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# "GMAT - Modern Mathematics"

Graduate Management Admission Test - Modern Mathematics Questions & Solutions with Explanation. Practice Questions with Solutions presented to you by MBA Rendezvous.

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## Modern Mathematics

1. Find the graph of y = |x| + 4.

A.

B.

C.

D.

2.

2.    Let f(x) = |x|, g(x) = [x] (greatest integer function), then

A. f(g(x)) > g(f(x))

B. f(g(x)) = g(f(x))

C. f(g(x)) < g(f(x))

D. Either A or B

E. None of these

3. If f(x) = |x|, h(x) = x – 3.5, p(x) = x2, q(x) =   (only positive square roots) g(x) = [x], then

A. p(1 – h(3)) > q(f(–9) –1)

B. h(q(16)) = h(g(p(2)))

C. h(q(16)) = h(p(3))

D. Both A and B

E. None of these

4.Real numbers x and y are related as y = (x – 2) (x – 3). When y is plotted against x: we get the following graph :

If another two real numbers, P and Q follow the relation P – 3 = |(Q – 5) (Q – 6)| then which of the following is the correct graph of P against Q?

A.

B.

C.

D. Both B and C

E. None of these

5. In the following graphs all inclined lines make an angle 45o with positive x-axis.

What best describes their relationship?

A. f(x) = 1 – g(x)

B. 1 – g(x) = h(x)

C. f(x) = h(x)+2

D. f(x) = h(x)

E. None of these

6.  If   , then

A. f(x) = –f(x)

B.

C.

D. f(x) = f(x2)

E. None of these

7. If  , the value of    will be

A. – 7

B. –4.5

C. 7

D. 4.5

E. None of these

8. Find out the number of permutation of the letters of the word FOLDER taking all the letters at a time?

A.6P6 = 6!

B.6C6 = 0

C. P6+1 = 6

D.Both A and B

E. none of these

9.  How many numbers greater than a million can be formed by number of the digits 4, 4, 5, 5, 6 and 0?

A. 460

B. 360

C. 300

D. 320

E. None of these

10. How many four digit numbers greater than 5000 can be formed by using the digits 4, 5, 6 and 7? If the repetition of the digits in the same number is not allowed

A. 18

B. 16

C. 19

D. 20

E. None of these

11. There are 10 points out of which not three are collinear. By joining the points how many straight lines can be formed.

A. 45

B. 44

C. 42

D. 39

E. None of these

12. There are 5 boys and 6 girls. A combination of 4 is to be selected so that it must consist of at least one boy and one girl?

A. 315

B. 310

C. 300

D. 298

E. None of these

13. In a school there are 400 candidates in an examination of three papers, 80% of the candidates passed in at least one subject. Two hundred passed in physics, 100 passed in chemistry, 120 passed in biology, 40 passed in both physics and biology, 60 passed in both physics and chemistry and 50 passed in both biology and chemistry. How many passed in all three subjects.

A. 45

B. 50

C. 53

D. 58

E. None of these

14. How many natural numbers are there not more than 100, which are not divisible by 2, 3, or 5

A. 26
B. 28
C. 30
D. 32
E. None of these

15. In a school there are 200 candidates, 120 study English and 180 study Hindi. How many of them study both subjects?

A. 2

B. 3

C. 4

D.Both A and B

E. None of these

16. There are 10 points in a plane of which no three points are collinear and 4 points are concyclic. The number of different circles that can be drawn through at least 3 points of these points is

A. 116
B. 120
C. 117
D.130
E. None of these

17. Let A be a set of n (> 3) distinct elements. The number of triplets (x, y, z) of the elements of A in which at least two coordinates are equal is

A. 3n2 + 2n

B. 3n2 – 2
C. 3n2 + 2
D. 3n2 – 2n
E. None of these

## Solutions

1. D
While looking at the graphs only, graph of D satisfies the conditions for both positive as well as negative values of x.

2. D
For x = 3.2

f(g(x)) = |[3.2]| = 3

g(f(x)) = [|3.2|] = 3

For x = –3.5

f(g(x)) = |[–3.5]| = 4

g(f(x)) = [|–3.5|] = 3

Thus, either A or B.

3. B

p(1 – h(3)) = 2.25

q(f(–9) – 1) = =

Clearly, (a) is false.

h(q(16)) = 0.5

h(g(p(2))) = 0.5

4. E
The variable P, Q can be replaced by y & x:
y – 3 = |(x – 5) (x – 6)|
Or, y – 3 = |(x – 3 – 2) (x – 3 – 3)|

we observe that the y against x graph of the above function can be obtained, simply by simultaneously translating the graph of y = |(x – 2) (x – 3)| by 3 units in the positive x – direction and 3 units in the +ve y direction. Option (a) shows the graph of y – 3 = |x(x + 1)|.

Option B shows the graph of y = |(x – 2) (x – 3)|

Option C shows the graph of y + 3 = |x(x + 1)|

Hence, option D is correct.

The correct graph of P – 3 = |(Q – 5) (Q – 6)| is as shown below :

5. C

for x < 0
for  x > 0

6. C

7. B

8.  A
Number of letters in the word FOLDER is 6

So the number of arrangements = 6P6 = 6!

9. B
Here 4 is occurring 2 times, 5 is occurring 3 times

Total number of arrangements =

Since a number beginning with 0 is not greater than a million, so total number of numbers beginning with ‘0’ is

So total number of numbers greater than a million is 420 – 60 = 360.

10. A

Total number of arrangements possible is 4P4 = 4!

Total number of arrangements by using the digits 5, 6 and 7 is = 3!

So the total number of required arrangements is 4! – 3! = 24 – 6 = 18

11. A
By joining any two points we will get one line.

So the total number of lines formed

= 10C2

12. B
The different possibilities are

I. 1 boys and 3 girls

II. 2 boys and 2 girl

III. 3 boy and 1 girl

In the first possibility total number of combination is 5C1 × 6C3

In the second possibility total number of combination is 5C2 × 6C2

In the third possibility total number of combination is 5C3 × 6C1

So the total number of combinations are

5C1 × 6C3 + 5C2 × 6C2 + 5C3 × 6C1 = 310

13. B

It is given that n(A U B U C) = 320 (80% of 400)

15. D
The answer of the question given cannot be determined because we don’t know whether every student study at lest one of these subject or not, there may be a case when certain students don’t read English or Hindi, they study physics.

16. C
The number of circles =(10C3 - 4C3)

17. C

Total number of triplets without restriction = n × n × n

The number of triplets with all different co-ordinates = nP3

The required number of triplets = n3 – n (n – 1) (n – 2) = 3n2 – 2n