A train running between two stations A and B arrives at its destination 15minutes late, when its speed is 45 km/hr. and 36 minutes late when its speed is 36 km/hr. find the distance between the stations A and B?

ANS:

Let ‘x’ km be the distance between the station A and B.

Speed of the train = 45 km/hr

Time taken =x/45 hours

Since the train is late by 15 minutes =1/4

Actual time =(x/45 -1/4)hr-------(1)

Time taken when the speed is 36 km/hr=x/36 hrs

Now, since the train is late by 36 min i.e.,36/60=3/5

actual time= (x/36-3/5)hrs ---------(2)

 Equating (1) and (2), we get x/45 -1/4 =x/36 -3/5 

∴ x/45 -1/4 =x/36 -3/5 

x=63km                       

 In a party of 80 people each person handshakes with the other. Find the total number of handshakes?

A)3160  B)3280 C)3260  D)2296

Answer:A)3160 

To calculate the number of handshakes, we can use the formula combination =nC2= n(n-1)/2, where n is the number of people. In this case, n = 80.

So, the number of handshakes would be 80(80-1)/2 = 3,160 handshakes.

 A sum of Rs.2000 is to be divided among three people so A,B and C.'A'receives thrice as much as 'B' and 'B' receives one fourth that of 'C'.Then A's share is?

A)Rs1000  B)Rs750 C)Rs250 D)Rs500

Ans:B)Rs750

Given:

Total sum to be divided: Rs. 2000

Three people: A, B, and C

A receives three times as much as B

B receives one fourth of C

Let's denote the amounts received by A, B, and C as follows:

Let B's share be x.

Then, A's share is 3x (three times as much as B).

And, C's share is 4x (since B receives one fourth of C).

Step 1: Setting up the Equations:

The total sum is divided among A, B, and C: A + B + C = 2000

Step 2: Expressing A, B, and C's Shares:

A = 3x

B = x

C = 4x

Step 3: Substituting into the Total Sum Equation:

3x + x + 4x = 2000

8x = 2000

x = 250

Step 4: Calculating A's Share:

A's share = 3x = 3 * 250 = Rs. 750

Therefore, A's share is Rs. 750 (Option B).

 40 men can do certain work in 25 days. After 10 days 25 men left. Find the number of days in which the remaining work is completed?

A)20 days      B)40 days C)15 days D)25 days

Ans:B)40 days

 Given:

40 men can complete a certain work in 25 days.

After 10 days, 25 men left.

Let's break down the problem into steps:

Step 1: Calculate the Total Work:

Let the total work be represented by ( W ).

40 men can complete the work in 25 days, so the total work is completed in ( 40 \times 25 = 1000 ) man-days.

Step 2: Work Done in 10 Days:

In 10 days, 40 men complete ( 40 \times 10 = 400 ) man-days of work.

Step 3: Work Remaining:

After 10 days, 25 men left, so the remaining men working are 40 - 25 = 15 men.

Work remaining after 10 days = Total work - Work done in 10 days = 1000 - 400 = 600 man-days.

Step 4: Calculate the Number of Days to Complete the Remaining Work:

The remaining work of 600 man-days will be completed by 15 men.

Number of days to complete the remaining work = Remaining work / (Men working * Days) = 600 / (15 * x), where x is the number of days.

Step 5: Solve for x:

( 600 = 15x )

( x = 600 / 15 = 40 )

Therefore, the number of days in which the remaining work is completed is 40 days (Option B).

 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.