MAH MBA CET Sample Paper 2026

Let the principal be Rs. X.

We have: $$8640=X\left(1.2\right)^3$$

Or, $$8640=1.728X$$

We get X = Rs. 5000

If Rs. 5000 is invested for 2 years at the rate of 40%,

Amount = $$5,000\times\ \left(1.4\right)^2$$

Rs. 9800

Let the total weight of her chocolate bar is 120X.

She distributed 60% of it among her cousins I.e. 60% of 120X = 72X

So, she received 120X - 72X = 48X

Now, if she distributed in equal ratio, then each one would have received 40X and Anisha would have received 150 grams less chocolate.

Or, 48X - 40X = 8X

8X = 160 grams

X = 20 grams

Hence, total weight = 120*20 = 2.4 Kg

Let's take the sock price on Monday start to be 100

Using the multiplication factor for all the increases and decreases through the week, we get:

$$100\times\ \frac{4}{5}\times\ \frac{1}{2}\times\ 0.97\times\ \frac{3}{2}\times\ \frac{6}{5}$$

$$97\times\ \frac{18}{25}$$

This would be less than 97. 

Hence, Option A would be the correct answer. 

Let the two numbers be Y and Z, such that

18% of Y = 45% of Z

Then, 63% of Y = ? % of Z

18% of Y = 45% Z

63% of Y = ? % of Z

∴ 18 x ? = 45 x 63

∴ ? = 45 x 63 = 157.5%

So answer is 157.5% of Z

We know a quadratic $$ax^2+bx+c$$ attains its minimum value at $$x=-\dfrac{b}{2a}$$

For the given quadratic $$x^2-10x+21$$

$$-\dfrac{b}{2a}=-\left(\dfrac{-10}{2}\right)=5$$

So, $$x=5$$ is the root of the quadratic equation $$x^2+kx-45=0$$

Putting $$x=5$$ we get

$$5^2+5k-45=0$$

$$5k=20$$

$$k=4$$

Hence, the answer is 4.

We see that this equation does not have rational roots, 
We will use the identity, 
$$a^2+b^2=\left(a+b\right)^2-2ab$$

We know that the sum of the roots of the equation is 13
And the product of the roots of the equation is 9

So, substituting these values we get, $$13^2-2\left(9\right)=169-18=151$$

The bottom part of the pool has a length of 12 m and the width of 6 m . So the area of the bottom ground = 12*6 = 72 m square.

Now, the pool has 2 opposite rectangular walls with the length of 12 m and the depth/width = 2 m so the area = 2*12*2 = 48 m square.

Further, it has another 2 opposite rectangular walls with length of 6 m and depth/width of 2 m so the area = 2*6*2 = 24 m square.

So, the total area to be painted = 72 + 48 + 24 = 144 m square.

Cost of painting = 144*25 = Rs. 3600

Number of diagonals can be found by using: $$^nC_2-n$$
$$\frac{n\left(n-1\right)}{2}-n=35$$
$$n^2-3n-70=0$$
$$\left(n-10\right)\left(n+7\right)$$

N cannot be negative so n=10

Internal angle can be calculated using, 
$$\frac{180\left(n-2\right)}{n}=\frac{180\left(8\right)}{10}=144$$

Hence the answer is 144. 

The area of a triangle formed by the medians of an equilateral triangle would be (3/4) of the area of the original triangle. 

The area of the original equilateral triangle can be calculated using the formula $$\frac{\sqrt{\ 3}}{4}a^2$$, where $$a$$ is the side length, giving the area of the original equilateral triangle to be $$\frac{\sqrt{\ 3}}{4}\times\ 12\times\ 12$$

This would give the area of the triangle formed by the medians to be $$\frac{\sqrt{\ 3}}{4}\times\ 12\times\ 12\times\ \frac{3}{4}=27\sqrt{\ 3}cm^2$$

Therefore, Option C is the correct answer. 

Let the common factor be K

∴ Length = 4K ; Breadth = 5K and Height = 7K

Tip:

Total Surface area of cuboid = 2(LB + BH + LH)

L = Length; B = Breadth/width; H = Height

Whole surface area of the rectangular plank = 2(4K x 5K + 5K x 7K + 7K x 4K)

∴ 20086 = 83*2*K^2

∴ K = 11

∴ Length = 4K = 44 sq.m