10th Grade > Mathematics
REAL NUMBERS MCQs
Total Questions : 58
| Page 2 of 6 pages
Answer: Option D. -> 4.8
:
D
√2=1.414213562373095..., is non-recurring and non-terminating. So it is an irrational number.
The value of πis 3.14159265358..., the digits neither terminate nor recur and hence it is an irrational number.
√3=1.732142857…, is non-recurring and non-terminating. So it is an irrational number..
4.8 = 4810satisfiesall the properties of a rational number. Therefore, it is a rational number.
:
D
√2=1.414213562373095..., is non-recurring and non-terminating. So it is an irrational number.
The value of πis 3.14159265358..., the digits neither terminate nor recur and hence it is an irrational number.
√3=1.732142857…, is non-recurring and non-terminating. So it is an irrational number..
4.8 = 4810satisfiesall the properties of a rational number. Therefore, it is a rational number.
Answer: Option B. -> 50
:
B
Given: L.C.M. (100,150) = 300
We know that product of two numbers is equal to the product of their HCF and LCM.
⇒ Product of numbers 100and 150 = HCF × LCM
⇒ 100×150 = HCF × 300
⇒ HCF = 100×150300
⇒ HCF = 15000300
⇒ HCF = 50
∴ HCF of 100 and 150 is 50.
:
B
Given: L.C.M. (100,150) = 300
We know that product of two numbers is equal to the product of their HCF and LCM.
⇒ Product of numbers 100and 150 = HCF × LCM
⇒ 100×150 = HCF × 300
⇒ HCF = 100×150300
⇒ HCF = 15000300
⇒ HCF = 50
∴ HCF of 100 and 150 is 50.
Answer: Option A. -> 9944
:
A
We have,
Largest 4-digit number = 9999
88) 9999 (113
9944
-----------
55
⇒ When 9999 is divided by 88, remainder is 55.
∴ Required number = (9999 - 55) = 9944
:
A
We have,
Largest 4-digit number = 9999
88) 9999 (113
9944
-----------
55
⇒ When 9999 is divided by 88, remainder is 55.
∴ Required number = (9999 - 55) = 9944
Question 14. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?
Answer: Option B. -> 36
:
B
Number of minutes = LCM of 18 and 12
Prime Factorisation:
12 = 2 ×2 × 3
18 = 2 × 3 × 3
L.C.M of the two numbers = 22×32=36
:
B
Number of minutes = LCM of 18 and 12
Prime Factorisation:
12 = 2 ×2 × 3
18 = 2 × 3 × 3
L.C.M of the two numbers = 22×32=36
Answer: Option C. -> 4
:
C
5005 = 5 × 7 × 11 × 13
∴ There are four prime factors of 5005.
:
C
5005 = 5 × 7 × 11 × 13
∴ There are four prime factors of 5005.
Answer: Option A. -> 3
:
A
Given rational number 141120
Here, 120=23×3×5
141=3×47
⇒ 141120 = 3×4723×3×5
=4723×5
Multiply and divide by 52.
=47×5223×5×52
=47×25(2×5)3
=11751000
= 1.175
Therefore, 141120will terminate afterthreedecimal places.
:
A
Given rational number 141120
Here, 120=23×3×5
141=3×47
⇒ 141120 = 3×4723×3×5
=4723×5
Multiply and divide by 52.
=47×5223×5×52
=47×25(2×5)3
=11751000
= 1.175
Therefore, 141120will terminate afterthreedecimal places.
Answer: Option B. -> 170 litres
:
B
Maximum capacity of the measuring vessel = HCF (850, 680)
We use Euclid's division algorithm to find the HCF of 850 and 680.
850 = 680 × 1 + 170
680 = 170 × 4 + 0
∴HCF (850, 680) = 170
So, the maximum capacity of the measuring vessel required is 170 litres.
:
B
Maximum capacity of the measuring vessel = HCF (850, 680)
We use Euclid's division algorithm to find the HCF of 850 and 680.
850 = 680 × 1 + 170
680 = 170 × 4 + 0
∴HCF (850, 680) = 170
So, the maximum capacity of the measuring vessel required is 170 litres.
Answer: Option A. -> 18 and 720
:
A
Prime factorisation of 90 and 144:
90=2×3×3×5
144=2×2×2×2×3×3
⇒HCF=2×32=18
LCM=24×32×5=720
∴ HCF and LCM of 90 and 144 are 18 and 720 respectively.
:
A
Prime factorisation of 90 and 144:
90=2×3×3×5
144=2×2×2×2×3×3
⇒HCF=2×32=18
LCM=24×32×5=720
∴ HCF and LCM of 90 and 144 are 18 and 720 respectively.
Answer: Option B. -> 11
:
B
Euclid's division algorithm to find HCF of 1650 and 847:
Step 1: 1650 = 847 × 1 + 803
Step 2: 847 = 803 × 1 + 44
Step 3: 803 = 44 × 18 + 11
Step 4: 44 = 11 × 4 + 0
Hence, 11 is the HCF of 1650 and 847.
:
B
Euclid's division algorithm to find HCF of 1650 and 847:
Step 1: 1650 = 847 × 1 + 803
Step 2: 847 = 803 × 1 + 44
Step 3: 803 = 44 × 18 + 11
Step 4: 44 = 11 × 4 + 0
Hence, 11 is the HCF of 1650 and 847.
Answer: Option C. -> 21
:
C
Number of rooms will be minimum if each room accomodates maximum number of participants.
In each room, the same number of participants are to be seated and all of them must be for the same subject.
∴ Number of participants in each room must be the HCF of 60, 84 and 108.
Prime factorisation of 60, 84 and 108:
60=22×3×5
84=22×3×7
108=22×33
∴ HCF =22×3=12
In each room, 12 participants can be accommodated.
Number of rooms=Total number of participants12
=60+84+10812
=25212
=21
∴ The total number of rooms is 21.
:
C
Number of rooms will be minimum if each room accomodates maximum number of participants.
In each room, the same number of participants are to be seated and all of them must be for the same subject.
∴ Number of participants in each room must be the HCF of 60, 84 and 108.
Prime factorisation of 60, 84 and 108:
60=22×3×5
84=22×3×7
108=22×33
∴ HCF =22×3=12
In each room, 12 participants can be accommodated.
Number of rooms=Total number of participants12
=60+84+10812
=25212
=21
∴ The total number of rooms is 21.
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