CAT 2024 Slot 1 Question Paper

Let us say the capacity of the glass is X, and it is completely filled with milk, 
If two-thirds of its content is poured out and replaced with water, the remaining fraction of the milk will be one third. 
And this is said to be done three more times, that means a total of 4 times. 
So the contents of Milk initially being X,
And after 4 times the contents will be, $$X\left(1-\frac{2}{3}\right)^4=\frac{X}{81}$$

Since the total contents is X, and the milk contents is X/81, the water contents will be 80X/81. 

Ratio of milk to water will be $$\frac{X}{81}:\ \frac{80X}{81}$$

Answer is $$1:80$$

The number of apples and mangoes are in the ratio 5 : 2.
Let us write the number of apples as 5X and number of mangoes as 2X
This means oranges will be 187-7X. 

After selling the remaining fruits, 
Apples: 5X-75
Mangoes: 2X-26
Oranges: (187-7X)/2

Unsold Apples to Unsold oranges is 3:2

$$\frac{2\left(5x-75\right)}{187-7x}=\frac{3}{2}$$

$$20x-300=561-21x$$
$$41x=861$$
$$x=21$$

Total number of unsold fruits will be, 
Apples: 30
Mangoes: 16
Oranges: 20

Total is 66. 

We can use the formula for when two people moving towards each other and meeting in a straight line. $$t^2=t_1\times\ t_2$$
Where $$t$$ is time taken for them to meet each other. $$t_1$$ is the time taken by person 1 to reach the destination after meeting and 
$$t_2$$ is the time taken by person 2 to reach the destination after meeting

We are told they meet each other in 1 and a half hours, that is 90 minutes. 
And if Sunil takes X minutes to reach A, Anil will take X+75 minutes
Since it is given that, Anil reaches B exactly 1 hour 15 minutes after Sunil reaches A

$$90^2=x\left(x+75\right)$$
$$8100=x^2+75x$$
$$x^2+75x-8100=0$$
$$\frac{\left(-75\pm\sqrt{5625+32400}\right)}{2}$$
$$\frac{\left(-75+195\right)}{2}$$
$$x=60$$

Total time of travel of Anil is, $$90\ \min\ +60\ \min\ +75\ \min$$
Total of 225 minutes. 
Time in hours will be 3.75 hours. 
Speed is $$\frac{40}{3.75}=12\ \frac{km}{hr}$$

Let us assume the four numbers to be a, b, c and d in ascending order. 

Average of first two numbers is 1 more than the first number
$$\frac{\left(a+b\right)}{2}=a+1$$
$$b-a=2$$
$$b=a+2$$

Average of first three numbers is 2 more than average of first two numbers
$$\frac{\left(a+b+c\right)}{3}=\frac{\left(a+b\right)}{2}+2$$
$$2c=a+b+12$$
Substituting the value for b
$$2c=a+a+2+12$$
$$2c=2a+14$$
$$c=a+7$$

Average of first four numbers is 3 more than average of first three numbers.
$$\frac{\left(a+b+c+d\right)}{4}=\frac{\left(a+b+c\right)}{3}+3$$
$$3d=a+b+c+36$$
Substituting the value of b and c
$$3d=a+a+2+a+7+36$$
$$3d=3a+45$$
$$d=a+15$$

d is the largest and a is the smallest and we know that d=a+15

Hence the difference between the smallest and the largest values is 15. 

We are told that, 10000 is deposited in bank A for a certain number of years at a simple interest of 5% per annum. 
Let us say that the number of years is x
Total value of the deposit after x years is, $$10000\left(1+x\left(0.05\right)\right)$$

On maturity, the total amount received is deposited in bank B for another 5 years at a simple interest of 6% per annum
Here we know the years and the interest rate, 
$$10000\left(1+x\left(0.05\right)\right)\left(1+5\left(0.06\right)\right)$$
$$10000\left(1+\left(0.05\right)x\right)\left(1.3\right)$$

Interest received from Bank A is $$\left(x\left(0.05\right)\right)10000$$
Interest received from Bank B is $$0.3\left(10000\left(1+x\left(0.05\right)\right)\right)$$

This ratio is given to be 10:13. 

$$\dfrac{x\left(0.05\right)}{0.3\left(1+x\left(0.05\right)\right)}=\dfrac{10}{13}$$

$$0.65x=3+0.15x$$
$$0.5x=3$$
$$x=6$$

Hence the number of years the money was invested in Bank A is 6 years. 

Let us say the quantity of grains is X
For the first customer he sells $$\frac{X}{2}+3$$
Remaining is $$\frac{X}{2}-3$$

Second customer he sells: $$\frac{X}{4}-\frac{3}{2}+3=\frac{X}{4}+\frac{3}{2}$$
Remaining will be $$\frac{X}{4}-\frac{9}{2}$$

Third customer he sells: $$\frac{X}{8}-\frac{9}{4}+3$$ = $$\frac{X}{8}+\frac{3}{4}$$
Remaining will be $$\frac{X}{8}-\frac{21}{4}$$
Now, this is said to be 0,
$$\frac{X}{8}-\frac{21}{4}=0$$

X=42

$$(a + b \sqrt{n})$$ is the positive square root of $$(29 - 12\sqrt{5})$$

So $$29-12\sqrt{5}=\left(a+b\sqrt{n}\right)^2$$
$$29-12\sqrt{5}=a^2+b^2n+2ab\sqrt{n}$$

$$a^2+b^2n=29$$ and
$$ab\sqrt{n}=-6\sqrt{5}$$
$$a^2b^2n=180$$
$$b^2n=\frac{180}{a^2}$$
Substituting this in the above equation, 
$$a^2+\frac{180}{a^2}=29$$
$$a^4-29a^2+180=0$$
$$a^2=\frac{\left(29\pm\sqrt{29^2-4\left(180\right)}\right)}{2}$$
$$a^2=\frac{\left(29+\sqrt{841-720}\right)}{2}$$
$$a^2=9\ or\ 20$$

That means, one of $$a^2\ or\ b^2n$$ is 9 and 20. 

We also have, $$ab\sqrt{n}=-6\sqrt{5}$$ that means one of a or b should be negative
And also the fact that this is a positive square root,
And we need to maximise the value of a, b and n. 

We can have a=-3, b=1 and n=20. 
This satisfies all the above equations, and the value of a+b+n=18. 

Given that, The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm
So, $$2(lb+bh+hl)=846$$. 

And $$4(l+b+h)=144$$
$$(l+b+h)=36$$
$$\left(l+b+h\right)^2=l^2+b^2+h^2+2\left(lb+bh+hl\right)$$
$$1296=\left(l^2+b^2+h^2\right)+846$$
$$450=l^2+b^2+h^2$$

We are told that this cuboid is inscribed in a sphere, the body diagonal of the cuboid equals the diameter of the sphere, this can be visualised as:

This is nothing but, $$\sqrt{l^2+b^2+h^2}=2R$$
$$l^2+b^2+h^2=4R^2$$
$$450=4R^2$$
$$R^2=\frac{225}{2}$$
$$R=\frac{15}{\sqrt{2}}$$

Volume of sphere will be $$\dfrac{4}{3}\times\ \pi\ \times\ \left(\dfrac{15}{\sqrt{2}}\right)^3$$

$$\dfrac{4}{3}\pi\ \left(\dfrac{3375}{2\sqrt{\ 2}}\right)$$

$$\pi\ \times\ 1125\sqrt{\ 2}$$