All Exams Test series for 1 year @ ₹349 only
Question

Each person in a group of teachers and students is given the same number of chocolates as the number of students. If 4 more students are added then in order to have the same number of chocolates per person as earlier, 28 more chocolates are needed. The total number of students now is

The correct answer is 11

This problem involves setting up algebraic equations based on the given information about teachers, students, and chocolates.

Students and Chocolates Problem Setup

Let's define our variables:

  • Let $T$ be the number of teachers.
  • Let $S$ be the initial number of students.
  • The total number of people initially is $T + S$.

Initially, each person (teachers and students) receives the same number of chocolates as the number of students. So, each person gets $S$ chocolates.

The initial total number of chocolates distributed is the total number of people multiplied by the number of chocolates per person:

\( \text{Initial Total Chocolates} = (T + S) \times S \)

Solving the Problem with More Students

Now, 4 more students are added to the group.

  • The new number of students is $S + 4$.
  • The total number of people now is $T + (S + 4) = T + S + 4$.

The number of chocolates per person remains the same as earlier, which is $S$ chocolates per person.

The new total number of chocolates distributed is the new total number of people multiplied by the number of chocolates per person:

\( \text{New Total Chocolates} = (T + S + 4) \times S \)

According to the problem, 28 more chocolates are needed in the new scenario compared to the initial one. This means:

\( \text{New Total Chocolates} = \text{Initial Total Chocolates} + 28 \)

Substitute the expressions for the total chocolates:

\( (T + S + 4) \times S = (T + S) \times S + 28 \)

Expand both sides of the equation:

\( TS + S^2 + 4S = TS + S^2 + 28 \)

Now, we can simplify the equation by subtracting $TS$ and $S^2$ from both sides:

\( 4S = 28 \)

To find the value of $S$, divide both sides by 4:

\( S = \frac{28}{4} \)

\( S = 7 \)

So, the initial number of students was 7.

Total Students Now

The question asks for the total number of students now. The number of students initially was 7, and 4 more students were added.

\( \text{Total Students Now} = \text{Initial Students} + 4 \)

\( \text{Total Students Now} = 7 + 4 \)

\( \text{Total Students Now} = 11 \)

Therefore, the total number of students now is 11.

Was this answer helpful?

Important Questions from Numerical Ability

  1. Arrange in the correct sequence the Percentage method of recipe adjustment.

    (i) Determine the total weight of the product needed.
    (ii) Calculate the percentage of each ingredient in relation to total weight.
    (iii) Convert all ingredients in recipe to gram unit and determine edible portion.
    (iv) Check the ratio of ingredients which should be in proper balance.
    (v) Multiply each percentage number by the total weight to give exact amount of the ingredient needed.
    (vi) Total the weight of ingredients using edible portion.

    Choose the correct answer from the code given below :
  2. In an examination 10,000 students appeared. The result revealed the number of students who have:

    passed in all five subjects = 5583

    passed in three subjects only = 1400

    passed in two subjects only = 1200

    passed in one subject only = 735

    failed in English only = 75

    failed in Physics only = 145

    failed in Chemistry only = 140

    failed in Mathematics only = 200

    failed in Bio-science only = 157

    The number of students passed in at least four subjects is:

  3. Four identical cones with base diameter of 10 cm are compactly placed inside a box in upright position. What will be the area of square (in cm2) formed by connecting tips of the cones?
  4. How many hollow spheres having inner radius of 1 cm can be completely filled by transferring water from a completely filled hollow sphere having inner diameter of 20 cm ?

  5. The period of a pendulum is given as T = 2 π (l/g)1/2 where g = 9.81 m/s2 and π = 3.1416. The period of a pendulum of length 1 m correct to the first place of decimal in seconds is

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App