Number System - Study Mode
[#426] If x : y be the ratio of two whole numbers and z be their HCF, then the LCM of those two numbers is:
Correct Answer
(D) $${ ext{xyz}}$$
Explanation
Solution: Ratio of the numbers = x : y
HCF of the numbers = z So, z is the common factor of the numbers Then, First number = xz Second Number = yz Now, First Number × Second Number = HCF and LCM of the numbers xzyz = z × LCM LCM = xyz
[#427] The sum of the digits of a two-digits numbers is 12 and when the digits of the to digit number are interchanged, the new number is 36 more than the original. What is the Original number?
Correct Answer
(B) 48
Explanation
Solution: Let the Original number is (X + 10Y) According to question, Sum of the digit of the number = 12 X + Y = 12 ----------- (1)
When digit interchanged, number become 36 more than original number, so, 10X + Y = 10Y + X + 36 X - Y = 4 ------------------- (2) On solving equation (1) and (2), we get X = 8 Y = 4 So, Original number, = X + 10Y = 8 + 10 × 4 = 48
[#428] A certain number of the capsule were purchased for Rs. 176. Six more capsule could have been purchased in the same amount if each capsule were cheaper by Rs. 3. What was the number of capsules purchased?
Correct Answer
(B) 16
Explanation
Solution: Let Price of each capsule were Rs. X, originally. So, Number of capsule purchased originally, = $$ frac{{176}}{{ ext{X}}}$$ If price of capsule is Rs. 3 less then 6 more capsule were purchased. In this case, the number of capsules, = $$ frac{{176}}{{{ ext{X}} - 3}}$$ According to question, $$eqalign{
& frac{{176}}{{{ ext{X}} - 3}} - frac{{176}}{{ ext{X}}} = 6 cr
& frac{{176{ ext{X}} - 176{ ext{X}} + 528}}{{{ ext{X}} imes left( {{ ext{X}} - 3}
ight)}} = 6 cr
& 6{{ ext{X}}^2} - 18{ ext{X}} - 528 = 0 cr
& {{ ext{X}}^2} - 3{ ext{X}} - 88 = 0 cr
& {{ ext{X}}^2} - 11{ ext{X}} + 8{ ext{X}} - 88 = 0 cr
& left( {{ ext{X}} - 11}
ight)left( {{ ext{X}} + 8}
ight) = 0 cr} $$ Either, X = 11 Or, X = -8 (Price cannot be negative) Thus, X = Rs. 11 Hence, Number of the capsule purchased = $$frac{{176}}{{11}}$$ = 16
[#429] An even number can be expressed as the square of an integer as well as cube of another integer. Then the number has to be necessarily divisible by?
Correct Answer
(B) 64
Explanation
Solution: Check the option. 32 and 128 are nither of a square and nor cube root. 64 is the square of 8 and a cube of 4
[#430] If n is an integer, then (n 3 - n) is always divisible by :
Correct Answer
(C) 6
Explanation
Solution: $$eqalign{
& left( {{n^3} - n}
ight){ ext{ and n is any integer}} cr
& { ext{put n = 2 so, }}{{ ext{2}}^3} - 2 = 6 cr
& { ext{It will be always divisible by 6}} cr
& { ext{(put n = 2, 3, 4}}....{ ext{)}} cr
& { ext{(n = 2, 3, 4}}.....{ ext{)}} cr} $$