Maths Olympiad Previous Year Paper -1 | Olympiad Preparation for Class 10 PDF Download

Ans: (c)
 The region common to all three shapes is 9.

Ans: (b)
The pattern followed is:

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

Ans: (a)
There can be two possible sitting arrangements. 

So, A and F are not a neighbour pair.

Ans: (b)
So distance between post office and post man = 40 + 50 = 90 m

Ans: (c)

Ans: (a)
Number of face with no face colored = 27

Ans: (a)

• First, we need to replace the symbols with their meanings: R = ×, Q = +, P = ÷, S = –.
• For option (a): 5 R 6 Q 8 P 2 S 9 becomes 5 × 6 + 8 ÷ 2 – 9.
• Calculating step-by-step: 5 × 6 = 30, then 30 + 8 = 38, next 8 ÷ 2 = 4, so 38 – 4 = 34, and finally 34 – 9 = 25.
• This confirms that option (a) is correct as it equals 25.

Ans: (c)

Ans: (c)

• First, we interchange the first and last digits of each number:
• 246 becomes 642, 525 becomes 525, 432 becomes 234, 726 becomes 627, and 283 becomes 382.
• Next, we add 1 to the middle digit of each number: 642 becomes 652, 525 becomes 535, 234 becomes 244, 627 becomes 637, and 382 becomes 392.
• Now, we arrange these numbers in descending order: 652, 637, 535, 392, 244.
• The middle number in this arrangement is 535, and the sum of its digits (5 + 3 + 5) is 13.

Ans: (a)

Ans: (c)

Ans: (b)

Ans: (d)

Ans: (b)
G ×× H + K – L means G is the father of H, who is mother of K. 
Also, K is the daughter of L. 
So, L is the husband of H. Hence, G is the father-in-law of L. 

Ans: (b)

• To find the value of m, we first need to solve the given linear equations.
• By solving (2x - 3y = 13) and (7x - 2y = 20), we can find the values of x and y.
• After finding y, we substitute it into the equation (y = mx + 7) to express m in terms of x.
• Through this process, we determine that the value of m is -5.

Ans: (b)

• In an arithmetic progression (A.P.), the terms can be expressed as: a₁, a₁ + d, a₁ + 2d, a₁ + 3d, a₁ + 4d, where d is the common difference.
• From the equation a₁ + a₃ + a₅ = –12, substituting the terms gives: a₁ + (a₁ + 2d) + (a₁ + 4d) = –12, simplifying to 3a₁ + 6d = –12.
• From a₁·a₂·a₃ = 8, substituting gives: a₁(a₁ + d)(a₁ + 2d) = 8.
• Solving these equations leads to the common difference d = -3.

Ans: (b)
480 = 25 x 31 x 51
Number of factors of 480 = (5 + 1) (1 + 1) (1 + 1) = 24.
Number of ways in which 480 can be resolved into two factors =  24/2 = 12 

Ans: (b)

Ans: (d)

• For a quadratic equation to have equal roots, the discriminant must be zero.
• The discriminant is calculated as b² - 4ac. Here, a = (a - 4), b = 2(a - 4), and c = 4.
• Setting the discriminant to zero gives us the equation: (2(a - 4))² - 4(a - 4)(4) = 0.
• Solving this leads to the values of a being 4 and 8, which means both options A and B are correct.

Ans: (a)
The required number is HCF of 612 and 342.
This number gives the maximum number of pencils in each box and the number of boxes with them be the least.
By using Euclid’s division algorithm, we have 612 = 342 × 1 + 270
342 = 270 × 1 + 72
270 = 72 × 3 + 54
72 = 54 × 1 + 18
54 = 18 × 3 + 0
Here we notice that the remainder is zero, and the divisor at this stage is 18. 
∴ HCF of 612 and 342 is 18.
So, the shopkeeper can pack 18 pencils per box.

Ans: (c)
For the number of rows to be the least, the number of students in each row must be the highest.
The number of students in each row is a common factor of 78 and 45. The highest value is the HCF of 78 and 45, which is 3.
Minimum number of rows = 78/3 + 45/3 = 41

Ans: (a)

• The total possible outcomes when rolling two dice is 36 (6 sides on the first die multiplied by 6 sides on the second die).
• The favorable outcomes where the difference is 1 are: (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5). This gives us a total of 10 outcomes.
• The probability is calculated as the number of favorable outcomes divided by the total outcomes: 10/36, which simplifies to 5/18.
• Thus, the probability that the difference of the numbers shown on the dice is 1 is 5/18.

Ans: (b)

• The volume of water displaced by the balls equals the volume of the balls themselves.
• The volume of the cylinder can be calculated using the formula: Volume = π × (radius)² × height. Here, the radius is 2 cm (half of the diameter) and the height is 0.8 cm.
• Calculating the volume gives: Volume = π × (2 cm)² × 0.8 cm = 3.2π cm³.
• Since there are 300 balls, the volume of one ball is: Volume of one ball = Total volume / Number of balls = (3.2π cm³) / 300.
• Using the formula for the volume of a sphere (V = 4/3 × π × (radius)³), we can find the radius of one ball and then double it to get the diameter, which results in a diameter of 0.4 cm.

Ans: (a)
Let f(x) = ax² + bx +c.
Given, f (− 1) = 7, f (− 2) = 12, f (−3) = 19 
a −b + c = 7 .... (1) 
4a − 2b + c = 12 ......(2) 
9a − 3b + c = 19 ..........(3) 
Solving these equations, we get, a = 1, b = -2, c = 4 
Therefore, a + b + c = 3

Ans: (a)

• The question asks for a rational number, which is a number that can be expressed as a fraction of two integers.
• Option (a) involves the sum of 2 and √3, which is not a rational number, but its inverse can be rational.
• Options (b) and (c) involve square roots of non-perfect squares, which are also irrational numbers.
• Thus, the correct answer is (a) because it includes the concept of an inverse that can yield a rational result.

Ans: (d)
Let the three terms of an AP be (a - d), a and (a + d).
The sum of these terms is 3a. 
3a = 69 
a = 23 
Product of these three terms is 
(a + d) a (a - d) = 11339
(23 + d) (23 - d) = 493 
529 - d² = 493 ⇒ d = ±6.
Taking d = 8, we get the terms as 17, 23 and 29.

Ans: (a)

• First, calculate p(2): p(2) = 2³ + 3(2)² – 2(2) + 4 = 8 + 12 - 4 + 4 = 20.
• Next, calculate p(–2): p(–2) = (–2)³ + 3(–2)² – 2(–2) + 4 = -8 + 12 + 4 + 4 = 12.
• Then, calculate p(0): p(0) = 0³ + 3(0)² – 2(0) + 4 = 4.
• Now, combine these results: p(2) + p(–2) – p(0) = 20 + 12 - 4 = 28.

Ans: (c)
If the numbers a, b and c are in AP, then 
b - a = c - b 
2b = a + c 
The given three numbers (2p - 1), (3p + 1) and 11 are in AP.
Then, 2(3p + 1) = (2p - 1) + 11 
6p + 2 = 2p - 1 + 11 
6p + 2 = 2p + 10 
6p - 2p = 10 - 2 
4p = 8 
p = 2 
So, the value of p is 2.

Ans: (a)
Let the two arithmetic series be a, a + d, a + 2d .... and A, A + D, A + 2D, .... Given that,

Ratio of the 10th terms of these series =
Putting n = 19 in (i), we have

Ans: (b)

Ans: (a)
The equations are: 
x + y + z = 100 
4x - 2y - z = 114 
4x - y - 2z = 136 
On solving the equations 
x = 50, y = 36, z = 14

Ans: (b)

• To find the values of p and q, we substitute x = -2 into both polynomials since x + 2 is a common factor.
• For p(x): p(-2) = (-2)² - 3(-2) + p = 4 + 6 + p = 10 + p. Setting this to 0 gives p = -10.
• For q(x): q(-2) = 2(-2)² + q(-2) + 2 = 8 + q(-2) + 2 = 10 + q(-2). Setting this to 0 gives q = 5.
• Thus, the values of p and q are -10 and 5 respectively.

Ans: (c)
The equation is: x² − 24 − 2tx + 11t = 0 
For equal roots, discriminant must be 0 
4t² − 4(11t − 24) = 0 
t² − 11t + 24 = 0 
(t − 3)(t − 8) = 0

Ans: (b)

• Let the initial number of students be x.
• The cost per student before 5 students dropped out is 500/x.
• After 5 students leave, the number of students becomes x - 5, and the new cost per student is 500/(x - 5).
• According to the problem, the new cost per student is ₹5 more than the old cost: 500/(x - 5) = 500/x + 5.
• Solving this equation gives x = 20, meaning 20 students initially planned for the picnic.

Ans: (c)

• Let A's investment be represented as 2x, B's as 3y, and C's as y.
• From the problem, we know that 2A = 3B and B = 4C, which gives us the relationships between their investments.
• By substituting the values, we find that B's share of the total profit is calculated based on the ratio of their investments.
• After calculating, B's share from the total profit of ₹ 29,700 comes out to be ₹ 10,800.

Ans: (c)

• Let the total number of workers be N.
• The total salary of all workers is N × 8,500.
• The total salary of 7 mechanics is 7 × 10,000 = 70,000.
• The total salary of the remaining workers is (N - 7) × 7,800.
• Setting up the equation: N × 8,500 = 70,000 + (N - 7) × 7,800.
• Solving this gives N = 22.

Ans: (c)

• Let the cost of one cap be C and one bat be B.
• From the first statement: 3C + 4B = 257.
• From the second statement: 4C + 3B = 324.
• By solving these two equations, we find C and B.
• After calculating, we find the cost of one cap and ten bats: C + 10B = 155.

Ans: (b)

• To find the speed of the second train, we first calculate the total distance covered when both trains pass each other, which is the sum of their lengths: 150 m + 100 m = 250 m.
• The time taken to pass is 10 seconds, so we can find the relative speed of both trains combined: Speed = Distance/Time = 250 m / 10 s = 25 m/s.
• Now, we convert the speed of the first train from km/h to m/s: 30 km/h = 30 * (1000 m / 3600 s) = 8.33 m/s.
• Using the formula for relative speed: Relative Speed = Speed of Train 1 + Speed of Train 2, we have 25 m/s = 8.33 m/s + Speed of Train 2.
• Solving for Speed of Train 2 gives us: Speed of Train 2 = 25 m/s - 8.33 m/s = 16.67 m/s, which converts back to km/h as 16.67 * (3600 s / 1000 m) = 60 km/h.

Ans: (c)

• The formula for compound interest is used to calculate the interest earned on an investment over time, which in this case is ₹993 for 3 years at 10% per annum.
• The simple interest can be calculated using the formula: Simple Interest = (Principal × Rate × Time) / 100.
• Given that the compound interest is higher than simple interest, we can derive that the simple interest for the same principal, rate, and time will be ₹900.
• This shows that while compound interest accumulates more over time, the simple interest remains straightforward and is less than the compound interest in this scenario.

Ans: (b)

• To find the length of the arc between two spokes, we first need to calculate the circumference of the wheel using the formula: Circumference = π × diameter.
• Here, the diameter is 91 cm, so the circumference is approximately 3.14 × 91 cm = 285.34 cm.
• Since there are 22 spokes, the length of the arc between two adjoining spokes is the total circumference divided by the number of spokes: 285.34 cm / 22 ≈ 13 cm.
• Thus, the length of the arc between two spokes is 13 cm, which corresponds to option (b).

Ans: (d)

• The volume of the hemispherical bowl can be calculated using the formula: V = (2/3)πr³. Here, the radius (r) is half of the diameter, which is 18 cm.
• The volume of one cylindrical bottle is given by the formula: V = πr²h. For the bottle, the radius is 3 cm and the height is 9 cm.
• After calculating both volumes, divide the volume of the bowl by the volume of one bottle to find out how many bottles are needed.
• The final calculation shows that 48 bottles are required to empty the bowl.

Ans: (c)

• The roots 5 and 9 suggest a quadratic equation of the form (x - 5)(x - 9) = 0, which expands to x² - 14x + 45 = 0.
• The roots 12 and 4 suggest another equation of the form (x - 12)(x - 4) = 0, which expands to x² - 16x + 48 = 0.
• To find the correct quadratic equation, we need to combine the information from both mistakes.
• The correct quadratic equation that fits the roots derived from the mistakes is x² - 14x + 48 = 0.

Ans: (b)
Let the cost of each burger be p.
Let the cost of each chocolate be c.
Let the cost of each pizza be z.
Then 2p + 14c + 5z = 400 ……….(1) 
Also, 20p + 7c + 15z = 1270 ……….(2) 
Multiply Eq. (2) by 2, we have 
40p + 14c + 30z = 2540 ………(3) 
Subtracting Eq. (1) from Eq. (3), we have 
38p + 25z = 2140

Ans: (a)
Given, x² + 3x – 10 = 0
⇒ x² + 5x – 2x – 10 = 0
⇒ x(x + 5) -2(x + 5) = 0
⇒ (x – 2) (x + 5) = 0
Zeroes are x – 2 = 0 or x + 5 = 0
∴ x = 2 or x = -5

Ans: (c)
Every year in the first 5 years, the number of carrots that is eaten is y (say).
The number of carrots produced in the xth year = 11x - y
Number of carrots produced in 4th year = 36
⇒ 11(4) - y = 36 ⇒ y = 8
Number of fruits produced in xth year in the last 5 years = 11x - y/2
Number of fruits produced in the 9th year = 11(9) - 8/2 = 99 - 4 = 95

Ans: (c)

Ans: (d)

Ans: (c)
Given, x-intercept (a) = 5 and y-intercept (b) = 7
∴ The equation of the required line is x/a + y/b = 1
x/5 + y/7 = 1
7x + 5y - 35 = 0

Ans: (a)
Let the ordinate of the second point be y. 
So, the first point is (2, -4) and the second is (10, y). 
The distance between them must be 10.

64 + y² + 16 + 8y = 100
y² + 8y − 20 = 0
y² + 10y − 2y − 20 = 0
y · (y + 10) − 2 · (y + 10) = 0
(y + 10) · (y − 2) = 0
y = −10 or y = 2