DSSSB TGT Social Science Mock Test - 1 - DSSSB TGT/PGT/PRT MCQ

a) In oxidation, oxygen is added to a substance :This is correct. Oxidation can involve the addition of oxygen to a substance or the loss of electrons.

b) In reduction, hydrogen is added to a substance :This is correct. Reduction can involve the addition of hydrogen to a substance or the gain of electrons.

c) Oxidizing agent is oxidized :This is incorrect. The oxidizing agent is reduced because it accepts electrons from another substance during the oxidation-reduction process.

d) Reducing agent is oxidized :This is correct. The reducing agent is oxidized because it donates electrons to another substance during the reaction.

From the image:

• Some actors are dancers. Yes
• No singer is actor. No

Only (1) conclusion follows.

All except Vapour are forms of precipitation

As 335 - 216 = 119

Similarly, 987 - X = 119

Therefore, X = 987 - 119 = 868.

Neither (1) nor (2) follows.

Q part of the figure represents those girls who are players but not coach.

The given numbers are squares of consecutive natural numbers in decreasing order starting with 15,
i.e., the numbers 225, 196, 169, ____, 100, 81 can be written as 15², 14², 13² , ____, 11² , 10² , 9².

Hence, the missing number is 12² = 144.

Clearly, there are two possible movements of Anuj as shown below.

The first, second, third, fourth, fifth and sixth letters of th word are respectively moved two, three, four, five, six and seven steps forward to obtain the corresponding letters of the code.

Each letter in the word is moved one step forward to obtain the corresponding letter of the code.

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 * 3) = Rs. 117.

 Principal = Rs. (815 - 117) = Rs. 698.

 Total votes = 12000.
 Total Valid votes = 85% of 12000 = 10200.
 Walter gets 80% of 10200 votes = 8160 votes.
 Jesse Pinkman would get 10200 – 8160 = 2040 Votes. 

Simple interest = P * r * t /100, where, P is the Principal, r is the rate of interest and t is the time period such that the rate of interest and the time period have mutually compatible units like r % per annum and t years.

Here, in the first case, let the Principal be Rs.P and rate of interest be r % per annum. Time period is given as 8 years . Interest is given as 80 P / 100
So, 80 P / 100 = P * r * 8 / 100
Or, 80 = 8 r
So, r = 10 % per annum.

Now, in the second case, Principal is given as Rs. 14,000, rate of compound interest is 10 % per annum as determined above; and t is 3 years.
Compound interest = P * ( 1 + r/100)t - P 
= 14,000 * ( 1+ 10 /100)3 - 14,000
= 14,000 * ( 1 + 0.1)3 - 14,000
= 14,000 *(1.1)3 - 14,000
(14,000 * 1.331) - 14,000
= 18,634 - 14,000 = 4,634
So, the compound interest on Rs.14,000 at the same rate of interest as in case 1 in 3 years will be Rs. 4,634.

Correct Answer :- d

Explanation : A and B do (1/8)th of the work in a day.

B and C do (1/12)th of the work in a day.

A and C do (1/16)th of the work in a day.

2A, 2B and 2C do (1/8)+(1/12)+(1/16) = (18/144)+(12/144)+(9/144) = (39/144) or (13/48)th of the work in a day.

Or A, B and C do (13/96)th of the work in a day.

So A, B and C together will complete the whole work in 96/13

Average annual exports of Company Z during the given period
= 1/7 x (60 + 90 + 120 + 90 + 60 + 80 + 100)
= Rs. ( 600 /7)crores
= Rs. 85.71 crores.
From the analysis of graph the exports of Company Z are more than the average annual exports of Company Z (i.e., Rs. 85.71 crores) during the years 1994, 1995, 1996 and 1999, i.e., during 4 of the given years.

To calculate the answer we need to get 1 man per day work and 1 woman per day work.
Let 1 man 1 day work =x
and 1 woman 1 days work = y. 
=> 6x+5y = 1/6 
and 3x+4y = 1/10 
On solving, we get x = 1/54 and y = 1/90 
(9 men + 15 women)'s 1 days work = 
(9/54) + (15/90) = 1/3 
9 men and 15 women will finish the work in 3 days

The correct option is C.
To make a profit of 20%, he has to sell the article for Rs. 720.
When he sold it for 480 he had lost 20%. So 80% of the actual price is 480. Hence the actual price of the article is (480×100÷80) = 600. To make a profit of 20% we have to calculate 120×600÷100=720.

Total weight increased = (8 x 2.5) kg = 20 kg.

Weight of new person = (65 + 20) kg = 85 kg.

So, the correct option is 'C'

Required time = y(1-(t/x)) = 27(1-(8/24))= 18 minutes

38% of 1000 = 380
Valid votes of Female = 100
Valid votes of Male = 380 – 100 = 280
=> 280:100
=> 14:5

Explanation:
Profit(Calculated on SP) = 20% of 1800 = 360
Profit(calculated on CP)
x + x/5 = 1800
x = 1500
Profit = 300
Difference = 360 – 300 = 60

(√7744 * 66) = 5808;

5808/352 = 16.5

So option D is correct

(√3 + 1)² = ? + √12
3 + 2√3 + 1 = ? + 2√3 = 4

Distance travelled by them in first hour = 12 km
Distance travelled by them in second hour = 13 km and so on
In 9 hours both will cover exactly 144 km.
In 9 hours each will cover half the total distance.

L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.

A. 60%
Explanation: Men = 600x Women = 400x Total engineers = 480x Male engineers = 480x * 0.66 = 320x Women who are Engineers = 160x Women who are not Engineers = 400x – 160x = 240x Required percentage = 240/400 * 100 = 60%

► Expenses for 120 boys = 8400
► Expenses for 150 boys = 10000.
Thus, variable expenses are 1600 for 30 boys.
► If we add 180 more boys to make it 330 boys,
we will get an additional expense of 1600 * 6 = 9600.
► Total expenses are Rs 19600.