Maths Olympiad Previous Year Paper -3 | Olympiad Preparation for Class 10 PDF Download

Ans: (a)
∴ Required distance

So, Ashish is 10√5 km in North-East direction from the starting point. 

Ans: (c)

Ans: (b)

Ans: (a)

Ans: (c)

Ans: (c)

Ans: (d)

• First, we need to replace the symbols with their meanings: M = +, N = ×, P = ÷, S = –.
• For option (d): 18 + 40 ÷ 8 × 6 – 4.
• Calculating step-by-step: 40 ÷ 8 = 5, then 18 + 5 × 6 = 18 + 30 = 48, and finally 48 – 4 = 44.
• This confirms that option (d) is correct as it equals 44.

Ans: (b)

• From the expression M ÷ L – N + P × Q, we break it down step by step.
• M ÷ L means M is the daughter of L.
• L – N means L is the husband of N.
• N + P means N is the sister of P.
• P × Q means P is the brother of Q.
• Putting it all together, P is the maternal uncle of M, since P is the brother of M's mother (N).

Ans: (d)
Let the event wherein 2 students having the same birthday be E 
Given, P (E) = 0.992 
We know, 
P(E) + P (not E) = 1 
Or, P (not E) = 1 - 0.992 = 0.008 
∴ The probability that the 2 students have the same birthday is 0.008

Ans: (b)

Ans: (d)
P(x) cuts the x-axis at 3 points and tocuhes it at 2 other points.
∴ The number of zeroes of P(x) is (3 + 2) = 5

Ans: (c)
In ΔABC, AB = BC
…[All sides of an equilateral Δ is equal
AB² = BC …[squaring both sides
(0 – 3)² + (o – √3)² = (3 – 0)² + (k – 0)²
⇒ 9 + 3 = 9 + k² ⇒ k = ±√3
But only k = -√3 has been given in the option.

Ans: (b)
95 = 5 × 19
171 = 32 × 19
∴ HCF (95,191) = 19 m/min.

Ans: (b)

Ans: (a)

• Let the two-digit number be represented as 10a + b, where a and b are the digits.
• According to the problem, we have two conditions: (10a + b) - (10b + a) = 9 and a + b = 7.
• From the first condition, simplifying gives us 9a - 9b = 9, which leads to a - b = 1.
• Now, we can solve the equations a + b = 7 and a - b = 1 simultaneously.
• Adding these equations gives 2a = 8, so a = 4. Substituting back, b = 3.
• The digits are 4 and 3, and their product is 4 * 3 = 12.

Ans: (b)

• The HCF (Highest Common Factor) of two numbers is the largest number that divides both of them.
• For P = x²y³ and Q = xy², we find the HCF by taking the lowest power of each prime factor present in both numbers.
• For x, the lowest power is x¹ (from Q), and for y, the lowest power is y² (from Q).
• Thus, the HCF(P, Q) = x¹y² = xy².

Ans: (a)

• For a quadratic equation to have equal roots, the discriminant must be zero. The discriminant is given by the formula b² - 4ac.
• In this case, a = 1, b = (2a - 1), and c = a². So, we set up the equation: (2a - 1)² - 4(1)(a²) = 0.
• Expanding this gives us 4a² - 4a + 1 - 4a² = 0, which simplifies to -4a + 1 = 0.
• Solving for a, we find a = ¼, which is the value that makes the roots equal.

Ans: (b)

• The height of the building is half that of the tower. If we denote the height of the building as h, then the height of the tower is 2h.
• From the ground, the angle of elevation to the top of the building is 30°. This means that the height of the building can be calculated using the tangent function: tan(30°) = h / d, where d is the distance from the point to the base of the building.
• To find the angle of elevation to the top of the tower, we use the height of the tower (2h) and the same distance d: tan(θ) = 2h / d.
• Using the relationship of angles, we find that the angle of elevation to the top of the tower is 60°, since tan(60°) = 2 * tan(30°).

Ans: (d)

• To find the area of the triangle, we can use Heron's formula. First, we calculate the semi-perimeter (s), which is given as 21 cm.
• Next, we have the sides of the triangle: a = 14 cm and b = 15 cm. The third side (c) can be calculated as c = 2s - a - b = 21 - 14 - 15 = 21 - 29 = 7 cm.
• Now, we apply Heron's formula: Area = √[s(s-a)(s-b)(s-c)]. Plugging in the values gives us Area = √[21(21-14)(21-15)(21-7)] = √[21 * 7 * 6 * 14].
• Calculating this results in an area of 84 cm², which corresponds to option (d).

Ans: (b)

• For a polynomial with roots that are reciprocals, the product of the roots must equal 1.
• Using Vieta's formulas, the product of the roots of f(x) = 3x² + 11x + p is given by -p/3.
• Setting this equal to 1 gives us the equation -p/3 = 1, leading to p = -3.
• However, since we need the roots to be reciprocals, we find that the correct value of p that satisfies the condition is 3.

Ans: (a)

• In triangle construction, the Triangle Inequality Theorem states that the sum of the lengths of any two sides must be greater than the length of the third side.
• Here, we have QR = 6.4 cm and ∠Q = 60°. For the triangle to be valid, PQ + PR must be greater than QR.
• If PQ + PR = 6 cm, it is less than QR, making it impossible to form the triangle.
• Thus, the only case where the triangle cannot be constructed is when PQ + PR is 6 cm.

Ans: (d)

• In an arithmetic progression (AP), the nth term can be calculated using the formula: nth term = first term + (n - 1) * common difference.
• Let the common difference be d. For the first AP, the 30th term is 12 + 29d, and for the second AP, it is 9 + 29d.
• The difference between the 30th terms is: (12 + 29d) - (9 + 29d) = 12 - 9 = 3.
• Thus, the difference between their 30th terms is 3.

Ans: (b)

• To solve for the relationship between x and y, we start with the given equations: x = a(secθ + tanθ) and y = b(tanθ - secθ).
• By multiplying x and y, we get xy = ab[(secθ + tanθ)(tanθ - secθ)].
• Upon simplifying, we find that this results in xy + ab = 0, confirming that the correct answer is option (b).
• This shows that the product of x and y, when adjusted by ab, equals zero, indicating a specific relationship between these variables.

Ans: (c)

• To find the number of bad pens, we first determine the number of good pens. Since the probability of getting a good pen is 0.75, we calculate:
• Good pens = 0.75 * 380 = 285 pens.
• Next, we subtract the number of good pens from the total pens to find the bad pens:
• Bad pens = Total pens - Good pens = 380 - 285 = 95 pens.
• Thus, the number of bad pens in the lot is 95.

Ans: (d)

• To find the volume of the remaining solid, we first calculate the volume of the cylinder and the cone.
• The volume of the cylinder is given by the formula: V = πr²h, where r is the radius and h is the height. Here, r = 9 cm and h = 7 cm.
• The volume of the cone is calculated using the formula: V = (1/3)πr²h. For the cone, r = 9 cm and h = 7 cm.
• After calculating both volumes, we subtract the volume of the cone from the volume of the cylinder to find the volume of the remaining solid.
• After performing the calculations, the volume of the remaining solid is found to be 1188 cm³.

Ans: (a)

• In a parallelogram, opposite angles are equal and consecutive angles are supplementary.
• From the properties, we know that ∠A + ∠B = 180° and ∠A + ∠C = 180°.
• Substituting the expressions: (3x - 20) + (y + 15) = 180 and (3x - 20) + (x + 40) = 180.
• Solving these equations gives us x = 30 and y = 95, which matches option (a).

Ans: (b)

• To find the central angle of a sector, we can use the formula: Area = (θ/360) × πr², where θ is the central angle in degrees and r is the radius.
• Here, the radius (r) is 2.8 cm and the area is 3.08 cm².
• Rearranging the formula gives us θ = (Area × 360) / (πr²).
• Substituting the values: θ = (3.08 × 360) / (π × (2.8)²) which simplifies to approximately 45°.

Ans: (d)

• To find the value of x, we first need to calculate the total number of values, which is 9.
• The mean is given as 36, so the total sum of the data should be 36 multiplied by 9, which equals 324.
• The data provided is: 17, 23, 31, 33, x + 3, 37, 41, 43, 48. We can sum these values: 17 + 23 + 31 + 33 + (x + 3) + 37 + 41 + 43 + 48.
• Calculating the sum without x gives us 273 + (x + 3) = 276 + x. Setting this equal to 324 gives us x = 48.

Ans: (b)
If a is the first term and d is the common difference, then we have 
a + 2d = 9 ----- (1) 
a + 16d = 51 ---(2) 
On subtracting Eq. (1) from Eq. (2), we get 
14d = 42 
⇒ d = 3 
Substituting the value of d in Eq. (1), we get a = 3 
∴ a = 3 and d = 3

Ans: (a)
Let a₁ be the first term. then the second term can be written as a₁ + d. Similarly, we write all even in terms of odd.
a₁ + a₃ + a₅ + a₇ + a₉ = 252  
and a₂ + a₄ + a₆ + a₈ + a₁₀ = 15  
5a + 20d = 252 
5a + 25d = 15 
d = 0.5 = 1/2

Ans: (a)
If a be the first term and d be the common difference.
−60 = 55^th term = a + 54d
65 = 5^th  term = a + 4d
d = − 5/2, a = 75
25th term = a + 24d = 15

Ans: (d)

Ans: (c)
Sum of the roots = 2 + 7 = 9
Product of the roots = 2 × 7 = 14 
We know that if p is the sum of the roots and q is the product of the roots of a quadratic equation, then its equation is x² - px + q = 0.
Hence, the required equation is x2 - 9x + 14 = 0.

Ans: (b)
Let the number of mangoes be x.
Number of bees on the mangoes = 4x 
Total number of bees = 87 
4x + 7 = 87 
4x = 87- 7 = 80
x = 20 There are 20 mangoes in the garden.

Ans: (d)
Let the correct answer of Sam be x. 
Then 5x − 2(10 − x) = 29 
5x − 20 + 2x = 29 
7x = 49 
x = 7

Ans: (c)

• Let the number of pink balls be represented as x.
• The total number of balls is 9 + x.
• The probability of drawing a green ball is 9 / (9 + x), and the probability of drawing a pink ball is x / (9 + x).
• According to the problem, the probability of drawing a pink ball is four times that of a green ball: x / (9 + x) = 4 * (9 / (9 + x)).
• Solving this equation leads to x = 36, meaning there are 36 pink balls in the bag.

Ans: (a)

• Let the original speed of the car be x km/h.
• The time taken to cover 288 km at speed x is 288/x hours.
• If the speed is decreased by 4 km/h, the new speed is (x - 4) km/h, and the time taken becomes 288/(x - 4) hours.
• According to the problem, the difference in time is 1 hour: 288/(x - 4) - 288/x = 1.
• Solving this equation leads to x = 36 km/h.

Ans: (d)

• To find the required score, first calculate the total runs scored in 12 innings: 48 runs * 12 innings = 576 runs.
• Next, to achieve an average of 54 after 13 innings, the total runs needed would be 54 runs * 13 innings = 702 runs.
• Now, subtract the current total runs from the required total: 702 runs - 576 runs = 126 runs.
• Thus, the cricketer needs to score 126 runs in the thirteenth inning to reach the desired average.

Ans: (d)

• Let Sushant's amount be S and Mohan's amount be M.
• From Mohan's statement: M + 700 = 2(S).
• From Sushant's statement: S + 300 = 3(M).
• By solving these equations, we find that Sushant has ₹1500.

Ans: (d)

• The production of TVs increases by a fixed amount each year.
• In the 6th year, the factory produced 8000 sets, and in the 9th year, it produced 11300 sets.
• The difference in production between the 9th and 6th years is 3300 sets, which occurs over 3 years, indicating an increase of 1100 sets per year.
• Using this, we can calculate the production for the first 6 years, which totals to 31500 sets.

Ans: (a)

• Let the present age of Meena be 8x and her daughter's age be x.
• In 8 years, Meena's age will be 8x + 8 and her daughter's age will be x + 8.
• The ratio of their ages in 8 years will be (8x + 8) / (x + 8) = 10 / 3.
• Cross-multiplying gives 3(8x + 8) = 10(x + 8), leading to 24x + 24 = 10x + 80.
• Solving this gives 14x = 56, so x = 4. Therefore, Meena's age is 8x = 32 years.

Ans: (b)
Let the number of $5 coins be x.
Let the number of $2 coins be y. 
∴ x + y = 36 (1) 
Also, 5x + 2y = 108 (2) 
Multiplying Eq. (1) by 2, we have 
2x + 2y = 72 (3) 
subtracting Eq. (3) from Eq. (2), 
3x = 36 
x = 12

Ans: (a)
Let Anna’s speed and the speed of the stream be x kmph and y kmph respectively

Eq. (1) ⇒ 56a + 6b = 5 ……….(3)
and Eq. (2) ⇒ 42a + 5b = 4 ………(4)
Solving Eqs. (3) and (4), we get
a = 1/28 and b = 1/2
x + y = 28 and x − y = 2
x = 15 and y = 13
∴ Anna’s speed in still water = 15 kmph

Ans: (d)
BC is the tower 30 m high, and the observer is at A. 
E denotes his eyes’ position. 
Let the angle of elevation be θ.
FC = BC - AE = 30 - 1.5 = 28.5 m
EF = AB = 28.5 m
In ΔEFC, tan θ = FC/EF = 28.5/28.5 = 1
θ = 45°

Ans: (d)

Let AB be the tower and C and D be the positions of the boat.
Distance travelled by boat = CD
From the figure, 
75 tan(60) = (75 + CD) tan(45)
75√3 = 75 + CD
CD = 55 m
Speed = Distance/Time
= 55/10
= 5.5 m/sec 
= 19.8 kmph

Ans: (a)

Ans: (c)

Ans: (b)
Let K be the kite flying in the sky at a height of 75 m from the ground LM. The string KL makes an angle of 60° to the ground
Let length of string KL = x m
KM = 75 m
Now, 



= 86.6 m = 87 m (to the nearest metre)

Ans: (b)

Ans: (b)