CMAT 4th May 2023 Slot-1

Statement I:

It is given, profit percentage = 20%

This implies, $$SP=\frac{120}{100}CP$$

It is given, $$\frac{120}{100}CP-CP=240$$

$$\frac{20}{100}CP=240$$

$$CP=1200$$

SP = 1200+240 = Rs 1440

Therefore, statement I is correct.

Statement II:

It is given,

10CP = 7SP

Profit % = $$\ \frac{\ SP-CP}{CP}\times100=\frac{300}{7}=42\frac{6}{7}\%$$

Therefore, statement II is correct.

The answer is option A.

It is given,

$$\ \ \frac{\ P\times8\times13}{100}=6500$$

P = Rs 6250

Amount after 2 years = $$6250\left(1+\frac{8}{100}\right)^2=6250\times\frac{108}{100}\times\frac{108}{100}=7290$$

Compound interest = 7290 - 6250 = Rs 1040

The answer is option A.

Statement A:

x : y = 3 : 1

$$x^3-y^3=\left(3k\right)^3-k^3=26k^3$$

This implies value of $$x^3-y^3$$ cannot be determined from the given information.

Therefore, statement A is incorrect.

Statement B: 

It is given,

x = y + 12 and x:y = 3:2

Let x = 3k and y = 2k

3k = 2k + 12

k = 12

x = 36 and y = 24

It is given, z : y = 1 : 3

z = $$\frac{24}{3}=8$$

z + x = 36 + 8 = 44

Therefore, statement B is correct.

Statement C:

It is given, 3x = 8y and 5y = 9z

x : y = 8 : 3 = 24 : 9

y : z = 9 : 5

x : y : z = 24 : 9 : 5

x : z = 24 : 5 = 72 : 15

Therefore, statement C is correct.

The answer is option C.

It is given,

$$\ \frac{\ a+b+c}{3}=11$$

a + b + c = 33 ...... (1)

$$\ \frac{\ c+d+e}{3}=17$$

c + d + e = 51 ...... (2)

$$\ \frac{\ e+f}{2}=22$$

e + f = 44 ...... (3)

$$\ \frac{\ e+c}{2}=17$$

e + c = 34 ...... (4)

(1) + (2) + (3) - (4) we get the sum of a, b, c, d, e and f

a + b + c + d + e + f = (a+b+c) + (c+d+e) + (e+f) - (e+c) = 33 + 51 + 44 - 34 = 94

Average = $$\frac{94}{6}=\frac{47}{3}=15\frac{2}{3}$$

The answer is option A.

In the question, it is given that area of triangle ABC is 2 times the area of triangle ABD.

$$\frac{1}{2}x\left(y+z\right)=2\left(\frac{1}{2}xy\right)$$

y + z = 2y

y = z

Triangles ABC and ABD are right-angled triangles.

$$W^2=x^2+4y^2$$

$$W^2=x^2+y^2+3y^2$$

$$AD^2=x^2+y^2$$

$$AD^2=W^2-3y^2$$

$$AD=\sqrt{\ W^2-3y^2}$$

The answer is option C.

$$(x+2)(x+5) > 40$$

$$x^2+7x+10>40$$

$$x^2+7x-30>0$$

$$\left(x+10\right)\left(x-3\right)>0$$

Range of x is $$\left(-\infty\ ,-10\right)U\left(3,\infty\right)$$

As none of the given options is same as above, this question is removed from official CMAT exam.

Let the number of square tiles be n and the area of each square tile be $$x^2$$

Length = 1591 cm

Breadth = 946 cm

x is the H.C.F of 1591 and 946.

$$1591$$ = $$43\times37$$

$$946$$ = $$43\times2\times11$$

x = 43

n(43)(43) = (1591)(946)

n = 37*22 = 814

The answer is option A.

Let the speed of bus be b km/hr and the speed of taxi be t km/hr.

It is given,

$$\frac{300}{b}+\frac{200}{t}=\frac{11}{2}$$ ...... (1)

$$\frac{260}{b}+\frac{240}{t}=\frac{56}{10}$$ ...... (2)

(1)*6 => $$\frac{1800}{b}+\frac{1200}{t}=\frac{66}{2}$$ ...... (3)

(2)*5 => $$\frac{1300}{b}+\frac{1200}{t}=\frac{56}{2}$$ ...... (4)

(3)-(4), we get

$$\frac{500}{b}=5$$

$$b=100$$

The answer is option A.

Statement 1:

k$$(x-2)(x-3)$$ = k($$x^2-5x+6$$) is the quadratic equation whose roots are 2 and 3.

Therefore, statement 1 is incorrect.

Statement 2:

Given equation: $$4x^{2}+3kx + 9 = 0$$

Roots of the equation are real and distinct when the discriminant of equation is greater than zero.

$$\left(3k\right)^2-\left(4\right)\left(4\right)\left(9\right)>0$$

$$\left(k-4\right)\left(k+4\right)>0$$

k < -4 and k > 4

But in the given statement, equal to sign is also considered which is incorrect.

If discriminant is equal to zero, roots of the equation are equal.

Therefore, statement 2 is incorrect.

The answer is option B.

Statement I:

$$\left(243\right)^{0.16}\times\left(9\right)^{0.1}=3^{0.8}\times3^{0.2}=3$$

Therefore, statement I is incorrect.

Statement II:

$$3.105\times10^P=0.00239+0.000715$$

$$3.105\times10^P=0.003105$$

$$3105\times10^P=3.105$$

P = -3

Therefore, statement II is correct.

The answer is option D.