Number System - Study Mode

[#506] 1397 x 1397 = ?
Correct Answer

(A) 1951609

Explanation

Solution: $$eqalign{
& 1397 imes 1397 cr
& = {left( {1397}
ight)^2} cr
& = {left( {1400 - 3}
ight)^2} cr
& = {left( {1400}
ight)^2} + {left( 3
ight)^2} - left( {2 imes 1400 imes 3}
ight) cr
& = 1960000 + 9 - 8400 cr
& = 1960009 - 8400 cr
& = 1951609 cr} $$

[#507] How many of the following numbers are divisible by 132 ? 264, 396, 462, 792, 968, 2178, 5184, 6336
Correct Answer

(A) 4

Explanation

Solution: 132 = 4 x 3 x 11 So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also. 264 ⇒ 11, 3, 4 (/) 396 ⇒ 11, 3, 4 (/) 462 ⇒ 11, 3 (X) 792 ⇒ 11, 3, 4 (/) 968 ⇒ 11, 4 (X) 2178 ⇒ 11, 3 (X) 5184 ⇒ 3, 4 (X) 6336 ⇒ 11, 3, 4 (/) Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336. Required number of number = 4.

[#508] (935421 x 625) = ?
Correct Answer

(B) 584638125

Explanation

Solution: $$eqalign{
& 935421 imes 625 cr
& = 935421 imes {5^4} cr
& = 935421 imes {left( {frac{{10}}{2}}
ight)^4} cr
& = frac{{935421 imes {{10}^4}}}{{{2^4}}} cr
& = frac{{9354210000}}{{16}} cr
& = 584638125 cr} $$

[#509] The largest 4 digit number exactly divisible by 88 is:
Correct Answer

(A) 9944

Explanation

Solution: $$eqalign{
& { ext{Largest}},{ ext{4 - digit}},{ ext{number}} = 9999 cr
& 88),,,,9999,,,,(113 cr
& ,,,,,,,,,,,,,88 cr
& ,,,,,,,,,,, - - - - cr
& ,,,,,,,,,,,,,119 cr
& ,,,,,,,,,,,,,,,,88 cr
& ,,,,,,,,,,, - - - - cr
& ,,,,,,,,,,,,,,,,,319 cr
& ,,,,,,,,,,,,,,,,,264 cr
& ,,,,,,,,,,,,,,, - - - cr
& ,,,,,,,,,,,,,,,,,,,,55 cr
& ,,,,,,,,,,,,,,, - - - cr
& ext{Required number} cr
& = left( {9999 - 55}
ight) cr
& = 9944 cr} $$

[#510] Which of the following pairs of non-zero values of p and q make 6-digit number 674pq0 divisible by both 3 and 11?
Correct Answer

(D) p = 5 and q = 2

Explanation

Solution: Given: 674pq0 is divisible by both 3 and 11 Concept used: If the sum of digits of a number is a multiple of 3, the number will be completely divisible by 3 If the difference of the sum of alternative digits of a number is 0 or divisible by 11, then that number is divisible by 11 completely. Calculation: As 674pq0 divisible by 3 So, 6 + 7 + 4 + p + q + 0 ⇒ 17 + p + q also divisible by 3 Now, Possible values of p + q = 4, 7, 10, 13, 16 From the options Option A and Option D is satisfying Now, For Option A The number is 674220 So, 6 + 4 + 2 - 7 - 2 = 3 Not divisible by 11 For Option D The number is 674520 So, 6 + 4 + 2 - 7 - 5 = 0 divisible by 11 ∴ The required answer is Option D.