Number System - Study Mode

[#56] The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32, 35 is.
Correct Answer

(B) 4714

Explanation

Solution: LCM of 20, 28, 32, 35 will be the greatest number which is divisible by these numbers. Firstly, we find LCM of 20, 28, 32, 35 20 = 2 × 2 × 5 28 = 2 × 2 × 7 32 = 2 × 2 × 2 × 2 × 2 35 = 5 × 7 LCM = 2 × 2 × 2 × 2 × 2 × 5 × 7 = 1120 Required greatest number which subtract from 5834 are divide by 20, 28, 32 and 35 = 5834 - 1120 = 4714

[#57] Which among $${2^{frac{1}{2}}}$$, $${3^{frac{1}{3}}}$$, $${4^{frac{1}{4}}}$$, $${6^{frac{1}{6}}}$$ and $${12^{frac{1}{{12}}}}$$ is largest?
Correct Answer

(B) $${3^{frac{1}{3}}}$$

Explanation

Solution: LCM in power 2, 3, 4, 6, 12 is 12 We multiplied the LCM to the power of the numbers. $${2^{frac{{1 imes 12}}{2}}},{3^{frac{{1 imes 12}}{3}}},{4^{frac{{1 imes 12}}{4}}},{6^{frac{{1 imes 12}}{6}}}{kern 1pt} { ext{and}},{12^{frac{{1 imes 12}}{{12}}}}$$ We get, = 2 6 , 3 4 , 4 3 , 6 2 , 12 = 64, 84, 64, 36, 12 Hence, greatest number would be $${3^{frac{1}{3}}}$$

[#58] LCM of two numbers is 936. If their HCF is 4 and one of the numbers is 72, the other is
Correct Answer

(B) 52

Explanation

Solution: Let two numbers be N 1 and N 2 .
Now, we have
HCF of the numbers × LCM of the numbers = Multiplication of the numbers.
936 × 4 = 72 × N 1 Or, N 1 = $$frac{{936 imes 4}}{{72}}$$
Or, N 1 = 52

[#59] The rightmost non-zero digit of the number 30 2720 is
Correct Answer

(A) 1

Explanation

Solution: (30) 2720 , we can write it as[(30) 4 ] 680 Or,[(10 × 3) 4 ] 680 The right most non-zero digit depends on the unit digit of [(3) 4 ] 680 Unit digit of [(3) 4 ] 680 , Or, (81) 680 The unit digit of 81 is 1 so any power of 81 will always give its unit digit as 1 Thus, required unit digit is 1

[#60] Each member of a club contributes as much rupees and as much paise as the number of members of the club. If the total contribution is Rs. 2525, then the number of members of the club is:
Correct Answer

(D) 50

Explanation

Solution: Let the total member be x. x member contribute is Rs. x and each member also contributes x paise. So total rupees is x member Rs. x = Rs. x 2 x member Rs. x paise $$ = frac{{x.x}}{{100}}$$ Total rupees is $$eqalign{
& {x^2} + frac{{{x^2}}}{{100}} = 2525 cr
& 101{x^2} = 2525 imes 100 cr
& 101{x^2} = 101 imes 25 imes 100 cr
& x08oxed{x = 50} cr} $$