Math Q&A

Interview Math Problems and solutions

1. Mrs. Rodger got a weekly raise of $145. If she gets paid every other week, write an integer describing how the raise will affect her paycheck.

Solution:

Let the 1st paycheck be x (integer).

Mrs. Rodger got a weekly raise of $ 145.

So after completing the 1st week she will get $ (x+145).

Similarly after completing the 2nd week she will get $ (x + 145) + $ 145.

= $ (x + 145 + 145)

= $ (x + 290)

So in this way end of every week her salary will increase by $ 145.

2. The value of x + x(xx) when x = 2 is:

(a) 10, (b) 16, (c) 18, (d) 36, (e) 64

Solution:

x + x(xx)

Put the value of x = 2 in the above expression we get,

2 + 2(22)

= 2 + 2(2 × 2)

= 2 + 2(4)

= 2 + 8

= 10

Answer: (a)

3. Mr. Jones sold two pipes at $1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the sale of the pipes, he:

(a) broke even, (b) lost 4 cents, (c) gained 4 cents, (d) lost 10 cents, (e) gained 10 cents

Solution:

20 % profit on $ 1.20

= $ 20/100 × 1.20

= $ 0.20 × 1.20

= $ 0.24

Similarly, 20 % loss on $ 1.20

= $ 20/100 × 1.20

= $ 0.20 × 1.20

= $ 0.24

Therefore, in one pipe his profit is $ 0.24 and in the other pipe his loss is $ 0.24.

Since both profit and loss amount is same so, it’s broke even.

Answer: (a)

4. The distance light travels in one year is approximately 5,870,000,000,000 miles. The distance light travels in 100 years is:

(a) 587 × 108 miles, (b) 587 × 1010 miles, (c) 587 × 10-10 miles, (d) 587 × 1012 miles, (e) 587 × 10-12 miles

Solution:

The distance of the light travels in 100 years is:

5,870,000,000,000 × 100 miles.

= 587,000,000,000,000 miles.

= 587 × 1012 miles.

Answer: (d)

5. A man has $ 10,000 to invest. He invests $ 4000 at 5 % and $ 3500 at 4 %. In order to have a yearly income of $ 500, he must invest the remainder at:

(a) 6 % , (b) 6.1 %, (c) 6.2 %, (d) 6.3 %, (e) 6.4 %

Solution:

Income from $ 4000 at 5 % in one year = $ 4000 of 5 %.

= $ 4000 × 5/100.

= $ 4000 × 0.05.

= $ 200.

Income from $ 3500 at 4 % in one year = $ 3500 of 4 %.

= $ 3500 × 4/100.

= $ 3500 × 0.04.

= $ 140.

Total income from 4000 at 5 % and 3500 at 4 % = $ 200 + $ 140 = $ 340.

Remaining income amount in order to have a yearly income of $ 500 = $ 500 - $ 340.

= $ 160.

Total invested amount = $ 4000 + $ 3500 = $7500.

Remaining invest amount = $ 10000 - $ 7500 = $ 2500.

We know that, Interest = Principal × Rate × Time

Interest = $ 160,

Principal = $ 2500,

Rate = r [we need to find the value of r],

Time = 1 year.

160 = 2500 × r × 1.

160 = 2500r

160/2500 = 2500r/2500 [divide both sides by 2500]

0.064 = r

r = 0.064

Change it to a percent by moving the decimal to the right two places r = 6.4 %

Therefore, he invested the remaining amount $ 2500 at 6.4 % in order to get $ 500 income every year.

Answer: (e)

6. Jones covered a distance of 50 miles on his first trip. On a later trip he traveled 300 miles while going three times as fast. His new time compared with the old time was:

(a) three times as much, (b) twice as much, (c) the same, (d) half as much, (e) a third as much

Solution:

Let speed of the 1st trip x miles / hr. and speed of the 2nd trip 3x / hr.

We know that

Speed = Distance/Time.

Or, Time = Distance/Speed.

So, times taken to covered a distance of 50 miles on his first trip = 50/x hr.

And times taken to covered a distance of 300 miles on his later trip = 300/3x hr.

= 100/x hr.

So we can clearly see that his new time compared with the old time was: twice as much.

Answer: (b)

7. If (0.2)x = 2 and log 2 = 0.3010, then the value of x to the nearest tenth is:

(a) -10.0, (b) -0.5, (c) -0.4, (d) -0.2, (e) 10.0

Solution:

(0.2)x = 2.

Taking log on both sides

log (0.2)x = log 2.

x log (0.2) = 0.3010, [since log 2 = 0.3010].

x log (2/10) = 0.3010.

x [log 2 - log 10] = 0.3010.

x [log 2 - 1] = 0.3010,[since log 10=1].

x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].

x[-0.699] = 0.3010.

x = 0.3010/-0.699.

x = -0.4306….

x = -0.4 (nearest tenth)

Answer: (c)

8. If 102y = 25, then 10-y equals:

(a) -1/5, (b) 1/625, (c) 1/50, (d) 1/25, (e) 1/5

Solution:

102y = 25

(10y)2 = 52

10y = 5

1/10y = 1/5

10-y = 1/5

Answer: (e)

9. The fraction (5x-11)/(2x2 + x - 6) was obtained by adding the two fractions A/(x + 2) and B/(2x - 3). The values of A and B must be, respectively:

(a) 5x, -11, (b) -11, 5x, (c) -1, 3, (d) 3, -1, (e) 5, -11

Solution:

 (5x-11)/(2x2 + x-6)  = [A(2x-3) +B(x+2)]/(x+2)(2x-3)
5x - 11 = A(2x2-3) +B(x+2)
5x-11 = (2A +B)x + (-3A + 2B)
5 = 2A + B ...(i)
-11 = -3A + 2B ...(ii)
A = 3
B = -1

Partial Fraction

Save

Answer: (d)

10. The sum of three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the third is 5/8. The second number is:

(a) 15, (b) 20, (c) 30, (d) 32, (e) 33

Solution:

Let the three numbers be x, y and z.

Sum of the numbers is 98.

x + y + z = 98………………(i)

The ratio of the first to the second is 2/3.

x/y = 2/3.

x = 2/3 × y.

x = 2y/3.

The ratio of the second to the third is 5/8.

y/z = 5/8.

z/y = 8/5.

z = 8/5 × y.

z = 8y/5.

Put the value of x = 2y/3 and z = 8y/5 in (i).

2y/3 + y + 8y/5 = 98

49y/15 = 98.

49y = 98 × 15.

49y = 1470.

y = 1470/49.

y = 30 .

Therefore, the second number is 30.

Answer: (c)
Part 1

Question – 1- Let A = {a, b, c} and the relation R be defined on A as follows:
R = {(a, a), (b, c), (a, b)}.
Then, write minimum number of ordered pairs to be added in R to make R reflexive and transitive.

Solution:
In order to make R reflexive, (b, b) and (c, c) will be added to R.
And in order to make R transitive, (a, c) will be added to R.
Therefore, The minimum number of order pair to be added to R will be (b, b), (c, c) and (a, c) - Answer

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Question – 2 – Let D be the domain of real valued function f defined by then, write D.

Solution:
Here given D is the domain of
Therefore,

Therefore, D = [– 5, 5] - Answer

Question - 3 – be defined by respectively. Then find g o f.

Solution:

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Question 4 – be the function defined by

Solution:

Question – 5 – If A = {a, b, c, d} and the function f = {(a, b), (b, d), (c, a), (d, c)}, write f ^{– 1}

Solution:
Given, f = {(a, b), (b, d), (c, a), (d, c)}
Therefore, f ^{– 1} ={(b, a), (d, b), (c, a), (c, d)} Answer

Question – 6 – If is defined by

Solution:
Given,

Question – 7 – Is g = {(1, 1), (2, 3, (3, 5), (4, 7)} a function? If g is described by g(x) = αx + β, then what value should be assigned to α and β?

Solution:
Given, g = {(1, 1), (2, 3, (3, 5), (4, 7)}
Therefore, each of the element of domain will be have unique image.
Consequently, g is a function.


Now, after substituting the value of α in equation (i), we get

Question – 16 – If A = {1, 2, 3, 4}, define relations on A which have properties of being

(a) Reflexive, transitive but not symmetric

Solution:

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(b) Symmetric but neither reflexive nor transitive

Solution:

(c) Reflexive, symmetric and transitive.

Solution:

Question: 17 – Let R be relation defined on the set of natural number N as follows: Find the domain and range of the relation R. Also, verify whether R is reflexive, symmetric and transitive.

Solution:

Thus, R is neither reflexive, nor symmetric and nor transitive.

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Question – 18 – Given A {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:

(a) an injective mapping from A to B

Solution:
But this is an injective mapping.

(b) a mapping from A to B which is not injective

Solution:
Here it is clear that it is not an injective mapping.

(c) a mapping from B to A

Solution:
Here it is clear that every first component is from B and second component is from A, thus h is a mapping from B to A.

Question – 19: Give an example

(i) Which is one-one but not onto

Solution:
Let A be the set of all 100 students in a school in a particular class say ninth. be the mapping defined by
Here it is clear that f is one-one because no two students of the same class can have the same roll number.
Let roll number of student start from 1 and ends on 100.
This implies that 101 in N is not the roll number of any of the student of the class, so that 101 is not an image of any element of A under f.
Therefore, f is not onto.

(ii) Which is not one-one but onto

Solution:
This is onto but not one-one.

(iii) Which is neither one-one nor onto

Solution:

Question: 22 – Each of the following defines a relation on N:

Determine which of the above relations are reflexive, symmetric and transitive.

Solution:
Hence, the given relation is only transitive.

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Hence, the given relation is only transitive.
Therefore, R is not reflexive.
Therefore, R is symmetric.
Therefore, R is not transitive.
Thus, R is only symmetric.
Therefore, R is reflexive.
Therefore, it is clear that R is symmetric.
Therefore, R is transitive.
Therefore, R is not reflexive.
Therefore, R is not symmetric.
Since, there is no element which begins with
Therefore, R is a transitive.

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Question: 23 – Let A = {1, 2, 3, ………, 9} and R be the relation in A x A defined by (a, b) R (c, b) if a + d = b + c for (a, b), (c, d) in A × A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Solution:
Therefore, R is reflexive.
Let (a, b) R (c, d)
Therefore, R is symmetric.
Let (a, b) R (c, d) and (c, d) R (e, f)
Therefore, R is transitive.
Thus, R is reflexive, symmetric and transitive.
Therefore, R is an equivalence relation.
Equivalence class containing {(2, 5)} is {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.

Question: 24 Using the definition, prove that the function is invertible if and only if f is both one-one and onto.

Solution:
By the definition of an invertible function:
A function is defined to be and invertible function, if there exists a function
The function g is called the inverse of f and is denoted by f ^{– 1}.
has to one-one and onto.
Therefore, f(x) should be both one-one and onto.

Question: 25 - Functions are defined respectively, by find:

(i) fog

Solution:

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(ii) gof

Solution:

(iii) fof

Solution:

(iv) gog

Solution:

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Question: 26 – Let * be the binary operation defined on Q. Find which of the following binary operations are commutative.

Solution:

Question: 27 – Let * be binary operation defined on R by Then the operation is:

(i) Commutative but not associative

(ii) associative but not commutative

(iii) neither commutative nor associative

(iv) both commutative and associative.

Answer: (i) is commutative but not associative.
Explanation:
Therefore, * is commutative.
Now, from equation (i) and (ii) it is clear that

Here it is neither one-one nor onto.