MAH MBA CET Sample Paper 2026

We know that the volume of a cylinder with radius $$r$$ and height $$h$$ is $$\pi r^2h$$

Given that the radius of the cylinder is decreased by 10%, i.e. the new radius is $$r\left(1-\frac{10}{100}\right)=0.9r$$

Also, the height is increased by 33.33%, i.e. the new height is $$h\left(1+\frac{33.33}{100}\right)=\frac{4}{3}h$$

Hence, the new volume of the cylinder is $$\pi\left(0.9r\right)^2\frac{4}{3}h=1.08\pi r^2h$$

So, the percentage change is $$\frac{\left(1.08\pi r^2h-\pi r^2h\right)}{\pi r^2h}\times100=8\%$$

Hence, the answer is 8% increase.

We see that the number 3240 can be factorised into: $$3240=2^3\times\ 3^4\times\ 5$$

We use the sum of factors formula: 
$$\frac{\left(2^{3+1}-1\right)}{2-1}\times\ \frac{\left(3^{4+1}-1\right)}{3-1}\times\ \frac{\left(5^{1+1}-1\right)}{5-1}$$

$$15\times\ 121\times6$$
10890

24 = $$\ 2^3\times3$$

We need to see the maximum power of both.

Maximum power of 2 in 124! = $$\left[\dfrac{124}{2}\right]+\left[\dfrac{124}{4}\right]+\left[\dfrac{124}{8}\right]+\left[\dfrac{124}{16}\right]+\left[\dfrac{124}{32}\right]+\left[\dfrac{124}{64}\right]$$ = 119

So, the maximum power of $$2^3$$ in 124! is $$\left[\dfrac{119}{3}\right]=39$$

Now, let's check for 3: $$\left[\dfrac{124}{3}\right]+\left[\dfrac{124}{9}\right]+\left[\dfrac{124}{27}\right]+\left[\dfrac{124}{81}\right]=41+13+4+1=59$$

Since the maximum power that can be common for both $$2^3$$ and 3 is 39, the maximum power for 24 will also be 39.

The maximum sum is possible when two numbers equals to the LCM.

So, the maximum sum = 1296+1296 = 2592.

We can see that 729 is 9 times 81, meaning that $$729=9^3=3^6$$

We want to find the minimum sum of the two numbers, in order to get the minimum sum, we would want the numbers to be as close to each other a possible, which would be when both are equal to $$3^3=27$$, giving the sum to be 54

But in that case, 729 would not be the lowest common multiple; it would be 27 itself. 

We see that since the 729 is a prime number multiplied repeatedly, the only number that can have LCM as 729 are 1 and 729 itself. 

Therefore, the minimum sum would be 730. 

Hence, Option E is the correct answer. 

When finding the last two numbers, we would want to change the number so that it has 1 as its unit digit. 

This can be done if we take $$217^{432}=\left(217^4\right)^{108}$$

Since the 7 raised to power 4 ends in 1, following the cycle of 7, 9, 3, 1,7....

To calculate the value of $$217^4$$, we would only have to consider the value of $$17^4$$, since the 200 would not impact the last two digits. 

17 squared is 289. to find the last two digits of 17^4, we only have to find the last two digits of 89 times 89, which would be 21

So we essentially have a number of the form $$\left(....21\right)^{108}$$

When the number is of the form $$\left(...a1\right)^n$$, the last two digits of the value are $$\left(a\times\ n\right)1$$

Hence, the last two digits of $$\left(...21\right)^{108}$$ would be 61

Therefore, Option B is the correct answer. 

$$47^{121}$$ can be written as $$47^{30}\times\ 47^{30}\times\ 47^{30}\times\ 47^{30}\times\ 47^{1}\times\ 47$$

We know by Fermat's theorem that

when $$a^{p-1}$$ is divide by $$p$$ where $$p$$ is prime and both $$a$$ and $$p$$ are co-prime, the remainder is 1.

Using the theorem in our expression, we get

$$47^{30}mod31=1$$

So the remainder when $$47^{121}$$ is divided by 31 is

$$1\times\ 1\times\ 1\times\ 1\times\ 1\times\ 47mod31=16$$

Hence, the answer is 16.

Seeing that 11, when divided by 13, leaves neither a remainder of 1 nor -1, we can look for a possible power of 11, which would leave a remainder of 1 or -1 when divided by 13. 
This method might be lengthy and not feasible without a calculator. 

Another property we can use here is Fermat little theorem, which states that $$\left[\frac{a^{p-1}}{p}\right]_R=1$$, where $$p$$ is a prime number. 

Hence, $$\left[\frac{\left(11^{10}\right)^{12}}{13}\right]_R=1$$

This would give the expression $$\left[\frac{11^{121}}{13}\right]_R=\left[\frac{\left(11^{10}\right)^{12}}{13}\right]_R\times\ \left[\frac{11}{13}\right]_R=1\times\ 11$$

Therefore, Option C is the correct answer. 

The Rabbit covers the distance in a geometric progression, with common ratio i.e 'r'= 1/5.

Since it travelled 12000 meters on the first day, the first term of the series is 12000

Now, we know that $$0<r<1$$ and in that case, the geometric progression is infinite. 

Hence, maximum distance it can cover = Sum of infinite GP= $$\dfrac{\ a}{1-r}$$, where a = 12000 and r = 1/5

So, combined maximum distance the rabbit can travel on all the days = $$\dfrac{12000}{1-\dfrac{1}{5}}\ =\ \dfrac{12000}{\dfrac{4}{5}}=12000\times\dfrac{5}{4}=15000$$ meters

The terms of series 1 is of the form 3n+2. The terms of series 2 is of the form 20+2m

3n+2 = 20+2m

when n = 6 and m = 1, the above equation satisfies. So, the first common term is 20.