the selling price of B be Q, then $$Q\left(1+\ \frac{GST\ \%}{100}\right)$$ = 1265

=> $$Q\left(1+\ \frac{15\ }{100}\right)$$ = 1265

=> $$Q\ =\ 1265\times\frac{100\ }{115}$$

Therefore, Q = Rs. 1100 and GST on item B = Rs. 165

Hence, the amount of GST not paid by the shopkeeper to the Government = 180+165 = Rs.345

The answer is option C.