What Might Be Next In The cuet general test mock test

What Might Be Next In The cuet general test mock test

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The CUET General Test is important for admission to numerous prominent university courses. With about 50 questions to solve in just an hour, speed and precision are key. Quantitative aptitude, Reasoning, and GK are the main sections, with Quant using up the biggest share.

Quant questions can be time-consuming, so being quick and accurate is important. Preferably, you ought to intend to solve around 1 question per minute. However, this can be challenging for students used to the more unwinded pace of school examinations, where they normally have around 3 minutes per question.

Imagine all of a sudden being thrust into a high-pressure environment where quick problem-solving is needed. Without practice and understanding shortcuts, it's easy to battle.

Recognizing the prospective mistakes, ExamDeewana uses free shortcuts for solving the most typical kinds of questions in the General Test. Let's get started and cut straight to the chase!

How to solve Questions on Solutions (General Test CUET 2022/2023).

Q) A jar contains a mixture of 2 liquids A and B in the ratio 4:1. When 10 litres of the mixture is replaced with liquid B, the ratio becomes 2:3. The volume of liquid A present in the jar previously was?

Solving such questions can be difficult in a timed environment if you have not practiced before. The key to solving such questions is to discover one liquid and develop an equation for it. For creating an equation use the below formula,.

M1V1 + M2V2 = M(V1+V2)
M1 = Concentration % of liquid in first mixture
V1 = Volume of first mixture
M2 = Concentration % of liquid in second mixture
V2 = Volume of second mixture
M = concentration % of liquid after mixing

Let’s solve the above question using this,
Taking B liquid as base,

First see which 2 liquids are getting mixed. We had a liquid with initial volume as V. This liquid’s (V-10) volume is getting mixed with pure 10l of B.

Concentration of B in liquid first jar = 4:1 so 1/5 or 0.2
Now, 10 l of this same liquid was replaced by 10 pure l of B
Concentration of B in the liquid that was added = 1 (Because the liquid added is pure B)
New concentration of B = 2:3 = 3/5 = 0.6

0.2 *(v-10) + 1*10 = 0.6* (v-10+10)
0.2v – 2 cuet general test mock test + 10 = 0.6v
8 = 0.4v
So, v = 20
This means initial volume was 20 l
So, volume of A earlier was 20*4/5 = 16

The formula works even if you change the base. Let’s solve the same question using liquid A as base,
Again, the liquids mixed are, mixture 1 with 10l replaced with pure B of 10l
M1 = 4/5 = 0.8
V1 = V-10 (As 10 l is replaced from initial volume)
M2 = 0 (As the mixed volume doesn’t have any A liquid in it)
V2 = 10
M = 2/5 = 0.4

0.8(v-10) + 0*10 = 0.4*(v)
0.8v – 8 = 0.4v
v= 20
Concentration of v in original liquid = 4/5 * 20 = 16

See, this formula works like magic! Can be pretty handy in CUET

Let’s solve another question and see similar application of this formula

A solution of 60 litre of acid and water contains 65% acid. How much water must be added to get a solution of 52% acid?

The 2 liquids merged are a solution with 65% acid and 35% water and some more pure water. The resultant is 52% acid.

With acid as base,
M1 = 65% or 0.65
V1 = 60
M2 = 0
V2 = x
M = 52% or 0.52

0.65*60 + 0*x = 0.52(60+x)
39 = 31.2 + 0.52x
X = 15l
I am sure you can solve the question yourself taking water as base now ????

Type II
Questions related to work. Let’s see how to solve such questions through an example,
In 8 days 50 men can do a job working 10 hr per day. If 20 boys and 15 girls work 15 hr per day, in how many days can the job be completed. Given boys and girls work equally hard and each of them works half as hard as a man

The key to solving such questions lies with using unit work/hr. Let one man do m work/hr, one boy do b work/hr and one girl do g work/hour. You must be wondering why take work/hour and not work/day or /week. If you see closely, the question itself talks about work hours in a day therefore taking unit as work/hour is the logical choice. If the lowest unit of time had been days I would have taken my variable as work/day. Always take the lowest unit of time as per question.

In 8 days 50 men working 10 hours will do
50*10m*8 = 4000m work
By question we can say, this is also the total work

Now 20 boys and 15 girls would give total work in a day as,
20b + 15g
Their 15 hour work would be
300b + 225g
As boy and girl are equally good and hard working it means b=g
Total work in a day by boy and girl
= 300*b + 225*b = 525b

Now lets say that boys and girls together work for y days to complete the total work. So,
525b*y = 4000m
Y = 4000m/525b
Y= 160/21 (m/b)

The question also tells that m/b =2 (Each works half as hard as a man)
So, y = 320/21 = 15.24
Therefore, they have to work till 15.24 days

For more such amazing questions and tricks do visit our site ExamDeewana.com. You can solve many such wonderful questions and can discuss their answers in detail on our community page

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