Tag: business maths

Questions Related to business maths

A businessman marks his goods in such a way that even after allowing $12.5 \%$ discount on a cash purchase he gains $20 \%$. If the cost price is Rs. $140$, the marked price is

business maths commission,brokerage and discount advantages of bill of exchange definition, characteristics and parties of bills of exchange simple transactions related to bills of exchange

  1. Rs. $162$

  2. Rs. $172$

  3. Rs. $193$

  4. Rs. $192$

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Correct Option: D
Explanation:

Business man gains $20\%$ on his cost price, that means, he has sold his goods,
$140 + \dfrac{140\times20}{100} =$ Rs. $168$
Selling price $= 168$

Now, let marked price was Rs. $x$
According to question,
$x - \dfrac{12.5 x}{100} = 168$
$\Rightarrow\dfrac{100x - 12.5 x}{100} = 168$
$\Rightarrow 100x - 12.5 x = 16800$
$\Rightarrow 87.5 x = 16800$
$\Rightarrow x = \dfrac{16800}{87.5}$
$\Rightarrow x = 192$
So, marked price was Rs. $192$

A money lender finds that due to reducing the interest rate decrease from $8\%$ to $7\dfrac{3}{4}\%$, his yearly income diminishes by Rs. 61.50. His capital is 

business maths commission,brokerage and discount advantages of bill of exchange definition, characteristics and parties of bills of exchange simple transactions related to bills of exchange

  1. Rs. 22400

  2. Rs. 23800

  3. Rs. 24600

  4. Rs. 26000

Show answer
Correct Option: C
Explanation:
Difference in rate$=\left(8-7\dfrac{3}{4}\right)\%=\dfrac{1}{4}\%$
Let the capital be rate Rs.$x$
$\therefore  \dfrac{1}{4}\%$ of $x=61.50$
$\therefore x = 61.50\times 100\times 4$
     Rs$=24600$

A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then $\displaystyle P(A \cup B)$ is

business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

  1. $\displaystyle \frac{2}{5}$

  2. $\displaystyle \frac{3}{5}$

  3. 0

  4. 1

Show answer
Correct Option: D
Explanation:

Here, $\displaystyle A=\left{ 4,5,6\right} $
and  $\displaystyle B=\left{ 1,2,3,4,\right} $
We have $\displaystyle A\cup B=\left{1,2,3,4,5,6\right} =S$
Where S is the sample space of the experiment of throwing a die.
$\displaystyle P(S) =1$ since it is a sure event.Hence $\displaystyle P(A\cup B)=1$

Twenty bags or sugar, each marked 10 kg actually give the following data:

The weight of a bag (in kg) No. of bags
     $9.5-9.8$          1
     $9.8-9.9$          2
     $9.9-10.0$          5
     $10.0-10.1$         12

The lower limits of the classes are inclusive and the upper limits are exclusive.

What is the probability that the bag selected at random (without any reference) weighs 10 kg or more?

business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

  1. $0.6$

  2. $0.8$

  3. $0.5$

  4. $0.3$

Show answer
Correct Option: A
Explanation:

Total number of bags = $12 + 5 + 2 + 1 = 20$
Bags which weigh more than $10 kg = 12$
Thus Probability (bag weighs more than 10 kg) = $\dfrac{12}{20} = 0.6$

In an experiment, there are exactly three elementary events. The probability of two of them are $\displaystyle\frac{2}{7}$ and $\displaystyle\frac{1}{7}$. What is the probability of third event?

business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

  1. $\displaystyle\frac{4}{7}$

  2. $\displaystyle\frac{3}{7}$

  3. $\displaystyle\frac{2}{7}$

  4. $\displaystyle\frac{1}{7}$

Show answer
Correct Option: A
Explanation:

We know that the sum of all the elementary events = 1
$\displaystyle\frac{2}{7} + \displaystyle\frac{1}{7} + P _3 = 1$ 

$\Rightarrow \quad P _3 = 1 - \left(\displaystyle\frac{2}{7} + \displaystyle\frac{1}{7}\right) = \displaystyle\frac{4}{7}$ 

If $n(A)=6,n(B)=8$ and $n(A\cup B)=12$, then $n(A\cap B)=$

business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

  1. $6$

  2. $2$

  3. $8$

  4. $12$

Show answer
Correct Option: B
Explanation:
We know that $n(A)+n(B)=n(A\cup B)+n(A\cap B)$
$n(A)=6,$ $n(B)=8$ and $n(A\cup B)=12$
$n(A)+n(B)-n(A\cup B)=n(A\cap B)$
$n(A\cap B)=6+8-12=2$

Let A be a set of $4$ elements. From the set of all functions from A to A, the probability that it is an into function is?

business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

  1. $\dfrac{3}{32}$

  2. $0$

  3. $\dfrac{29}{32}$

  4. $1$

Show answer
Correct Option: A
Explanation:
Let set A be of n element and from A to A then no of onto element is n! and total function are $n^{2}$

$P(E)=\dfrac{n!}{n}=\dfrac{4!}{4\times 4\times 4\times 4}$

$=\dfrac{4\times 3\times 2}{4\times 4\times 4\times }=\dfrac{3}{32}$

From a well shuffled standard pack of $52$ playing cards, one card is drawn. What is the probability that it is either a King of hearts or a Queen of diamonds.

business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

  1. $\dfrac {1}{2}$

  2. $\dfrac {1}{4}$

  3. $\dfrac {1}{8}$

  4. $\dfrac {1}{26}$

Show answer
Correct Option: D
Explanation:

No. of Kings of heart $=$ 1, No. of Queen of diamonds $=$1
Probability of king of hearts or queen of diamonds $ = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26}$

If a coin is tossed twice, then the events 'occurrence of one head',  'occurrence of $2$ heads' and 'occurrence of no head' are -

business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

  1. Independent

  2. Equally likely

  3. Not equally likely

  4. Both (A) and (B)

Show answer
Correct Option: C
Explanation:
Possible outcomes when a coin is tossed $=$ H or T
Possible outcomes when a coin is tossed twice$=$ HH,TT,TH,HT
$=4$ outcomes
P(occurrence of $1H$) $=$ TH,HT
$=2/4=1/2$
P(occurrence of $2H$) $=$ HH
$=1/4$
P(occurrence of no H) $=$ TT
$=1/4$
Therefore, the probability of occurrence of no head and occurrence of no head and occurrence of two head is same but probability of occurrence of one Head is $1/2$. Thus the events are:-
Not equally likely.

If events A and B are independent and P(A) $=$ 0.15, P(A $\cup $ B) $=$ 0.45, Then P(B) $=$ ..............

business maths probability - i addition theorems of probability statistics and probability descriptive statistics and probability

  1. $\displaystyle \frac{6}{13}$

  2. $\displaystyle \frac{6}{17}$

  3. $0.315$

  4. $0.352$

Show answer
Correct Option: B,D
Explanation:
Events A and B are independent then $P\left( A\bigcap { B }  \right) =0$
$P\left( A\bigcup { B }  \right) =P(A)+P(B)-P(A).P(B)$
Given: $P(A)=0.15$
$P\left( A\bigcup { B }  \right) =0.45$
$\therefore P(B)=P\left( A\bigcup { B }  \right) -P(A)+P\left( A \right) P(B)$
$P(B)=0.45-0.15+1.15P(B)$
$P(B)=0.3+0.15P(B)$
$P(B)(1-0.15)=0.3$
$P(B)(0.85)=0.3$
$P(B)=\cfrac { 0.3 }{ 0.85 } =0.352=\cfrac { 6 }{ 17 } $