Previous Year Solved Questions 2022

Question 2: What is the greatest possible speed at which a girl can walk 95 m and 171 m in an exact number of minutes? [CBSE 2022 TERM 1]
(a) 17 m/min (b) 19 m/min (c) 23 m/min (d) 13 m/min

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Ans: (b) 19 m/min
Find HCF of 95 and 171:
95 = 5 × 19
171 = 3² × 19
HCF = 19
Greatest possible speed = 19 m/min

Question 3: Three alarm clocks ring their alarms at regular intervals of 20 min, 25 min and 30 min respectively. If they first beep together at 12 noon, at what time will they beep again for the first time? [CBSE 2022 TERM 1]
(a) 4:00 pm (b) 4:30 pm (c) 5:00 pm (d) 5:30 pm

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Ans: (c) 5:00 pm
LCM of 20, 25, 30:
20 = 2² × 5
25 = 5²
30 = 2 × 3 × 5
LCM = 2² × 3 × 5² = 300 minutes
300 minutes = 5 hours
12:00 noon + 5 hours = 5:00 pm

Question 4: The greatest number which when divides 1251, 9377 and 15628 leaves remainder 1, 2 and 3 respectively is [CBSE 2022 TERM 1]
(a) 575 (b) 450 (c) 750 (d) 625

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Ans: (d) 625
1251 – 1 = 1250
9377 – 2 = 9375
15628 – 3 = 15625
Find HCF of 1250, 9375, 15625:
1250 = 2 × 5⁴
9375 = 3 × 5⁵
15625 = 5⁶
HCF = 5⁴ = 625

Question 5: Given a and b are coprime, whose H.C.F is 1. Then, a³ and b³ are also coprime numbers, whose H.C.F is 1. [CBSE 2022 TERM 1]

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True
Since a and b are coprime (HCF=1), they have no common prime factors.
a³ and b³ will have the same prime factors as a and b respectively, with exponents tripled.
Thus they also have no common prime factors, so HCF(a³, b³) = 1.

Question 6: The number 2(5ⁿ + 6ⁿ) always ends with the digit 2 for every n ∈ ℕ. [CBSE 2022 TERM 1]

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True
5ⁿ always ends with 5
6ⁿ always ends with 6
5ⁿ + 6ⁿ ends with 5+6=11 → units digit 1
2 × (5ⁿ + 6ⁿ) ends with 2×1=2

Question 7: The LCM of two numbers is 2400. Which of the following cannot be their HCF? [CBSE 2022 TERM 1]
(a) 300 (b) 400 (c) 500 (d) 600

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Ans: (c) 500
HCF must divide LCM.
Check divisibility:
2400 ÷ 300 = 8 ✓
2400 ÷ 400 = 6 ✓
2400 ÷ 500 = 4.8 ✗ (not integer)
2400 ÷ 600 = 4 ✓
500 does not divide 2400.

Question 8: Two positive numbers have their HCF as 12 and their product as 6336. The number of pairs possible for the numbers is: [CBSE 2022]
(a) 2 (b) 3 (c) 4 (d) 1

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Ans: (a) 2
Let numbers be 12a and 12b, where a,b coprime.
12a × 12b = 6336
144ab = 6336
ab = 44
Coprime pairs (a,b): (1,44), (4,11)
Numbers: (12,528) and (48,132)
Total 2 pairs.

Question 9: If ‘n’ is any natural number, then 12ⁿ cannot end with the digit: [CBSE 2022]
(a) 2 (b) 4 (c) 8 (d) 0

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Ans: (d) 0
12ⁿ = (2² × 3)ⁿ = 2²ⁿ × 3ⁿ
Contains only prime factors 2 and 3, no factor 5.
To end with 0, number must have factor 10 = 2×5.
Since no factor 5, cannot end with 0.

Question 11: The number 385 can be expressed as the product of prime factors as: [CBSE 2022]
(a) 5 × 11 × 13
(b) 5 × 7 × 11
(c) 5 × 7 × 13
(d) 5 × 11 × 17

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Ans: (b) 5 × 7 × 11
Prime factorisation of 385:
385 ÷ 5 = 77
77 ÷ 7 = 11
11 ÷ 11 = 1
385 = 5 × 7 × 11

Real NumberPrevious Year Solved Questions 2022

Real Number
Previous Year Solved Questions 2022