IIFT Quantitative Ability MCQ Quiz - 1 - CAT MCQ

Draw perpendiculars from O on to AC and BC. Let them meet AC and BC at D and E respectively. D and E lie on the circumference of the semicircle, since the radius and the tangent at the point of contact are perpendicular. Draw OC.
Let the radius of the semicircle be R.

Hence, option 3.

The ratio of the speeds of the fastest and the slowest runner is 2 : 1. i.e. They will meet for the first time, when the fastest runner completes 2 rounds while the slowest runner completes 1 round. Now, they meet for the first time after 5 min. The fastest runner completes 2 rounds in 5 min. i.e. he covers 2000 metres in 5 min.
Thus, the fastest runner completes the race i.e. covers 4000 metres in 10 min. Hence option 1.

Refer the diagram, 

The shaded region represents the maximum area that can be grazed by the cow. 

Let the cow be tethered to the top right comer (point B) of the rectangular house ABCD.
From the figure, it can be seen that: (i) The cow can graze 3/4th of the area of the circle of radius 20 ft. (ii) Moving to the left, along the length of the house the rope can go up to 16 ft along the boundary of the house. The remaining 4 ft of the rope would enable the cow to graze a small quarter circle of radius 4 ft. (iii) Moving down, along the breath of the house, the rope can go up to 10 ft along the boundary of the house. The remaining 10 ft of the rope would enable the cow to graze another quarter circle of radius 10 ft. The maximum total area that the cow can graze.

Given series: {xⁿ, x^{n + 1,} . . x²ⁿ}.
It can be seen that this series is a geometric progression with the first term = a = xⁿ and the common ratio = r = x
The number of terms of the series = 2n - n + 1 = (n + 1) terms

Year 2007:
In a stock split, the number of shares increases decreasing the face value of the stock. Since it was a 2 for 1 stock split, Mani’s shares increase to 12000 and the face value of each share reduces to Rs. 5.

Hence, option 2.

It will take a total of 20 minutes for one tank to be depleted.
Hence, option 2.

Quantity of Marcus wine = 15.6 - 10.4 = 5.2 litres.
Hence the ratio = 14.4 : 10.4 : 5.2 = 36 : 26 : 13. Hence, option 2.

The long pin forms the axis of rotation of the given triangle, and since it passes through the centroid; when the triangle is rotated, it rotates along its centroid thus creating a circumcircle around the triangle. The radius of this circumcircle is equal to line joining the centroid to one of the vertex of the triangle. In an equilateral triangle, the altitudes and the medians are the same and the centroid is also the circumcentre. The centroid divides the altitude in the ratio of 2 : 1.

Bacteria required to decompose 256 x 10²⁷ kg of waste = 16000 x 10⁵⁴.

Bacteria required to decompose 144 x 10⁹ kg of waste

Hence, option 1.

Let E = Event of Monica taking a Pink hair band from Phoebe E’ = Event of Monica taking a blue hair band from Phoebe B = Event of Rachel picking a Pink hair band from Monica’s hair band bag. 

g(f(x )) = ln[sin(x³ - 2x² - 5x + 6)]

x³ - 2x² - 5x + 6 = (x - 1)(x + 2)(x - 3)

As the argument of a natural logarithm has to be positive, g(f(x)) cannot defined for all the values of x, for which sin(x³ - 2x² - 5x + 6) gives a non positive value.
The range of sin x is -1 to 1 but it cannot take 0 or negative values.
Hence, the range of ln(x), for all x, such that 0 < x 1 is -∞ to 0. Hence, option 4.

Since the railway line runs parallel to the y-axis, its equation is of the form x = c.
Since it runs at a distance of 4 km to the right of the centre of the town, equation of the railway line is x = 4.
Equation o f the highway = (2x - y + 1) + k(x + 3y + 2) = 0. x(2 + k) + y(3k - 1) + 2k + 1 = 0 Since the highway is perpendicular to the railway line, the highway is parallel to the x-axis.

Hence, the co-efficient of x is 0.  Therefore 2 + k = 0; k = - 2 Hence, option 1.

Let x and y be the number of almond and coffee flavored chocolates.
She bought chocolates worth Rs. 358.

3.7x + 4.8y = 358 Since there is only one equation in two variables, the only way to solve it is by substituting answer options.
Only the values in option 2 satisfy this equation.
Hence, option 2.

Without the sale, Mr. Mickey gets a 20% discount and then a 25% discount.
During the sale, the shopkeeper gives Mr. Mickey the 20% discount as well as the 25% voucher (which is also valid).
Mr. Mickey also gets a further reduction of 30% due to the sale.
Thus, the percentage saving of Mr. Mickey is nothing but the additional percentage discount that he gets apart from the 20% and 25% that he was already getting.
Since he gets a 30% reduction, that is his savings.
Hence, option 2.

Alternatively, It is given that the marked price of the suit is Rs. 2,000.
Mr. Mickey originally gets 20% off from the shopkeeper and a further discount of 25% on the reduced price due to his voucher.
Mr. Mickey's cost price = 0.75 x 0.8 x 2000 = Rs. 1,200.

During the sale, Mr. Mickey gets a 30% reduction in the marked price, then a 20% discount from the shopkeeper and finally a 25% discount due to the voucher.
Mr. Mickey's cost price = 0.7 x 0.75 x 0.8 x 2000 = Rs. 840.

The percentage saving of Mr. Mickey = [(1200 - 840)/1200)] x 100
= 30%

Hence, option 2.

Let log₅(2x + 5y) = Z

From the first given equation we get, Z² - 6 Z + 5 = 0. 
Z = 1 or 5 i.e. log₅(2x + 5y) = 1 or 5 i.e. 2x + 5y = 5 or 5⁵ .

However, it is given that (2x + 5+) < 10.

i.e. log₃(3x + 2y) = 2 or 3. 

i.e. 3x + 2y = 9 or 27 However, it is given that (3x + 2y) < 10.

3x + 2y = 9 ... (ii)

Odd numbers between 300 and 500 divisible by 11 are 319, 341, 363, 385, 407,429, 451,473 and 495.
There are 9 such numbers.
Count of odd numbers between 300 and 500 = 100 Count of odd numbers between 300 and 500 not divisible by 11 = 100- 9 = 91
Even numbers between 100 and 300 divisible by 5 are 100, 110, 120, 130, 140,..., 280, 290, 300
There are 21 such numbers.
Count of even numbers between 100 and 300 = 101

Count of even numbers between 100 and 300 not divisible by 5 = 101-21 = 80
Required difference = 91 - 80 =11 Hence, option 2.  

Let the length of the trains be l.
Since the speed of 2nd train is 50% more, let the speeds of the 1st and 2nd train be 2x kmph and 3x kmph respectively.

It has been shown that p( p + 1 ) = 2(N + 16). Both p and N are integers.
Now, the options represent potential values of N.
Use the values of N from the options and solve for p. See if you get integral value of p.
Option 1: N= 5033.

p(p + 1) = 2(5033 + 16) = 10098 Now, 100 x 101 = 10100

Thus, no integral value of p satisfies this equation.
Option 2: N = 20085

p(p+ 1) = 2(20085 + 16) = 40202 Now, 200 x 201 = 40200 Thus, no integral value ofp satisfies this equation.
Option 3: N = 80183.

p(p + 1) = 2(80183 + 16) = 160398 Now, 400 x 401 = 160400

Thus, no integral value of p satisfies this equation.
Hence, option 4.
Note: You can verify that N = 320384 gets satisfied for p = 800, but you need not calculate it in the exam as the other options have been eliminated.

Let s be the total distance by the ball before coming to rest. 

Hence, option 3.

Assume that 100 kids were surveyed.

Hence, 90% of these i.e. 90 kids like GTA.
Now, the randomly selected kid likes GTA. Hence, this kid can be chosen in ⁹⁰C₁ = 90 ways
Now, number of kids who like GTA and Assassin’s creed but not FIFA = 0.90 x 90 = 81. Similarly, number of kids who like GTA and FIFA but not Assasin's Creed = 0.067 x 90 ~ 6.

Since there is no kid who likes only GTA, the number of kids who like all three games is: 90 - (81 + 6) = 3

6 out of 10 consonants can be selected in ¹⁰C₆ ways while 4 out of 5 vowels can be selected in 5C4 ways.
Let C1, C2, C3, C4, C5, and C6 be the 6 consonants. Since no two vowels are adjacent to each other, place the consonants first as shown below. _C1_C2_C 3_C 4_C 5_C 6_
Now, the 4 vowels can occupy any four places out of the seven blank spaces shown above. This can be done in ⁷C₄ ways.

Since all the consonants as well as all the vowels are distinct, the consonants and vowels can be respectively arranged in 6! and 4! ways.
Total number of words = ¹⁰C₆ x ⁵C₄ x ⁷C₄ x 6! x 4! = ¹⁰C₆ x ⁵C₄ x ⁷C₄ x 17280
Hence, option 1.