Practice Test for NMAT - 12 - CAT MCQ

Without losing generality we can assume that shopkeeper has 100 articles with marked price of 100.
Now, for first 70 articles he offered discount of Rs. 40 on each of the articles.
Hence selling price of these 70 articles will be Rs. 60 each.
∴ Total selling price of these 70 articles will be,
70 x 60 = 4200
And he offers additional 30% discount on remaining articles.
Hence selling price of these 30 articles will be 70% of 60 i.e. Rs. 42 each.
∴ Total selling price of these 30 articles will be,
30 x 42 = 1260
Total selling price of 100 articles is,
1260 + 4200 = 5460
This is 20% more than cost price. Hence,



Hence, option 1.

Note th a t (x - 2), (x - 1), x and (x + 1) are four con secutive integers.
The magnitude of no set of four consecutive integers adds up to 7.
Hence, no integral values of x satisfy the given equation.
Hence, option 1.

Only students who passed could apply for jobs and all such students applied for jobs.

∴ Number of students applying for a job in ‘Others’ = (0.33 x 800) + (0.12 x 650) + (0.13 x 1100) + (0.28 x 1200) + (0.16 x 1000)
= 264 + 78 + 143 + 336+ 160
= 981
Since the number of students applying for Systems jobs has to be mimimized, the number of students applying for HR should be the highest possible.
The number of students applying for HR jobs was the same each year.
Hence, this number is maximized when there are 78 applicants for HR each year (as number of students each year is equal and there cannot be more than 78 applicants in 2009).
∴ Maximum number of students applying for an HR job = 78 x 5 = 390
∴ Minimum number of students applying for a Systems job = 981 - 390 = 591 
Hence, option 2.

In 2009, students applying to Nefertiti = 25% of 48% of 650
Also, students selected in Nefertiti = 50% of 25% of 48% of 650
= 0.5 x 0.25 x 0.48 x 650
= 39
Each got a salary that was 10% above the starting salary paid in Marketing.
Total monthly salary outflow for Nefertiti = 39 x 1.1 x 6390 = Rs. 2,74131. 
Hence, option 3.

Since the average salary for Others is not known, the average salary for the B- school cannot be found for any year.
Hence, option 5.

The total amount earned by students applying for Finance jobs in 2010 = = 0.23 * 1100 x 7550 = Rs. 19,10,150
The total amount earned by students applying for Operations jobs in 2012 = 0.2 x 1000 x 8640 = Rs. 17,28,000
Thus, the required percentage is definitely greater than 100.
Hence, options 1 and 3 can be directly eliminated.
Required percentage = (1910150/1728000) x 100 110%
Hence, option 4.

Hence, option 4.

Out of 12 nuts, 4 are defective while out of 24 bolts, 8 are defective.
So, there are 12 defective items and 24 non-defective items.
So, the probability that both items are non-defective (when picked without replacement one after the other)

Hence, option 3.

A(minor segment MCNO) = A(minor sector MCNO) - A(ΔMON)

Since MN subtends a right angle at the centre of the circle, MNO is an isosceles right triangle with base = height = 14 cm

∴ A(minor segment MCNO) = 154 - 98 = 56 sq.cm
Hence, option 2.

A perfect square is of the form n^2a
∴ Number of factors = (2a + 1), which is odd
Hence, statement (a) is correct.
Hence, options 2 and 5 can be eliminated.
Since statement (c) is easier to check using a direct formula, check for statement (c) first.
7² = 2³ x 3²
Number of factors = (3 + 1)(2 + 1) = 12
Thus, statement (c) is incorrect.
Hence, options 3 and 4 can be eliminated.
Hence, option 1.
Note: Statement (a) can also be checked by considering 2 or 3 random perfect squares and finding their number of factors.
Statement (b) can be verified as under:
N is a perfect square and has 5 factors between 1 and √n.
Half of the factors of a number which is a perfect square are lesser than its square root and half of them are greater than its square root.
e.g., 36 has the factors 1, 2, 3, 4, 6, 9, 12, 18 and 36 and √36 = 6
1,2, 3 and 4 are less than 6 while 9, 12, 18 and 36 are greater than 6
∴ If the number of factors of n between 1 and √n = 5, then the number of factors between √n and n is also 5.
Apart from this, 1, √n and n are also factors of n.
Total number of factors = 5 + 5 + 3=13

For the index to ‘decline’, only one stock should go up on that particular day.
This is satisfied only on days 4, 8 and 9.
Hence, the index ‘declined’ on three days.
Hence, option 3.

Value of index = 3A + 2B + C
The maximum rise in index would occur when the index has its least possible value on day 1 and highest possible value on day 10.
In the index above, A is multiplied by the largest number (i.e. 3) and B is multiplied by the second largest number (i.e 2).
Hence, to get the minimum value on day 1, take A = 12, B = 15.5 and C = 23.5
∴ Index on day 1 = (3 * 12) + (2 * 15.5) + 23.5 = 36 + 31 + 23.5 = 90.5
Similarly, index on day 10 = (3 * 14.6) + (2 * 16.7) + 24 = 43.8 + 33.4 + 24 = 101.2
∴ Rise in index = 101.2 - 90.5 = 10.7
∴ Required percentage = (10.7/90.5) x 100 = 11.82%
Hence, option 3.

Consider the solution to the previous question.
Observe that each graph has its least possible value on day 1.
Since the index = 3A + 2B + C, the least value of the index is when A = 12, B = 1.5 and C = 23.5
Hence, as calculate earlier, least value of the index = 90.5
Hence, option 5.

Index = 3A + 2B + C
The value of the index on day 1 is 105.5
Because there are three graphs, there are 3! = 6 possibilities.
1) X = C, Y = B, Graph 3 = A ⇒ Here, index = 90.5 (as calculated earlier)
2) X = C, Y = A, Graph 3 = B ⇒ Here, index = 3(15.5) + 2(12) + 23.5 = 46.5 + 24 + 23.5 = 94
3) X = B, Y = C, Graph 3 = A ⇒ Here, index = 3(12) + 2(23.5) + 15.5 = 36 + 47 + 15.5 = 98.5
4) X = B, Y = A, Graph 3 = C ⇒ Here, index = 3(15.5) + 2(23.5) + 12 = 46.5 + 47 + 12 = 105.5
Since the correct value of the index has been obtained, you do not need to calculate the remaining two casese.
Using these values, index on day 3 = 3(17.1) + 2(27.2) + 13 = 51.3 + 54.4 + 13 = 118.8
Hence, option 1.

Let the usual speed of Mohsin be x.
∴ Mohsin’s speed on Monday = 1.25x
Assume that usually Mohsin takes t minutes to reach the office.
∴  Time taken on M onday = t - 16
∴ Distance tra velled = xt =1.25x (t - 16) Solving this, we get
t = 80 minutes
On Tuesday, Mohsin reached office 20 minutes late, travelling with his usual speed,
i.e., he travelled for (20/80 =) 25% more.
As the speed with which he travelled was same, distance a time.
As time taken is 25% more, the distance travelled will also be 25% more.
Hence, option 3.

Since one internal angle of the parallelogram is 150°, the other has to be 30°
Let the adjacent sides of the parallelogram measure a cm and b cm respectively.
Perimeter = 2(a + b) = 40
i.e. a + b = 20 ... (i)
Area = ab * sin 30 = 48
∴ ab = 48 x 2 = 96 ... (ii)
Dividing (i) by (ii)


The only values of a and b (from the options) that satisfy this equation are 12 and 8.
Hence, option 1.

|3x² - 5| > 5x + 3
∴ ±(3x² - 5) > 5x + 3

Case I:

3x² - 5 > 5x + 3
∴ 3x² - 5x - 8 > 0
∴ 3x² - 8x + 3x - 8 > 0
∴ x(3x - 8) + 1 (3x - 8) > 0
∴(x+ 1)(3 x -8 )> 0
i.e. x < - 1 or x > 8/3 ...(i)

Case II:

-(3x² - 5) > 5x + 3
∴ -3x² + 5 > 5x + 3
∴ 3x² + 5x - 2 < 0
∴ 3x² + 6 x - x - 2 < 0
∴ 3x(x + 2) - 1 ( x +2) <0
∴ (3 x - 1)(x + 2) < 0
i.e. -2 < x < 1/3 ...(ii)
Combining the two ranges in (i) and (ii)
(-∞,-1) U (1/3, ∞)
Hence, option 5.

Per capita income = total income / total population
Since the question asks for the average of the annual per capita income, first find the per capita income for each year (as shown below).

Average of this per capita income (in Rs.) = 4124.19

Hence, option 2.

Consider the solution to the first question.
From the table, the highest per capita income is in 2014-15.
The national income for this year is Rs. 4,33,500 crore.
Average national income = Rs. 322321.33 crore
∴ Required difference = 433500 - 322321.33 = 111178.67
Hence, option 4.

Consider the solution to the first question.
First calculate the absolute change in the per capita income for each and see if some options can be eliminated.

2010 - 11: Change = 3482.32 - 3097.64 = 384.68 and base = 3097.64 i.e. % increase > 10%
2011 - 12: 3786.44 - 3482.32 = 304.12 and base = 3482.32 i.e. % increase < 10%
2012 - 13: 4202.98 - 3786.44 = 416.54 and base = 3786.44 i.e. % increase > 10%
2013 - 14: 4856.74 - 4202.98 = 653.76 and base = 4202.98 i.e. % increase > 10%
2014 - 15: 5319.02 - 4856.74 = 462.28 and base = 4856.74 i.e. increase < 10%

Hence, the least percentage growth is in 2011-12 or 2014-15
% growth in 2011-12 = (304.12/3482.32) x 100 = 8.73%
% growth in 2014-15 = (462.28/4856.74) x 100 = 9.52%

Hence, option 4.

Consider the solutions to the previous questions.
National income in 2015-16 = 433500 * 1.25 = 541875
Per capita income in 2015-16 = 5319.02 x 1.35 = 7180.68
∴ Population in 2015-16 = 541875/7180.68 = 75.46
Hence, in ascending order of population, 2015-16 would be ranked third.
Hence, option 5.

Option 3 can be directly eliminated as B can finish the complete work in 9 days in that case, whereas B takes 10 days to complete partial work.
Let B and A take x and x + 6 days to complete the work individually.
Hence,

7x + 10x + 60 = x² + 6x
∴ x² - 11 x - 60 = 0
∴ x = 15 or x = - 4
Since number of days cannot be negative, x = 15
Thus, B and A can do the work in 15 and 15 + 6 = 21 days respectively.
Hence, option 1

Original quantity of crude oil in tank = 84 x 0.7 = 58.8 litres
Let x litres be the quantity of crude oil replaced with cheaper oil of 20% purity (or 80% impurity).
Though the overall amount of oil in the tank is still 84 litres, the purity has gone down by 10 percentage points.

∴ 5(58.8 - 0.5x) = 3(84) = 252
∴ 294 - 2.5x = 252
∴ x= 16.8
Hence, option 2.

33 w hen divided by 17 gives a re m aind er of - 1.
∴ 33¹⁷ w ould give a rem ainder of (- 1)¹⁷ = -1 = 16
Hence, option 2.

Using statement A alone:

There is no information on the number of runs scored or the number of matches played by Pratap and Jagtap.
Hence, the total runs scored by each player cannot be found.
Thus, the question cannot be answered using statement A alone.

Using statement B alone:

Let the average runs scored by Pratap be x runs per match.
∴ Average runs scored by Jagtap = 1 .1x per match.
Let the number of matches played by Jagtap be y.
∴ Number of matches played by Pratap = 1.1y
∴ Total runs scored by Pratap = x * 1.1y = 1.1xy ... (1)
and, total runs scored by Jagtap = 1.1 x * y = 1.1xy... (2)
Thus, Pratap and Jagtap have scored the same number of runs i.e. Pratap has not scored more runs than Jagtap at the end of the tournament.
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.

Let the length and the breadth of the rectangle be denoted as / and b respectively.
From the given statement, (b + 2) * (l - 3) = lb
∴ 2l - 3b = 6 ...(i)

Using statement A alone:

(b -1)x (l+ 3) = lb
∴ 3b - l = 3 ...(ii)
From (i) and (ii), l = 9 and b = 4.
Thus, the question can be answered using statement A alone.

Using statement B alone:

(l + 2) x (b+ 1) = 2lb
∴ Ib + 2b + l + 2 = 2lb
∴ 2b+ l+2 = lb ...(iii)
Now from (i),
I = (3b+ 6)/2 ...(iv)
Substituting from (iv) in (iii) we get,
2b + [{3b + 6)/2] + 2 = [(3b + 6)/2] x b
∴ 4b + 3b + 6 + 4 = 3 x b² + 6b
∴ 3b² - b - 10 = 0
∴ 3b² - 6b + 5 b - 10 = 0
∴ 3b(b - 2) + 5(b - 2) = 0
∴ (3b + 5) x (b - 2) = 0
b = 2
Since b is known, / and consequently the area can be found.
Thus, the question can be answered using statement B alone.
Thus, the question can be answered using either of the statements alone.
Hence, option 3.

Let the six digits of the telephone number be ABCDEF.

Using statement A alone:

Since the number is divisible by 22, its units digit should be a multiple of 2 and the difference of digits at even and odd places must be a multiple of 11.
However, multiple six-digit numbers satisfy this condition.
So, a unique telephone number cannot be found.
Thus, the question cannot be answered using statement A alone.

Using statement B alone:

A + C + E = 27 and B + D + F = 5
Since A, C and E are single digit numbers, A + C + E = 27 only when A = C = E = 9
Since D = F = 0, B = 5
So, the telephone number is 959090.
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 I alone:

From the information available, we have the number of students those do not have anything (i.e., neither chocolate nor ice-cream) Thus, statement I alone is not sufficient to answer the question.

Using statement II alone:

Let x and y be the number of students those had only chocolates and those had only ice-creams respectively.
It is given that fifty students had both chocolates and ice-creams. Also, x and y are prime numbers. We do not have unique values for x and y.
Thus, statement II alone is also not sufficient to answer the question.

Using both the statements together:

The number of students who had both chocolates and ice-creams = 50
∴ x + y + 50 = 75
∴ x + y = 75 - 50 = 25
We need to find the value of x
Now we have two possibilities:
x = 2 , y = 23 OR x = 23, y = 2
Unique value of x cannot be found.
Thus, the question cannot be answered.
Hence, option 5.

Using statement A alone:

Hraday went to a movie 10 days ago. Since there is no other reference point, the day corresponding to today cannot be found.
Thus, the question cannot be answered using statement A alone.

Using Statement B alone:

Hraday goes to the movie only on Sunday. Since there is no other reference point, the day corresponding to today cannot be found.
Thus, the question cannot be answered using statement B alone.

Using statements A and B together:

Hraday went to a movie 10 days ago and he goes to the movie only on a Sunday.
So, today has to be 10 days after Sunday, i.e., Wednesday.
Thus, the question can be answered using both statements together but not by using either statement alone.
Hence, option 4.

Using statement A alone:

y²= xz
This is the only conclusion that can be drawn.
Hence, A alone is not sufficient to answer the question.

Using statement B alone:

Multiplying the above two equations, we get,


Hence, 2, log_y x, log_z y and log_x z do not form an AP.
Hence, statement B alone is sufficient to answer the question.
Hence, option 2.