[Mission 2023] Insta–DART (Daily Aptitude and Reasoning Test) 11 August 2022

Answer : a

Sol. LCM of 10, 15, 21 and 28 = 420 , Largest number of 4 – digits = 9999

Now, (9999/420) = 23 ´ (339/420) \ Remainder = 339

\ Four – digit number divisible by 10, 15, 21 and 28 = 9999 – 339 = 9660

Also, k = 10 – 4 = 15 – 9 = 21 – 15, 28 – 22 = 6

\ Required number = ( 9660 – k )    = ( 9660 – 6 )  = 9654

Sol. (d) Let the numbers be 21 a and 21b, where a and b are coprimes.

Then, 21a x 21 b = (21 X 4641) , ab = 221

Two coprimes with product 221 are 13 and 17.

.: Required number = (21 x 13, 21 x 17) = (273,357)

Answer : a

Sol: Given expression = + + + …. +

= ( – ) + ( – ) + ( ) +….+( – ) = – =  = = 0.16

Answer : a

Sol: Since the given number is divisible by 5, so 0 or 5 must come in place of #. But, a number ending with 5 is never divisible by 8. So, 0 will replace #. Now, the number formed by the last three digits is 4@0, which becomes divisible by 8, if @ is replaced by 4. Hence, digits in place of @ and # are 4 and 0 respectively.

Answer : a

Sol. If the number is divisible by 5 and 11 it must be divisible by 55. The numbers are less than 660. Hence, dividing 659 by 55 gives the number of multiples of 55 = 11 (ignoring fraction part).The 11 multiples of 55 which are less than 560, but of these 11 multiples some can be multiples of 3. The numbers of such, multiples is the quotient of 11 by 3. Quotient of 113=3113=3. Out of 11 multiples of 55, 3 are multiples of 3. Hence, numbers less than 660 and divisible by 5 and 11 but not by 3=11−3=8