I travelled a certain distance at a speed of $40 \mathrm{~km} / \mathrm{h}$ and the remaining distance at $60 \mathrm{~km} / \mathrm{h}$. The total distance of 240 km was covered in 5 hours. Find the distance covered at the speed of $40 \mathrm{~km} / \mathrm{h}$.
Answer
Let distance covered at the speed of $40 \mathrm{~km} / \mathrm{h}$ be $\text{x km}$ $\therefore$ Time taken $=\frac{x}{40}$ hrs Distance covered at the speed of $60 \mathrm{~km} / \mathrm{h}=(240-\mathrm{x}) \mathrm{km}$ $\therefore$ Time taken $=\frac{240-x}{60}$ hrs $\therefore \frac{x}{40}+\frac{240-x}{60}=5 \Rightarrow 3 \mathrm{x}+480-2 \mathrm{x}=600 \Rightarrow \mathrm{x}=120$ $\therefore$ Distance covered $=120 \mathrm{~km}$
There are 210 members in a club. 100 of them drink tea and 65 drink tea but not coffee, each member drinks tea or coffee. Find how many drink coffee? How many drink coffee but not tea?
Are the following pair of sets equal? Give reason. $A=\{2,3\} ; B=\left\{x \mid x\right.$ is a solution of $\left.x^{2}-5 x+6=0\right\}$
Answer
Given set $A=\{2,3\}\ldots\text{(i)}$ and set $B=\left\{x: x\right.$ is a solution of $\left.x^{2}-5 x+6=0\right\}$ For solution of $x^{2}-5 x+6=0$ $(\mathrm{x}-2)(\mathrm{x}-3)=0 \Rightarrow \mathrm{x}-2=0$ or $\mathrm{x}-3=0 \Rightarrow \mathrm{x}=2,3$ $\therefore$ set $B=\{2,3\} \ldots\text{(ii)}$ From (i) and (ii), we notice sets $A$ and $B$ are equal.
The average marks of 15 students are 45 . If average marks of the first 8 students are 48 and that of the last 8 students is 42 , find the marks obtained by 8th student.
Answer
Given, average of marks of 15 students is 45 So sum of marks obtained by 15 students $=15 \times 45=675$ Also, average of first 8 students is 48 $\Rightarrow$ sum of marks obtained by first 8 students $=8 \times 48=384$ and average of last 8 students is 42 $\Rightarrow$ sum of marks obtained by last 8 students $=8 \times 42=336$ The marks obtained by 8th student $=(384+336)-675=720-675=45$ Hence, the marks obtained by 8th student are 45