SSC Stenographer Practice Test - 6 - SSC Stenographer MCQ

The boy born in class B will marry a girl of class A. So, the daughter will belong to class A. Thus, (1) is false.
Since the boy's daughter will belong to class A, she would marry a boy of class B and so the son-in-law would be from class B. So, (2) is false.
The boy, born in class B, will have mother born in class B and father born in class A. So, his uncle (i.e. mother's brother or father's brother) can be from class A or B.
So, (3) is true.
The boy in class B gets married to a girl in class A and himself becomes of class A.
Their son also belongs to class A, who will marry a girl of class B and himself becomes of class B.
When the son divorces, he reverts to his original class, i.e. A; hence, (4) is false.

The mother of class B female belongs to class B. So, her father was born in class A. Thus, her father's mother belonged to class A, and father's father was born in class B. So, the grandfather belongs to class B and grandmother belongs to class A.

(30 - 24) x 8 = 48

and (23 - 12) x 8 = 88

Therefore, (92 - 86) x 8 = 48.

(4 x 7) % 4 = 7

and (6 x 2) % 3 = 4

Therefore, (6 x 2) % 2 = 6.

(1)² + (5)² + (4)² + (3)² = 51 x 10 = 510

and (3)² + (4)² + (6)² + (2)² = 65 x 10 = 650

Similarly (0)² + (1)² + (2)² + (8)² = 69 x 10 = 690.

= 9 / 5 × 120+ 32 = 248°F

J – This is only an individual opinion.
F – They have talked about statistics. Thus, this is a fact.
J – This is only an individual opinion.
J – This is only an individual opinion.

Local time varies only with longitudes.
The flight travelled from 10E to 70W longitudes i.e. 80° longitudes.
Time change because of longitude change = 80 × 4 = 320 minutes = 5 hours and 20 minutes
Time at 18N 10E after 10 hour journey = 2 am + 10 hours = 12 pm
Because the flight travelled towards West from East, the local time at 36N 70W
= 12 pm - 5 hrs 20 m = 6:40 am

Time from 3 pm on a day to 6 am after 4 days is 87 hours. Now, 23 hr 51 min on this clock are the same as 24 hr on the correct clock.
That is, on this clock = 24 hr on the correct clock
∴ 87 hr on this clock = hr on the correct clock
= 87 hr and 33 minutes approx. on the correct clock
So, the correct time will be 6:33 am.

The watch loses 5 minutes in 3 hours.
It loses 1 minute in (3/5) hours.
To show the correct time again, watch must create a 24-hour difference. (Since in one round, an hour-hand covers 24 hours.)
So, it loses 60 minutes in (3/5) × 60 hours or 36 hours.
It loses 24 hours in 36 × 24 hours or 864 hours or 36 days
Therefore, after exactly 36 days, the clock will show the correct time.

Hour hand:
1 day = 36 hours and the hour hand covers the dial twice every day
So, the dial covers 36/2 = 18 hours in one complete round
One complete angle in the clock dial is 360°.
So, in 1 hour, hour hand moves 360/18 = 20°
So, for 12 hours, hand moves 12 × 20 = 240°
Also, 1 hour = 90 minutes
Now, for 1 minute, hour hand moves (20/90)°
Thus, for 18 minutes, it moves 20/90 × 18 = 4°
So, for 12:18 am, hour hand moves = 240 + 4 = 244°
Minute hand:
Now, minute hand moves 90 minutes to complete one hour.
Thus, for 1 minute, it moves 360/90 = 4°
For 18 minutes, it moves 4 × 18 = 72°
So, the angle between hour hand and minute hand at 12.18 am is 244 - 72 = 172°

As North-West is 135^o clockwise from South in the same were North-East is 135^o clockwise from the West.

CEDH -- Reverse it ---> HDEC.

VRNP -- Reverse it ---> PNRV.

Therefore VRNP is the answer.

As, 123 → 13²

As, 235 → 25³

The middle digit of first term becomes power to the next term.

Sure, let's find the correct answer for the given problem.

144 : 10 :: 169 : ?

- 144 is a perfect square (12²) and its square root is 12.
- 12 is then reduced to 10 by subtracting 2 (12 - 2 = 10).
- Similarly, 169 is a perfect square (13²) and its square root is 13.
- Following the same pattern, subtract 2 from 13 (13 - 2 = 11).

Thus, the correct answer is:

B: 11

- 169 : 11

As Word is a group of letters similarly Book is a group of papers.

Second one had one more side than the first.

These are all the languages of computer.

As 'Indolence' and 'Work' are opposite to each other in the same way 'Taciturn' and 'speak' are opposite to each other.

As the young one of a 'cat' is 'kitten' in the same way the young one of 'woman' is 'baby'.

x > 1 means x is greater than 1 in this equation the values of x can be 2,3,4,5.............∞
x < -1 means is less than -1 therefore values of x can be -2, -3, -4, -5............ -∞
So no common point for x
Hence there will be no solution

To solve the inequality 3(a - 6) < 4 + a, we need to simplify and isolate the variable a on one side of the inequality. Here are the steps:
1. Distribute the 3 on the left side of the inequality: 3a - 18 < 4 + a
2. Subtract a from both sides of the inequality: 3a - a - 18 < 4
3. Combine like terms on the left side of the inequality: 2a - 18 < 4
4. Add 18 to both sides of the inequality: 2a < 22
5. Divide both sides of the inequality by 2: a < 11
Therefore, the solution to the inequality 3(a - 6) < 4 + a is a < 11. This means that a is less than 11. Therefore, the correct answer to the question is D: a < 11
 

Given inequality is, P > Q ≤ R = S ≥ T

1. Q > S → Q ≤ R = S, Q < S → False.

2. Q = T → Q ≤ R = S ≥ T, data is not sufficient to determine the relation → False.

3. R < T → R = S ≥ T, R ≥ T → False.

So, none of the options are correct.

Take x = 2, then option (1), (2) will become true and hence those are not the answers.
Take x = – 2, option (4) will be true and hence not the answer.
But for any value of x, 2x > x.
So option (3) is not true at all. It is always false.

At x = 0, inequality is satisfied.
Thus, options, (c) and (d) are rejected.
At x = 1, inequality is satisfied
Hence, we choose option (a). 

As (a - b)² is never negative.
a² + b² - 2ab is greater than or equal to zero.
Therefore, we can say that a² + b² ≥ 2ab.

m cannot be < -3 because that will result in (-ve)(-ve)/(-ve), which will be negative and thus < 0.
m cannot be -2 or -3 because that will result in 0, which is not > 0.
m cannot be -1, 0 or 1 because that will result in (+ve)(+ve)/(-ve), which is < 0.
m cannot be 2 because that will result in dividing by 0.
Thus, m can only be 3, 4, 5, 6, ......., 99.
So, required sum is