d Question 9 & 10 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

Question 9 & 10

Question 9: A total of Rs 71 is made up of 324 coins of 20 paisa and 25 paisa. What is the number of coins of 25 paisa ?

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Let there be ‘x’ 20 paisas coins and ‘y’ be the number of 25 paisas coin

So, by given condition all 20 paisas and 25 paisas coins are 324 in total; i-e

x+y=324 ---(1)

But All coins together make Rs 71  and Rs71 make 7100 paisas.

Now all 20 paisas coins will make 20x paisas and all 25 paisas coins will make 25y paisas. But by given condition the sum of all paisas is 7100; so,

20x+25y=7100 ----(2)

From eq (1)  we get;


Put this value of x in eq (2);

\Rightarrow 20 (324-y)+25y=7100

\Rightarrow 6480-20y+25y=7100




So, the sum of 25 paisas coins is 124

Question 10: A printer numbers a book’s pages from 1 to 3189 digits in total. How many pages are there for the book ?

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The book is having total pages 3189 and first page is starting from number 1.

So, here we can see that;

Case 1:

For digits 1 to 9 there are 9 pages, which are numbered with single digits which are 9 ( i-e 1,2,3,4,5,6,7,8,9)

Case 2:

For pages starting from 10 to 99 there are 90 pages  and these 90 pages are numbered with double digits(i-e 10, 11, 12,……up to 99), so,

Total digits for pages starting from 10 to 90  will be ’90 pages  multiplied by two digits’ i-e  90×2=180 digits

Case 3: 

Now after  99 th page it starts with case 3; i-e now onward the pages are numbered with three digit number  (i-e  from 100, 101. 102, …….. up to 999th page) and these will be 900 pages . So the number of digits for nine hundred triple digit numbered 900 pages = 900×3=2700 digits

Now we add digits used in case 1, case 2 and case 3 so, 9+180+2700 = 2889 digits,But we have been given that total digits are 3189. So, now we will count pages for 3189-2889=300 digits.

Case 4: 

Now after 999th page comes 1000th page which is numbered with 4 digits. We need 75 tetra digit pages to use 300 digits. So we see that 1000 to 1074 will make 75 pages having 4 digit numbering, and digits used by 75 four digit numbers = 75×4=300

Now add pages of all cases i-e 9+90+900+75 =1074 pages

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