Maths

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.

f (t) = t³ – 6t² + 9t + 3

$\Rightarrow f\text{'}\left(t\right)=3t^2-12t+9=3\left(t-1\right)\left(t-3\right)$

$\therefore f\text{'}\left(t\right)=0\Rightarrow t=1,3$

$\Rightarrow f\left(1\right)=1,f\left(3\right)=-3$

$g\left(x\right)=\{\begin{array}{l}f\left(x\right),0\le x<1\\ 1,1\le x\le 3g\left(x\right)iscontinuous\\ 4-x,3<x\le 4\end{array}$

Hence g (x) is not differentiable at only x = 3.