Maths Olympiad Previous Year Paper -1 | Mathematics Olympiad for Class 9 PDF Download

Ans: (b)
The number common to rectangle and triangle only is 8.

Ans: (d)

• The pattern in the series involves multiplying the previous number by 3 and then adding 3.
• Starting with 4: 4 * 3 + 3 = 15.
• Next, 15: 15 * 3 + 3 = 48.
• Then, 48: 48 * 3 + 3 = 147.
• Continuing this, 147: 147 * 3 + 3 = 444, which is the missing number.
• Finally, 444: 444 * 3 + 3 = 1335, confirming the pattern.

Ans: (a)  
From figure (i) to (ii), the number of arrows is increased by two on each side and the direction of arrows get reversed.

Ans: (d)

Ans: (c)

• Vikrant starts by walking 70 m East.
• Then he turns left (to the North) and walks 60 m.
• Next, he turns left again (to the West) and walks 20 m.
• Finally, he turns right (to the North) and walks 60 m.
• After all these movements, he is 130 m away from the starting point in the North-East direction.

Ans: (b)

Ans: (d)
Hence, the distance from this original position i. e. A to E is (40 + 20) m = 60 m.

Ans: (c)

Ans: (b)

Ans: (c)

• The code language involves a specific pattern of shifting letters. Each letter in the word is replaced by another letter based on a fixed rule.
• For example, in the word GUIDELINES, each letter is shifted to create GFKWIUGPKN.
• Applying the same shifting pattern to SEPARATELY, we find that it translates to TCRGUANGVC.
• This means that the correct answer is TCRGUANGVC, which corresponds to option (c).

Ans: (a)

• First, replace the symbols with their respective operations: 24 – 65 ÷ 13 + 16 × 5.
• Next, perform the division: 65 ÷ 13 = 5.
• Now, substitute back: 24 – 5 + 16 × 5.
• Then, perform the multiplication: 16 × 5 = 80.
• Finally, calculate: 24 – 5 + 80 = 19 + 80 = 99.

Ans: (d)
Triangles formed are : a, b, c, d, e, f, g, h, i, j, k, ab, bc, cd, gf, fe, hi, ij, jk, abc, bcd, gfe, hij, ijk, abcd and hijk i.e., 26 in number.

Ans: (d)

• To find the pairs of letters in the word APPOINTMENT, we need to check each letter and see if the number of letters between them in the word matches the number of letters between them in the alphabet.
• For example, the letters A and C have 1 letter between them in the alphabet, and they also have 1 letter between them in the word.
• After checking all possible pairs, we find that there are four such pairs: AP, IN, OT, and MT.
• Thus, the answer is Four pairs of letters.

Ans: (b)

• First, let's break down the symbols: H % E means H is the father of E.
• Next, E * D indicates that E is the sister of D.
• Then, D $ G shows that D is the brother of G.
• Putting it all together, H is the father of E, who is the sister of D, and D is the brother of G. Therefore, H is the father of G.

Ans: (d)

Ans: (b)
The numbers on opposite faces are: (1, 2), (3, 5) and (4, 6).

Ans: (d)

• To find the value of k, substitute x = 3 and y = –2 into the equation 3x – ky = 1.
• This gives us 3(3) – k(-2) = 1, which simplifies to 9 + 2k = 1.
• Rearranging this, we get 2k = 1 - 9, leading to 2k = -8.
• Dividing both sides by 2, we find k = -4.

Ans: (c)
Let the coordinates of the point of internal division A be (x, y). Then 

Ans: (a)
Let the line 2x + 3y - 30 = 0 divide the join of A (4,5) and B (6,7) at point C (p, q) in the ratio k: 1. Then,

As the point C lies on the line 2x + 3y - 30 = 0, it satisfies the given equation, i.e,

Ans: (a)
The perpendicular distance from the x-axis is y co-ordinate and vice versa So, x = 7, y = 5
In the III quadrant, both x and y are negative. Hence coordinates are (- 7, - 5)

Ans: (c)

• To find the smallest number to add to 454189 to make it a perfect square, we first need to determine the nearest perfect square greater than 454189.
• The square root of 454189 is approximately 673.5. The next whole number is 674.
• Calculating 674 squared gives us 454276. Now, we subtract 454189 from 454276.
• This results in 87, which is the smallest number that needs to be added to 454189 to achieve a perfect square.

Ans: (d)
Given, 2x + 3y = 6 
kx + 6y = 15 
The system has a unique solution, 
i.e., 2/k ≠ 3/6
⇒ k ≠ 4
The value of k is any real number except 4.

Ans: (c)
Let the speed of the man in still water be x kmph and speed of the stream be y kmph 

⇒ x + y = 10    (1)
⇒ x - y = 2    (2)
Eq. (1) + Eq. (2) ⇒  2x = 12 ⇒  x = 6

Ans: (b)

• To find the value of x³ + y³ + z³, we can use the identity: x³ + y³ + z³ = (x + y + z)(x²+ y² + z² - xy - yz - zx) + 3xyz.
• Given that x + y + z = 1 and xyz = -1, we need to calculate x² + y² + z².
• Using the formula x² + y² + z² = (x + y + z)² - 2(xy + yz + zx), we find x² + y² + z² = 1² - 2(-1) = 1 + 2 = 3.
• Now substituting these values into the identity gives us: x³ + y³ + z³ = 1(3 - (-1)) + 3(-1) = 1(4) - 3 = 4 - 3 = 1.

Ans: (d)

• We know that a polyhedron has 5 faces.
• Let the number of edges be E and the number of vertices be V.
• According to the problem, V = (2/3)E.
• Using Euler's formula for polyhedra, which states V - E + F = 2, we can substitute F = 5.
• This gives us the equation: V - E + 5 = 2, or V - E = -3.
• Now, substituting V from the earlier equation, we get: (2/3)E - E = -3.
• This simplifies to (-1/3)E = -3, leading to E = 9.
• Now substituting E back to find V: V = (2/3) * 9 = 6.
• Thus, the total number of vertices is 6.

Ans: (d)
Using the basic proportionality theorem, 

⇒ 9p - 6 = 4p + 14 
⇒ p = 4

Ans: (c)

• According to the transitive property in mathematics, if two things are equal to a third thing, they are also equal to each other.
• This means that if A = C and B = C, then A must equal B.
• Thus, the correct answer is that they are equal to one another.
• Options like "perpendicular" and "not equal" do not apply in this context.

Ans: (d)

• Let the two numbers be 7x and 5x based on the ratio 7:5.
• When we decrease both numbers by 3, we have (7x - 3) and (5x - 3).
• The new ratio is given as 3:2, which leads to the equation: (7x - 3)/(5x - 3) = 3/2.
• Cross-multiplying gives us 2(7x - 3) = 3(5x - 3), simplifying to 14x - 6 = 15x - 9.
• Solving for x, we find x = 3. Therefore, the numbers are 21 and 15, and their sum is 36.

Ans: (a)

• To find the area of the triangle, we first need to determine the length of the third side. The perimeter is 32 cm, so the third side is 32 - (8 + 11) = 13 cm.
• Now, we can use Heron's formula to calculate the area. The semi-perimeter (s) is (32 / 2) = 16 cm.
• Using Heron's formula: Area = √[s(s-a)(s-b)(s-c)], where a, b, and c are the sides of the triangle. Here, Area = √[16(16-8)(16-11)(16-13)] = √[16 × 8 × 5 × 3] = √(2400) = 20√(6) cm².
• Since the area is given as k × 30 cm², we can set 20√(6) = k × 30. Solving for k gives k = (20√(6) / 30) = 8.

Ans: (c)

• The incorrect step in constructing quadrilateral PQRS is Step 3.
• In this step, it states to draw an arc from Q with a radius of 5 cm to find point P.
• However, the correct construction should ensure that the angle and lengths are consistent with the given dimensions.
• Thus, the arc may not intersect correctly, leading to an inaccurate quadrilateral.

Ans: (b)

• In a parallelogram, opposite sides are equal. Here, AB is extended to E.
• Since ED bisects BC at O, it means BO is equal to OC.
• From the properties of the parallelogram, we can conclude that AB is equal to BE because E is just an extension of AB.
• Thus, the correct statement is that AB = BE.

Ans: (b)

• We start with the equations 2/3x – 4/5y = 4 and xy = 60.
• From xy = 60, we can express y in terms of x: y = 60/x.
• Substituting this into the first equation allows us to solve for x and then find y.
• After finding x and y, we can calculate 4/9x² + 16/25y² to get the final result, which is 80.

Ans: (c)

Ans: (c)
We have ∆APQ ~ ∆ABC 
So,
Let AQ = 3x, QC = 2x (AQ : QC = 3 :2) 
∴ AC = AQ + QC = 5x 

⇒ BC = 25 cm

Ans: (b)

• The volume of a cone is calculated as (1/3)πr²h.
• The volume of a hemisphere is (2/3)πr³.
• The volume of a cylinder is πr²h.
• When comparing these volumes, we find that the ratio of the volumes of the cone, hemisphere, and cylinder simplifies to 1 : 2 : 3.

Ans: (c)

• First, calculate the volume of the original sphere using the formula: Volume = (4/3)πr³. Here, r = 12 cm.
• Next, find the volume of the original sphere: Volume = (4/3)π(12)³ = 576π cm³.
• Since this volume is divided into 8 smaller spheres, the volume of each small sphere is: 576π cm³ / 8 = 72π cm³.
• Now, use the volume formula for a sphere to find the radius of the small spheres: Volume = (4/3)πr³, leading to r³ = 54. Thus, the radius of each small sphere is 3√6 cm.
• Finally, calculate the surface area of each small sphere using the formula: Surface Area = 4πr². Plugging in the radius gives Surface Area = 4π(3√6)² = 144π cm².

Ans: (d)

• Let the cost price (C.P.) of the article be 100.
• The marked price (M.P.) is then 125 (which is 25% above C.P.).
• After applying a 10% discount, the selling price (S.P.) becomes 112.5 (which is 90% of M.P.).
• The profit is calculated as S.P. - C.P. = 112.5 - 100 = 12.5.
• The profit percentage is then (Profit/C.P.) * 100 = (12.5/100) * 100 = 12.5%.

Ans: (b)

• To find the area of a triangle with sides 120 m, 80 m, and 50 m, we can use Heron's formula.
• First, calculate the semi-perimeter (s): s = (120 + 80 + 50) / 2 = 125 m.
• Then, 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 gives us the area as 375√15 m², which is the area available for planting.

Ans: (d)

• The total amount of food can be calculated as 300 students * 24 days = 7200 student-days of food.
• After some students left, the food lasts for 36 days. Let the number of remaining students be x.
• Now, the food can also be expressed as x students * 36 days = 7200 student-days of food.
• Setting the equations equal gives us 7200 = x * 36, leading to x = 200 students remaining.
• Thus, the number of students who left is 300 - 200 = 100.

Ans: (b)

• The problem states that the cost of a notebook is twice the cost of a pen.
• This means if the pen costs ₹y, then the notebook costs ₹2y.
• We can express this relationship as x = 2y, which can be rearranged to x - 2y = 0.
• Thus, the correct equation that represents this condition is x – 2y = 0.

Ans: (d)

• To find the time taken by A, B, and C together, we first calculate their individual rates of filling the tank.
• A and B together fill the tank in 8 hours, so their combined rate is 1/8 of the tank per hour.
• B and C together fill it in 12 hours, giving a rate of 1/12 of the tank per hour.
• A and C together fill it in 16 hours, resulting in a rate of 1/16 of the tank per hour.
• By adding these rates and solving for the combined rate of A, B, and C, we find that they can fill the tank together in hours.

Ans: (a)

• The man walks at 5/7 of his usual speed, which means he is slower than normal.
• This slower speed causes him to be 14 minutes late.
• To find his usual time, we can set up a relationship between speed, time, and distance.
• Since speed and time are inversely related, if speed decreases, time increases. Thus, we can calculate that his usual time is 35 minutes.

Ans: (d)

• Let the number of people be represented as x.
• Each person contributed 2x rupees.
• The total contribution can be expressed as x * 2x = 2x².
• Setting this equal to the total contribution, we have 2x² = 4418.
• Solving for x, we find x² = 2209, leading to x = 47.

Ans: (d)

• To solve this, we first determine the work done by men and children. If a man completes the work in 1 day, then a child takes 2 days.
• In 1 day, 12 men can do 12 units of work, while 8 children can do 4 units (since 8 children do 4 units in 1 day).
• Thus, together they complete 16 units of work in 1 day.
• Over 9 days, they finish 144 units of work (16 units/day * 9 days).
• Now, if only 12 men work, they can complete 12 units of work per day, so to finish 144 units, it will take them 12 days (144 units / 12 units/day).

Ans: (b)

• The area of the rectangle is given as (14x² – 11x – 15) m².
• To find the length and breadth, we need to factor this expression.
• Factoring gives us (7x + 5) and (2x – 3), which are valid dimensions.
• Option (a) does not yield the correct area when multiplied, confirming that option (b) is the correct answer.

Ans: (b)

In ΔABC, AB = BC.
⇒ ∠B=90°
∠AIC = 90° + ∠B/2
= 90° + 45° =135°

Ans: (b)
PRST is a rhombus. All sides are equal. 
= ∠PRS = 60° = ∠RST =120°

Ans: (a)
The sides are 40 cm and 50 cm. 

Ans: (c)
Opposite angles are supplementary. 180° - 50° = 130°

Ans: (a)

Let the diagonals meet at O as shown in the figure. 
∠POS = ∠ROQ = 90° 
Also, OP = OQ = OS = OR, i.e., the diagonals are equal and bisect at right angles. Clearly, PRQS is a square.