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.