CMAT 4th May 2023 Slot-2

Statement I:

Bank A: r = 6.2% and compounded anually

Compound interest at the end of an year = $$P\left(1+\frac{6.2}{100}\right)-P=\frac{6.2P}{100}$$

Bank B: r = 6% and compounded semi-anually

C.I = $$P\left(1+\frac{3}{100}\right)^2-P=\frac{P\left(6.09\right)}{100}$$

Interest is less in bank B.

Therefore, statement I is correct.

Statement II:

Let the sum of money be 'S'

It is given,

$$S\left(1+\frac{r}{100}\right)^3=2S$$

$$\left(1+\frac{r}{100}\right)^3=2$$

$$\left(1+\frac{r}{100}\right)^9=4$$

$$S\left(1+\frac{r}{100}\right)^9=4S$$

Therefore, P = 4

Statement II is incorrect.

The answer is option C.

F(x) = |2x| + 4

Option A: G(x) = x + 3

If x > 0 or x = 0

2x + 4 = x + 3

x = -1 < 0

Therefore, this case is not possible.

If x < 0

-2x + 4 = x + 3

3x = 1

x = 1/3 > 0

Therefore, this case is not possible.

Option B: G(x) = 2x - 2

If x > 0 or x = 0

2x + 4 = 2x - 2

This case is not possible.

If x < 0

-2x + 4 = 2x - 2

x = 1.5 > 0

This case is not possible.

Option C: G(x) = 2x + 3

If x > 0 or x = 0

2x + 4 = 2x + 3

This case is not possible.

If x < 0

-2x + 4 = 2x + 3

x = 0.25 > 0

This case is not possible.

Option D: G(x) = 3x - 2

If x > 0 or x = 0

2x + 4 = 3x - 2

x = 6

This case is possible.

Therefore, the answer is option D.

It is given,

320SP = 400CP

4SP = 5CP

Profit percentage = $$\ \frac{\ SP-CP}{CP}\times100=\ \frac{\ \frac{5CP}{4}-CP}{CP}\times100=25\%$$

The answer is option D.

In 2020:

Cost of component A = $$\frac{90}{360}\times12000=3000$$

Cost of component B = $$\frac{120}{360}\times12000=4000$$

Cost of component C = $$\frac{150}{360}\times12000=5000$$

In 2021:

Cost of component A = 1.1*3000 = Rs 3300

Cost of component B = 1.2*4000 = Rs 4800

Cost of component C = 1.1*5000 = Rs 5500

The answer is option B.

A - 2 terms

B - 4 terms

C - 6 terms

..

2 + 4 + 6 + ..... 2n terms => this sum should be greater than or equal to 95.

$$2\left(1+2+3+...+n\right)\ge\ 95$$

$$n\left(n+1\right)\ge\ 95$$

If n = 9, we get only 90 terms.

95th term occurs when n = 10, i.e. J.

The answer is option C.

Let the amount invested be 'P' and the rate of interest be 'r'.

Simple interest =$$\ \frac{\ P\times r\times2}{100}$$, which is equal to 800

=> $$\frac{\Pr}{100}=400$$

Compound interest = $$P\left(1+\frac{r}{100}\right)^2-P\ $$ = 900

$$P\left(\frac{r}{100}\right)^2+\frac{2\Pr}{100}=900$$

$$P\left(\frac{r}{100}\right)^2=100$$

$$400\left(\frac{r}{100}\right)=100$$

r = 25%

P = $$\frac{400\left(100\right)}{25}$$ = Rs 1600

The answer is option B.

It is given, x > 0

3.6% of $$\frac{5x}{12}$$ = $$\frac{3.6}{100}\times\frac{5x}{12}=\frac{36}{1000}\times\frac{5x}{12}=\frac{3x}{200}$$

A) 3 percent of 20x = $$\frac{3\left(20x\right)}{100}=\frac{3x}{5}\ne\ \frac{3x}{200}$$

B) x percent of $$\frac{3}{2}$$ = $$\frac{x}{100}\left(\frac{3}{2}\right)=\frac{3x}{200}$$

C) 3x percent of 0.2 = $$\frac{3x}{100}\left(0.2\right)=\frac{6x}{1000}=\frac{3x}{500}\ne\ \frac{3x}{200}$$

D) 0.05 percent of 3x = $$\frac{0.05}{100}\left(3x\right)=\frac{3x\left(5\right)}{10000}=\frac{3x}{2000}\ne\ \frac{3x}{200}$$

E) $$\frac{3x}{200}$$

Only B and E are equal to the calculated value.

Therefore, the answer is option D.

Statement I:

Let the actual time taken by Ravi to reach the college be 't'.

If Ravi travels at a speed of 5 km/hr, time taken is t + $$\frac{10}{60}$$

If Ravi travels at a speed of 6 km/hr, time taken is t - $$\frac{4}{60}$$

Distance = speed*time = P = $$5\left(t+\frac{10}{60}\right)=6\left(t-\frac{4}{60}\right)$$

$$5\left(t+\frac{10}{60}\right)=6\left(t-\frac{4}{60}\right)$$

$$5t+\frac{50}{60}=6t-\frac{24}{60}$$

$$t=\frac{74}{60}$$

Distance(P) = $$5\left(\frac{74}{60}+\frac{10}{60}\right)=5\left(\frac{84}{60}\right)=7$$ km

Therefore, statement I is incorrect.

Statement II:

Let the speed of Babita be 3x.

The speed of Amit is 7x.

B covers distance y. 

In the same time, A should cover 80+y.

$$\frac{y}{3x}=\ \frac{\ 80+y}{7x}$$

7y = 240 + 3y

4y = 240

y = 60

Total distance = 80 + 60 = 140m

Therefore, statement II is correct.

The answer is option D.

It is given,

a + b + c = 96 and $$a\le\ b\le\ c$$

It is also given that age of Q is 6 years more than the age of P, i.e. b = a + 6

Minimum value c can take is equal to b, i.e. a + 6

a + a + 6 + a + 6 = 96

3a = 84

a = 28

Minimum possible value of c = 28 + 6 = 34 years

The answer is option A.

Total number of 8-letter words possible using letters S, Y, L, L, A, B, U, S is $$\dfrac{8!}{2!2!}$$

Statement I:

We can find total number of 8-letter words by considering 2 S's as a single unit and arranging 7 units, i.e. (SS), Y, L, L, A, B, U

Total number of ways = $$\dfrac{7!}{2!}$$

Required probability = $$\ \dfrac{\ \dfrac{7!}{2!}}{\dfrac{8!}{2!2!}}=\dfrac{2}{8}=\dfrac{1}{4}$$

Therefore, statement I is correct.

Statement II:

We can find total number of 8-letter words such that L's are at ends by arranging remaining 6 letters in the middle.

Total number of ways = $$\dfrac{6!}{2!}$$

Required probability = $$\ \dfrac{\ \dfrac{6!}{2!}}{\dfrac{8!}{2!2!}}=\dfrac{2}{56}=\dfrac{1}{28}$$

Therefore, statement II is correct.

The answer is option A.