IIFT Quantitative Ability MCQ Quiz - 2 - CAT MCQ

S (in ascending order) = {10, 13, 30, 40, 42, 66, 80, 99}

the median of the series  = (40+42)/2= 41

Now, x andy are to be added to this series. This can be added in the following ways:

(i) To get the minimum possible value of the median, x and y have to be as small as possible, since they are both two digit numbers, consider the following values for x and y:

x = 11 andy = 12

S" = {10, x, y, 13,30,40,42, 66, 80, 99}

Median = (30+40)/2 = 35

(ii) To get the maximum possible value of the median, x andy have to be as big as possible, since they are both two digit numbers, consider the following values for x and y:

x = 97 andy = 98

S"= {10, 13,30, 40, 42, 66, 80,x,y, 99}

Median = (42+66)/2 = 54

∴ The difference between the maximum and minimum possible values of the new median = 54 - 35 = 19

Hence, option 1.

Since the two children are given ropes of the same length, and there is no wastage in making the two figures, the two triangles have the same perimeter.

Let the congruent sides of the isosceles right triangle be x cm.

∴ Perimeter of isosceles right triangle = x + x + 

Also, area  of isosceles right triangle = 

∴ Perimeter of isosceles right triangle =  

This is the same as the perimeter of the equilateral triangle.

∴ Side of equilateral triangle = 

Hence, option 1.

Let the interest be compounded n times a year. Hence, the man will get a return of;

Here, r is the nominal interest rate (as a decimal) and time period is 2 years.

Now, if the compounding is done once a year, amount = (1 + 1/10)² = Rs. 1.21
If the compounding is done every six months, then amount = (1 + 1/20)⁴ = Rs. 1.2155

If the compounding is done every three months, then amount = (1 + 1/40)8 = Rs. 1.2184

So, we get the maximum return if the interest is calculated every instant (i.e. n —> ∞).

Hence, A_max = e^2r= e^2x0.1 = e⁰²= (1.1051)² = 1.2212 

Hence, option 1.

Sum of first 35 natural numbers = (35 x 36)/2 = 630, but the sum given is

∴ The wrong number is (648 - 630) = 18 more than the number that is added twice.
Also, as the wrong number is added twice, it has to be less than 35.
The only such number is 13.
Hence, option 1.

Let us assume he buys n goods. 

Total CP = 20n 

Total SP = 2 + 4 + 6 + 8 ….n terms 

Total SP should be at least 40% more than total CP 

2 + 4 + 6 + 8 ….n terms ≥ 1.4 * 20 n 

2 (1 + 2 + 3 + ….n terms) ≥ 28n 

n(n + 1) ≥ 28n 

n2 + n ≥ 28n 

n2 - 27n ≥ 0 

n ≥ 27

Let the height of the bird from the top of the pole be x.

As shown in the figure and using the properties of 30-60-90 and 45-45-90 triangles,

√3x = x + 8

∴ x = 10.92 m ~ 11 m

Hence, option 2.

Let the brothers receive Rs. 6x, Rs. 7x, Rs. 8x and Rs. 9x respectively.

∴ 36x² + 49x² + 64x² + 8lx² = 5750 ∴ 230x² = 5750

∴ x² = 25

∴x = 5
∴ Amount received by the three elder brothers = 6x + 7x + 8x = 2 lx = Rs. 105
Hence, option 1.

Profit from x units of the product is P = S - C = 156x² - 150x - x³

Average profit = f[x) = P/x = 156x - 150 - x²

Now,f(x) = 156 - 2x

f"' ( x ) = - 2

Since, f'(x) is negative, function f(x) assumes maximum value at the value of x corresponding to f ( x ) = ∴ 156 - 2x = 0

i.e. x = 78

Hence, option 2.

Since the caps are shaped as a right circular cone and 13 such caps are needed,

Total area of cardboard required = 13  curved surface area of one cap

=13πrl

Hence, option 1.

if x + y + z is constant, the product xyz takes its maximum value when x = y =z

The speed of the boat when there is only one person on board is 50 kmph.

∴ c₁ + c₂ — 50 ... (1)

The speed of the boat when there are 4 people on board is 35 kmph.

∴ c1+ (c₂/4) = 35 ... (2)

On solving (1) and (2), c1= 30 and c2 = 20

The current of the sea water is 10 kmph away from the island and towards Aksa beach.

Thus, the effective speed while going to the island will be 40 - 10 = 30 kmph and the effective speed while returning will be 34 + 10 = 44 kmph.

Since the same distance is travelled in both parts of the journey, the average speed will be

Hence, option 3.

The given data could be represented as shown in the above diagram.

Let R and r denote the radii of the outer and the inner circle respectively.
PQ is a tangent to the inner circle, hence it is perpendicular to the line joining the centre (C) of the inner circle and the point of intersection (T) of the circle and the tangent. Here, PT = TQ = 7

Also, ∠CTP =∠CTQ = 90°

In ΔCTP, CP₂ = CT₂ + PT₂  

R² = r² + 7²

Thus, 7, r and R form a Pythagorean triple. 7 is part of only one such triplet, 7, 24 and 25.

r = 24,R = 25

Sum of the radii of both the circles = r + R = 24 + 25 = 49

The sum of the diameters of both the circles = 2 x 49 = 98

Hence, option 1.

Let log₁₀x = y

∴ y²+ y - 2 = 0 

∴( x+ 2)(y - 1) = 0

∴ y = 1 or -2
 

For, y = log₁₀ x= 1, x = 10

Hence, option 3.

∴ a ³ - 1 6 a = 42 + 1 8 a - 1 6 a = 42 + 2 a

Hence, option 3.

As nothing is stated otherwise, we can assume that each of them contributed the same amount of work on a day when he was present at work. Let this amount of work be k. Further, let the number of days John, Tom and Harry remained present at work be (x - 20), x and y respectively. We have:

Simplifying, we gety = x - 10

∴The number of days for which Harry worked more than John = y — (x - 20) = (x - 10) - (x - 20) = 10 days.

Hence, option 2.

Area of the outermost hexagon ABCDEF =

As shown in the figure infinite number of hexagons is drawn.

Draw perpendicular AM on side UP.

Since m∠UAP = 120° and AM is the perpendicular bisector, Δ AMU is a 30- 60-90 triangle

UP — 2UM — side of the second hexagon =

Area of the second hexagon PQRSTU =

∴ A rea o f th e second hexagon is 3/4tim es th a t o f th e first

∴ Similarly , area of the third hexa gon is 3/4 times  that of  the  seco nd and  so on, 

∴ The areas of all the inner hexagons form an infinite decreasing G.P.

Com m on ratio (r) 3/4 nd first term (a )

 ∴ Sum of the are as  of  all  the  hexagons = 

Hence, option 2.

Unit’s digit of 7⁴⁶ = Units digit of 7² = 9

Unit’s digit of 3²¹ = Units digit of 3¹ = 3

Unit’s digit of 2² = 4

∴ U nit’s digit o f 7⁴⁶ x 3²¹ x 2² = units digit of  9 x 3 x 4 = 8

Thus, the given term is of the form abcd... efgh... 8.

When this term is divided by 100, the outcome is of the form (abed.. .efgh..) + 0.yS

∴  d= 8

Hence, option 1.

Product of abc and the sum of its digits is 2005.

∴ (100a + 10 b + c){a + b + c) = 2005

Now, either the number should have its units digit as 5 or the sum of its digits should have its units digits as 5, or both.

Hence, try to express 2005 as a product of two numbers such that at least one of the two has its units digit as 5 and one of the two numbers is a 3-digit number.

2005 = 401 x 5

Since 5 and 401 are both prime numbers, there is no other way of expressing 2005 in the manner required.

∴ abc = 401

∴ b = 0

Hence, option 4.

Given that the king takes every ten of his hundred sons to the jungle with no ten being repeated

∴ The number of times that the king goes to the forest is given by  i-e. X = 
Also, each son goes with every two of his brothers selected from the remaining 6 brothers.

∴ The number of times that each son goes to the forest is given by  i.e. Y = 

Hence, option 4.

Circumference of the track = Distance that has to be travelled by both the motorcycles to complete one round = 2 x 22/7 x 105 = 660 m.

Time taken by the first motorcycle to complete one round = 660/33 = 20 seconds

Time taken by the second motorcycle to complete one round = 660/44 = 15 seconds

∴ The two motorcycles meet for the first time at the starting point at LCM (15, 20) = 60 seconds

So, they meet for the fifth time at 60 x 5 = 300 seconds = 5 minutes.

Hence, option 4.

Probability of drawing a queen = 4/52

Probability of drawing any card other than the queen = 48/52

∴ Probability of drawing a queen exactly 4 times out of 15

=  

Let t be the time taken, d be the distance, TV be the number of coaches and U be the number of electricity units utilized per km. Let k be the constant of proportionality.

In the first case

Now, in the second case, N= 20 and t = 3

Hence, option 3.