Maths Olympiad Previous Year Paper -2 | Mathematics Olympiad for Class 9 PDF Download

Ans: (c)
In each step, the figure rotates 45° anti-clockwise. 
Also, shaded circles becomes unshaded and unshaded circle becomes shaded.

Ans: (d)

• First, replace the letters with their corresponding operations: 15 + 30 ÷ 6 × 20 - 18.
• Next, follow the order of operations (BODMAS/BIDMAS): Start with division and multiplication from left to right.
• Calculate 30 ÷ 6 = 5, then 5 × 20 = 100.
• Now, the expression is 15 + 100 - 18, which equals 97.

Ans: (a)

• The relationship between the numbers is that the second number is the square of the first number plus a certain value.
• For the pair 11 : 144, we see that 11 squared is 121, and 121 plus 23 equals 144.
• Applying the same logic to 14, we calculate 14 squared, which is 196.
• Thus, the missing number is 14 squared minus 23, which gives us 225.

Ans: (c)
Length of each side of a rhombus =

Length of each side of a rhombus =
Length of each side of a rhombus =
Length of each side of a rhombus = 15 cm

Ans: (d)
R + r > d > R - r 
6 + 3 > 5 > 6 - 3 
9 > 5 > 3 
This condition is satisfying so the circles are intersecting.

Ans: (c)
In a cyclic quadrilateral, opposite angles are supplementary. 

Ans: (b)
d = 7, R = 5, and r = 2 
d = R + r 
Circles are touching externally. Number of common tangents is 3

Ans: (c)
We have HE × HD = HG × HF 
⇒ 6 × (HE + ED) = 5 × HF 
⇒ 6 × (6 + 9) = 5 × HF 
⇒ HF = 18 cm 
∴ GF = HF - HG = 18 - 5 = 13 cm

Ans: (c)

∠ ADB = 1/2 × (Reflex ∠ACB)
1/2 × 240° = 120°

Ans: (a)
Given, ∠PQR = 55°. 
An angle subtended by an arc at the centre of the circle is double the angle subtended by the same arc at any point on the remaining part of the circle. 
∠PCR = 2(∠PQR) =110° 
Reflex ∠PCR = 360° - 110° = 250°

Ans: (a)

• The machine rearranges the input by first sorting the numbers in ascending order, followed by the words in alphabetical order.
• For the input "61 place time 32 cat 76 july", the first step will sort the numbers: 32, 61, 76.
• Then, it will arrange the words: "cat", "july", "place", "time".
• Following the same pattern as the previous example, the last step will be Step IV, which is the final arrangement of numbers and words.

Ans: (d)
Semi perimeter of triangle s =
By Heron’s formula: The area of the triangle
Cost of painting = 30 x 9 = 270 cents = $2.70

Ans: (d)

The line of division is along the diagonal of the rhombus-shaped field.
So, each son gets an equal area.
By Heron's formula: Area of triangle is

Given: a = 100 m, b = 100 m, and c =160 m
Area of ΔMBC =

Ans: (b)
area of ∆PGQ = area of ∆QGR = area of ∆PGR

Ans: (d)
Given LM // NO and the area of the parallelogram, LMQO is 48cm²
ΔLQM, ΔLPM and ΔLNM and parallelogram LMQO are on the same base and between the same parallels
Area of ΔLQM = Area of ΔLPM = Area of ΔLNM
= 1/2 × Area of parallelogram LMQO
= 1/2 × 48 = 24 cm²

Ans: (c)

• When the diagonals of a quadrilateral bisect each other at right angles, it indicates a special property.
• This property is characteristic of a rhombus, which has diagonals that not only bisect each other but also intersect at 90 degrees.
• While a parallelogram has bisecting diagonals, they do not necessarily meet at right angles.
• A rectangle has equal angles but does not guarantee that the diagonals intersect at right angles.

Ans: (b)

• To find the value of k, substitute x = –1 and y = –1 into the equation: 9k(-1) + 12k(-1) = 63.
• This simplifies to -9k - 12k = 63, which combines to -21k = 63.
• Now, divide both sides by -21 to isolate k: k = 63 / -21.
• This results in k = -3, which is the correct answer.

Ans: (c)

• To solve for x² + (1/x²), we start with the equation x - (1/x) = √7.
• We can square both sides: (x - (1/x))² = (√7)², which gives us x² - 2 + (1/x²) = 7.
• This simplifies to x² + (1/x²) = 7 + 2 = 9.
• Thus, the value of x² + (1/x²) is 9.

Ans: (b)

• To solve the equation, we first simplify the expression 3 + 2√5 / 3 - 5√5.
• By multiplying the numerator and denominator by the conjugate of the denominator, we can eliminate the square root in the denominator.
• This results in a form where we can identify p and q, which are the coefficients of the rational and irrational parts respectively.
• After finding p and q, we calculate 11(p + q) and find that it equals –41.

Ans: (c)

• To find the volume of one cylindrical pillar, we use the formula: Volume = π × radius² × height.
• The radius is 20 cm, which is 0.2 m (since 100 cm = 1 m). The height is 10 m.
• Calculating the volume of one pillar: Volume = π × (0.2)² × 10 = π × 0.04 × 10 = 0.4π m³.
• Now, for 14 pillars, the total volume is 14 × 0.4π m³ ≈ 17.6 m³ (using π ≈ 3.14).

Ans: (c)

• To construct a quadrilateral, certain conditions must be met.
• When you have two adjacent sides and two angles, it is not enough to determine a unique quadrilateral.
• This is because the angles can lead to different shapes, making it impossible to construct a single quadrilateral.
• In contrast, the other options provide sufficient information to create a quadrilateral.

Ans: (a)
Dimensions of a room = 12 m × 8 m × 9 m
Length of the diagonal =

Ans: (d)

• To determine the value of p for which the polynomial is divisible by x + 2, we can use the Remainder Theorem.
• According to this theorem, if a polynomial f(x) is divisible by (x - a), then f(a) = 0.
• Here, we substitute x = -2 into the polynomial: f(-2) = 2(-2)⁴ + 3(-2)³ + 2p(-2)² + 3(-2) + 6.
• Calculating this gives us: 32 - 24 + 8p - 6 + 6 = 0, which simplifies to 8p + 8 = 0.
• Solving for p, we find p = -1, confirming that the correct answer is (d).

Ans: (b)

• The rule for divisibility by 11 states that you subtract the sum of the digits in odd positions from the sum of the digits in even positions.
• For 841763, the sums are: odd positions (8 + 1 + 6) = 15 and even positions (4 + 7 + 3) = 14. The difference is 1, which is not divisible by 11.
• In contrast, the other numbers can be divided evenly by 11 based on this rule.
• Thus, 841763 is the only number that does not meet the divisibility condition.

Ans: (d) 
Curved surface area of a cylinder = 2πrh 
h = 7 × 2 + 1 = 15 cm 
r = 7 cm 
Curved surface area of a cylinder = 660 cm²

Ans: (b)

• First, simplify the fractions: a = 15/18 simplifies to 5/6 and b = 28/36 simplifies to 7/9.
• Next, calculate a + b: 5/6 + 7/9. To add these, find a common denominator, which is 18. This gives us (15/18 + 14/18) = 29/18.
• Now, calculate a - b: 5/6 - 7/9. Using the same common denominator, we get (15/18 - 14/18) = 1/18.
• Finally, divide the results: (29/18) / (1/18) = 29, since dividing by a fraction is the same as multiplying by its reciprocal.

Ans: (b)

• To find the abscissa of point P, we look at its coordinates, which are (–3, 4). The abscissa is the first number, so for P, it is –3.
• For point Q, with coordinates (–4, 5), the abscissa is –4.
• Now, we calculate the difference: abscissa of P (–3) minus abscissa of Q (–4).
• This gives us: –3 - (–4) = –3 + 4 = 1.

Ans: (a)

• To find the number of persons from Patna city, we use the formula: (Central Angle / 360) * Total Persons.
• Here, the central angle is 72° and the total number of persons is 2500.
• Calculating: (72 / 360) * 2500 = (1/5) * 2500 = 500.
• Thus, the number of persons from Patna city is 500.

Ans: (b)

• To find the area of the triangle, we first need to determine the length of the third side. Since the perimeter is 64 cm, the third side is 64 - (16 + 22) = 26 cm.
• Next, we can use Heron's formula to calculate the area. First, we find the semi-perimeter, which is (16 + 22 + 26) / 2 = 32 cm.
• Now, applying Heron's formula: Area = √[s(s-a)(s-b)(s-c)], where s is the semi-perimeter and a, b, c are the sides of the triangle.
• Substituting the values: Area = √[32(32-16)(32-22)(32-26)] = √[32 * 16 * 10 * 6] = 32√30 cm².

Ans: (c)

• To find the volume of a sphere, we use the formula: V = (4/3)πr³.
• The radius (r) is half of the diameter, so r = 3.5 mm / 2 = 1.75 mm.
• Now, substituting the radius into the formula: V = (4/3)π(1.75)³.
• Calculating this gives approximately 22.46 mm³, which is the volume of the tablet.

Ans: (d)

• Euclid’s axioms are fundamental truths in geometry.
• The statement in option (d) is incorrect because it does not reflect a basic principle of equality.
• In contrast, options (a), (b), and (c) are all valid axioms that describe relationships between parts and wholes.
• Thus, option (d) is the only one that does not belong to Euclid’s axioms.

Ans: (d)

• To solve the equation, we first simplify the left side: (5/7) - 4X * (7/5) - 7.
• Next, we can rewrite the equation as (5/7) - 4X * (7/5) = (7/5) 9 + 7.
• After simplifying, we isolate x and find that x equals 4.
• Thus, the correct answer is 4, which is option (d).

Ans: (a)

• Let the C.P. of the chair be x.
• With an 18% profit, the selling price (S.P.) is 1.18x.
• If sold for ₹172 more, the S.P. becomes 1.18x + 172, which gives a 22% profit, so it equals 1.22x.
• Setting the two expressions for S.P. equal: 1.18x + 172 = 1.22x.
• Solving this gives x = ₹4300, which is the C.P. of the chair.

Ans: (d)
We know that:
Curved surface area = 2πr² sq. units 
Total surface area = 3πr² sq. units
Curved surface area =

Curved surface area = 420 cm²

Ans: (a)
Volume of clay = 10m³
Area of land = 100 acres
= 100 × 100 m² = 10000 m²
Rise in the level of ground = 10/10000 = 0.001 m

Ans: (c)

• First, calculate the total volume of the tank: 40 m × 25 m × 15 m = 15,000 m³.
• Convert the volume to litres: 15,000 m³ = 15,000,000 litres (since 1 m³ = 1,000 litres).
• Next, find the daily water requirement for the village: 5000 people × 75 litres = 375,000 litres per day.
• Finally, divide the total water in the tank by the daily requirement: 15,000,000 litres ÷ 375,000 litres/day = 40 days.

Ans: (d)

• To find Mr. Raj's salary after 3 years with a 20% annual increase, we can use the formula for compound interest: Final Amount = Principal × (1 + Rate)^Time.
• Here, the Principal is ₹ 500000, the Rate is 20% (or 0.20), and the Time is 3 years.
• Calculating it: Final Amount = 500000 × (1 + 0.20)³ = 500000 × (1.20)³ = 500000 × 1.728 = ₹ 864000.
• Thus, after 3 years, Mr. Raj's salary will be ₹ 864000.

Ans: (a)

• To find the area of the triangular wall, we can use Heron's formula. First, we calculate the semi-perimeter (s) of the triangle: s = (18 + 12 + 10) / 2 = 20 m.
• Next, we apply Heron's formula: Area = √[s(s-a)(s-b)(s-c)], where a, b, and c are the sides of the triangle.
• Substituting the values: Area = √[20(20-18)(20-12)(20-10)] = √[20 * 2 * 8 * 10] = √[3200] = 40√2 m².
• Thus, the area that has been painted is 40√2 m².

Ans: (c)

• Let Sumit's current age be represented as S.
• Then, Manish's current age is S - 7.
• In 6 years, Manish will be (S - 7 + 6) = S - 1 years old.
• At that time, Sumit will be (S + 6) years old.
• The equation states that Manish's age will be 7 years more than half of Sumit's age: S - 1 = 0.5(S + 6) + 7.
• Solving this gives S = 22, which means Sumit's present age is 22 years.

Ans: (d)

• First, calculate how much work B does in 8 days. B completes 1/12 of the work in a day, so in 8 days, B finishes 8/12 or 2/3 of the work.
• This means 1/3 of the work is still left. A can complete 1/15 of the work in a day.
• To find out how many days A needs to finish the remaining 1/3 of the work, divide 1/3 by 1/15, which equals 5 days.
• Thus, A will take 5 days to complete the remaining work.

Ans: (d)

• The distance between A and B is 90 km.
• Both cars travel for 9 hours before meeting.
• If they are moving in the same direction, the distance covered by the car from A plus the distance covered by the car from B must equal 90 km.
• The equation can be set up as: 9x - 9y = 90, which simplifies to x - y = 10. This does not match any of the provided options, hence the answer is None of these.

Ans: (b)

• To find the number of rows, we first calculate how many students can be arranged in a square formation. This means we need to find the largest perfect square less than or equal to 6000.
• The largest perfect square less than 6000 is 5929, which is 77 squared (77 x 77 = 5929).
• Since 71 students are left out, we subtract 71 from 6000, giving us 5929, which confirms that 77 rows and 77 columns can be formed.
• Thus, the number of rows is 77, as it matches the number of columns.

Ans: (d)

• To find out how much water to add, we first calculate the total cost of the milk: 45 litres x ₹4 = ₹180.
• Next, we want the new cost per litre to be ₹3. If we add 'x' litres of water, the total volume becomes (45 + x) litres.
• We set up the equation: ₹180 / (45 + x) = ₹3.
• Solving this gives us x = 15 litres. Therefore, we need to add 15 litres of water to achieve the desired cost.

Ans: (b)

• Pipe A fills the cistern in 9 hours, so it fills 1/9 of the cistern in 1 hour.
• Pipe B fills the cistern in 12 hours, so it fills 1/12 of the cistern in 1 hour.
• When both pipes are used alternately for 1 hour each, in 2 hours, they fill 1/9 + 1/12 of the cistern.
• Calculating this gives a total of 5/36 of the cistern filled in 2 hours.
• To fill the entire cistern, it will take 10(1/4) hours when both pipes are used alternately.

Ans: (d)

• First, calculate the total number of students who like singing and dancing: 21 + 17 = 38.
• Next, find the number of students who like other activities: 60 - 38 = 22.
• The probability of selecting a student who likes other activities is the number of students who like other activities divided by the total number of students: 22/60.
• Simplifying 22/60 gives us 11/30, which is the correct answer.

Ans: (b)

• To solve for p³ + q³ + r³ - 3pqr, we can use the identity: p³ + q³ + r³ - 3pqr = (p + q + r)((p + q + r)² - 3(pq + qr + rp)).
• Substituting the values: p + q + r = 6 and pq + qr + rp = 5, we get: 6(6² - 3*5) = 6(36 - 15) = 6*21 = 126.
• Thus, the value of p³ + q³ + r³ - 3pqr is 126.
• Therefore, the correct answer is option (b) 126.

Ans: (b)
Volume all the containers = Volume of one container × Number of containers
Volume all the containers = 1/3 πr²h × 100
Volume all the containers = 6600 cm³

Ans: (c)
From the figure, PR = 5 cm [Pythagoras theorem] 
Perimeter of the rectangle = 7 + 7 = 14 cm 
Perimeter of the rhombus = 4 × 5 = 20 cm
Ratio of their perimeters = 14 : 20 
= 7 : 10

Ans: (a)
Let the endpoint of the diagonal be (x, y)

x = 1 and y = 6
Length of diagonal =

Ans: (c)
Let ON = x units
In ΔMNO ~ ΔLNP

N is (-4,0)
Area of trapezium. LMOP = 1/2 (OM + LP) · OP = 1/2 (3 + 6) · 4 = 18 sq units