CUET 2026 General Aptitude Test May 22 Shift 2 Question Paper- Download Answer Key and Solution PDF

Concept:
Speed is defined as: \[ Speed = \frac{Distance}{Time} \]
To find the ratio of speeds: \[ Ratio = \frac{Speed\ of\ car}{Speed\ of\ train} \]
Step 1: Calculate the speed of the car.
Given: \[ Distance = 105\,km \] \[ Time = 3\,hours \]
Using: \[ Speed = \frac{Distance}{Time} \]
\[ Speed of car = \frac{105}{3} \]
\[ = 35\,km/h \]
Step 2: Calculate the speed of the train.
Given: \[ Distance = 252\,km \] \[ Time = 4\,hours \]
\[ Speed of train = \frac{252}{4} \]
\[ = 63\,km/h \]
Step 3: Find the ratio of their speeds.
\[ 35 : 63 \]
Divide both terms by \(7\): \[ 5 : 9 \]
Therefore, \[ \boxed{5:9} \] Quick Tip: Remember: \[ Speed = \frac{Distance}{Time} \] For ratios: First calculate individual speeds Simplify the ratio to lowest terms

Concept:
Area of a circle: \[ A = \pi r^2 \]
For a rectangle whose sides are in ratio \(11:7\), let the sides be: \[ 11x \quad and \quad 7x \]
Then: \[ Area = 11x \times 7x \]
Step 1: Find the area of the circular garden.
Given: \[ Diameter = 140\,m \]
Hence radius: \[ r = \frac{140}{2} = 70\,m \]
Using: \[ A = \pi r^2 \]
\[ A = \frac{22}{7}\times 70 \times 70 \]
\[ A = 22 \times 10 \times 70 \]
\[ A = 15400\,m^2 \]
Step 2: Form the area equation for the rectangle.
Sides of rectangle are in ratio: \[ 11:7 \]
Let sides be: \[ 11x \quad and \quad 7x \]
Area: \[ 11x \times 7x = 77x^2 \]
Since areas are equal: \[ 77x^2 = 15400 \]
\[ x^2 = \frac{15400}{77} \]
\[ x^2 = 200 \]
\[ x = 10\sqrt{2} \]
Step 3: Calculate the perimeter of rectangle.
Perimeter: \[ P = 2(l+b) \]
\[ P = 2(11x + 7x) \]
\[ P = 2(18x) \]
\[ P = 36x \]
Substituting: \[ x = 10\sqrt{2} \]
\[ P = 36 \times 10\sqrt{2} \]
\[ P = 360\sqrt{2} \]
Therefore, \[ \boxed{360\sqrt{2}} \] Quick Tip: Remember: Area of circle: \[ \pi r^2 \] If rectangle sides are in ratio \(a:b\): \[ Sides = ax,\; bx \] Perimeter: \[ 2(l+b) \]

Concept:
Work done is calculated using: \[ Work = Efficiency \times Time \]
If two groups complete the same work in the same time, then their total efficiencies are equal.
Step 1: Find the relation between efficiency of men and women.
Given: \[ 12\ men\ complete\ work\ in\ 54\ days \]
Also: \[ 20\ women\ complete\ same\ work\ in\ 54\ days \]
Hence: \[ 12\ men = 20\ women \]
Dividing by \(4\): \[ 3\ men = 5\ women \]
Therefore: \[ 1\ man = \frac{5}{3} women \]
Step 2: Convert all workers into women-equivalent.
Given: \[ 9\ men\ and\ 12\ women \]
Since: \[ 1\ man = \frac{5}{3} women \]
\[ 9 men = 9 \times \frac{5}{3} = 15 women \]
Total equivalent women: \[ 15 + 12 = 27\ women \]
Step 3: Calculate the required number of days.
We know: \[ 20\ women\ complete\ work\ in\ 54\ days \]
Using inverse proportion: \[ Days = \frac{20 \times 54}{27} \]
\[ = 40 \]
Therefore, \[ \boxed{40 days} \] Quick Tip: Remember: Work and time are inversely proportional to efficiency If: \[ a\ men = b\ women \] then efficiencies can be converted easily More workers \(\Rightarrow\) fewer days

Concept:
For problems involving angles of elevation, use: \[ \tan\theta = \frac{Perpendicular}{Base} \]
If two objects are observed from each other’s foot, the horizontal distance between them remains the same.
Step 1: Assume the height of the hill and horizontal distance.
Let: \[ Height of hill = h \]
Let the horizontal distance between the tower and hill be: \[ x \]
Height of tower: \[ 50\,m \]
Step 2: Use angle of elevation from foot of tower to top of hill.
Given: \[ \theta = 60^\circ \]
Using: \[ \tan60^\circ = \frac{h}{x} \]
\[ \sqrt{3} = \frac{h}{x} \]
\[ h = \sqrt{3}x \quad \cdots (1) \]
Step 3: Use angle of elevation from foot of hill to top of tower.
Given: \[ \theta = 30^\circ \]
Using: \[ \tan30^\circ = \frac{50}{x} \]
\[ \frac{1}{\sqrt{3}} = \frac{50}{x} \]
\[ x = 50\sqrt{3} \]
Step 4: Substitute value of \(x\) into equation (1).
\[ h = \sqrt{3}(50\sqrt{3}) \]
\[ h = 50 \times 3 \]
\[ h = 150\,m \]
Therefore, \[ \boxed{150\,m} \] Quick Tip: Remember: \[ \tan\theta = \frac{Height}{Distance} \] \[ \tan60^\circ = \sqrt{3} \] \[ \tan30^\circ = \frac{1}{\sqrt{3}} \]

Concept:
Work done per day is called efficiency.
\[ Efficiency = \frac{Work}{Time} \]
Combined efficiency: \[ Total work done = (Combined efficiency) \times Time \]
Step 1: Find efficiency of A.
A completes: \[ \frac{1}{3} \]
of the work in \(30\) days.
So efficiency of A: \[ A = \frac{1/3}{30} \]
\[ A = \frac{1}{90} \]
Thus: \[ A's one day work = \frac{1}{90} \]
Step 2: Find efficiency of B.
B completes: \[ \frac{4}{5} \]
of the work in \(24\) days.
So efficiency of B: \[ B = \frac{4/5}{24} \]
\[ B = \frac{4}{120} \]
\[ B = \frac{1}{30} \]
Thus: \[ B's one day work = \frac{1}{30} \]
Step 3: Calculate work done by A and B together in \(20\) days.
Combined one day work: \[ \frac{1}{90} + \frac{1}{30} \]
Taking LCM: \[ = \frac{1+3}{90} \]
\[ = \frac{4}{90} = \frac{2}{45} \]
Work done in \(20\) days: \[ 20 \times \frac{2}{45} \]
\[ = \frac{40}{45} = \frac{8}{9} \]
Thus remaining work: \[ 1 - \frac{8}{9} = \frac{1}{9} \]
Step 4: Find efficiency of C.
C completes remaining: \[ \frac{1}{9} \]
work in \(8\) days.
So: \[ C = \frac{1/9}{8} \]
\[ C = \frac{1}{72} \]
Step 5: Find combined efficiency of A, B and C.
\[ \frac{1}{90} + \frac{1}{30} + \frac{1}{72} \]
LCM of \(90,30,72 = 360\)
\[ = \frac{4+12+5}{360} \]
\[ = \frac{21}{360} = \frac{7}{120} \]
Thus total one day work: \[ \frac{7}{120} \]
Required time: \[ \frac{1}{7/120} = \frac{120}{7} \approx 17.14 \]
Closest option: \[ 18 days \]
Therefore, \[ \boxed{18 days} \] Quick Tip: Remember: \[ Efficiency = \frac{Work}{Time} \] Remaining work: \[ 1\ -\ completed\ work \] Total time: \[ \frac{Total work}{Combined efficiency} \]

Concept:
Difference between Compound Interest (CI) and Simple Interest (SI) for \(2\) years is: \[ CI - SI = P\left(\frac{R}{100}\right)^2 \]
where:
\(P\) = Principal
\(R\) = Rate of interest
Step 1: Write the given values.
Principal: \[ P = 1250 \]
Rate: \[ R = 4% \]
Time: \[ T = 2 years \]
Step 2: Use the formula for difference between CI and SI.
\[ Difference = 1250\left(\frac{4}{100}\right)^2 \]
\[ = 1250 \times \frac{16}{10000} \]
\[ = 1250 \times 0.0016 \]
\[ = 2 \]
Thus, the extra amount earned in compound interest over simple interest is: \[ 2 rupees \]
Hence, if the amount had been given at simple interest, the loss would be: \[ \boxed{2} \] Quick Tip: Remember: \[ CI - SI = P\left(\frac{R}{100}\right)^2 \] for \(2\) years. This shortcut saves time in competitive exams.

Concept:
For boats and streams: \[ Upstream speed = b-s \] \[ Downstream speed = b+s \]
where:
\(b\) = speed of boat in still water
\(s\) = speed of stream
Also: \[ Time = \frac{Distance}{Speed} \]
Step 1: Write the given information.
Speed in still water: \[ b = 6 km/h \]
Let speed of stream be: \[ s km/h \]
Thus: \[ Upstream speed = 6-s \] \[ Downstream speed = 6+s \]
Step 2: Use the condition on time.
Given: \[ Time upstream = 2 \times Time downstream \]
For equal distances: \[ \frac{1}{6-s} = 2\left(\frac{1}{6+s}\right) \]
Step 3: Solve the equation.
\[ \frac{1}{6-s} = \frac{2}{6+s} \]
Cross-multiplying: \[ 6+s = 2(6-s) \]
\[ 6+s = 12-2s \]
\[ 3s = 6 \]
\[ s = 2 \]
Therefore, \[ \boxed{2 km/h} \] Quick Tip: Remember: Upstream speed: \[ b-s \] Downstream speed: \[ b+s \] For same distance: \[ Time \propto \frac{1}{Speed} \]

Concept:
Time taken: \[ Time = \frac{Distance}{Speed} \]
Total time includes:
Walking time
Rest time
Step 1: Calculate walking time for \(6\) km.
Given: \[ Speed = 8 km/h \]
Distance: \[ 6 km \]
Using: \[ Time = \frac{6}{8} \]
\[ = \frac{3}{4} hour \]
Converting into minutes: \[ \frac{3}{4}\times 60 = 45 minutes \]
Step 2: Calculate total rest time.
He rests after every kilometre.
For \(6\) km journey, rest is taken after: \[ 1,2,3,4,5 km \]
Thus total rests: \[ 5 \]
Each rest: \[ 4 minutes \]
Total rest time: \[ 5 \times 4 = 20 minutes \]
Step 3: Find total time.
\[ Total time = 45 + 20 \]
\[ = 65 minutes \]
Therefore, \[ \boxed{65 minutes} \] Quick Tip: Remember: Convert hours into minutes when needed Number of rests: \[ Distance\ in\ km\ -\ 1 \] if no rest is taken at the destination

Concept:
For a tower of height \(h\) and shadow length \(x\): \[ \tan\theta = \frac{h}{x} \]
Thus: \[ x = \frac{h}{\tan\theta} \]
As the angle of elevation decreases, the shadow becomes longer.
Step 1: Find the shadow length when angle is \(60^\circ\).
Let height of tower be: \[ h \]
Let shadow length at \(60^\circ\) be: \[ x_1 \]
Using: \[ \tan60^\circ = \frac{h}{x_1} \]
\[ \sqrt{3} = \frac{h}{x_1} \]
\[ x_1 = \frac{h}{\sqrt{3}} \]
Step 2: Find the shadow length when angle is \(30^\circ\).
Let shadow length at \(30^\circ\) be: \[ x_2 \]
Using: \[ \tan30^\circ = \frac{h}{x_2} \]
\[ \frac{1}{\sqrt{3}} = \frac{h}{x_2} \]
\[ x_2 = h\sqrt{3} \]
Step 3: Use the condition on difference of shadows.
Given: \[ x_2 - x_1 = 30 \]
Substituting values: \[ h\sqrt{3} - \frac{h}{\sqrt{3}} = 30 \]
Taking LCM: \[ \frac{3h-h}{\sqrt{3}} = 30 \]
\[ \frac{2h}{\sqrt{3}} = 30 \]
\[ 2h = 30\sqrt{3} \]
\[ h = 15\sqrt{3} \]
Therefore, \[ \boxed{15\sqrt{3} meter} \] Quick Tip: Remember: \[ \tan\theta = \frac{Height}{Shadow} \] \[ \tan60^\circ = \sqrt{3} \] \[ \tan30^\circ = \frac{1}{\sqrt{3}} \]

Concept:
Mean is given by: \[ Mean = \frac{Sum of observations}{Number of observations} \]
If one observation is wrongly recorded, first correct the total sum and then calculate the corrected mean.
Step 1: Calculate the incorrect total sum.
Given: \[ Mean = 40 \]
Number of observations: \[ n = 10 \]
Thus: \[ Incorrect sum = 40 \times 10 \]
\[ = 400 \]
Step 2: Correct the wrongly copied observation.
Actual observation: \[ 45 \]
Wrongly copied as: \[ 15 \]
Difference: \[ 45-15 = 30 \]
Correct sum: \[ 400 + 30 \]
\[ = 430 \]
Step 3: Calculate the correct mean.
\[ Correct mean = \frac{430}{10} \]
\[ = 43 \]
Therefore, \[ \boxed{43} \] Quick Tip: Remember: \[ Mean = \frac{Sum}{n} \] If an observation is misread: \[ Correct Sum = Wrong Sum + (Correct value - Wrong value) \]