Sample IIFT Quantitative Ability Quiz - CAT MCQ

In the available 6 slots, 5 subjects can be organized in ⁶P₅ = 720 ways.
The remaining one slot can be filled with any one of the 5 subjects.
Hence total number of ways = 720 x 5 But since one subject is repeated, the total number of distinct possibilities = 

We simplify the equation x² - xy - 101 = 0 as, x(x- y) = 101 = 1 * 101. As x and y are integers, we have, x = 1 and x - y = 101, or x = 101 and x - y = 1.
If x  = 1 , y = 1 - 101 = -100.
If x = 101 , y = 101 - 1 = 100.

As x and y are natural numbers, we must disregard the solution (x, y) = (1, - 100).
There is only one solution: x = 101 and y = 100.
For this set, we have (x + y) = 101 + 100 = 201.
Hence, option 4.

Ratio of the shares of Amav, Anubha and Astha = 1,00,000 x 4 : 50,000 x 6 : 60,000 x 7 = 20: 15 : 21
Profit to be divided = 80% of Rs. 14,000 = Rs. 11,200. 

Hence, option 2.

log (4^x + 1) + 2x = log 15 + x log 4

Therefore log (4^x + 1) = log 15 + log 4^x - 2x.

log (4^x + 1) - log 4^x = log 15 - 2x log 10 ...(since log10¹⁰ = 1)

We try to equate each factor on the L.H.S. to a corresponding factor on the R.H.S. and find that, for x = 1/2, we have (4^x + 1) = 3 and (5)^2x = 5 Thus, x = 1/2. For this value of x, we have 2x = 1, 4x = 2 and 8x = 4.
Only 4x = 2 is a prime.
Hence, option 3.

The number of outcomes that the BPO would close down is given by:

The number of outcomes that the BPO would close down is given by:

Consider a rectangle of length l cm and breadth b cm.  Its area = A = l x b.

A threefold increase in its sides will make the new length equal to 3l and the new breadth equal to 3b. New area of the rectangle = 9(l x b) = 9A. Similarly, consider a triangle of sides a cm, b cm and c cm. 

A threefold increase in its sides will make the new sides equal to 3a, 3b and 3c.

Area of the triangle will be equal to the area of the rectangle. Hence, option 4.

Let there be 100 items, each costing Re. 1.
Thus, total cost price = Rs. 100.
For an overall profit percentage of 25%, total selling price = 100 x 1.25 = Rs. 125
Since one-fifth of the goods are sold at a 10% loss, it means that 20 items are sold at Rs. 0.9 each.
So, amount received from this sale = 20 x 0.9 = Rs. 18. 20% of the remaining goods are sold at neither profit nor loss. Hence, 0.2 x 80 i.e 16 items are sold at Re. 1 each. 

So, amount received from this sale = 16 x 1= R s . 16 Now, there are 100 - (20 + 16) = 64 items left to be sold.
Hence, amount that should be received from these 64 goods = 125 - (18 + 16) = Rs. 91.

Hence, each of these items is sold at Rs. 91/64.

Profit percentage = {[(91/64) - 1]/1} x 100 = 42.19% Hence option 3.

As the first bucket is filled at a rate of 2x, water leaks from the first bucket at a rate of x.
The second bucket is filled at a rate of x, therefore, water leaks from this bucket at a rate of 0.5x.
The third bucket is filled at a rate of 0.5x, therefore, water leaks from this bucket at a rate of 0.25x

Continuing in this manner we get an infinite geometric progression with a = x and r = 1/2 

So, sum of the amount of water leaking at any point of time 

Since 65% of the villagers own neither a car nor a phone, 35% of the villagers own atleast one of the two.
Let the percentage of villagers who own a car as well as a phone be x%. 25% of the villagers own a phone and 15% of the villagers own a car. (25 - x) + x + (15 - x) = 35. So  x = 5
Hence, 25 - 5 i.e. 20% of the villagers own only a phone.

Since 2000 villagers own both a phone and a car, 5% of the total number of villagers corresponds to 2000.
Hence, the number of villagers who own only a phone (20%) is 4 * 2000 = 8000
Hence, option 3.

Let the two particular boys be A and B.
They can be arranged in 2 ways after keeping a gap of 3 seats between them.
Three boys who are to be seated in between A and B can be selected from remaining 8 boys and arranged in 8P3 ways.
The remaining 5 boys can be arranged in 5! ways.
The group of 5 boys including A and B and the ones sitting between them can be seated in the arrangement in 6 ways.
The total number of ways in which all the 10 boys can be seated is 2 x ⁸P₃ x 5! x 6 = ⁸P₃ x 6! x 2 = ⁸P₃ x 1440 Hence, option 2.

Let the cost of the banana, cake, cola and book be b, c, a, d respectively.
So,
2b + 3c + 4a + 5d= 59.5 ... (i)

2b + 4c + 5a = 39.5 ... (ii)

Subtracting (i) by (ii) we get,

- c - a + 5d = 20 ... (iii)

We are given that, d - c = 2 ... (iv) And d or c has 50 paisa in it.
So we check all the combination where the difference between the cake and the book.

Also as none of them come free, the cost of anything cannot be greater than 7 and the cost of book is greater than the cost of cake, so using equation (iv) the combinations of the cost of a book and a cake are (2.5, 0.5), (3.5, 1.5), (4.5, 2.5), (5.5, 3.5), (6.5, 4.5).
By substituting each combination of the book and the cake in equation in (iii) we get, the value of the book and the cake to be (5.5, 3.5) and the corresponding value of the cola is Rs. 4 Hence, option 4.

The petrol pump has a gain of 10%. i.e., if CP = Rs. 100, SP = Rs. 110. It is given that SP = Rs. 66

Petrol in 1st bottle = 20 litres.

Petrol in 2nd bottle = 20 x (100/120) = 20 x (5/6) litres.

Petrol in 3rd bottle = 20 x (5/6)² litres.

Petrol in 5th bottle = 20 x (5/6)⁴ = 3125/324 litres.

 Kerosene in the 5th bottle = 20 - (3125/324) = 3355/324 litres Hence, option 2.

Since the largest digit in Sheetal’s number is 6, her base system has to be atleast 8. It has to be an even number as it is twice the number system that Manju used.
Similarly, Manju’s base system has to be atleast 4.
If Sheetal wrote the number in base system x, 

Thus, Sheetal wrote her number in base 8 while Manju wrote it in base 4.
Since Komal’s base system was the average of these two, Komal wrote her number in the base system 6.
Hence, option 3 can be directly eliminated as 6 cannot be a digit in base system 6.

Now, consider options 1 and 2; and check if they give 419.

Option 1: 2423

2(6)³ + 4(6)² + 2(6)¹ + 3 = 2(216) + 4(36) + 12 + 3 = (591)₁₀
Hence, option 1 can be eliminated.

Option 2: 1535

1(6)³ + 5(6)² + 3(6)¹ + 5 = 216 + 180+ 18 + 5 = (419)₁₀
Hence, option 2.

Consider the given equation,

ax² + bx + c = 0 
Since roots are in the ratio p : q, let the roots be pk and qk.

On substituting the value of k from equation (ii) to equation (i), we get, b²pq = ac(p + q)²
Hence, option 2.

Let the total capital be 12x.
Then Gopi contributes 3x.
Hari contributes twice as much as Gopi and Sufi together contribute. If Sufi contributes y, then Hari contributes 2(3x + y). Therefore 3x + y + 2(3x+ y) = 12x.

y = x. 

Sufi contributes x and Hari contributes 8x.
Gopi’s, Hari’s and Sufi’s contributions are in the ratio 3 : 8 : 1. Their profits also would be in the same proportion and would be Rs. 1200, Rs. 3200 and Rs. 400.
Hence, option 2.

The rectangular cube is pasted lengthwise symmetrically over the diameter.

Area of the disc attached to the rectangular cube = 2 x (area of the sector + area of the triangle) = 2 x (13.1+ 10.82) = 47.84 cm² Hence, option 4.

101010.10101 x 10^-7 x 10ⁿ < 1000000 < 1010101.0101 x 10^-8 x 10ⁿ. Therefore 
-7 + n  0.  n  7. Thus, for 1000 < k < 1000000; n = 5, 6, 7. The sum of possible values of n = 5 + 6 + 7 = 18. Hence, option 3.

Let x and y be the length of diagonals of Ravi’s rhombus.  Area of rhombus = 24 = (product of diagonals)/2 = xy/2. xy = 48 Now, x and y also correspond to the sides of Raju’s rectangle.
Area of rectangle = xy = 48 sq.cm. Area of Ronak’s square = 48(3) = 144 sq.cm. Length of side of square = 12 cm Hence, option 2.

The following table lists the different probabilities: 

From, the above table, it can be seen that the probability of answering the random question correctly is 0.65. The probability of answering the random question incorrectly = 1 - 0.65 = 0.35
Hence, option 4.

i.e. Shiva left after 2.2 days.
Hence, option 2.

Total sales this year = 116.8 + (59.3 x 100)/108.2 + 70.4 ~ 242 million dollars
Expected sales next year = (116.8 x 1.0725) + 59.3 + (70.4 x 1.0999) ~ 262 million dollars.

Expected sales growth: