Business Mathematics Test

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Business Mathematics Test - 1

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Question 1
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The number of ways in which 8 different books can be arranged on a shelf so that 3 particular books shall not be together:

A
11! – 3!

B
361000

C
8! × 3! - 5!

D
36000

Solution

Number of ways in which 8 books can be arranged = 8!

Number of ways when three particular books are together = 6! × 3!

Therefore Number of ways when three particular books are not together = 8! × 6! - 3!

= 6!(7 × 8 - 3 × 2)

= 6! × 50

= 720 × 50 = 36000 ✅
Question 2
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Find the interval in which the function f(x) = x² − 2x is strictly increasing?

A
(1, ∞)

B
(1, ∞)

C
(0, ∞)

D
(-∞ , 1)

Solution
Question 3
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In any group, the number of improper subgroups is:

A
2

B
3

C
Depends on the group`

D
1

Solution

If G is a group, then the subgroup consisting of G itself is the improper subgroup of G. All other subgroups are proper subgroups

In any group, the number of improper subgroups is 2.
Question 4
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If A = {x : x is a multiple of 3}, B = {x : x is a multiple of 4} and C = {x : x is a multiple of 12}, then A ∩ (B ∩ C) is equal to:

A
A

B
B

C
C

D
{ x : x is a multiple of 144}

Solution
Question 5
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Out of 85 children playing badminton or table tennis or both, the total number of girls in the group is 70% of the total number of boys in the group. The number of boys playing only badminton is 50% of the number of boys and the total number of boys playing badminton is 60% of the total number of boys. The number of children playing only table tennis is 40% of the total number of children and a total of 12 children play badminton and table tennis both. The number of girls playing only badminton is:

A
14

B
16

C
17

D
35

Solution
Question 6
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There are 12 points in a plane out of which 5 are collinear. The number of triangles formed by the points as vertices is:

A
185

B
210

C
220

D
175

Solution

Total number of triangles that can be formed with 12 points (if none of them are collinear).

= ¹²C₃

(this is because we can select any three points and form the triangle if they are not collinear).

With collinear points, we cannot make any triangle (as they are in straight line). Here 5 points are collinear. Therefore we need to subtract ⁵C₃ triangles from the above count.
Question 7
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Find the sum of all numbers divisible by 6 in between 100 to 400.

A
12,550

B
12,450

C
11,450

D
11,550

Solution

Given,

1^st term = a = 102

(Which is the 1^st term greater than 100 that is divisible by 6.)

The last term less than 400, Which is divisible by 6 is 396.

Terms in the AP; 102, 108, 114 … 396

Now

First term = a = 102

Common difference = d = 108 - 102 = 6

n^th term = last term (l) = 396

As we know, nth term of AP = a_n

= a + (n – 1) d

⇒ 396 = 102 + (n - 1) × 6

⇒ 294 = (n - 1) × 6

⇒ (n - 1) = 49

∴ n = 50
Question 8
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Suman has Rs. 265 consisting of only denominations of Rs. 2 and Rs. 5 coins and thus total number of coins is 80. Find the number of coins in Rs. 5 denomination.

A
30

B
45

C
35

D
25

Solution

Given:

Total number of coins = 80

Calculations:

Suppose Suman has ‘x’ and ‘y’ number of coins of denominations of Rs. 2 and Rs. 5 respectively.

∴ 2x + 5y = 265 ---- (1)

And

x + y = 80 ------(2)

Multiplying equation (2) by 2 and subtracting it from equation (1)

3y = 265 - 160

⇒ y = 35

∴ Number of coins in Rs. 5 denomination = 35
Question 9
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If in an A.P. pth term is q and th qth term is p, then which term will be zero?

A
(p + 1)^thterm

B
(q + 1)^thterm

C
(p - q)^thterm

D
( p+q)^thterm

Solution

Let a be the first term and d be the common. difference

Then T_p = a + (p - 1)d

⇒ a + (p - 1)d = q ---(i)

T_q = a + (q - 1)d

⇒ a + (q - 1)d = q ---(ii)

Solving equation (i) and (ii), we get

a = p + q - 1 and d = -1

∴ T_p+q = (p + q - 1) + (p + q - 1)(-1)

= (p + q - 1) - (p + q - 1) = 0
Question 10
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The value of ordinate of the graph of y = 2 + cos⁡ x lies in the interval:

A
[0, 1]

B
[0, 3]

C
[-1, 1]

D
[1, 3]

Solution