NMAT Quantitative Skills MCQ Quiz - 3 - CAT

Since 10 is not a prime number, we have to first find the maximum power of 2 and 5 in 125!.
Maximum power of 2 = Multiples of 2 in 125! + Multiples of 4 in 125! + Multiples of 8 in 125! + Multiples of 16 in 125! + Multiples of 32 in 125! + Multiples of 64 in 125!

Where [x] = Maximum integer less than or equal to x. = 62 + 31 + 15 + 7 + 3 + 1 = 119 
Similarly the maximum power of 5

= 25 + 5 + 1 = 31
To get the number 10, we need one 2 and one 5.
As the number of 5's are lesser, the highest power of 10 in 125! will be 31.
Hence, option 5.

A can take 4 places - first to fourth - to have a P after it.
Case 1: A is first i.e. A

There are 4 letters, out of which P is present twice.

∴ No. of permutations possible = 

Case 2: A is second i . e . __A _______ Within this, there are two possibilities: 1) Both P’s are after A.
Here, one of L and E is also after A.
So, there are three letters after A i.e. P, P and E or L

∴ No. of permutations possible =

2) Exactly one P is after A.
Here, the letters after A are P, E and L, and there is no repetition.
∴ No. of permutations possible = 3! = 6
∴ Total no. of permutations when A is second = 6 + 6=12

Case 3: A is third i . e . ____ A _____ Again, within this, there are two possibilities: 1) Both P’s are after A.
Hence, L and E are before A, without any repetition.
These L and A can be arranged in 2! ways No. of permutations possible = 2! = 2 2) Exactly one P is after A.
So, one of L and E is before A and one is after A.
The two letters before A can be arranged in 2! ways.
Similarly, the two letters after A can also be arranged in 2! ways. Finally, L or E can be after A in another 2 ways.
∴No. of permutations possible = 2! *2! x 2 = 8

∴Total no. of permutations when A is third = 2 + 8 = 10 Case 4: A is fourth i . e . _______ A __

Now, exactly one P is after A.
So, the letters before A are P, L and E, without any repetition.
∴No. of permutations possible = 3! = 6

∴Total number of permutations = 12 + 12 + 10 + 6 = 40 Hence, option 5.

The natural growth rate gives the growth in population for every 1000 people. Natural growth rate for Maharashtra in 2009 = 17.4

∴ Population at the end of 2009 =

Natural growth rate for Maharashtra in 2010 = 17.3 

∴ Population at the end of 2010 =

Hence, option 4.

In 2009, the growth rate in Maharashtra =17.4 per 1000

Birth rate = 26 per 1000
∴ Overall death rate = 26 - 17.4 = 8.6 per 1000
Infant mortality in Maharashtra in 2009 = 70 ∴ Infant mortality for 26 births on 1000 = (70 x 26)/1000 = 1.82

∴ Required percentage = (1.82/8.6) x 100 = 21.16%
Hence, option 3.

The death rate for each state for each year is as shown below:

The death rate for Karnataka is the same in 2007 and 2010.
Hence, the percentage decrease for Karnataka is 0.

Observe that the death rate in West Bengal is greater than 10% while in the remaining states, it is less than 10%.

Hence, the greatest death rate is in West Bengal over the given period.

Hence, option 4.

The birth rate in Tamil Nadu increased from 2007 to 2008 and then remained constant throughout. Hence, it never decreased.

The birth rate in each of the other states decreased atleast once.

Hence, option 2.

Total profit made by firm = profit made in Q1 + profit made in Q2 + interest income
 Profit made in Q1 = 4.3 - 3.1 = Rs. 1.2 crores
Profit made in Q2 = 5.2 - 4.4 = Rs. 0.8 crores
Rs. 1.2 crores is invested at 8% per annum for 3 months

∴ Interestin come = 

∴ Total profit = 1.2 + 0.8 + 0.024 = Rs. 2.024 crores
Hence, option 1

Average quarterly income of previous year = 

Average quarterly  expenditure of previous year = 

∴ Profit in Q1 of next year = 4.375 - 3.525 = Rs. 0.85 crores

Hence, option 2.

This question can be solved faster by observation.
Note that the income is less than expenditure in Q3. Hence, there is a percentage loss in
Q3. Hence, Q3 can be ignored in calculations.
Now, compare Q1 and Q4.
The income in Q4 is more than the income in Ql, while the expenditure in Q4 is less than the expenditure in Ql.
Hence, the percentage profit in Q4 is more than that in Ql and so, Ql can be ignored in calculations.

For Q2: percentage profit = 

For Q4: percentage profit =

Thus, the maximum percentage profit is approximately 61%.
Hence, option 4.

Expenditure in Ql of next year = expenditure in Q4 of this year = Rs. 2.8 crores From the solution to the previous question, percentage profit in Q4 this year = 60.71%
∴ Percentage profit in Ql next year = (60.71 + 5)% i.e. 65.71%.

Hence, option 2.

The given data can be represented as shown below:

Let the area of the square, circle and triangle be S, C and T respectively.
Area o f cardboard thrown away when the circle is cut = A - C
And, area o f cardboard thrown away when the triangle is cut = C - T
∴ Total area o f cardboard thrown away = A - C + C - T = A - T

Area of square =A = a² sq.cm

Since all three vertices of the right triangle lie on the circle, the hypotenuse of the triangle is the same as the diameter of the circle (which in turn is the same as the side of the square i.e. a cm).

If the centre of the circle is joined to the third vertex of the triangle, it becomes the height of the triangle (as shown in the figure). The hypotenuse becomes the base of the triangle.

∴  Area of triangle =

∴ Required ratio = 

Hence, option 4.

Let w, r and t be the weight, radius and thickness of the circular disc respectively. 
∴ w ∝ r₂ x t

∴ w = k x r₂ x t

∴ W₁ = k x r1₂ x t₁ 

_{Hence, ratio of radius of two discs is 4 : 3.
Hence, option 2.}

Instead of trying to solve this equation, substitute the value of x from the options and check which value satisfies this equation.

Only x = 81 satisfies this equation.
Hence, option 3.

Let the age of the members in the row be A, B, C, D, E, F, G, H and I.

∴  A + B + C + D + E + F + G + H + I = 1 0 * 9 = 630 ...(i)

Since the average age of the first four members is 64 years,

A + B + C + D = 6 4 x 4 = 256 ...(ii)

§jnce the average age of the last seven members is 80 years,

C + D + E + F + G + H + I = 80 x 7 = 560 ...(iii)

Subtracting (iii) from (i),

A + B = 70 ... (iv)
Subtracting (iv) from (ii),
C + D= 186

∴ Averageage of third and four the member = 

Hence, option 3.

Using the given data, the number of matches played, won, drawn and lost can be found as shown below.

 ∴ Total number of matches played by Indore, Chennai, Ludhiana, Banglore and Kolkata = 80 + 80 + 40 + 20 + 60 = 280 
Hence, option 1.

Correct Answer :- b

Explanation : Number of matches drawn by Bangalore = 4

Number of matches drawn by Chandigarh = 18

% = (18-4)/2 * 100

= 77.77%

Consider the solution to the first question.
Number of matches lost by Pune =14 Number of matches lost by Ludhiana = 2

∴Required difference = 1 4 - 2 = 12

Hence, option 1.

Consider the solution to the first question.

Total number of matches won by all the teams = 16 + 30 + 28 + 2+ 12 + 26 + 56 + 8 = 178
Total number of matches drawn by all the teams = 10 + 9 + 12 + 4 + 10 + 12 + 20 + 3 = 80
∴ Difference = 178 - 80 = 98

Hence, option 5

The maximum length of the match stick will be equal to the biggest possible diagonal of the cube (match box).
Dimensions of the box are given as 10 x 6 x 2

∴ The biggest possible diagonal = 

Hence, option 4.

Let A alone do a units of work per day and B alone do b units of work per day.

When A starts, A works for 5 of the 9 days and B works for the remaining 4.

∴ Work done in 9 days when A starts = 5a + 4b

Let the total work be the LCM of 10 and 6 i.e. 30 units.

When B starts, B works for 5 of the 10 days and A works for the remaining 5.

∴  Work done in 10 days when A starts = 5a + 5b

(ii) - (i) gives b = 4

Hence, option 2.

The numerator is equal to:

Similarly, the denominator is equal to:

Hence, option 1.

The boy picked 11 mangoes in the remaining 3 days.
Using statement A alone: The number of mangoes plucked could be (1, 1, 9), (1, 3, 7), (1, 5, 5) and (3, 3, 5).
Thus, the question cannot be answered using statement A alone.
Using statement B alone: The number of mangoes plucked could be (1, 3, 7), (1,2, 8) and so on.
Thus, the question cannot be answered using statement B alone.
Using both the statements together: When conditions are combined, (1, 3, 7) is the only combination satisfying both conditions.
Thus, the boy picked 1, 3 and 7 mangoes (in no particular order) on the remaining 3 days.
Thus, the question can be answered using both the statements together but not by using either statement alone.
Hence, option 4.

Let Jai and Veeru’s current age be 2x and jc respectively.
Using Statement A alone: Though the ratio of Veeru and Gabbar’s age is known, there is no additional information on Jai. So, none of the ages can be found.
Thus, the question cannot be answered using statement A alone.
Using Statement B alone:
This equation can be solved for x. Once x is found, Veeru’s age is also found.
Thus, the question can be answered using statement B alone.
Thus, the question can be answered using statement B alone but not by using statement A alone.
Hence, option 2.

Using statement A alone: Let the cost price of each pen be Rs. y.

∴ Profit = (x x x) - (x x y) = x2 - xy = Rs. 150 

Since we don’t know the value of x, we cannot find the profit percent.
Thus, statement A alone is not sufficient to answer the question.
Using statement B alone: Cost price of 12 pens = Rs.

∴Cost price of one pen = Rs. x/12

Selling price of one pen = Rs. x/8

Thus, statement B alone is sufficient to answer the question.
Hence, option 2.

Let the speed of the cruise ship in still water = s m/s and length of the cruise ship = ℓ m Using Statement A alone:

Since the length of the cruise ship is not known, the speed cannot be determined Thus, the question cannot be answered using statement A alone.
Using Statement B alone:

Since the length of the cruise ship is not known, the speed cannot be determined.
Thus, the question cannot be answered using statement B alone.
Using Statements A and B together: Solving equations i and ii, 5 = 30 m/s.
Thus, the question can be answered using both the statements together but not by using either statement alone.
Hence, option 4.

(P/m)< 1 is possible when p < m .

Using statement A alone:

∴  p - m > 0 

∴  p > m

∴  p/m > 1

Thus, p/m is not less than one.
Thus, the question can be answered using statement A alone. Using statement B alone:

p < 8 and m - 8 > 0 

∴  m > 8 

∴  p < m

∴ p/m < 1

Thus, the question can be answered using statement B alone.

Thus, the question can be answered using either statement alone.
Hence, option 3.

Statements A and B alone cannot give a unique solution.

Using both the statements, the number is in the form of (a¹ x b¹) or a³, where a and b are prime.

If the number is of the form a3, its factors will be 1, a, a² and a³.

Therefore the number cannot be in the form of a³ since a² + a + 1 = 56 does not have any integer solutions.

∴ The number is of the form a¹x b¹
Its factors are 1, a, b and ab.
Sum of the factors excluding N is given to be 56.

a + b = 56 - 1 = 55

∴ a and b should be two primes such that,

a + b = 55

∴ a = 2, b = 53

∴ N= 53 x 2 = 106

∴ Statements A and B together are together sufficient to answer the question.
Hence, option 4.

Remainder when (a + b + c) is divided by n = rem (a/n) + rem (bln) + rem (dn) x! = X x (x - 1) X (x - 2) x ... X 3 X 2 x 1 Thus, every term from 12! to 55! will have a ‘12’ in it. Thus, every term from 12! to 55! is divisible by 12.
Since 12 = 4 x 3 , any x! that has a term (4 x 3) in it will also be divisible by 12. 4! =4x 3x 2xl
Since every subsequent term from 4! to 11! will have a 4! in it, each term from 4! to 11! is divisible by 12.

∴ Required remainder = rem(l !/12) + rem (2!/12) + rem (3!/12) = rem(l/12) + rem (2/12) + rem (6/12) = 1 + 2 + 6 
= 9
Hence, option 5.

∴ x +y > 1

Even then x + y > 1

∴ x + y is always greater than 1.
Hence, option 2.

Let the present age of Esha and Ankita be E and A years respectively.
Number of years between Esha’s current age and the age when she was as old as Ankita’s current age = E - A Ankita’s age when her age was reduced by this number = A - ( E - A ) = 2 A - E

∴ E = 2(2A-E)
i.e. 28 = 4A - 56

∴ 4A = 84
∴ A = 21

Hence, option 3.

When the hour hand is ahead of the minute hand, the angle is positive and when the hour hand is behind the minute hand, the angle is negative.

∴ Difference between Raju and Shyam’s viewing = 36.72 - 6.91 30 minutes

Hence, option 2.

Let the average daily expenditure per student be Rs. a.

∴ Total weekly expenditure for 35 students = 35 x a x 7 = Rs. 245a Also, average expenditure per head per week = Rs. la

When the number of students increases by 7 to 42, average expenditure per head per week reduces by Rs. 7

∴ Total weekly expenditure for 42 students = 42 x (la - 7) = Rs. 294(a - 1)

However, it is also known that in the same case of 42 students , total expenditure increases compared to the previous case by Rs. 42 per day.

∴ Total weekly expenditure for 42 students as per this data = 245a + 42(7) = Rs. 49(5a + 6)
Equating the two, 294(a - 1) = 49(5a + 6)

∴  5a + 6 = 6(a - 1) 

∴  a = 12

∴  Original weekly expenditure of the mess = 245 x 12 = Rs. 2,940

Hence, option 4.

Since AQ = QP = PC and AC = 12, AQ = QP = PC = 4 cm

Let the radius of the circle be jc cm as shown above.
Since OSBR forms a square, BR = BS = x cm 

Also, AR² = AQ.AP 

∴ AR² = 4(8) = 32

Since ΔAEC is a right isosceles triangle, AE = EC = x + 4√2

Hence, option 3.

The data given in the question can be represented as shown below.

The order of B, P, Q and C is because B-P-Q and P-Q-C.
Based on the figure,

BP = 2x, PC = 5x, BQ =y and QC = 2y

BC = BP + PC = BQ + QC
∴ 2x + 5 x = y + 2y

∴3y-7x = 0 ...(i)

Since the two lines are drawn parallel to each other, PQ = 2 units ... QB - PB = 2

∴ y - 2x = 2

∴ 3 y - 6 x = 6 ...(ii)

Solving (i) and (ii), x = 6 and y = 14

∴ BC = lx = 42 units

Hence, option 1.

Let the price of 1 apple, 1 orange and 1 guava be x, y and z respectively.
∴ 3x + 5y + 7z = 317 ...(i)

2x + 3y + 4z=191 (ii)

We need to find the value of 7x + 11y + 15z

Multiplying equation (ii) by 2 and then adding to equation (i),

∴ 7x + Uy + 15z = 317 + 191 x 2 = 699

Hence, option 1.

Average units elling price of a Lakme product = 

Hence, option 3.

Note that the number of companies in the Others category is not known. Also, the breakup of sales volume and sales value within this category is not known.
Hence, the company that has realized the lowest average unit sales price in 2010 cannot be found.
Hence, option 5.

Sales of all companies, except Lakme and L’Oreal, in 2010 (by value)
= (6 + 6 + 8 + 30 + 8)% of 120000 = 58% of 120000 = USD 69,600

Sales of Lakme in 2011 (by value) = 1.5 x 0.28 x 120000 = USD 50,400
Sales of L’Oreal in 2011 (by value) = 0.84 x 0.14 x 120000 = USD 14,112

∴Total size of US market (by value) in 2011 = 69600 + 50400 + 14112 = USD 134112

∴ Percentage growth in market size (b y value) = 

Hence, option 3.

AveragB e unit selling price o f an Elle IS Kp roduct = 

Average unit selling price o f a Nivea product =

∴ Required ratio = 44 : 54 = 22 : 27 

Hence, option 5.

Let the cost price of each saree be Rs. 100. ∴ Marked price = Rs. 160

We can say that,
160(0.9)ⁿ ≥ 100, where n is the number of successive discounts.

∴ (0.9)ⁿ ≥ 0.625 

0.9² = 0.81 
0.9³ = 0.729 
0.9⁴ = 0.6561
0. 9⁵ = 0.59 < 0.625 
∴ n = 4 Hence, option 3.

In a chessboard of dimensions 6x6, Number of squares of dimension 1 x 1 = 6² Number of squares of dimension 2 x 2 = 5² Number of squares of dimension 3 x 3 = 4² Number of squares of dimension 4 x 4 = 3² 

^{and so on...
∴Total number o f squares =}

^{∴ The required probability is:}

i. e. 0.37

Hence, option 5.

There are 8 shirts out of which 4 are white, 2 are red and 2 are black. No two shirts of the same colour are adjacent.
So, the white shirts should be alternately placed at either the 1^st, 3^rd, 5^th and 7^th positions or the 2^nd, 4^th, 6^th and 8^th positions from the left.

Thus, the white shirts can be arranged in 2 ways.
Now, the remaining 4 shirts can be placed in the remaining 4 positions in 4! ways. However, there are 2 red and 2 black shirts.

Therefore number of ways of arranging remaining 4 shirts = 

∴ Total number o f ways o f arranging the shirts = 2 x 6 = 1 2

Hence, option 1.

231 = 7 x 11 x3

Similarly, it can be shown that the remainder when 256^874! is divided by 11 = 1 and the remainder when 256^874! is divided by 3 = 1 ∴ The remainder is of the form:

11/7+1= 3q + 1 = 7r + 1 where p, q, r are integers The smallest value which satisfies all three is 1.
Hence, option 2.

Total capacity of coffee industry in India =

 Total capacity of Brooke Bond =

Total capacity of Nestle = 

Total capacity of Lipton = 

Total capacity of  MAC =

All the figures above are in ‘000 tonnes.

∴ Total capacity of companies under Others (in '000 tonnes) = 18.92 - (3.88 + 3.48 + 3

2.53 + 2.59) = 6.44

Hence, option 4.

Unutilised capacity (in ’000 tonnes) = Total capacity - Actual production 

Total capacity of coffee industry in India =

 Total capacity of Brooke Bon d =

Total capacity of Nestle = 

Total capacity of Lipton = 

Total capacity of  MAC =

All the figures above are in ‘000 tonnes.

∴ Total capacity of companies under Others (in '000 tonnes) = 18.92 - (3.88 + 3.48 + 3

2.53 + 2.59) = 6.44

Actual production of Others (in ’000 tonnes) = 11.6- (2.97 + 2.48 + 1.64 + 1.54) = 2.97

Hence, the unutilised capacity for each company is:

Lipton: 2.53 - 1.64 = 0.89
Nestle: 3.48 - 2.48 = 1
Brooke Bond: 3.88 - 2.97 = 0.91
MAC: 2.59- 1.54= 1.05 
Others: 6.44 - 2.97 = 3.47

Thus, the minimum unutilised production capacity is for Lipton.

Hence, option 1.

The cost of coffee (in Rs. per tonne) can be found for the top four producers.

However, within Others, there may exist some producer that sells coffee at a higher rate.

Hence, the most expensive coffee cannot be identified.

Hence, option 5.

Sales volume of Others (in ‘000 tonnes) = 10.67 - (2.55 + 2.03 + 1.26 + 1.47) = 3.36

Sales value of Others (in Rs. Cr.) = 132.8 - (31.15+ 26.75 + 15.25 + 17.45) = 42.2

CoffeeBean contributes to 25% of the sales volume and 20% of the sales value.

∴ Cost of CoffeeBean (in Rs. pertonne ) =

Hence, option 2.

Length of the diameter of the circle = 2 x 3.5 = 7 cm

Since the chord of maximum length is the diameter, QR passes through the centre of the circle.

Let O be the centre of the circle and the triangle intersects the circle at the points S and T.

Draw OS and OT.

In ΔQOS,

OQ = OS and m∠OQS = 60°

∴  ΔQOS is an equilateral triangle and OQ = 3.5 cm

Similarly, we can prove that ΔSOT and ΔTOR are equilateral triangles and since OQ = OS = OT = OR = 3.5 cm, ΔQOS, ASOT and ATOR are congruent.

A(n QRTS) = 3 x A(ΔOQS)

Area of shaded region = Area of semicircle - A(ΔQRTS)

Area of the circle 

Hence, option 1