Question 12 Marks
Convert the decimal number 437 into the binary number.
Answer
View full question & answer→The given decimal number is 437
Hence, the required binary number is 110110101
Hence, the required binary number is 110110101
Questions
2 Marks Questions
🎯
Test yourself on this topic
7 questions · timed · auto-graded
Question 22 Marks
Find the derivative of the given function: $\left(x^2+1\right)(x-2)$
Answer
View full question & answer→Let $f(x)=\left(x^2+1\right)(x-2)=x^3-2 x^2+x-2$, diff. w.r.t. $x$, we get $f^{\prime}(x)=3 x^2-2 \cdot 2 x^1+1-0=3 x^2-4 x+1$
Altermatively
$\begin{array}{l} f ( x )=\left( x ^2+1\right)( x -2) \text {, diff. w.r.t. } x \text {, we get } \\ f ^{\prime}( x )=\left( x ^2-1\right) \cdot \frac{d}{d x}( x -2)+( x -2) \cdot \frac{d}{d x}\left( x ^2+1\right) \text { (product rule) }\end{array}$
$\begin{array}{l}=\left(x^2+1\right)(1-0)+(x-2)(2 x+0) \\ =x^2+1+2 x^2-4 x=3 x^2-4 x+1\end{array}$
Altermatively
$\begin{array}{l} f ( x )=\left( x ^2+1\right)( x -2) \text {, diff. w.r.t. } x \text {, we get } \\ f ^{\prime}( x )=\left( x ^2-1\right) \cdot \frac{d}{d x}( x -2)+( x -2) \cdot \frac{d}{d x}\left( x ^2+1\right) \text { (product rule) }\end{array}$
$\begin{array}{l}=\left(x^2+1\right)(1-0)+(x-2)(2 x+0) \\ =x^2+1+2 x^2-4 x=3 x^2-4 x+1\end{array}$
Question 32 Marks
If $xy ^2=1$, prove that $2 \frac{d y}{d x}+ y ^3=0$.
Answer
View full question & answer→Given as $xy ^2=1$
Differentiating with respect to x,
$\frac{d}{d x}\left( xy ^2\right)=\frac{d}{d x}(1)$
$\begin{array}{l}\Rightarrow x \frac{d}{d x}\left( y ^2\right)= y ^2 \frac{d}{d x}( x )=0 \\ \Rightarrow x (2 y ) \frac{d y}{d x}+ y ^2(1)=0 \\ \Rightarrow 2 xy \frac{d y}{d x}=- y ^2 \\ \Rightarrow \frac{d y}{d x}=\frac{-y^2}{2 x y} \\ \Rightarrow \frac{d y}{d x}=\frac{-y}{2 x}\end{array}$
On putting the value of $x=\frac{1}{y^2}$ in above equation
$\begin{array}{l}\Rightarrow \frac{d y}{d x}=\frac{-y}{2\left(\frac{1}{y^2}\right)} \\ \Rightarrow \frac{d y}{d x}=\frac{-y}{2\left(\frac{1}{y^2}\right)} \\ \Rightarrow 2 \frac{d y}{d x}=- y ^3 \\ \Rightarrow 2 \frac{d y}{d x}+ y ^3=0\end{array}$
Differentiating with respect to x,
$\frac{d}{d x}\left( xy ^2\right)=\frac{d}{d x}(1)$
$\begin{array}{l}\Rightarrow x \frac{d}{d x}\left( y ^2\right)= y ^2 \frac{d}{d x}( x )=0 \\ \Rightarrow x (2 y ) \frac{d y}{d x}+ y ^2(1)=0 \\ \Rightarrow 2 xy \frac{d y}{d x}=- y ^2 \\ \Rightarrow \frac{d y}{d x}=\frac{-y^2}{2 x y} \\ \Rightarrow \frac{d y}{d x}=\frac{-y}{2 x}\end{array}$
On putting the value of $x=\frac{1}{y^2}$ in above equation
$\begin{array}{l}\Rightarrow \frac{d y}{d x}=\frac{-y}{2\left(\frac{1}{y^2}\right)} \\ \Rightarrow \frac{d y}{d x}=\frac{-y}{2\left(\frac{1}{y^2}\right)} \\ \Rightarrow 2 \frac{d y}{d x}=- y ^3 \\ \Rightarrow 2 \frac{d y}{d x}+ y ^3=0\end{array}$
Question 42 Marks
A clock gains 4 seconds in 3 minutes and was set right at 9:00 a.m. What time will it show at 11:00 p.m. on the same day?
Answer
View full question & answer→Given that the clock gains 4 seconds in 3 minutes
$\Rightarrow$ it gains 20 × 4 = 80 seconds in 20 × 3 = 60 minutes i.e. 1 hour
Now, from 9:00 a.m. to 11:00 p.m. on the same day time passed is 14 hours.
So, in 14 hours, the clock will gain 14 × 80 seconds = 1120 seconds
= 18 min. 40 sec.
Hence, the clock will show 11:18:40 p.m. at 11:00 p.m.
$\Rightarrow$ it gains 20 × 4 = 80 seconds in 20 × 3 = 60 minutes i.e. 1 hour
Now, from 9:00 a.m. to 11:00 p.m. on the same day time passed is 14 hours.
So, in 14 hours, the clock will gain 14 × 80 seconds = 1120 seconds
= 18 min. 40 sec.
Hence, the clock will show 11:18:40 p.m. at 11:00 p.m.
Question 52 Marks
In a certain language if LUCKNOW is coded as NWEMPQY, how is DELHI coded?
Answer
View full question & answer→Clearly, each letter of the word LUCKNOW is moved two steps forward to obtain the corresponding code
$\therefore$ 'DELHI' is coded as 'FGNJK
$\therefore$ 'DELHI' is coded as 'FGNJK
Question 62 Marks
If 'men are very busy' means, '1234', 'Busy person need encouragement' means '4567', 'encouragement is very important' means '3589' and 'Important persons are rare' means, '2680', then
i. What is the code for encouragement.
ii. According to given code, what is the code of Men need encouragement?
i. What is the code for encouragement.
ii. According to given code, what is the code of Men need encouragement?
Answer
View full question & answer→Given statements and their codes are as follows:
(I) Men are very busy - 1234
(II) Busy persons need encouragement - 4567
(III) Encouragement is very important - 3589
(IV) Important persons are rare - 2680
i. In second (II) and third (III) sentences, common word is 'encouragement' and common code no. is '5'. Hence, number 5 stands for encouragement.
ii. In first (I) and fourth (IV) sentences, the common word is 'are', hence 'are' stands for '2' from first (I) and third (III) sentences, 'very' stand for '3' from first (I) and second (II) sentences, 'busy' stands for '4'. Hence, we get from first sentence 'Men stand for '1' similarly 'needs' stand for '7'. From this, we conclude that 'Men need encouragement' will be coded as '1 5 7'.
(I) Men are very busy - 1234
(II) Busy persons need encouragement - 4567
(III) Encouragement is very important - 3589
(IV) Important persons are rare - 2680
i. In second (II) and third (III) sentences, common word is 'encouragement' and common code no. is '5'. Hence, number 5 stands for encouragement.
ii. In first (I) and fourth (IV) sentences, the common word is 'are', hence 'are' stands for '2' from first (I) and third (III) sentences, 'very' stand for '3' from first (I) and second (II) sentences, 'busy' stands for '4'. Hence, we get from first sentence 'Men stand for '1' similarly 'needs' stand for '7'. From this, we conclude that 'Men need encouragement' will be coded as '1 5 7'.
Question 72 Marks
A is thrice as good a workman as B and so takes 60 days less than B for doing a job. In how many days A and B together will complete the same job?
Answer
View full question & answer→. Let A and B separately complete the job in $n _{ A }$ and $n _{ B }$ days respectively. It is given that A is thrice as good a workman as B. This means A takes one third of the time taken by B to complete the job.
$\therefore n _{ A }=\frac{1}{3} n _{ B } \Rightarrow n _{ B }=3 n _{ A } \ldots$ (i)
It is also given that A takes 60 days less than B to complete the job.
$\therefore n _{ A }= n _{ B }-60 \ldots$ (ii)
Solving (i) and (ii), we obtain
$n _{ A }=30$ and $n _{ B }=90$
Suppose A and B together can complete the job in $n _{ AB }$ days. Then,
$\frac{1}{n_{A B}}=\frac{1}{n_A}+\frac{1}{n_B} \Rightarrow \frac{1}{n_{A B}}=\frac{1}{30}+\frac{1}{90} \Rightarrow \frac{1}{n_{A B}}=\frac{4}{90} \Rightarrow n _{ AB }=\frac{90}{4}=22 \frac{1}{2}$
Hence, A and B together can complete the job in $22 \frac{1}{2}$ days
$\therefore n _{ A }=\frac{1}{3} n _{ B } \Rightarrow n _{ B }=3 n _{ A } \ldots$ (i)
It is also given that A takes 60 days less than B to complete the job.
$\therefore n _{ A }= n _{ B }-60 \ldots$ (ii)
Solving (i) and (ii), we obtain
$n _{ A }=30$ and $n _{ B }=90$
Suppose A and B together can complete the job in $n _{ AB }$ days. Then,
$\frac{1}{n_{A B}}=\frac{1}{n_A}+\frac{1}{n_B} \Rightarrow \frac{1}{n_{A B}}=\frac{1}{30}+\frac{1}{90} \Rightarrow \frac{1}{n_{A B}}=\frac{4}{90} \Rightarrow n _{ AB }=\frac{90}{4}=22 \frac{1}{2}$
Hence, A and B together can complete the job in $22 \frac{1}{2}$ days