Ncert solutions

Ncert solutions

Grade 8

Playing with Numbers | Exercise 16.2

Question 1

Q1) If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

Solution 1:

2\times1000+1\times100+y\times10+5\ =\ 21y5

for 21y5 to be divisible by 9

2+1+y+5 = no. divisible by 9

8+y = 9,18,27....

therefore, y = 9-8

y =1

No = 2115

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subject-cta

Question 1

Q1) If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

Solution 1:

2\times1000+1\times100+y\times10+5\ =\ 21y5

for 21y5 to be divisible by 9

2+1+y+5 = no. divisible by 9

8+y = 9,18,27....

therefore, y = 9-8

y =1

No = 2115

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

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Book a free class now
subject-cta
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