Test: CAT Quantitative Aptitude- 6 (October 22) - CAT MCQ

Answer: 4

Let x = 200,000,000
Then, the factors become (x – 3)(x + 3) = x² – 9.
So, the product is 40,000,000,000,000,000 – 9 = 39,999,999,999,999,991, whose digit sum is 3 + 15(9) + 1 = 139.

Let the unit of work done by 1 man in 1 hour and 1 day be 1 MDH unit (Man, Day, Hour).
According to question,
Work done by N workers working 8 hours per day in 20 days = N × 8 × 20 MDH units.
Since this work is equal to 70% of the total work,
i.e. 70% of the total work = 160N MDH units
Therefore, total work = 
So, remaining 30% of the total work = 
This work is completed by (N - 15) workers working 12 hours in 10 days = (N - 15) (10)(12) = 120(N - 15)

So, 

 

 

So, N = 35

When two numbers divided by a common number N leave the same remainder, then their difference must be divisible by N.
Thus must be divisible by N.
194 = 97 x 2
Since N is of two digits, the number is 97.

Number of zeros can be found by calculating highest power of 10 in the number.

As 10 = 2 x 5? so the number of zeros will be equal to the highest power of 5 in the number.

Highest power of 5 in 60! = (60 / 5) + (60 / 25) = 12 + 2 = 14 The number of zeros in 60! = 14 Answer: 14

Option 2

12^th term of an A.P. = Sum of first 12 terms - Sum of first 11 terms = 555 - 420= 135

4^thterm of G.P. = 12^th term of an A.P. =135

Let the first term and common ratio of G.P. be a and r.

4^th term pf G.P. = ar³ = 135 .

∴ r = 135/5 = 27

∴ r = 3 Answer: 3

Option 3

The digits at thousands place can be anything from 1 to 9 and the digit at hundreds place can be anything from 0 to 9.

The sum of the two digits will be either a single or double digit.

∴ The number of required numbers = 9 x 10 = 90

Hence, option 4.

Let T be the position of tree.

m∠TPR = 2 x m∠TPS

∴ PS is the angle bisector.

 ∴  x = 6 meters

Hence, option 2.

From the given conditions;

We can see that P_m increases exponentially as m increases.

Option 3 is the only possible answer.

Hence, option 3.

Let g(x) = a and g(x - 1 ) = b

Let x = x + 1

Similarly, g(x + 3) = l/a, g(x + 4) = bla, g(x + 5) = b, g(x + 6) = a .. .and so on.
∴ g(x + 6) = g(x)

Answer: 6 optin 3

Let the current population of the village be x.

The population of the village just after the third batch of migrants entre village X = x + 3(0.2x) = 1.6x

The total sum of the ages of the migrants just after the third batch of migrants entre village X = 0.2x x 27 + 0.2x x 26 + 0.2x x 25 = 15.6x

The average age of the village after 3 years = (33x + 15.6x )/1.6x = 30.375 years

Hence, option 4.

When 3¹,3², 3³ and 3⁴ divided by 12 leaves remainder 3, 9, 3 and 9 respectively.

As the pattern repeats itself after 2 steps, 3 odd number leaves remainder 3 and 3even number leaves remainder 9.

∴ R em ainder of 3¹³⁶ when divided by 12 = 9

Hence, option 3.

Since each user receive files in multiple of three, let us assume that they receive 3a, 3b, 3c, 3d and 3e number of files, respectively, where a, b, c, d and e are natural numbers.
Therefore, 3a + 3b + 3c + 3d + 3e = 36
a + b + c + d + e = 12
So, we have to find in how many ways the sum of five numbers can be 12.
Therefore, number of ways in which the sum of five numbers is equal to 12 = ^{(12 - 1)}C_{(5 - 1)} = 330

As shown in the figure the square PQRS cuts the axes at A(2,0), B(0,2), C (-2,0) and D(0, - 2).
When rotated by 60° the axes will cut the square at A', B', C' and D' respectively.

PD’ = AP - AD'

Hence, option 3.

Let free cab service is for a fixed distance for z km.
Let per kilometer charge = Rs. y
Let the distance covered by Biswas and Amit is 'a' km and 'b' km, respectively.
So,
(a + b - 2z )y = 110 ...(1)
(2a - z)y = 200 …(2)
(2b - z )y = 100 …(3)
(3) + (2) - (1)
We get
(a + b) y = 190
Also, if we add (1) and (3)
(a + b - z ) y = 150 ..(4)
Simplifying (1) and (4), we get
19z = 4a + 4b.. (5)
Also,
2a - 4b = - z …(6)
From (1) and (4)zy = 40
Also,
a = 3z
Biswas initially paid (3a - z)y, if he travelled thrice the distance travelled initially by him which is 8zy = Rs. 320

Given: Speed of bus B = 75% of speed of A = 3/4 of speed of A
i.e. Ratio of speeds of A and B = 4 : 3
Let their speeds be 4V and 3V.

Also, the distance between S₁ and S₃ is 60% of that between S₁ and S₂.
i.e. S₁S₃ = 60/100 of S₁S₂ = 3/5 of S₁S₂
So, S₁S₃ = 3d, S₂S₃ = 2d and S₁S₂ = 5d

Now, time taken by bus A to reach S₃ = 3d/4v hours ---(1)
Time taken by bus B to reach S₃ = 2d/3v hours ---(2)

We need to find the time taken by A to travel from S₁ to S₂, i.e. 5d/4v.

Now, bus B started one hour later than bus A.

i.e. The time taken by bus A to reach S₃ is one hour more than the time taken by bus B to reach S_3.
i.e. 

Let d/v = k


=> k = 12

So,  hours

Total purchase price = 45.45 x 100 = Rs. 4545 Let the tax paid be Rs. x

∴ (6817.5 - 4545 - x)/(4545 + x) = 0.2

∴ 2272.5 - x = 909 + 0.2x

∴ x = 1136.25
The tax percentage levied over cost price = (1136.25/4545) x 100 = 25% 
Answer: 25

The correct option is Option B.

Let the capacity of Container 1 be ‘a’ litres and that of Container 2 be ‘6’ litres.

∴ Petrol in container 1 = 0.3a 

Petrol in container 2 = 0.456

Similarly, petrol in the new container = 0.35(a + b)

∴ 035 a + 0.45 b = 0.35(a + b)

∴ a = 2b

Since a = 8, b = 4.

Hence, the capacity of container 2 is 8 litres.

Hence, the answer is 4

The speed of the train = 270 km/hr = 75 m/s The time after which the difference in position of A is 1800 m = 1800/(75 - 3) = 25 seconds Now, in these 25 seconds,

• A walks 25 x 3 = 75 m towards rearmost part of the train.
• B walks 25 x 4 = 100 m towards foremost point of the train.

At his stage, there is distance of 2100 m between them.

∴ The length of the train = 75 + 2100+100 = 2275 m

Hence, option 4.

log₂(x² - y²) - log₂(x + y) = 1

x - y = 2 ....(1)
log₃x + log₃y = 1
log₃(x . y) = 1
xy = 3¹ = 3
Put the value of x from equation (1) in (2):
(2 + y)y = 3
y² + 2y - 3 = 0
y = 1, -3
y = -3 rejected
So, the value of y is 1.

The correct option is Option D.

The number of chocolates received by Siya is multiple of 3. 

Among the given options, only 15 is multiple of 3.

So, Siya gets 15 chocolates.

As per the given condition,
hog(x) + fog(x) = 30
x² + 13|x| - 30 = 0
(For x > 0, |x| = x)
When x = +ve,
x² + 13x - 30 = 0
(x + 15)(x - 2) = 0
x = 2, x = -15 (rejected as x > 0)
When x = -ve,
x² - 13x - 30 = 0
(x + 2)(x - 15) = 0
x = -2, x = 15 (rejected because x < 0)
Sum = -2 + 2 = 0