Aptitude Discussion

Q. |
If we write down all the natural numbers from 259 to 492 side by side get a very large natural number $259260261....491492$. How many 8's will be used to write this large natural number? |

✖ A. |
32 |

✖ B. |
43 |

✖ C. |
52 |

✔ D. |
53 |

**Solution:**

Option(**D**) is correct

From 259 to 458, there are 200 natural numbers so there will be $2×20 =40$ 8's

From 459 to 492 we have 13 more 8's and so answer is $40+13 =$ **53**

**Ratishchandran Nair**

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**Abhishek**

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268,278,288,298,308,318,328,338,348,358,368,378,388,398,408,418,428,438,448,458,468,478,488,--280,281,282,283,284,285,286,287,289,380,381,382,383,384,385,386,387,389,480,481,482,483,484,485,486,487,489

**Jasmine**

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plz explain clearly ????? anyone tell me the concept

Follow the comment of 'RATISHCHANDRANNAIR'. It is same as q.2

**Krunal**

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what is the calculation of 2*20 and how comes 13 ????

please explain m

In the span of 100 natural numbers (from 259 to 358 OR 359 to 458) there are 20, 8's. So in the span of 200 natural numbers (from 259 to 458) total 8's will be ($2 \times 200$).

After this remaining numbers are from (459 to 492), if you count there will be 13 more 8's.

Thus it will account to total of $40+13=\textbf{53, 8's.}$

**A Thankful User**

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Thank you for providing such a good question. More detailed solution will be highly appreciated. thank you

Hi,

Please see Gopal Jha's answer or reply to Ajay's comment.

Hope this helps

**Ajay**

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please give a clear idea about this.

Gopal Jha has nailed it for us. Reposting his answer here.

from 260 to 460 there are 200 numbers in which 8 can appears at unit, tens or hundred place. Lets calculate the no of 8's between 260 to 360:

261-270

271-280

281-290

.

.

.

.

351 - 360

Here you can observe, at unit's place 10 8's occurred and at ten's place also 10 8's occurred. so, total no of 8 is 20.

similarly from 361 to 460 there are 20 8's.

so, total 8's are $2*20=40$...... (A)

again, in the remaining numbers only from 461 to 490, 8 can appears at unit's or ten's place.

Let's calculate:

461-470

471-480

481-490

here u can observe 3 8's appeared at unit place and 10 8's appeared at ten's place.

so, total no of 8's appeared: $10+3=13$..... (B)

$A+B=53$

**Gopal Singh**

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we get 200 natural number 208, 218, 228..... like this we get 8's 10 times

similarly 280, 281 ,,,,,289 another 8's get 10 times

so in 100 natural number's we get 20 8's

so in 200 natural number's = 40

**Sree**

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pls tell... clearly i cant get the concepts if any know the explanation pls say....

we get 200 natural number 208,218,228..... like this we get 8's 10 times

similarly 280,281 ,,,,,289 another 8's get 10 times

so in 100 natural number's we get 20 times 8's

so in 200 natural number;s = 40

**Dhinesh**

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Y dont u count 10th digit 8's in 280 to 289 & 380 to 389???

Count the number of 8 in Tens and Units Place:

Ten place : 30 (i.e. from 80 to 89 in 200, 300 and 400 series)

Units place: 23 (i.e. from 269 to 488)

Therefore the total number of 8 used is 53