Solution

To find the total amount Seema paid to the shopkeeper, we calculate the total cost of each item type and then sum these costs.

Cost of Pens

There are 20 pens at ₹7 each, so the total cost for pens is:

\[
20 \times 7 = ₹140
\]

Cost of Wax Colours

There are 8 packets of wax colours at ₹22 each, so the total cost for wax colours is:

\[
8 \times 22 = ₹176
\]

Cost of Calculators

There are 6 calculators at ₹175 each, so the total cost for calculators is:

\[
6 \times 175 = ₹1050
\]

Cost of Pencil Boxes

The price of one pencil box is ₹14 more than the combined price of one pen and one packet of wax colours. Therefore, the price of one pencil box is:

\[
7 (pen) + 22 (wax colour) + 14 = ₹43
\]

Since there are 7 pencil boxes, the total cost for pencil boxes is:

\[
7 \times 43 = ₹301
\]

Total Amount Paid

Adding up all the costs:

\[
140 (pens) + 176 (wax colours) + 1050 (calculators) + 301 (pencil boxes) = ₹1667
\]

Therefore, Seema paid a total of ₹1667 to the shopkeeper.

The correct answer is (c) ₹1,667.

Solution

The hostel initially has provisions for 250 students for 35 days. Let’s calculate how the provisions are affected by changes in the number of students.

Initial Provisions Consumption

For the first 5 days, 250 students consume the provisions. This consumption rate leaves provisions for 30 days for 250 students.

After 5 Days

After 5 days, 25 more students are admitted, increasing the total to 275 students. These 275 students will consume the provisions faster.

Calculation of Remaining Provisions

The total provisions can be seen as “student-days” of food. Initially, this is \(250 \times 35 = 8750\) student-days.

After 5 days of consumption by 250 students, \(250 \times 5 = 1250\) student-days of food are consumed, leaving \(8750 – 1250 = 7500\) student-days of food.

Provisions for 275 Students

For the next 10 days, there are 275 students, consuming \(275 \times 10 = 2750\) student-days of food.

After this consumption, \(7500 – 2750 = 4750\) student-days of food remain.

Adjustment for Student Departure

Following the departure of 25 students after these 10 days, 250 students remain. This is 15 days into the original 35 days.

Remaining Provisions Duration

To find out for how many more days the remaining provisions can last for 250 students:

\[
\text{Remaining days} = \frac{\text{Remaining student-days of food}}{\text{Number of students}} = \frac{4750}{250} = 19 \text{ days}
\]

Therefore, the remaining provisions will last for 19 days after the initial 5 days and the adjustments in student numbers.

The correct answer is (b) 19 days.

Given the equation \(\frac{x^{2}+y^{2}+z^{2}-64}{xy-yz-zx}=-2\) and the condition that \(x + y = 3z\), we proceed to find the value of \(z\) as follows:

First, we note that:

\[
x^2 + y^2 + z^2 – 64 = -2(xy – yz – zx)
\]

Additionally, we have the identity that can be used:

\[
[x + y + (-z)]^2 = x^2 + y^2 + z^2 + 2(xy – yz – zx)
\]

Substituting \(x + y = 3z\) into this identity, we get:

\[
(3z – z)^2 = x^2 + y^2 + z^2 + 2(xy – yz – zx)
\]

Which simplifies to:

\[
(2z)^2 = x^2 + y^2 + z^2 + 2(xy – yz – zx)
\]

Using the given equation and substituting \(-2(xy – yz – zx)\) for \(x^2 + y^2 + z^2 – 64\), we equate this to \((2z)^2\), resulting in:

\[
(2z)^2 = 64
\]

This simplifies to:

\[
4z^2 = 64
\]

Therefore, solving for \(z\):

\[
z^2 = 16
\]

Given \(z\) is positive (implied by the context), we find:

\[
z = 4
\]

The value of \(z\) is 4.

The answer is (c) 4.

Solution

Let the two numbers be \(x\) and \(y\). According to the given conditions:

• The sum of the two numbers is \(12\), which means \(x + y = 12\).
• Their product is \(35\), which means \(xy = 35\).

We are asked to find the sum of the reciprocals of these numbers, which is \(\frac{1}{x} + \frac{1}{y}\).

Using the properties of fractions, the sum of the reciprocals can be rewritten as:

\[
\frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy}
\]

Substituting the given values for \(x + y\) and \(xy\), we get:

\[
\frac{x + y}{xy} = \frac{12}{35}
\]

Therefore, the sum of the reciprocals of the two numbers is \(\frac{12}{35}\).

The correct answer is (a) \(\frac{12}{35}\).

Solution

To find the value of \(x\) given the equation \(\frac{2 x}{1+\frac{1}{1+\frac{x}{1-x}}}=1\), we can start by simplifying the complex fraction:

\[
\frac{2x}{1 + \frac{1}{1 + \frac{x}{1 – x}}} = 1
\]

Step 1: Simplify the Innermost Fraction

First, simplify the fraction inside:

\[
1 + \frac{x}{1 – x}
\]

Getting a common denominator:

\[
\frac{1 – x + x}{1 – x} = \frac{1}{1 – x}
\]

Step 2: Simplify the Next Fraction

Now, plug this back into the original equation:

\[
\frac{2x}{1 + \frac{1}{\frac{1}{1 – x}}} = 1
\]

Simplify the denominator further:

\[
\frac{2x}{1 + (1 – x)} = 1
\]

\[
\frac{2x}{2 – x} = 1
\]

Step 3: Solve for \(x\)

Multiply both sides by \(2 – x\) to get rid of the denominator:

\[
2x = 2 – x
\]

Add \(x\) to both sides:

\[
3x = 2
\]

Divide by 3:

\[
x = \frac{2}{3}
\]

Therefore, the value of \(x\) is \(\frac{2}{3}\).

The correct answer is (a) \(\frac{2}{3}\).