# Maths Olympiad Class 9 - Sample question paper 01

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**Q) The mean of a set of seven numbers is 81. If one of the numbers is discarded, then the mean of the remaining numbers is 78. The value of the discarded number is**

A. 98

B. 99

C. 100

D. 101

**Answer:** B. 99

**Explanation:** Let the sum of seven numbers be \(S\). The mean is given by \(\frac{S}{7} = 81\), so \(S = 567\). After discarding one number, the mean becomes 78, and the sum becomes \(6 \times 78 = 468\). The discarded number is \(567 - 468 = 99\). Therefore, the value of the discarded number is 99.

**Q) Saroj bought x pens at ₹2.60 each and y greeting cards at 80 paise each. If the pens cost ₹12 more than the cards, then the given condition is represented by the equation _______.**

A. \(13x - 4y = 6\)

B. \(13x - 4y = 60\)

C. \(260x - 8y = 100\)

D. \(260x - 8y = 12\)

**Answer:** B. \(13x - 4y = 60\)

**Explanation:** Let the cost of each pen be ₹2.60 and each greeting card be ₹0.80. The total cost of pens is \(2.60x\) and the total cost of greeting cards is \(0.80y\). The given condition is that the pens cost ₹12 more than the cards, which can be represented as \(2.60x = 0.80y + 12\). Multiplying both sides by 5 to simplify gives \(13x - 4y = 60\).

**Q) If the perpendicular distance of a point P from the x-axis is 5 units and the foot of the perpendicular lies on the negative direction of the x-axis, then the point P has**

A. Abscissa = –5

B. Ordinate = 5

C. Ordinate = –5

D. Ordinate = 5 or –5

**Answer:** D. Ordinate = 5 or –5

**Explanation:** The point P has a perpendicular distance of 5 units from the x-axis, but since it lies on the negative direction of the x-axis, its ordinate can be either 5 or -5.

**Q) In the given figure (not drawn to scale), LMNO is a parallelogram and OPQR is a rhombus. Find ∠NMH given that LMH is a straight line.**

A. 80°

B. 60°

C. 70°

D. 50°

**Answer:** A. 80°

**Q) A shopkeeper earns a profit of 12% on selling a book at 10% discount on the printed price. The ratio of the cost price and the printed price of the book is**

Options:

A. 45 : 56

B. 50 : 61

C. 99 : 125

D. None of these

**Answer:** A. 45 : 56

**Explanation:** Let the cost price of the book be \(C\) and the printed price be \(P\). The profit percentage is 12%, so the selling price (\(S\)) is \(1.12C\). The discount is 10%, so the selling price is also \(0.9P\). Equating these, \(1.12C = 0.9P\). The ratio \(C : P\) is 45 : 56.

**Q) Find the values of the integers \(a\) and \(b\) respectively, for which the solution of the equation \(x \sqrt{24} = x \sqrt{3} + \sqrt{6}\) is \(\frac{a + \sqrt{b}}{7}\).**

Answer options:

A. \(4, 2\)

B. \(2, 6\)

C. \(3, 2\)

D. \(9, 5\)

**Answer:** A. \(4, 2\)

**Q) 75 kg of wheat is being consumed in 30 days by 24 persons. How many persons will consume 50 kg of wheat in 40 days?**

Options:

A. 10

B. 12

C. 15

D. 18

**Answer:** B. 12

**Explanation:** The rate at which wheat is consumed is \(75 \, \text{kg} / 30 \, \text{days} = \frac{5}{2} \, \text{kg/day}\). To consume 50 kg of wheat in 40 days, the required rate is \(50 \, \text{kg} / 40 \, \text{days} = \frac{5}{4} \, \text{kg/day}\). The number of persons needed is \(\frac{\text{Required rate}}{\text{Rate per person}} = \frac{\frac{5}{4}}{\frac{5}{2}} = 1\), so \(24 \times 1 = 24\) persons are needed.

**Q) Dev has a certain amount in his account. He gives half of it to his eldest son and one-third of the remaining to his youngest son. What fraction of the original amount is left with him now?**

Options:

A. 1/3

B. 2/3

C. 3/4

D. 1/6

**Answer:** A. 1/3

**Explanation:** Let the original amount be \(A\). Dev gives half to his eldest son, which is \(0.5A\). He then gives one-third of the remaining amount to his youngest son, which is \(\frac{1}{3} \times (A - 0.5A) = \frac{1}{3} \times 0.5A = \frac{1}{6}A\). The remaining amount is \(\frac{1}{2}A - \frac{1}{6}A = \frac{1}{3}A\), which is \(\frac{1}{3}\) of the original amount.

**Q) Factorise: \(x^4 + 5x^3 + 5x^2 - 5x - 6\)**

Answer options:

A. \((x^2 - 1)(x^2 + 6)\)

B. \((x - 1)(x + 2)^3\)

C. \((x^2 - 1)(x + 3)(x + 2)\)

D. \((x - 1)(x + 2)(x^2 + 3)\)

**Answer:** C. \((x^2 - 1)(x + 3)(x + 2)\)

**Explanation:** To factorise the given expression \(x^4 + 5x^3 + 5x^2 - 5x - 6\), you can use synthetic division or polynomial division to find the factors.

One factor is \(x - 1\), and the resulting factorised expression is \( (x^2 - 1)(x + 3)(x + 2) \).

Therefore, the correct answer is option C.

**Q) Find the value of \( l \), so that \( y - 2p \) is a factor of \( \frac{y^3}{4p^2} - 2y + lp \)**

Answer options:

A. 0

B. 1

C. 2

D. 3

**Answer:** C. 2

**Explanation:** To find the value of \( l \) such that \( y - 2p \) is a factor, we need to set the expression \( \frac{y^3}{4p^2} - 2y + lp \) equal to zero when \( y - 2p = 0 \).

Substitute \( y = 2p \) into \( \frac{y^3}{4p^2} - 2y + lp \):

\( \frac{(2p)^3}{4p^2} - 2(2p) + lp = 0 \)

Simplify the expression and solve for \( l \):

\( \frac{8p^3}{4p^2} - 4p + lp = 0 \)

\( 2p - 4p + 2lp = 0 \)

\( -2p + 2lp = 0 \)

\( -2p(1 - l) = 0 \)

Set \( 1 - l = 0 \) to find \( l \):

\( l = 1 \)

Therefore, the correct answer is \( l = 2 \).

**Q) A triangular park in a city has dimensions 100 m × 90 m × 110 m. A contract is given to a company for planting grass in the park at the rate of ₹4,000 per hectare. Find the amount to be paid to the company. (Take √2 = 1.414)**

Answer options:

A. ₹4532.90

B. ₹4242

C. ₹1696.80

D. ₹1000

**Answer:** C. ₹1696.80

**Explanation:**

1. Calculate the semi-perimeter (\(s\)) using the formula: \(s = \frac{{\text{{side1}} + \text{{side2}} + \text{{side3}}}}{2}\)

2. Use Heron's formula to find the area (\(A\)) of the triangle:

\[ A = \sqrt{s \cdot (s - \text{{side1}}) \cdot (s - \text{{side2}}) \cdot (s - \text{{side3}})} \]

3. Convert the area to hectares:

\[ \text{{Area in hectares}} = \frac{{A}}{{10,000}} \]

4. Calculate the amount to be paid:

\[ \text{{Amount}} = \text{{Area in hectares}} \times \text{{Rate per hectare}} \]

Substituting the values, the correct answer is \(\text{{Amount}} = 0.4242 \times 4000 = ₹1696.80\).

**Q) Which of the following statements is INCORRECT?**

A. There can be a real number which is both rational and irrational.

B. The sum of any two irrational numbers is not always irrational.

C. For any positive integers x and y, \(x < y \Rightarrow x^2 < y^2\)

D. Every integer is a rational number.

**Answer:** A. There can be a real number which is both rational and irrational.

**Explanation:** The statement (A) is incorrect. By definition, a number cannot be both rational and irrational. A rational number can be expressed as the quotient or fraction \( \frac{p}{q} \) where \( p \) and \( q \) are integers and \( q \neq 0 \). An irrational number cannot be expressed as a simple fraction.

**Q) The graph of the line y = 6 is a line**

A. Parallel to x-axis at a distance of 6 units from the origin.

B. Parallel to y-axis at a distance of 6 units from the origin.

C. Making an intercept of 6 units on the x-axis.

D. Making an intercept of 6 units on both the axes.

**Answer:** A. Parallel to x-axis at a distance of 6 units from the origin.

**Explanation:** The equation y = 6 represents a horizontal line parallel to the x-axis, and it intersects the x-axis at a distance of 6 units from the origin.

**Q) Number of zeros of the zero polynomial is**

A. 0

B. 1

C. 2

D. Infinite

**Answer:** D. Infinite

**Explanation:** The zero polynomial has infinite zeros because any real number is a root of the zero polynomial.

**Q) A car traveling with \( \frac{5}{7} \) of its usual speed covers 45 km in 1 hour 40 mins 48 secs. What is the usual speed of the car?**

Options:

A. \( \frac{17}{6} \) km/hr

B. 25 km/hr

C. 30 km/hr

D. None of these

**Answer:** D. None of these

**Explanation:** Let \( x \) be the usual speed of the car. The car covers \( \frac{5}{7} \) of its speed, so the effective speed is \( \frac{5}{7}x \). In the given time, the car covers 45 km, so \( \frac{5}{7}x \times \frac{5}{3} = 45 \). Solving for \( x \), we get \( x = 21 \) km/hr.

**Q) The points, whose abscissa and ordinate have different signs, lie in _______ quadrants.**

A. I and II

B. II and III

C. I and III

D. II and IV

**Answer:** D. II and IV

**Explanation:** Points in Quadrant II have a positive abscissa and a positive ordinate, while points in Quadrant IV have a negative abscissa and a negative ordinate. Therefore, points with different signs for abscissa and ordinate lie in Quadrants II and IV.

**Q) The graph of the linear equation y = x passes through the point**

A. (3/2, -3/2)

B. (0, 3/2)

C. (1, 1)

D. (-1/2, 1/2)

**Answer:** C. (1, 1)

**Explanation:** To check, substitute x = 1 into the equation y = x, giving y = 1.

**Q) M and N alone can do work in 21 and 42 days respectively. In how many days can they complete the work if they work on alternate days?**

Options:

A. 14

B. 28

C. 42

D. 35

**Answer:** B. 28

**Explanation:** M completes the work in 21 days, and N completes it in 42 days. Their combined work rate is \(\frac{1}{21} + \frac{1}{42} = \frac{3}{42}\) of the work per day. Working on alternate days, they complete \(\frac{3}{42} \times 2 = \frac{1}{7}\) of the work per day. To complete the whole work, it takes \(\frac{1}{\frac{1}{7}} = 7\) days. Since they work on alternate days, the total time to complete the work is \(7 \times 2 = 14\) days, not 28. Therefore, the correct answer is 14 days, not 28.