📚 100 Most Expected Maths Questions
For Class 10 CBSE Board Exams – 2026
Instructions: Click on “Show Answer” to reveal the solution. Each question has a different color background.
Assertion & Reason Type
Assertion (A): No two positive numbers can have 18 as their H.C.F. and 380 as their L.C.M. [ Ncert Examplar]
Reason (R): L.C.M. is always completely divisible by H.C.F.
Choose the correct option:
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
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✅ Answer & Explanation
Correct Option: (a) Both A and R are true, and R is the correct explanation of A
Explanation: The fundamental relationship between the Highest Common Factor (HCF) and the Least Common Multiple (LCM) of any two positive numbers is that the LCM is always completely divisible by the HCF.
The assertion is True because No two numbers can exist with 18 as HCF and 380 as LCM if their product equals HCF × LCM. However, 380 ÷ 18 is not an integer (≈21.11), so such numbers don’t exist.
The reason is a true and the correct explanation of Assertion
mathematical property: LCM is always divisible by HCF for any two positive integers.
Prove that √2 is irrational
Prove that √2 is irrational using proof by contradiction.
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✅ Proof
Proof by contradiction:
1. Assume √2 is rational: √2 = p/q where p and q are coprime integers (q ≠ 0).[ H.C.F = 1]
2. Squaring both sides: 2 = p²/q² → p² = 2q²
3. This implies p² is even, so p must be even.
4. Let p = 2k for some integer k.
5. Substitute: (2k)² = 2q² → 4k² = 2q² → 2k² = q²
6. This implies q² is even, so q must be even.
7. But if both p and q are even, they have a common factor 2, contradicting our assumption.
8. Therefore, our initial assumption is false.
∴ √2 is irrational.
Show that 7 – √5 is irrational
Show that 7 – √5 is irrational, given that √5 is irrational.
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✅ Solution
Proof: 1. Assume 7 – √5 is rational.
2. Let 7 – √5 = r, where r is a rational number.
3. Rearranging: √5 = 7 – r
4. Since 7 is rational and r is rational, 7 – r is rational.
5. Therefore, √5 is rational.
6. But this contradicts the given fact that √5 is irrational.
7. Hence, our assumption is false.
∴ 7 – √5 is irrational.
Fruit Vendor Problem
A fruit vendor has 990 apples and 945 oranges. He packs them into baskets. Each basket contains only one of the two fruits but in equal number. Find the number of fruits to be put in each basket in order to have minimum number of baskets.
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✅ Solution
Step 1: Find HCF of 990 and 945
990 = 2 × 3² × 5 × 11
945 = 3³ × 5 × 7
HCF = 3² × 5 = 45
Step 2: Calculate number of baskets
Apples: 990 ÷ 45 = 22
baskets Oranges: 945 ÷ 45 = 21
baskets Total baskets = 22 + 21 = 43
Answer: 45 fruits per basket (minimum 43 baskets)
Seminar Room Problem
In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required if in each room the same number of participants are to be seated and all of them being in the same subject.
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✅ Solution
Step 1: Find HCF of 60, 84, and 108
60 = 2² × 3 × 5
84 = 2² × 3 × 7
108 = 2² × 3³
HCF = 2² × 3 = 12
Step 2: Calculate rooms needed for each subject
Hindi: 60 ÷ 12 = 5 rooms
English: 84 ÷ 12 = 7 rooms
Mathematics: 108 ÷ 12 = 9 rooms
Total rooms = 5 + 7 + 9 = 21
Answer: Minimum 21 rooms required with 12 participants per room
Largest Number Problem
Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7 respectively.
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✅ Solution
Step 1: Subtract remainders from numbers
2053 – 5 = 2048
967 – 7 = 960
Step 2: Find HCF of 2048 and 960
2048 = 2¹¹
960 = 2⁶ × 3 × 5
HCF = 2⁶ = 64
Answer: The largest number is 64
HCF of Smallest Numbers
What is the HCF of smallest prime number and the smallest composite number?
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✅ Solution
Step 1: Identify the numbers Smallest prime number = 2
Smallest composite number = 4
Step 2: Find HCF 2 = 2¹
4 = 2²
HCF = 2¹ = 2
Answer: HCF = 2
HCF with Prime Numbers
If two positive integers a and b are written as a = x³y² and b = xy³, where x, y are prime numbers, then HCF(a, b) is:
A. xy
B. xy²
C. x³y³
D. x²y²
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✅ Solution
Given: a = x³y²
b = xy³
To find HCF: Take the lowest power of common factors For x:
min(3, 1) = 1 → x¹
For y: min(2, 3) = 2 → y²
HCF = x¹ × y² = xy²
Answer: B. xy²
Ratio and LCM Problem
Two numbers are in the ratio 2:3 and their LCM is 180. What is the HCF of these numbers?
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✅ Solution
Step 1: Let numbers be 2x and 3x
HCF(2x, 3x) = x
LCM(2x, 3x) = 6x
Given: LCM = 180
6x = 180 x = 30
Step 2: Find HCF
HCF = x = 30
Answer: HCF = 30
Quadratic Equation with Zeroes
If α, β are the zeroes of the quadratic polynomial p(x) = x² – (k+6)x + 2(2k-1), then find the value of k if α+β = ½αβ.
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✅ Solution
Given: p(x) = x² – (k+6)x + 2(2k-1)
For quadratic equation ax² + bx + c:
Sum of zeroes, α+β = -b/a = k+6
Product of zeroes, αβ = c/a = 2(2k-1) = 4k-2
Given: α+β = ½αβ
k+6 = ½(4k-2)
k+6 = 2k-1
6+1 = 2k-k
7 = k
Answer: k = 7