CMAT 2024 Slot 1 Question Paper

A) $$16\frac{2}{3}$$ % = 0.1667
B) $$\frac{2}{15}$$ = 0.1333
C) 0.17 = 0.17
D) 0.25 % of 64 = 0.16

Hence, 

B < D < A < C

We can rewrite the following as:

A) $$\sqrt[4]{3}$$ = $$3^{\dfrac{1}{4}}$$

B) $$\sqrt[3]{2}$$ = $$2^{\dfrac{1}{3}}$$

C) $$\sqrt[6]{5}$$ = $$5^{\dfrac{1}{6}}$$

D) $$\sqrt[6]{7}$$ = $$7^{\dfrac{1}{6}}$$

Now, we will make the denominator of all equal:

A)$$3^{\dfrac{3}{12}}$$

B) $$2^{\dfrac{4}{12}}$$

C) $$5^{\dfrac{2}{12}}$$

D) $$7^{\dfrac{2}{12}}$$

Simplifying further, we get, 

A)$$3^{\dfrac{3}{12}}$$ = $$\left(3^3\right)^{\dfrac{1}{12}}=\left(27\right)^{\dfrac{1}{12}}$$

B) $$2^{\dfrac{4}{12}}$$ = $$\left(2^4\right)^{\dfrac{1}{12}}=\left(16\right)^{\dfrac{1}{12}}$$

C) $$5^{\dfrac{2}{12}}$$ = $$\left(5^2\right)^{\dfrac{1}{12}}=\left(25\right)^{\dfrac{1}{12}}$$

D) $$7^{\dfrac{2}{12}}$$ = $$\left(7^2\right)^{\dfrac{1}{12}}=\left(49\right)^{\dfrac{1}{12}}$$

Arranging them in sequence: We get: B < C < A < D

Let the cost price be Rs. 1000 per kg

So, Selling price = Rs. 900 per kg

But he sells only 70% of the quantity claiming it to be 100% quanitity.

So, the actual cost price of 700 grams = Rs. 700

Actual selling price = Rs. 900

Profit = Rs. 200

Profit % = (Profit/Cost Price)*100 = 200/700 = 200/7 % = $$28\dfrac{4}{7}$$%

If roots are reciprocal to each other, the product of roots will be 1. 

From the given equation, the product of roots will be $$\frac{r}{p}$$. That wil be equal to 1.

So, r = p

Statement I :

If Shyam can finish the task by working alone in 8 days, so he will do $$\dfrac{6}{8}\times100=75\%$$ os the work in 6 days.

The remaining 25% is done by Ram is 6 days.

So, if Ram does 25% of the work in 6 days, he will take $$\dfrac{100}{25}\times6=24$$ days to complete the work alone.

Hence, the given statement is true.

Statement II :

Assume 1 person does 1 man hour work in one hour

So, in one week, working 8 hours a day, 

The total work done by 6 person is $$6\times8\times7=336$$ man hours

Assume the payment to be Rs. $$x$$ per man-hour

So, $$336x=8400$$

or, $$x=25$$

The total man hours for 9 persons working 6 hours a day for a week will be

$$9\times6\times7=378$$

So, the money earned is $$378x=378\times25=9450$$

Hence, the given statement is true.

If abc is positive, it can have two cases:

Case 1: when all three integers are postive

Case 2: when any two are negative and the third is positive (as the two negative will cancel while multiplying)

We need to find the option which must be true:

Option 1: $$\dfrac{a}{b} < 0$$

It is not necessary that one of a and b is negative and the other one is positive. It can be possible that but both a and b are negative and hence the fraction can be greater than 0. Therefore, this option is not necessarily true. 

Option 2: $$\dfrac{ab}{c} > 0$$

Here, if all the three integers are positive then the expression will definitely be greater than 0. On the other hand, if any two are negative the according to the expression, the negatives will cancel out and the expression will again be greater than 0. Hence, this option must be true. 

Option 3: $$bc < 0$$

Same as option 1, it is not necessary that one of  b and c is negative and the other one is positive for the expression to be true. There can also be the possibility when both are negative and the product can be greater than 0 in that case.

Option 4: $$a > bc$$

We don't know the numeric values so we can't say that a will be greater than the product of b and c necessarily. 

The volume of the sphere = $$\dfrac{4}{3}\times\ \pi\ \times\ \left(3\right)^3=36\pi\ $$

Now, the water level will increase equal to the volume of the sphere. 

Let the water level increasing by h cm.

Therefore, $$36\pi\ =\pi\ \times\ \left(4\right)^2\times h$$

$$h=\dfrac{9}{4}$$

"OMEGA" comprises 5 letters: O, M, E, G, A.

Vowels: O, E, A (3 vowels)

Consonants: M, G (2 consonants)

Odd positions in a 5-letter word are: 1st, 3rd, and 5th.

The vowels O, E, and A can be placed in these positions in 3! = 6 ways.

The even positions are 2 and 4.

The consonants M and G can be arranged in these positions in 2! = 2 ways.

The total number of arrangements is the product of the arrangements of vowels and consonants = 3! x 2! = 6*2 = 12 ways.

Let A be his additional investement. The interest obtained on A will be 0.08A.

The interest obtained on Rs. 1,200 will be Rs. 60.

60+0.08A = 0.06(1200+A)

0.02A = 12

A = 600

Assume the length of the first train to be $$x$$ m

So, the length of the second train is $$0.5x$$ m

The total length to be covered = $$1.5x$$ m

As they are travelling in the opposite directions, their speed will add in relative frame

So, the combined speed is $$48+42\ =\ 90\ $$kmph or $$\ 90\times\dfrac{5}{18}=25$$ m/s

So, $$\dfrac{1.5x}{25}=12$$

$$1.5x=300$$

or, $$x=200$$ m

Given it passes the station in 45 sec

Let the length of the station be $$l$$ m

So, $$\dfrac{l+x}{48\times\dfrac{5}{18}}=45$$

$$l+x=600$$

$$l=400$$ m

Hence, the answer is 400 m