Practice Test for NMAT - 13 - CAT MCQ

The abscissa and ordinate stand for the X-axis and Y-axis respectively.
Thus, figures on the X-axis stand for the total exports for a given year while those on the Y-axis stand for the export of herbal products in that year.
Thus, in 2006, the total exports were Rs. 5,500 lakhs, whereas the exports of herbal products were Rs. 750 lakhs.

Hence, option 2.

In 2006, the herbal exports were Rs. 750 lakhs, whereas in 2007, they were Rs. 650 lakhs.

Hence, option 3.

In 2004, the herbal exports were Rs. 500 lakhs while at the end of the period, i.e. in 2007, they were Rs. 650 lakhs.

Hence, option 2.

The total export in 2004 were Rs. 5100 lakhs, while in 2007, they were Rs. 6000 lakhs.

This rise is over a period of 3 years.
The average percentage increase in export = 17.65/3 = 5.88% Hence, option 4.

Cost price : Marked price = 2 : 3 and Profit % : Discount% = 3 : 2

Let cost price, (C.P.) = 2x; marked price, (M.P.) = 3x; Profit % = 3y and Discount % = 2y

Solving this, we get

Hence, option 1.

Let the time taken to build the compound when both Jai and Veeru are working together be x.

Time taken by Jai to complete the job alone is x + 8 and the time taken by Veeru to complete the job alone is x + 12.5 Thus we get the following equation:

2x² + 20.5x — x² + 20.5x +100

x² = 100

x = 10
Thus Jai will take 18 hours to complete the job alone.
Hence, option 2.

a, b and c are in continued proportion. 

b² = ac ...(i)

b, c and a are in continued proportion. c² = ab •••(ii)

Multiplying (i) and (ii), we get,

a²bc = c²b²

a² = be ...(iii)

All three will be satisfied only when a = b = c = k (say)

= 45.83
Hence, option 1.

A 4-digit palindrome will be of the form abba where a > 0 and b can take integer values from 0 to 9.

a can take 9 values and b can take 10 values.

The total number of 4-digit palindromes is 9 x 10 = 90 Hence, option 2.

Let the original two digit number be xy. 10x + y + l0y + x = 40 + 2(10x + y)

11x + 1 ly = 45 + 20x + 2y

9y - 9x = 45

y-x = 5 ... (I)

Also, y + x = 9 ... (II)

Solving (I) and (II), y = 7 and x = 2 Hence, the original number is 27.
Hence, option 5.

Number of males in North 2 = 0.756b And, number of males in North = 0.6(a + b)

0.4a + 0.756 = 0.6(a + 6)

0.4a + 0.756 = 0.6a + 0.66

0.2a = 0.156

Hence, option 1.

Number of male employees in West = 55% of 500 = 275

Let there be a employees in West 1 and (500 - a) employees in West 2.

Number of male employees in West 1 = 0.75a

And, number of male employees in West 2 = 0.5(500 - a)

0.75a+ 0 .5(500 - a) = 275

0.75a + 250 - 0.5a = 275

0.25a = 25

 a = 100
Number of male employees in West 2 = 0.5(500 - 100) = 200

Hence, option 1.

Let the total employees in South be a.
Since there are 60 male employees in South 2,

Total employees in South 1 = a - 150. Total male employees in South 1 = 0.2(a - 150) = 0.2a - 30

Also, total male employees in South = 30% of a = 0.3a

0.2a - 30 + 60 = 0.3a

0.1a = 30

a -300

Hence, option 4

Let there be 100 employees in East Number of female employees in East = 55% of 100 = 55

Now, let a employees from East be in East 1 and the remaining (100 - a) be in East 2.

Number of female employees in East 1 = 50% of a = 0.5a

And, number of female employees in East 2 = 60% of (100 - a) = 0.6(100 - a)

Now, 0.5a + 0.6(100 - a) = 55 0.5a + 60 - 0.6a = 55

0.1a = 5

a = 50

Hence, option 2.

Total girls in the university = 25 + 23 + 25 + 12 + 25 + 20 + 1 2 + 3 = 145.
Total boys in the university = 45 + 186 + 120 + 100 + 65 + 32 + 58 + 5 = 611.
Total students = 145 + 611 = 756

Number of students who passed = 611 + 0.8(145) = 727

Hence, option 5.

Business Management has 50 girls and 110 boys i.e. 50 girls in 160 students Computers has 43 girls and 218 boys i.e. 43 girls in 261 students Finance has 37 girls and 178 boys i.e. 37 girls in 215 students Statistics has 15 girls and 105 boys i.e. 15 girls in 120 students It can be directly observed that the number of girls as a percentage of total number of students in that course is the highest in Business Management (31.25%)
Hence, option 1.

The number of students pursuing Computers = 23 + 186 + 20 + 32 = 261 The number of students pursuing Finance = 25 + 120 + 12 + 58 = 215

Hence, option 5.

The Business Management course has 50 girl students Number of successful girl students in the Business Management course = 46% of 50 = 23

Number of successful students in the Computers course = 3 x 23 = 69 The Computers course has 261 students.
Required percentage = 69/261 x 100 = 26.43  26%

Hence, option 4.

where a, b, c are the sides of the triangle and A is the area of the triangle.
D, E, F are the mid points of the sides BC, CA and AB respectively of a triangle ABC.
Hence, from the mid-point theorem, the sides of ΔDEF are half the sides of Δ ABC and A( ΔDEF) is one-fourth the Δ(AABC)

Hence, option 2.

In all there are 15 points in the plane. Number of ways in which four points can be selected from these 15 points is ¹⁵C₄ = 1365.
But to form a quadrilateral, not more than 2 points can be on the same line.
So the cases when a selection of 4 out of 15 points, made in the above manner, will not result in a quadrilateral are:

Case (i):

When all the 4 points selected are on line a.
This selection can be made in ⁴C₄ = 1 way.
Case (ii):

When all the 4 points selected lie on line b.
This selection can be made in ⁵C₄ = 5 ways.
Case (iii):

When 3 points lie on line a, while the 4^th point lies on either line b or is one of the 6 points in the plane.
This selection can be made in ⁴C₃ x ⁵C₃ + ⁴C₃ x ⁶C₁ = 44 ways.
Case (iv): When 3 points lie on line b, while the 4^th point lies on either line a or is one of the 6 points in the plane.
This selection can be made in 5C3 x ⁴C₃ + ⁵C₃ x ⁶C₁ = 100 ways.
So the total number o f quadrilaterals that can be formed is 1 3 6 5 - (1 + 5 + 44 + 100) =1215.

Hence, option 3.

If the side opposite to ∠x is 3p and the hypotenuse is 5p , the adjacent side has to be 4p .
Hence, using the standard definitions of the given trigonometric ratios,

Hence, option 4.

In a 200 m race, Raman gives Suman a head start of 35 m but still beats him by 45 m.
So, Raman starts when Suman has covered 35 m and when Raman completes the race, Suman is yet to cover 45 m i.e. Suman has covered 155 m.
So, in the same time that Raman covers 200 m, Suman only covers 155 - 35 = 120 m.

So, Raman covers 100 m in the same time that Suman covers 60 m.
Therefore, when both start together in a 100 m race, Suman has only covered 60 m when Raman completes the race.
So, Raman beats Suman by 100 - 60 = 40 m.
Hence, option 1.

Expected winning = probability of winning x amount at stake.
P (A winning) = p + (1 - p )( 1 - p )p + (1 - p )(1 - P )(1 -p)(1- p)p + •••

= p + (1 - p )²p + (1 - p)⁴p + ...

The expression in the brackets is an infinite G.R with a = 1 and r = (1 - p )²

Hence, option 5.

Since the same wire is bent from a circle to a square, the circumference of the circle is equal to the perimeter of the square.
Let the radius of the circle measure r cm.

πr² = 471 

r = 2 cm

Perimeter of square = circumference of circle = 2πr = 4π

Side of square = 4π/4= π cm

Area of square = π² sq.cm

Hence, option 3.

a + b + c = 339 ...(i)

Now, c is as much more than the average as much as a is less than the average

 c - 113 = 113 + a

a + c = 226 ...(ii)

Subtracting equation (ii) from equation (i), we get, 
b = 113

Hence, option 3.
Alternatively,
The simplest way to solve this question is to understand that for an arithmetic progression, arithmetic mean of three numbers is equal to the middle term and the difference between the middle term and the first number is equal to the difference between the second term and the middle number.
b=113
Hence, option 3.

Number of days required by Amar to complete the job alone = 540/36 = 15 days

Number of days required by Akbar = 324/18 = 18 days

Number of day srequired by Anthony = 320/32 = 10 days

The fraction of job that will be completed in a day when all three are working together is

4.5 days will be required to finish the job when all the three are working together.

Hence, option 3.

Let log_x² = a

From the above equation we get a = 1/2.

x = 4

Hence, option 3.

Area of original sheet = π x 14² = 19671 sq.cm.
Area of the five circles = 5 x π x 6² = 180π sq.cm.
Area of the remaining portion of the sheet = 16π sq.cm.

Required ratio = 16π : 196π = 4 : 49 Hence, option 4.

Let x be the number of 4 rupee coins that Rajesh paid andy be the number of 5 rupee coins that he paid.

4x + 5y - 150... (i)
Dividing (i) by 4 we get;

Now, as x andy are number of coins, they should be integers.

y - 2 = 4p

y = 4p + 2 ...(ii)

Substituting (ii) in (i) we get,

4x + 5(4p + 2) = 150

4x = 150 - 10 - 20p

4x = 140 - 20p

x = 35 - 5p ...(iii)

For p > 7 , x will be negative and for p < 0, y will be negative.

Now for  will be greater than x and for p < 4, y will be less than x. 

In no case will y be equal to x.
Probability that x > y = 4/8 = 0.5 and probability that x < y = 4/8 = 0.5.
Hence, both the quantities are equal. Hence, option 3.

Alternatively,

Let x be the number of 4 rupee coins that Rajesh paid and y be the number of 5 rupee coins that he paid.

4x + 5y = 150 ... (i)

5y = 150 - 4x

Since the LHS is a multiple of 5, the RHS has to be divisible by 5. Hence, 4x has to be a multiple of 5.
Hence, x can take 8 values i.e. 0, 5, 10, 15, 20, 25, 30 and 35.
The corresponding values ofy are 30, 36, 22, 18, 14, 10, 6 and 2.
Observe that for the first four values, x <y, and for the last four values, x >y

Probability that x > y = 4/8 = 0.5 and probability that x < y = 4/8 = 0.5.
Hence, both the quantities are equal.
Hence, option 3.

Solution: The roots of x² + x - 6 = 0 are x = 2, -3

Let the common root of both equations be x = 2 & let the other root of 2x² - 7x + k = 0 be a.

a = 3/2

Also, product of roots = 

Let the common root of both equations be x = -3 & let the other root of 2x² - 7x + k = 0 be b.

Also, product of roots  = -3b = K/2

Using statement A alone:

Since k is a positive integer, k = 6.
Thus, the question can be answered using statement A alone.

Using statement B alone:

k + 39 is a natural number

For k = 6 or - 39 , k + 39 = 45 or 0

0 is not a natural number.

k = 6.
Thus, the question can be answered using statement B alone.
Thus, the question can be answered using either statement alone.
Hence, option 3.

Let at present, number of workers in unit A and unit B be x andy respectively.
Using statement I alone:

Five days ago;

Number of workers in unit A = x - 5 Number of workers in unit B = y - 5

It is given that, five days ago, the number of workers in Unit A was twice the number of workers in Unit B. That means five days ago, the number of workers in Unit A is greater than the number of workers in Unit B. Hence, we can conclude that even now the number of workers in Unit A is more than the number of workers in Unit B (as equal number of workers have been added over the last 5 days to both the units).
Hence, statement I alone is sufficient to answer the question.
Using statement II alone:

Ten days ago;

Number of workers in unit A = x - 10

Number of workers in unit B = y - 10

It is given that, ten days ago, the number of workers in Unit B was thrice the number of workers in Unit A then even now the number of workers in Unit B must be greater than the number of workers in Unit A (as equal number of workers have been added over the last 10 days to both the units).
Hence, Statement II alone is sufficient to answer the question.
The question can be answered by using either statement alone.
Hence, option 3.