Quantitative Aptitude Prep Tips for MBA

We have to find last non zero digit by 4 so we will eliminate 0 in question so now it becomes 104925 = 4 as 4odd = 4

It is given that they were paid in the ratio 4 : 5 : 6 s for doing the job.

So, maximum amount would have been got by z and the minimum amount obtained by x.

Amount obtained by x = $\frac{}{}$ $=4*\frac{T}{15}$

where T is the total amount that all of them got together.

Amount obtained by z   $=6*\frac{T}{15}$ $\frac{}{}$

Now as per the problem,

$=6*\frac{T}{15}-4*\frac{T}{15}=300$

$=2*\frac{T}{15}=300$

T = Rs. 2250

Share of x and z $=6*\frac{T}{15}+4*\frac{T}{15}=10*\frac{T}{15}=$ $\frac{}{}\frac{}{}\frac{}{}$ Rs. 1500

2¹⁸⁹ – 2¹⁸⁸ – 2¹⁸⁸ = 2¹⁸⁸ (2² – 2 – 2) = 0

$\frac{100}{3}\mathrm{\%}\text{?}\text{?}=\text{?}\text{?}\frac{x}{1000\text{?}\text{?}?\text{?}\text{?}x}\text{?}\text{?}*\text{?}\text{?}100\mathrm{\%}$

x = 250

Hence, he uses weight, 1000 – 250 = 750 g

Let other number be x.

6 * 84 = 12 * x

x = 42

the highest power =   $\frac{80}{9}$  = 8

HCF = $\frac{HCFof(81,27,9)}{LCMof(16,20,40)}=\frac{9}{80}$  

Quantitative Aptitude Prep Tips for MBA

$\begin{array}{l}=\sqrt[]{(l^2+b^2+h^2)}\\ =\sqrt[]{100+100+225}\\ =\sqrt[]{425}=5\sqrt[]{17} m\end{array}$

$\frac{417}{4}$ gives a remainder of 1.

For all numbers which are in the form  x^1/x and all x is greater than 3, the one with the greatest base is the smallest number.