Solution:

No. of A problems = 50
No. of B problems = 200 - 50 = 150

Let X be the time taken to solve B problems
Then the time taken to solve A problems = 2X
Maximum time available = 3 hours = 180 minutes

According to question,
150X + 50(2X) = 180
150X + 100X = 180
250X = 180min

X = 180/250
X = 0.72min

Or if we convert it into seconds,

X = 0.72*60seconds
X = 43.2sec

Thus, time taken for A problems is = 43.2*2*50 = 4320 seconds
= 4320/60sec = 72minutes

Solution:

If you are habitual of solving series problems, this might not take much of your time.
First digit = 8
Second digit = 24 (8*3)
Third digit = 12 (24/2)
Fourth digit = 36 (12*3)
Fifth digit = 18 (36/2)
Sixth digit = 52 (18*3)

Thus seventh digit will be 52/2 = 17

Solution:

We are given a fact that we have to mix equal quantities of b and c
Thus we may as well take average of them both and consider them as a single type

Let us call that type as d
The price will be 170/2 = Rs.85/Kg

Now we have to mix only a which is Rs.95/Kg and d which is Rs.85/Kg

If you compare with the target price, a is Rs.5 more than it and d is Rs.5 less than it

Thus we may mix 50% of each into the final mixture

Required a = 50 Kg
Required d = 50 Kg
D is an equal mixture of b and c thus the final mixture will have 25 kg each of b and c.

So in order to obtain the required results, we will have to mix 50 Kg of a and 25 Kg of each b and c.

Solution:

Original are of the rectangle = l*b where l= length and b = breadth
After the sides are increased by 100%, the area will increase by
= (((2l*2b)-(a*b))*100) / (a*b) = 300%

Solution:

Let the simple interest be x
According to the question
1120 - 4x = Amount ----- eq (1)
1200 - 5x = Amount ----- eq (2)
Equating eq (1) and eq (2)
x=80
Thus the Amount is 1200 - 400=800

Option b is correct

A simpler way to solve will be

Interest for one year = Interest for five years - Interest for four years
= 1200 - 1120 = Rs.80

Interest for four years = 4*80 = Rs.320
Thus sum = 1120 - 320 = Rs. 800

Solution:

If the seller is buying at Rs.12/liter and mixing it with 20% of water and selling it for Rs.15/liter, he is selling the thus formed 1.2 liters of mixture at Rs.18

Thus the percentage of his gain is 50%.

Solution:

Let the number of people required originally = X
10*X = Work --- eq (1)
(10+10)*(X-10) = Work --- eq (2)
Equating eq (1) and eq (2)
10X=200
=>X=20
Therefore originally 20 people were tasked at the work
Thus option d is the correct answer.

Solution:

In GP, a/r+a+ar....=38
a (1+r+r^2) = 38r --- eq1
Product a^3 = 1728
a = 12
Putting the value in eq1
12 (1+r+r^2) = 38r
r=2/3 and 3/2

Thus the smallest required number is 8

Solution:

Let us assume that the initial price was Rs.100
Then after reducing it 30% the price will be Rs.70

To make it back to Rs.100, the price must be increased by
100*30/70 = 300/7 % = 42.86%

Solution:

Total games won by the team = 30 percent of 60 = 18
Now, the team wins continuous matches and raises winning % to 50
Thus, ((18 + x) / (60 + x)) * 100 = 50
(18 + x) / (60 + x) = 1/2
x = 24
Thus the team must have won 24 matches to raise the winning percentage to 50 percent.