Test: CAT Quantitative Aptitude- 8 (October 29) - CAT MCQ

Let's assume the capacity of the tank is 1.

As tap A fills the tank in 16 hours

So, tap A filling capacity per hour, ie. The volume of water tap A can fill in one hour. 

=> capacity of tank / hours tap A take to fill tank    =  1 / 16

Let  tap B filling capacity per hour = x

 As given; 

2 * ( tap A filling capacity per hour) + 4*

( tap A  and tap B together filling capacity per hour)   = capacity of tank 

=> 2 * 1/16 + 4(1/16 +x)  = 1

=> 2 + 4 + 64 x = 16

=>  64 x = 10

=>  x = 5/32

ie. The filling capacity of  tap B per hour

So, the time tap B will take to fill the tank;-

=> capacity of tank/ filling capacity of B per hour  =  32/5 hours 

=>  32/5 * 60  

=> 384 min

The effective depth and radius after brick layering would be 14 m and 5 m respectively.
The volume of water in it would be = π x 5 x 5 x l 4 = 1100 cu. m Hence, option 1.

For a triangle to be formed, 7 < x < 17 
Case I: x ≥ 12 i.e. x is the largest side
x² < 13², for an acute-angled triangle.
Thus, the only possible integer value of x is 12.
Case II: x <12 i.e. 12 is the largest side 
x² > 12² - 5², for an acute-angled triangle.
Thus, the only possible integer value of x is 11.
1 + 1 = 2 acute-angled triangles can be formed.
Hence, option 1.

A = { 3 , 6 , 9 ....... 99}
B = {2, 4 , 6 ........100}
C = {7, 1 4 ,2 1 ..... 98}
D = {1,4, 9, 16, 2 5 ..... 100}
∴ (A ∩ C) = {21, 42, 63, 84} and (B ∩ D) = {4, 16, 21, 36, 64, 100}
∴ (A ∩ C) ∪ (B ∩ D) = {4, 16, 21, 36, 42, 63, 64, 84, 100}
The sum of the numbers of (A ∩ C) ∪ (B ∩ D) = 430 
Hence, option 3.

In 50 gm alloy 40 gm is gold.
Consider x gm gold to be added in the alloy to make gold 90%. (40 + x)/(50 + x) = 90/100

Solving the equation we get, x = 50
50 gm of gold should be added.
Answer: 50

The cost price per box = 40 x 20 = Rs. 800 The profit percentage = (160/800) x 100 = 20%

The required profit percentage = 25%

The number of chocolates now = 1.15 x 40 = 46

The cost price per box = 46 x 20 = Rs. 920 Selling price per box = 920 x 1.25 = Rs. 1150

Hence, the increase in the selling price per box = 1150- 960 = Rs. 190

Answer: 190

Exterior angle of a polygon = 360/n

Since, the sum of interior and exterior angle = 180°

(360/n) + 4 x (360/n)= 180

n = 10

Hence, option 3.

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= (3 x 9⁸) + (5 x 9⁷) + (a x 9⁶) + ... + (2 x 9⁰)

= [3 x (8 + 1)⁸] + [5 x (8 + 1)⁷] + [a x (8 + 1)⁶] + ... + [2 x (8 + 1)⁶]

When the above expression is expanded, only the last term of each binomial expression is not divisible by 8.
Thus, the number will be divisible when sum of its digits is divisible by 8. Sum of digits = 3 + 5 + <z + 2 + 4 + 6 + 7 + 7 + 2 = 36 + a The sum will be divisible by 8 when a = 4.
Answer: 4

4/5 of 40 = 32 matches played out of which 29 won.

8 more matches are to be played.

75% of 40 = 30

To maintain more than 75% win record, it has to win at least 31 matches.
Team India has to win at least two out of the remaining eight matches. Hence, team India can afford to lose at most six matches.
Answer: 6

For a real x, x² is greater than zero.

x² < -16 has no solution.
Answer: 0

Let the units digit and tens digit be 'a' and 'b' respectively.

The number = 10b + a

Required difference = |(10b + a) - (10a +b)|= |9b - 9a| = 9 x |b - a|

The difference will always be divisible by 9.
Hence, option 3.

Polex is set 38 minutes behind the original time and Tastfrack is set 57 minutes ahead of original time.
Thus, the total time difference = 95 minutes Polex gains 2 minutes in every hour and Tastfrack loses 3 minutes in every hour.
Thus, the relative speed of the watches = 5 minutes/hour They will cover 95 minutes in 95/5 = 19 hours i.e. at 2:00 am on 2^nd January 2016.
Hence, option 1.

Interest received by Meena after 2 years = 5,000(1.08)² - 5,000 = Rs. 832
Suppose Seema invested Rs. y in the investment scheme.
According to the given condition:
0.24y = 1.3077 × 832
y = Rs. 4,533 (approx.)

Let the total number of students be x.
The total number of chocolates = 14x

Four students amongst them will now distribute 4 x 14 = 56 chocolates

This is equal to the number of chocolates which are recieved by all of them in this distribution = 4x

4x = 56 i.e. x = 14 Thus, there are 14 x 14 = 196 chocolates.

Answer: 196

As we do not know in which direction the other train is running.
Hence cannot determine the time taken.
Hence, option 4.

Since, the polygon is regular the diagonal passing through the centre of the polygon will bisect the internal angle.

The angle formed at the centre of the triangle = π-(2 x (7π/16)) = π/8

Hence, the given polygon has 16 sides.
The number of distinct lines between sixteen vertices = ¹⁶C₂ =120 
Hence, option 3

The number of ways of choosing the 3 non-zero and non-unity digits out of 8 digits = ⁸C₃ = 56
These can be arranged in (6!)/(2! × 2! × 2!) = 90 ways and the digit 1 can be placed in any of the 7 possible locations with respect to the 6 digits (7 × 56 × 90 = 35,280).
There are 35,280 trials that are needed.

Selling price after successive discount = 700 x 0.8 x 0.75 x 0.7 = Rs. 294 
Answer: 294

Interior angle of an n-sided regular polygon = [(n - 2) × 180]/n

∠BAF = (6 - 2) × 180° ×  = 120°
∠BAR = (5 - 2) × 180° ×  = 108°
∠FAR + ∠BAF + ∠BAR = 360° (angles around the point)

⇒ ∠FAR = 360° - 120° - 108° = 132°

As m(AF) and m(AR) are equal, angle opposite to the equal side is equal.

⇒ ∠AFR =  = 24°
Therefore, 24 is the correct answer.

I is the incentre => mzBIC = 120°

m(arc BDC) = 240° ... (Inscribed angle theorem)

So, m(arc BIC) = 120° => mzBOC = 120° Therefore, mZlOC = mzOIC = 60°

Thus, ΔIOC is an equilateral triangle.
ΔIEC is 30°-60°-90° triangle.

Let IO = IC = OC = r
∴ EC = (r√3) / 2
∴ AC = (r√3)
∴ (√3 / 4) x (r√3)² = 8√3
r² = 32/3
πr² = 32π/3 
Hence, option 2.

As the number is divisible by 18, it is a even number divisible by 9.
For a number to be divisible by 9, sum of the digits should be a multiple of 9.

Sum of digits = 2 + 4 + 2 + a + 2 + 7 + b=17+ a + b

a + b can either be 1 or 10.
Considering the options only valid option is (4, 6).
Hence, option 2.

According to the first race, in the time Car-1 covers a distance of 100 km, Car-2 covers a distance of 90 km and Car-3 covers a distance of 80 km.
So, in the time Car-1 covers a distance of 1 km, Car-2 covers a distance of 0.9 km and Car-3 covers a distance of 0.80 km.
Now, for the second race, Car-1 has to cover a distance of 550 km, Car-2 has to cover a distance of 500 km and Car-3 has to cover a distance of 450 km.
In the time Car-1 covers a distance of 550 km, Car-2 covers a distance of 0.9 × 550 km = 495 km and Car-3 covers a distance of 0.80 × 550 km = 440 km
Thus, in the second race, it is Car-1 that wins the race.