Quantitative Aptitude - Numbers and Algebra
- Option : D
- Explanation : Let HCF be h and LCM be l.
I. Let, numbers be ah and bh. Then abh = and (a + b)h = m
=> (a — b)h = n
Using these ah and bh can be uniquely determined. Thus, I is true II.
II. If HCF = LCM, then two numbers are equal and same as HCF or LCM. Thus, II is true.
III. LCM/HCF = prime i.e. l/h =P Then one of the numbers is equal to h and other is equal to E. Thus, III is true.
13. What is the remainder when 2050 x 2071 x 2095 is divided by 23?
- Option : C
- Explanation : 2050 x 2071 x 2095
=(23x89+3)x(23x90+1)x(23x91+2)
=(23[89 x 90+ 89 + 90 x 3]+ 3)x(23 x 91+2)
=(23 *K+ 3)* (23 * 91 + 2) )
=(23 *[K* 91 + K* 2 + 91 * 3]+ 6)
= 23 * L + 6 (where K and L are constants).
Hence remainder is 6.
- Option : A
- Explanation :
Total distance travelled by the ball before coming to rest
= 12 + 2 (6 + 3 + 1.5 +...to?)
= 12 + 2. ( 6/ (1-1/2))
= 12 + 2(12)
= 36m
