10th Maths 1.2

NCERT Class 10th solution of Exercise 1.1

NCERT Class 10th solution of Exercise 1.3

NCERT Class 10th solution of Exercise 1.4

Exercise 1.2

Q1. Express each number as a product of its prime factors:
i) 140    ii) 156    iii) 3825    iv) 5005    v) 7429
Sol.
i) 
140=2×2×5×7
140=22×51×71
Answer:
=22×51×71.
ii)
156=2×2×3×13
156=22×3×13
Answer:
=22×3×13.
iii)
3825=3×3×5×5×17
3825=32×52×17.
Answer:
=32×52×17.
iv)
5005=5×7×11×13
5005=51×71×111×131.
Answer:
=51×71×111×131.
v)
7429=17×19×23
7429=171×191×23
Answer:
=171×191×23


Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM×HCF=porduct of the two numbers.
i) 26 and 91    ii) 510 and 92    iii) 336 and 54
Sol.
26=2×13
91=7×13
HCF=13
and
LCM=2×7×13=182
Verify
LCM×HCF=Product of the two numbers
182×13=26×91
2366=2366
Answer:
HCF=13
LCM=182
ii)
510=2×3×5×17
92=2×2×23
HCF=2
and
LCM=2×2×3×5×17×23×=23460
Verify 
LCM×HCF=Product of the two numbers
2×23460=510×92
46920=46920
Answer :
HCF=2
LCM=23460
iii)
336=2×2×2×2×3×7
54=2×3×3×3
HCF=2×3=6
LCM=2×2×2×2×3×3×3×7=3024
Verify
LCM×HCF=Product of the two numbers
3024×6=336×54
18144=18144
Answer :
HCF=6
LCM=3024
Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
i) 12,15 and 21    ii) 17,23 and 29    iii) 8,9 and 25
Sol.
12=2×2×3
15=3×5
21=3×7
HCF=3
LCM=2×2×3×5×7=420
Answer :
HCF=3
LCM=420
ii)
17=1×17
23=1×23
29=1×29
HCF=1
LCM=17×23×29=11339
Answer :
HCF=1
LCM=11339
iii) 
8=1×2×2×2
9=1×3×3
25=1×5×5
HCF=1
LCM=2×2×2×3×3×5×5=1800
Answer :
HCF=1
LCM=1800
Q4. Given that HCF(306,657)=9, find LCM(306,657).
Sol. :
LCM×HCF=Product of two numbers
LCM×9=306×657
LCM=306×6579
LCM=2010429
LCM=22338
Answer :
LCM=22338
Q5. Check whether 6n can end with the digit 0 for any natural number n.
Sol. :
We know that 6n=2n×3n in which 5 is not a prime factor and the number which end with the digit 0 is divisible by 5, has 5 as prime factor.
Answer :
Thus, the number 6n cannot end with the digit 0 for any natural number.
Q6. Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite numbers.
Sol. :
=7×11×13+13
=13(7×11+1)
=13×78
Which is a composite number
and 
=7×6×5×4×3×2×1+5
=5(7×6×4×3×2×1+1)
=5(1008+1)
=5×1009
which is a composite number.
Answer :
Thus, the given numbers are composite numbers.
Q7. 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?
Sol.
18=2×3×3
12=2×2×3
LCM(18,12)=2×2×3×3×=36
Answer :
Thus, they will meet again at the starting point after 36 minutes.



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