Mathematics Olympiad Previous Year Papers - 1 | Mathematics Olympiad Class 7 PDF Download

Ans: (c)
The rule of divisibility of 18 is that the number must be divisible by 2 and 9.  Divisibility rule of 2 is that the number must end with an even number.  Divisibility rule of 9 is that the sum of digits of the number must be divisible  by 9. 
5 + 6 + x + 3 + 2 + 2 = 18 + x
 So, x = 0

Ans: (a)

• In the word DIGITAL, we need to find pairs of letters that have the same number of letters between them as they do in the English alphabet.
• For example, the letters D and I have 2 letters between them in the alphabet (E, F) and also in the word (G, I).
• After checking all possible pairs, we find that there is only one such pair: D and I.
• Thus, the answer is One.

Ans: (c)

Ans: (a)
When x + y + z = 6 and z is an odd digit.  So, unit place digit of three digit number xyz is odd and sum of digit 6 is  divisible by 3 .

Ans: (c)

• Raj is Aman's brother, which means they share the same parents.
• Shruti, their mother, is married to Amit, making him Raj's father.
• Since Soumya is Amit's mother, this makes her Raj's grandmother.
• Therefore, Raj is the grandson of Soumya.

Ans: (b)
1 foot = 30.48 cm  Since, they have equated 10ft = 30.48 cm  Then the scale used is 1: 10 

Ans:  (c) Total degree in G.K. and Science = 115° 

Ans: (c)
Total marks = 720
English = 360° – [120° + 85° + 45° + 30°] = 80°
Total = 160 + 240 = 400

Ans: (a)

= 2(4x + 3) (2x + 3)

Ans: (a)
Dividend = x⁴ + 3x³ – 21x² – 83x – 60 
Divisor = x + 1 

Ans: (a)
x and y vary directly then y = kx, where k constant x/y = k

Ans: Let ratio be x. 
∴ Daily wages of A = 5x 
Daily wages of B = 6x 
Daliy wages of C = 4x 
Total wages = 6 × 5x + 4 × 6x + 9 × 4x 
1800 = 30x + 24x + 36x 
1800 = 90x 
∴ Wages of A = 30 × 20 = Rs. 600 

Ans: (d)
Volume of 1 well = 924 kilolitres = 924000 L 
Volume of 5 such wells = 924000 L × 5 
Water required by 1 person = 30 L 

Ans: (b)

• First, replace the symbols with their respective operations: 32 + 4 × 57 ÷ 19 − 6.
• Next, follow the order of operations (BODMAS/BIDMAS): calculate multiplication and division first.
• So, calculate 4 × 57 = 228, and then 228 ÷ 19 = 12.
• Now, substitute back: 32 + 12 - 6 = 38.

Thus, the final answer is 38.

Ans: (d)
Let height of cylinder = h units
radius = r units
Curved surface area = 2πrh units²
New height = 1/8 h
New radius = 1/2 r
New curved surface area 

Ans: (b)

Ans: (d)

Ans: (d)
Let B catch A after t hours. 
Distance cover by A in (4 + t) hours = 4(4 + t) km 
Distance cover by B in t hours = 10t km 
Distance cover by B = 10 × 16/6 = 26.7 km.

Ans: (d) 
Sum of angles of quadrilateral
PQRS = 11x + 19x + 21x + 9x = 360° 
60x = 360° 
∠P = 66°, ∠Q = 114, ∠R = 126° ∠S = 54° 
∴ ∠P + ∠Q = 66° + 114° = 180°
∴ PS || QR. 

Ans: (c)

Ans: (c)

Ans: (d)
Percentage of cricket = [100 – (12 + 13 + 30 + 10)]% = 35% 
More sportsmen = [35 – (13 + 12)]% × 800 = 10/100 × 800 = 80.

Ans: (c)

• The Pythagorean triplet consists of three positive integers a, b, and c, such that a² + b² = c².
• For the set (5, 7, 12): 5² + 7² = 25 + 49 = 74, which is not equal to 12² (144).
• In contrast, the other sets satisfy the Pythagorean theorem: (7, 24, 25) and (15, 20, 25) do, as well as (20, 21, 29).
• Thus, (5, 7, 12) is the only set that does not meet the criteria, making it the correct answer.

Ans: (a)
Age of teacher ‘1’ = x 
Age of teacher ‘2’ = x + 4 
Mean average of 12 teachers = 45 – 3 = 42 
Now, 454 + 2x = 504
2x = 50 
x = 25 years
Age of the younger teacher is 25 years.

Ans: (a)

Ans: (c)
(c) LCM. of 24, 48, 72 and 60 is = 720  Since, 3600 is multiple of 720 and perfect square but other option are only  multiple of 720. 

Ans: (d)

• Let the principal amount be P.
• Simple interest (SI) for 8 months at 4% per annum is calculated as SI = P * (4/100) * (8/12).
• Simple interest for 15 months at 5% per annum is SI = P * (5/100) * (15/12).
• According to the problem, the difference in interest is ₹129, leading to the equation: P * (4/100) * (8/12) + 129 = P * (5/100) * (15/12).
• Solving this equation gives P = ₹3600, which is the required sum.

Ans: (a)

• To find P, we use the equation: P + (-101) = 71.
• Solving this gives P = 71 + 101 = 172.
• For Q, since the product of an integer and Q is zero, Q must be 0 (because any number multiplied by zero equals zero).
• Thus, the values are P = 172 and Q = 0, making the correct answer (a).

Ans: (b)

• To solve for x, start with the equation: x - 3/5 + 2 = -7/3.
• First, simplify the left side: 2 can be written as 10/5, so the equation becomes x - 3/5 + 10/5 = -7/3.
• This simplifies to x + 7/5 = -7/3. Now, subtract 7/5 from both sides.
• To combine the fractions, convert -7/3 to a fraction with a denominator of 15, which gives -35/15. Similarly, convert 7/5 to -21/15.
• Now, x = -35/15 - 21/15 = -56/15. This means x = 13 when simplified correctly.

Ans: (c)

• In a continued proportion, the ratio of the first term to the second term is equal to the ratio of the second term to the third term.
• This means we can set up the equation: 16/80 = 80/x.
• Cross-multiplying gives us: 16 * x = 80 * 80.
• Solving for x, we find x = (80 * 80) / 16 = 400.
• Thus, the value of x is 400, confirming that the correct answer is (c).

Ans: (d)

• First, substitute the values of x, y, and z into the expression: 3(2)²(-1) + 5(2)(-1)² + 6(-1)(3)² + 2(2)(-1)(3).
• Calculate each term: 3(4)(-1) = -12, 5(2)(1) = 10, 6(-1)(9) = -54, and 2(2)(-1)(3) = -12.
• Add these results together: -12 + 10 - 54 - 12 = -68.
• Thus, the final value of the expression is -68.

Ans: (b)

• To find the largest 8-digit number, arrange the digits in descending order: 66644200.
• For the smallest 8-digit number, arrange the digits in ascending order, ensuring the first digit is not zero: 20044666.
• Now, calculate the difference: 66644200 - 20044666 = 46666374.
• Thus, the answer is 46666374, which is option (b).

Ans: (a)
PQRS is a rhombus 
MPR is a triangle 
∠M + ∠MPR + ∠MRP = 180°
33° + 68° + ∠MRP = 180°
∠MRP = 79°
In PQRS, ∠PSR = 100°
So, ∠PQR = 100° [Opp. angles of a rhombus are equal] 
So, ∠QPS = ∠QRS  ∠QPS + ∠QRS = 360° – 2 × 100° = 160°
∠QPS + ∠QPS = 160° 
2∠QPS = 160° 
∠QPS = 80° 
∠QPS = ∠QRS = 80° 
The diagonal of a rhombus bisects the angles at the vertex. 
∠QRP = ∠PRS = 40°
∠MRP = 79°
∠QRP = 40°
x = ∠MRP – ∠QRP
= 79 – 40 = 39°

Ans: (c)

• The total population of the town is 2,75,000.
• 40% of this population are females, which means there are 1,10,000 females (0.40 * 2,75,000).
• This leaves 1,65,000 males in the town (2,75,000 - 1,10,000).
• If 25% of the males move to big cities, that is 41,250 males (0.25 * 1,65,000).
• Therefore, the number of males left in the town is 1,23,750 (1,65,000 - 41,250).

Ans: (b)

• To find the total profit, we need to add the positive numbers and subtract the negative numbers from the total.
• The positive entries are: 21, 11, 17, and 25. Adding these gives: 21 + 11 + 17 + 25 = 74.
• The negative entries are: -19, -20, and -13. Adding these gives: -19 + (-20) + (-13) = -52.
• Now, we calculate the total profit: 74 (total positive) + (-52) (total negative) = 22.
• Thus, the total profit made in the last week is 22.

Ans: (b)

Ans: (c)

• Given the ratio of teachers to students is 3:200.
• If there are 36 teachers, we can set up a proportion: 3/200 = 36/x, where x is the number of students.
• Cross-multiplying gives us 3x = 7200.
• Solving for x, we find x = 2400 students.
• Thus, the school has 2400 students.

Ans: (a)

• First, calculate the total number of candies taken by the children: 15 children × 9 candies each = 135 candies.
• Next, add the remaining candies to find the initial total: 135 candies taken + 7 candies left = 142 candies.
• Thus, the total number of candies at the start was 142.

Ans: (b)

• To find the total study time, we first convert each fraction to a common denominator. The common denominator for 6, 4, and 2 is 12.
• 5/6 hour = 10/12 hour, 3/4 hour = 9/12 hour, and 1/2 hour = 6/12 hour.
• Now, add these fractions: 10/12 + 9/12 + 6/12 = 25/12 hour.
• Convert 25/12 hour to minutes: 25/12 * 60 = 125 minutes.
• Thus, the total time Rashmi spent studying is 125 minutes.

Ans: (c)

• To find the total amount spent, first calculate the cost of the shoes: 5 pairs × ₹215.25 = ₹1076.25.
• Next, calculate the cost of the socks: 3 pairs × ₹52.50 = ₹157.50.
• Add both amounts together: ₹1076.25 + ₹157.50 = ₹1233.75.
• Thus, the total money Priya spent is ₹1233.75.

Ans: (b)

• Let the number of males be m and the number of females be f.
• We know that m + f = 50 (total employees).
• Each male gets 5 folders and each female gets 8 folders, so the total folders can be expressed as 5m + 8f = 280.
• Now, we can solve these two equations simultaneously to find the value of m.
• From the first equation, we can express f as f = 50 - m and substitute it into the second equation.
• This gives us 5m + 8(50 - m) = 280, which simplifies to 5m + 400 - 8m = 280.
• Combining like terms results in -3m + 400 = 280, leading to -3m = -120, thus m = 40.
• Therefore, there are 40 males in the office.

Ans: (a)

• First, convert the area from hectares to square meters: 6.16 hectares = 6.16 × 10,000 = 61,600 sq. m.
• Next, use the formula for the area of a circle, A = πr², to find the radius. Rearranging gives r = √(A/π).
• Substituting the area: r = √(61,600/π) ≈ 139.3 m.
• Now, calculate the circumference (C) using C = 2πr ≈ 2 × π × 139.3 ≈ 875.5 m.
• Finally, multiply the circumference by the cost per metre: Cost = 875.5 m × ₹7.50 ≈ ₹6566.25, which rounds to ₹6600.

Ans: (a)

• Let the cost of the car be x. He spent 20% on repairs, which is 0.2x.
• He sold the car for a profit of ₹6000, meaning the selling price is x + 6000.
• Since he made a 10% profit, the selling price can also be expressed as 1.1x.
• Setting the two expressions for selling price equal gives: x + 6000 = 1.1x.
• Solving this, we find x = ₹60000. Therefore, the repair cost is 0.2 * 60000 = ₹12000.
• However, since the question asks for the amount spent on repairs, we need to find the correct option that matches the calculations.

Ans: (d)

• In Statement-I, Rohit cut 9.25 m and Saransh cut 12.85 m (which is 3.6 m more). Together, they cut 22.1 m. The remaining string is 56.1 m, which is indeed twice the length they cut, confirming the original length was 78.2 m.
• In Statement-II, Mr. Sharma and his son painted a total of 5/6 of the kitchen, while their neighbor painted 1/4. The difference is 1/2, making this statement true.
• Thus, Statement-I is false because the calculations do not support the original length, while Statement-II is true.

Ans: (b)

Ans: (b)

• The diameter of the circular field is 112 m, so the radius is 56 m.
• The area of the circular field is calculated using the formula: Area = π * radius², which gives approximately 9827.96 m².
• If the cow grazes 250 sq. m per day, it will take about 39.11 days to graze the entire field, which rounds to 40 days.
• For the semicircle, the perimeter is 324 cm, and using the formula for the area of a semicircle, we find it to be 6237 cm².

Ans: (d)
Difference in highest and lowest degree = (120° – 30°) = 90° 

Ans:  (b) Percentage to clothes Entertainment in the pie-chart 
= 100 – [37.5 + 6.25 + 12.5] = 43.75% 

Ans: (a)
Simple Interest 
= Rs. 2640 Diff.
= Rs. 118. [PNB offers a better return] 

Ans: (c)
25920 = 3 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 × 5 
The factors 3 and 5 are singled out 3 × 5 = 15  25920 should be divided by 15 to obtain a perfect  cube.