CMAT 2018 Slot 1 - Quantitative Aptitude

Given AB + C = D................................ (1)
If B = 6 and D = 30, then from (1),
6A + C = 30........................ (2)
And If B = 8 and D = 36,
8A + C = 36........................ (3)
(3) - (2) => 2A = 6
=> A= 3
Substituting the value of A in equation (2) we get, 18 + C = 30
=> C = 12.

y² + 3y − 18 ≥ 0
⇒ y² + 6y − 3y − 18 ≥ 0
⇒ y(y + 6) − 3(y + 6) ≥ 0
⇒ (y − 3)(y + 6) ≥ 0
⇒ y ≥ 3 and y ≤ −6

Since balls of same color should come together, let us consider 3 red balls as one unit and 2 blue balls as one unit and 4 yellow balls as another unit.
So, we have a total of 3 different units which can be arranged in '3!' ways.
These 3 red balls can internally be arranged in '3!' ways.
Similarly the blue balls and yellow balls can be arranged internally in '2!' and '4!' ways respectively.
So, total number ways = 3! × 3! × 2! × 4!
= 6 × 6 × 2 × 24 =1728 ways

By carefully observing the given data and by observing the given options, The charges will be least for the parcel from Kolkata to Mumbai which is Rs. 7

Given X, Y and Z are in the ratio 12: 15: 25
Let say X = 12k, Y = 15k and Z = 25k, where k is any constant.
Difference between Y and X = 15k - 12k = 3k.............................. (1)
And Difference between Z and Y = 25k - 15k = 10k..................... (2)
Ratio of (1) and (2)
= 3k ÷ 10k = 3:10.

Given speed of Ankush = 3 rounds per hour = 1 round per 20 min
Whereas speed of Babulal = 5 rounds per hour = 1 round per 12 min.
Let us consider Ankush as stationery at the starting point.
The relative speed of Babulal with respect to Ankush = 3 rounds per hour + 5 rounds per hour = 8 rounds per hour
This implies, In one hour Babulal crosses Ankush 8 times.
So, starting from 9.00 am to 10.30 am i.e., 1.5 hours or 90 min, Babulal crosses Ankush 12 times.

Given monthly savings of Amit and Bharat are in the ration 37:18. Let the savings of Amit and Bharat be 37k and 18k, where 'k' is a constant.
=> Total monthly savings of Amit and Bharat = 55k.
Given Total annual savings of Amit and Bharat = 1,32,000.
=> Total monthly savings = 11,000.
=> 55k = 11000
=> k = 200.
Hence monthly savings of Amit and Bharat are 7400 and 3600 respectively.
From the given data,
Let the monthly incomes of Amit and Bharat be 5x and 4x respectively, where 'x' is a constant.
Similalrly, let the monthly expenditures of Amit and Bharat be 19y and 21y respectively, where 'y' is a
constant.
Savings = Income - Expenditure
=> 5x - 19y = 7400.......................... (1)
And 4x - 21y = 3600............................... (2)
Solving both the equations we get x = 3000 and y = 400.
Therefore, the monthly income of Amit is 5x i.e., 15000.

As can be seen from the above venn diagram distribution

Number of households reading only one newspaper = 22 + 26 + 16 = 64.

Hence, option C.

Let say f(x) = x³ + 2x² + bx + 68 and given (x + 4) is a factor of f(x).
=> f(x) = (x + 4) × k…………. (1)
Where ‘k’ is the quotient when f(x) is divided by (x + 4).
Substituting ‘x = -4’ in the equation (1), we get f(-4) = 0
=> -4³ + 2(- 4)² + b (-4) + 68 = 0
=> -64 + 32 – 4b + 68 = 0
=> 4b = 36
=> b = 9.

Average speed = Total distance covered ÷ Total time taken.
Total distance covered = 12 + 36 + 32 = 80 km.
Total time taken = 12 ÷ 6 + 36 ÷ 9 + 32 ÷ 4
= 2+ 4 + 8 = 14 hrs.
=> Average speed = 80 ÷ 14 = 5.71 km/hr.

Let the capacity of the cistern be 600 units.
From the given data, the efficiencies of pipes A and B are 5 units/ min and 4 units/min respectively.
Let the efficiency of outlet pipe C be 'k' units/min.
Given the time taken to fill the cistern when all the three pipes are open = 100 minutes
⇒ Efficiency of pipes × time taken = Capacity of cistern
⇒ (5 + 4 − k) × 100 = 600
⇒ 9 − k = 6
⇒ k = 3
Therefore the time taken (t) by pipe C to empty the cistern
= Capacity of the cistern ÷ efficiency of pipe C
=> t = 600 ÷ 3 = 200 minutes = 3 hrs. 20 min.

As per the given conditions, weeks 1 and 3 will have similar pay structure and weeks 2 and 4 will have similar pay structure.
For weeks 1 and 3
Wage/hr for working = 60
Wage/hr during rest = 30
Total payment per day = (60 × 6) + (30 × 3) = 450
Total payment for weeks 1 and 3 = 450 × 12 = 5400
For weeks 2 and 4
Wage/hr for working = 20
Wage/hr during rest = 10
Total payment per day = (20 × 3) + (10 × 6) = 120
Total payment for weeks 2 and 4 = 120 × 12 = 1440
∴Total monthly salary = 5400 + 1440 = 6840

From 32 litres, 4 litres is drawn out and is replaced with water.
Proportion of wine drawn = 4/32 = 1/8th of the total volume of the wine in the mixture.
When wine is drawn out for the second time, (1/8) x (28) = 3.5 litres of wine will be drawn out.
Wine remaining in the jar = 28-3.5 = 24.5 litres
When wine is drawn out for the third time, (1/8) x (24.5) = 3.0625 litres of wine will be drawn out.
Wine remaining in the jar = 24.5 - 3.0625 = 21.4375 litres
Water in the jar = 10.5625 litres
Ratio of wine remaining in the jar to water remaining in the jar = 343:169