SSC Stenographer Practice Test - 3 - SSC Stenographer MCQ

Tennis is to be shown on friday only. This implies that we can not show football a day after i.e. on Saturday as swimming is to be shown before football.
Moreover as it is given that we can not show a football a day before friday i.e. Thursday. Thus we can not even place it on wednesday as if we will do so then we will not be able to place kho-kho and cricket together as no consecutive days will be left with us.
Also football can not be shown on the first day i.e. Monday because swimming is to be shown before football.
Therefore to maintain a proper order we can place football only on Tuesday.

It is clear from the table.

it is clear from the table. Tennis is played on friday

(5 + 3)/2 = 4

and (6 + 8)/2 = 7

Therefore (8 + 4)/2 = 6.

20 + 273.15 < K < 25 + 273.15

⇒ 293.15 < K < 298.15

J – A tutorial being the best cannot be independently verified as best is a subjective term. It may have different meanings for different people.
I – The first half is a fact, whereas the second half is an inference derived from the fact.
F – This is a verifiable fact.
I – The first half is a fact from which the second half of the statement is inferred.

Total holidays are 6 and the number of days in January is 31. Number of working days = 31 - 6 = 25. Thus, this option is correct.

At 5 o'clock, the hands are 25 minute space apart.
To be at right angle, that too between 5:30 and 6:00, the minute hand has to gain (25 + 15) or 40 minute space.
55 minute space is gained in 60 minutes.
40 minute space is gained in 40 ×
Therefore, required time =  

To be together between 9 and 10 o'clock, the minute hand has to gain 45 minute spaces.
55 minute space is gained in 60 minutes.
45 minute space will be gained in 
At 49minutes past 9, the hands of the clock will be together.

As there is 1 odd day in 1900 years
In 49 years from 1900 to 1949, there are 49 ÷ 4 = 12 leap years and 37 ordinary years. There are (12 × 2 = 24) + 37 = 61 odd days or 5 odd days
From 1st January, 1950 till 26th January, 1950, there are 5 odd days
So, in total there are 0 + 1 + 5 + 5 = 11 odd days or 4 odd days
Hence, From 1st January, 1600 till 26th January, 1950, there are 4 odd days
So 26th January 1950 is Thursday.

As on melting, liquid is formed similarly on freezing solid is formed.

A Priest wears cassock (a full-length garment worn by certain Christian clergy, members of church choirs, and others having an office or role in a church.) while Graduate wears gown.

As Peacock is the national bird of India, similarly Bear is the national animal of Russia.

As 335 - 216 = 119

Similarly, 987 - X = 119

Therefore, X = 987 - 119 = 868.

27 → 3³; 125 → 5³; 64 → 4³

Therefore ? = 6³ = 216.

As Wick is a part of Candle similarly Wheel is a part of Bicycle.

As all terms given in the question are medical terms and Hematology is also medical term.

Carpenter, Plumber, Electrician and Blacksmith are in same working category.

'Jade' is a 'green' precious stone in the same way 'Garnet' is a 'red' precious stone.

The given inequality is 5x– 3 < 7
=> 5x – 3 + 3 < 7 + 3                             [3 is added both sides]
=> 5x < 10
=> x < 10/5
=> x < 2
When x is a real number, the solutions of the given inequality are given by x < 2, i.e., all real numbers x which are less than 2.

Statements:

U > H = G ≥ N; E < N > W ≥ O

Conclusion:

I. G > E → True → According to the statement G ≥ N > E

II. H > N → False → According to the statement U > H = G ≥ N

Hence, Option (b) is correct.
Mistake Points:
As from the equation H = G ≥ N.
There is a difference in (≥, >) both the sign.

When H ≥ N means H is greater or equal to N.

When H > N means H is greater than N.

As in conclusion 2, it is said H = G ≥ N from this we cannot say H is greater than N.

Thus conclusion 2 is false.

Given: A > B, B > C and C > D

After combining the operators,

we get: A > B > C > D

Now, checking the options:

Option A: A > D → True (As, A > B > C > D gives A > D)

Option B: A > C → True (As, A > B > C > D gives A > C)

Option C: D > A → False (As, A > B > C > D gives A > D)

Option D: B > D → True (As, A > B > C > D gives B > D)

So, Option C, D > A is definitely wrong.

Hence, the correct answer is "Option c".

At x = 0, inequality is not satisfied. Thus, option (c) is rejected. Also x = 0 is not a solution of the equation. Since, this is a continuous function, the solution cannot start from 0. Thus options (a) and (b) are not right. Further, we see that the given function is quadratic with real roots. Hence, option (d) is also rejected.

At x  = 0 inequality is satisfied, option (b) is rejected.
At x  = 2, inequality is satisfied, option (c) is rejected.
At x = 5, LHS = RHS.
At x = –1, LHS = RHS.
Thus, option (d) is correct. 

So, we can't find the value of N 

So, the answer can't be determined.