A chemist has two alcohol solutions of different strength, 30% alcohol and 45% alcohol solutions, respectively. How many cubic centimeters of each must he use so as to make a mixture of 30 cubic centimeters which will contain 39% alcohol.

(a)10 and 20 

(b)6 and 24

(c)12 and 18 

(d)15 and 15

Ans:

Given:

Two alcohol solutions: 30% and 45% alcohol solutions.

Mixture volume: 30 cubic centimeters.

Target alcohol concentration in the mixture: 39%.

Let:

( x ) be the volume (in cubic centimeters) of the 30% alcohol solution.

( y ) be the volume (in cubic centimeters) of the 45% alcohol solution.

Step 1: Write the Equations based on the Mixture:

The total volume of the mixture is 30 cubic centimeters: ( x + y = 30 ) (Equation 1).

The total amount of alcohol in the mixture is 39% of 30 cubic centimeters: ( 0.30x + 0.45y = 0.39 \times 30 ) (Equation 2).

Step 2: Solve the Equations:

From Equation 1, we have ( x = 30 - y ).

Substitute ( x ) in Equation 2: ( 0.30(30 - y) + 0.45y = 11.7 ).

Simplify the equation: ( 9 - 0.30y + 0.45y = 11.7 ).

Further simplify: ( 0.15y = 2.7 ) and ( y = 18 ).

Substitute ( y = 18 ) in ( x = 30 - y ): ( x = 12 ).

Step 3: Check the Solution:

The chemist should use 12 cubic centimeters of the 30% alcohol solution and 18 cubic centimeters of the 45% alcohol solution to make a 30 cubic centimeter mixture with 39% alcohol content.

Therefore, the correct answer is (c) 12 and 18.

 The average marks of 28 students in mathematics were 50.Eight students left the school and the average increased by 5.

What is the average marks obtained by the students  who left the school?

A)45 B)42.5 C)37.5 D)50.5

ANS:37.5

Let's solve this step by step considering you are in school:

Given:

The average marks of 28 students in mathematics were 50.

Eight students left the school and the average increased by 5.

Step 1: Initial Average Calculation

Initially, the average marks of 28 students were 50.

Total marks of 28 students = 28 * 50 = 1400.

Step 2: Average After Students Leave

After 8 students leave, there are 20 students left.

The average increases by 5, so the new average is 50 + 5 = 55.

Step 3: New Total Marks Calculation

Total marks of 20 students with the new average = 20 * 55 = 1100.

Step 4: Total Marks of Students Who Left

Total marks of the 8 students who left = Total marks before - Total marks after they left

Total marks of the 8 students who left = 1400 - 1100 = 300.

Step 5: Average Marks of Students Who Left

Average marks = Total marks of students who left / Number of students who left

Average marks = 300 / 8 = 37.5.

Therefore, the average marks obtained by the students who left the school is 37.5 (Option C).

Permutation and combination

 If there are 30 cans out of them one is poisoned if a person tastes very little he will die within 14 hours so if there are mice to test and 24 hours to test, what is the minimum no. of mice’s required to find poisoned can?

  a) 3                   b) 2                

  c) 5                   d) 1

ANS:

with 5 mice we can make 2^5 =32 possibilities ,In that first possibility no mice eat a can would be avoid other 31 combinations are used for test.

let A,B ,C,D and E be the mice,

0-indicate mice not eat that can

1-indicate mice eat that can

A      B      C     D    E

0      0       0     0     1 ---1st can

0      0       0     1     0----2nd can

0      0       1    0      0------3rd can

.  

.

.

1     1       1     0     0......29th can

1      1       1    1      0......30th can

Above 30 combinatins are used for test  poissined 'can' can which combination of mice is die.

In a world cup cricket tournament two group are present ,in each group seven teams are present each team play match against all the other team in the group only once.First four teams in each group are qualified for next stage.The Second stage is a knockout stage.winners of the knockout stage will participate in semifinal,and the winners will participate in the final.Total how many matches are present in the tournament?

Ans:

In first stage there are   7C2   matches are present in each group.

Total matches in first stage=2* 7C2 =2*21=42. 

4 matches present in the second stage,2 in semi and one in final Total matches=42+4+2+1=49

A hollow cube of size 5 cm is taken, with a thickness of 1 cm. It is made of smaller cubes of size 1 cm. If 1 face of the outer surface of the cube are painted, totally how many faces of the smaller cubes remain unpainted?

a) 900   b) 488    c) 563   d) 800

Ans: Total cubes to form hollow cube ?

If it is cube it contains how many cubes =5^3=125

How many number of  small cubes  required to form a hollow cube= 125-3^3=98

Total number of faces in small cubes=98*6=588

number of painted faces=5*5=25

Number of unpainted faces=588-25=563

From a group of 6 men & 5 women, 4 persons are to be selected to form a committee such that at least 2 men are in committee. In how many ways it can be done?  

(A) 150        (B) 250    (C) 265    (D) 115

Ans:C)265

To find the number of ways to select 4 persons from a group of 6 men and 5 women to form a committee such that at least 2 men are in the committee, we can use combinations.

We need to consider the following cases:

Selecting 2 men and 2 women

Selecting 3 men and 1 woman

Selecting 4 men

Case 1: Selecting 2 men and 2 women Number of ways to select 2 men from 6 men = 6C2 Number of ways to select 2 women from 5 women = 5C2 Total ways for this case = 6C2 * 5C2

Case 2: Selecting 3 men and 1 woman Number of ways to select 3 men from 6 men = 6C3 Number of ways to select 1 woman from 5 women = 5C1 Total ways for this case = 6C3 * 5C1

Case 3: Selecting 4 men Number of ways to select 4 men from 6 men = 6C4

Total number of ways to form the committee = Total ways for Case 1 + Total ways for Case 2 + Total ways for Case 3 Total ways = (6C2 * 5C2) + (6C3 * 5C1) + 6C4

Calculating the values: 6C2 = 15 5C2 = 10 6C3 = 20 5C1 = 5 6C4 = 15

Substitute these values into the formula: Total ways = (15 * 10) + (20 * 5) + 15 Total ways = 150 + 100 + 15 Total ways = 265

Therefore, there are 265 ways to select 4 persons from the group of 6 men and 5 women to form a committee such that at least 2 men are in the committee.

4 men or 8 women can do a piece of work in 24 days. In how many days, will 12 men and 8 women do the same work?

A)6      B)8  C)5      D)None of these

Answer:6

The efficiency of men is twice that of women

8 women are equal to 4 men

So 12 men and 8 women = 16 men

4 men can do the work in 24 days

16 men need 24/4 =6 days