1. A box contains chocolates and the total weight of the box and the chocolates is 3 pounds. If 2/5^{th} of the chocolates are eaten, and now the box with the remaining chocolates weighs 2 pounds, then what is the weight of the empty box in pounds?

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**Correct answer is option - 4**

**Explanation:**

Let the weight of the box be = ‘b’ pounds.

Let the weight of the chocolates be = ‘c’ pounds.

Then given b + c = 3 pounds and (2/5^{th} * c) is eaten.

Remaining box with chocolates weigh = 2 pounds ==> chocolates eaten weigh = 3 – 2 = 1 pound.

So 2/5 * c = 1 pound==> c = (1 * 5)/ 2 ==> c = 2.5 pounds.

Since b + c = 3 pounds ==> b = w(3 – 2.5) = 0.5 pounds.

2. The midpoint of a line joining the point P and Q is (4, -1). If the point Q is (2, -3), then what is the coordinate of point P?

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**Correct answer is option - 4**

**Explanation:**

The midpoint of the line PQ is (4, -1) and the point Q is (2, -3)

Midpoint of a line segment joining (x_{1}, y_{1}) and (x_{2}, y_{2}) is ((x_{1} + x_{2})/2, (y_{1}+ y_{2})/ 2)

Let the point P = (x, y).

This gives: (4, -1) = ((x + 2)/2, (y – 3)/2)

This implies: 4 = (x + 2)/ 2 ==> x + 2 = 8 ==> x = 6.

Similarly, -1 = (y – 3)/2 ==> y – 3 = -2 ==> y = 1.

Hence the point P is (6, 1).

3. If log_{b}(2) = p and log_{b}(7) = q, then what is log_{b}(b/14)?

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**Correct answer is option - 5**

**Answer:**

log_{b}(b/14) can also be written as log_{b}(b) – log_{b}(14) which gives: 1 – (log_{b}(14)).

Now log_{b}(14) can also be written as: log_{b}(14) = log_{b}(2 * 7) ==> log_{b}2 + log_{b}7.

Hence we get: 1 – (log_{b}(14)) = 1 – (log_{b}2 + log_{b}7).

This implies: log_{b}(b/14) = 1 – (p + q) ==> 1 – p – q.

4. If n, n + 1 and n + 2 are three consecutive integers and if their sum is equal to ‘s’, then what is the value of (s/3) – 1 ?

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**Correct answer is option - 1**

**Explanation:**

The three consecutive integers are n, n + 1 and n + 2.

Sum of the three integers is: n + n + 1 + n + 2 = s and this gives: 3n + 3 = s

Now from the give question: (s/3) – 1 ==> [(3n + 3)/ 3] – 1 ==> 3(n + 1)/3 – 1.

The ‘3’ cancels and we get: n + 1 – 1 ==> n.

Therefore the answer is: n.

5. If f(x) = 3a – 2b and f(y) = 12a – 8b, then what is the value of f(y)/ f(x)?

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**Correct answer is option - 2**

**Explanation: **f(y)/ f(x) = (12a – 8b)/ (3a – 2b) ==> f(y)/ f(x) = 4(3a – 2b)/ (3a – 2b).

Therefore we get: f(y)/ f(x) = 4.

6. On a paper of length 12.5cm and width 9cm, a margin of 0.5cm is drawn all around and is cut from the paper. What is the area covered by the remaining paper in cm^{2}?

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**Correct answer is option - 3**

**Explanation: **When a picture is drawn, it can be clearly observed that if a margin of 0.5cm is cut from every side, then the length and the width is reduced by (0.5 * 2) = 1cm each.

This gives: length of the paper = 12.5cm – 1cm = 11.5cm

Similarly, width of the paper = 9cm – 1cm = 8cm.

Therefore the area of the remaining portion of the paper is = 11.5cm * 8cm = 92cm^{2}

7. As shown in the figure below, OABC is a rectangle where side BC measures 12m and side OC measures 9m. What is the length of the arc in meters, in the figure?

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**Correct answer is option - 2**

**Explanation: **OABC is a rectangle and OB is the radius of the given part of the circle.

OB is also the diagonal of the rectangle and we can use the Pythagorean Theorem.

In triangle OBC, OC^{2} + BC^{2} = OB^{2} ==> 9^{2} + 12^{2} = 81 + 144 = 225 = OB^{2}.

Therefore, OB = √225 ==> OB = 15m.

Arc length = (radius) * θ * (π/ 180°) where θ = angle in degrees = 90° (angle made at the center).

This gives: Arc length = (15m) * (90° * π/ 180°) ==> Arc length = 7.5π.