RS Aggarwal 2019,2020 solution class 7 chapter 7 Linear Equation In One Variable Exercise 7B

Exercise 7B

Page-114

Question 1:

Twice a number when decreased by 7 gives 45. Find the number.

Answer 1:

$\begin{array}{l}\text{ Let the number be }x. \\ \text{ Then, we have}:\\ \Rightarrow 2x-7=45\\ \Rightarrow 2x=45+7\\ \Rightarrow x=\frac{45+7}{2}\\ \Rightarrow x=\frac{\sout{5}{\sout{2}}^{26}}{{\sout{2}}_1}\\ \Rightarrow x=26\\ \therefore \text{The required number is 26}\text{.}\end{array}$

Question 2:

Thrice a number when increased by 5 gives 44. Find the number.

Answer 2:

$\begin{array}{l}\text{Let the number be }x.\\ \text{Then, we have:}\\ \Rightarrow 3x+5=44\\ \Rightarrow 3x=44-5\\ \Rightarrow x=\frac{44-5}{3}\\ \Rightarrow x=\frac{\sout{3}{\sout{9}}^{13}}{{\sout{3}}_1}\\ \Rightarrow x=13\\ \therefore \text{The required number is 13}\end{array}$

Question 3:

Four added to twice a number yields $\frac{26}{5}$. Find the fractions.

Answer 3:

$\begin{array}{l}\text{Let the number be }x.\\ \text{Then, we have:}\\ \Rightarrow 2x+4=\frac{26}{5}\\ \Rightarrow 2x=\frac{26}{5}-4\\ \Rightarrow 2x=\frac{26-20}{5}\\ \Rightarrow x=\frac{{\sout{6}}^3}{\sout{1}{\sout{0}}_5}\\ \Rightarrow x=\frac{3}{5}\\ \therefore \text{ The required fraction is }\frac{3}{5}\text{.}\end{array}$

Question 4:

A number when added to its half gives 72. Find the number.

Answer 4:

$\begin{array}{l}\text{ Let the required number be }x.\\ \text{Then, we have:}\\ \Rightarrow x+\frac{x}{2}=72\\ \Rightarrow \frac{2x+x}{2}=72\\ \Rightarrow \frac{3x}{2}=72\\ \Rightarrow 3x=72\times 2\\ \Rightarrow x=\frac{\sout{7}{\sout{2}}^{24}\times 2}{\begin{array}{l}{\sout{3}}_1\\ =48\end{array}}\\ \therefore \text{The required number is 48.}\end{array}$

Question 5:

A number added to its two-thirds is equal to 55. Find the number.

Answer 5:

$\begin{array}{l}\text{ Let the required number be x.}\\ \text{ Then, we have:}\\ \Rightarrow x+\frac{2x}{3}=55\\ \Rightarrow \frac{3x+2x}{3}=55\\ \Rightarrow 5x=55\times 3\\ \Rightarrow x=\frac{\sout{5}{\sout{5}}^{11}\times 3}{\begin{array}{l}{\sout{5}}_1\\ =33\end{array}}\\ \therefore \text{The required number is 33.}\end{array}$

Question 6:

A number when multiplied by 4, exceeds itself by 45. Find the number.

Answer 6:

$\begin{array}{l}\text{ Let the required number be }x.\\ \text{ Then, we have:}\\ \Rightarrow 4x-x=45\\ \Rightarrow 3x=\frac{45}{3}\\ \Rightarrow x=15\\ \therefore \text{The required number is }15.\end{array}$

Question 7:

A number is as much greater than 21 as it is less than 71. Find the number.

Answer 7:

$\begin{array}{l}\text{ Let the number be }x. \\ \text{Then, we have:}\\ (x-21)=(71-x)\\ \Rightarrow x+x=71+21\\ \Rightarrow 2x=92\\ \Rightarrow x=\frac{\sout{9}{\sout{2}}^{46}}{{\sout{2}}_1}\\ \Rightarrow x=46\\ \therefore \text{The required number is 46.}\end{array}$

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Question 8:

$\frac{2}{3}$ of a number is less than the original number by 20. Find the number.

Answer 8:

$\begin{array}{l}\text{ Let the original} \text{number be }x. \\ \text{Then, we have:}\\ \Rightarrow \frac{2}{3}x=x-20\\ \Rightarrow \frac{2x}{3}-x=-20\\ \Rightarrow \frac{2x-3x}{3}=-20\\ \Rightarrow -x=-20\times 3\\ \Rightarrow x=60\\ \therefore \text{The original number is 6}0.\end{array}$

Question 9:

A number is $\frac{2}{5}$ times another number. If their sum is 70, find the numbers.

Answer 9:

$\begin{array}{l}\text{ Let the number be }x. \\ \mathrm{Then}, \mathrm{the} \mathrm{other} \mathrm{number} \mathrm{will} \mathrm{be} \frac{2x}{5}.\\ \begin{array}{l}\text{Now, we have:}\\ \Rightarrow x+\frac{2x}{5}=70\\ \Rightarrow \frac{5x+2x}{5}=70\\ \Rightarrow \frac{7x}{5}=70\\ \Rightarrow x=\frac{\sout{7}{\sout{0}}^{10}\times 5}{{\sout{7}}_1}\\ \therefore \mathrm{Other} \mathrm{number}=50\times \frac{2}{5}=20\\ \mathrm{Hence}, \mathrm{the} \mathrm{numbers} \mathrm{are} 50 \mathrm{and} 20.\end{array}\end{array}$

Question 10:

Two-thirds of a number is greater than one-third of the number by 3. Find the number.

Answer 10:

$\begin{array}{l}\text{ Let the} \text{number be }x. \\ \text{ Then, we have:}\\ \frac{2}{3}x=\frac{1}{3}x+3\\ \Rightarrow \frac{1}{3}x=\frac{2x}{3}-3\\ \Rightarrow \frac{x}{3}-\frac{2x}{3}=-3\\ \Rightarrow \frac{x-2x}{3}=-3\\ \Rightarrow x-2x=3\times (-3)\\ \Rightarrow -x=-9\\ \therefore \text{The required number is 9.}\end{array}$

Question 11:

The fifth part of a number when increased by 5 equals its fourth part decreased by 5. Find the number.

Answer 11:

$\begin{array}{l}\text{ Let the number be }x. \\ \text{Then, we have:}\\ \Rightarrow \frac{x}{5}+5=\frac{x}{4}-5\\ \Rightarrow \frac{x}{5}-\frac{x}{4}=-5-5\\ \Rightarrow \frac{-x}{20}=-10\\ \Rightarrow x=200\\ \\ \therefore \text{The required number is 200.}\end{array}$

Question 12:

Find two consecutive natural numbers whose sum is 63.

Answer 12:

$\begin{array}{l}\text{ Let the two consecutive natural number be }x\text{ and }(x+\text{1}).\\ \text{Then, we have:}\\ x+(x+1)=63\\ \Rightarrow x+x+1=63\\ \Rightarrow 2x=63-1\\ \Rightarrow x=\frac{\sout{6}{\sout{2}}^{31}}{{\sout{2}}_1}\\ \Rightarrow x=31\\ \therefore \text{The required numbers are 31 and 32 (i.e., 31+1).}\end{array}$

Question 13:

Find two consecutive positive odd integers whose sum is 76.

Answer 13:

$\begin{array}{l}\text{ Let the two consecutive odd integers whose sum is 76 be }x\text{ and }(x+2).\\ \text{Then, }x+x+2=76\\ \Rightarrow \text{2x }+\text{2}=\text{76}\\ \Rightarrow \text{2x}=\text{76}-\text{2}\\ \Rightarrow \text{x}=\text{74}÷\text{2 }\\ \Rightarrow \text{x}=\text{37}\\ \therefore \text{The required integers are 37} \mathrm{and} 39 (\mathrm{i}.\mathrm{e}., 37+2).\\ \end{array}$

Question 14:

Find two consecutive positive even integers whose sum is 90.

Answer 14:

$\begin{array}{l}\text{Let the three consecutive positive even integers be }x, (x+2) \mathrm{and} (x+4).\\ \mathrm{Let} x\text{ be the even number.}\\ \text{Then, }x+x+2+x+4=90\\ \Rightarrow \text{3}x=\text{9}0-\text{6}\\ \Rightarrow \text{3}x=\text{84}\\ \Rightarrow x=\frac{84}{3}=\text{28}\\ \therefore \mathrm{The} \mathrm{r}\text{equired numbers are 28}, \text{3}0 \mathrm{and} \text{32.}\end{array}$

Question 15:

Divide 184 into two parts such that one-third of one part may exceed one-seventh of the other part by 8.

Answer 15:

$$\begin{gathered} \begin{array}{l}\text{Let the two parts be}\mathit{\text{ x }}\text{and }(184-x).\\ \text{Then, we have:}\\ \frac{\text{1}}{3}x=\frac{\text{1}}{7}(\text{184}-\text{x)+8}\\ \begin{array}{l}\text{⇒}\frac{\text{1}}{3}x-\frac{\text{1}}{7}(\text{184}-\text{x})=\text{8}\\ \text{⇒}\frac{\text{1}}{3}x-\frac{184}{7}+\frac{x}{7}=\text{8}\\ \text{ ⇒}\frac{1}{3}x+\frac{1}{7}x=\frac{184}{7}+8\\ \text{⇒}\frac{7x+3x}{21}=8+\frac{184}{7}\\ \text{⇒}\frac{10x}{21}=\frac{56+184}{7}\\ \text{⇒}\frac{10x}{21}=\frac{240}{7}\\ \text{⇒}x=\frac{240\times 21}{7\times 10}\\ =72\\ \mathrm{Now}, \mathrm{ot}\text{her part =184}-72=112\end{array}\end{array} \\ \therefore \mathrm{The} \mathrm{two} \mathrm{parts} \mathrm{are} 72 \mathrm{and} 112. \end{gathered}$$

Question 16:

A sum of 500 is in the form of denominations of 5 and 10. If the total number of notes is 90, find the number of notes of each type.

Answer 16:

$\begin{array}{l}\text{Let the number of five rupee notes be }x.\\ \text{Then, the number of ten rupee notes will be }(90-x).\\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \mathrm{we} \mathrm{have}:\\ \begin{array}{l} \text{ 5}x+10(90-x)=500\\ \text{⇒5}x+900-10x=500\\ \Rightarrow -5x=-400\\ \Rightarrow x=80\\ \mathrm{Number} \mathrm{of} \mathrm{ten} \mathrm{rupee} \mathrm{notes}=90-80=10\\ \therefore \mathrm{There} \mathrm{are} 80 \mathrm{five} \mathrm{rupee} \mathrm{notes} \mathrm{and} 10 \mathrm{ten} \mathrm{rupee} \mathrm{notes}.\end{array}\end{array}$

Question 17:

Sumitra has 34 in 50-paise and 25-paise coins. If the number of 25-paise coins is twice the number of 50-paise coins, how many coins of each kind does she have?

Answer 17:

$\begin{array}{l}\text{Let the numbers of 5}0\text{ paise coins and 25 paise coins be }x\text{ and 2}x,\text{ respectively.}\\ \mathrm{Then}, \mathrm{we} \mathrm{have}: \\ \text{5}0x+\text{25}\times \text{2}x=\text{34}00\\ \Rightarrow \text{5}0x+\text{5}0x=\text{34}00\\ \Rightarrow \text{1}00x=\text{34}00\\ \Rightarrow x=\text{34}\\ \therefore \mathrm{Number} \mathrm{of} \text{5}0\text{ paise coins }=\text{34}\\ \text{and number of 25 paise coins }=\text{68}\end{array}$

Question 18:

Raju is 19 years younger than his cousin. After 5 years, their ages will be in the ratio 2 : 3. Find their present ages.

Answer 18:

$\begin{array}{l}\text{Let the present ages of Raju and his cousin be (}\mathit{\text{x}}\text{-19) yrs and }x \mathrm{yrs}. \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \mathrm{we} \mathrm{have}:\\ \begin{array}{l}\frac{(x-19)+5}{x+5}=\frac{2}{3}\\ \text{⇒3}(x-14)=2x+10\\ \Rightarrow \text{3}x-42=2x+10\\ \Rightarrow x=52\\ \therefore \mathrm{Age} \mathrm{of} \mathrm{Raju}\text{'}\mathrm{s} \mathrm{c}\text{ousin = 52 yrs }\\ \text{ and age of Raju = }52-19=33 \mathrm{yrs}\end{array}\end{array}$

Question 19:

A father is 30 years older than his son. In 12 years, the man will be three times as old as his son. Find their present ages.

Answer 19:

$\begin{array}{l}\text{Let the age of the son and the father be }x \mathrm{yrs}\text{ and }(x+30) \mathrm{yrs}, \mathrm{respectively}.\\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \mathrm{we} \mathrm{have}:\\ \begin{array}{l}\text{3}\times (x+12)=x+30+12\\ \text{⇒3}x+36=x+42\\ \Rightarrow \text{3}x-x=42-36\\ \Rightarrow \text{2}x=6\\ \Rightarrow x=3\\ \therefore \mathrm{Son}\text{'}\mathrm{s} \mathrm{a}\text{ge}=3 \text{yrs}\\ \text{Father's age}=(x+30) \mathrm{yrs}=(3+30) \mathrm{yrs}=33\text{ yrs}\end{array}\end{array}$

Question 20:

The ages of Sonal and Manoj are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.

Answer 20:

$\begin{array}{l}\text{Given ratio of Sonal's and Manoj's ages }=\text{7 }: \text{5}\\ \text{Let the ages of Sonal and Manoj be 7}x \mathrm{yrs} \text{and 5}x \mathrm{yrs}.\\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \mathrm{we} \mathrm{have}:\\ \begin{array}{l}\frac{7x+10}{5x+10}=\frac{9}{7}\\ \Rightarrow \text{7}(7x+10)=9(5x+10)\\ \Rightarrow \text{49}x+70=45x+90\\ \Rightarrow \text{49}x-45x=90-70\\ \Rightarrow 4x=20\\ \Rightarrow x=5\\ \therefore \text{Sonal's present age is 7}\times \text{5=35 yrs}\end{array}\\ \text{ Manoj's present age is 5}\times \text{5=25 yrs}\end{array}$

Question 21:

Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Answer 21:

$$\begin{gathered} \begin{array}{l}\text{Let }\mathit{\text{x }}\text{yrs be the present age of son.}\\ \text{Then, the age of the son 5 years ago would be (}x-5) yrs \end{array} \\ \mathrm{Then}, \mathrm{Age} \mathrm{of} \mathrm{father}= 7(x-5) \mathrm{yrs} \\ \mathrm{After} \text{5 yrs, }\mathrm{the age of the son}\text{ will be }(x+5) yrs \\ \mathrm{Then}, \mathrm{Age} \mathrm{of} \mathrm{father}= 3(\mathrm{x}+5) \mathrm{yrs} \\ \begin{array}{l}\text{Now, we have 3}(x+5)=7(x-5)+10\\ \text{⇒ 3}x+15=7x-35+10\\ \Rightarrow \text{4}x=40\\ \Rightarrow x=10\\ \therefore \mathrm{P}\text{resent age of the father is = 3(}\mathit{\text{x}}\text{+5)-5}\end{array} \\ =\text{3}(10+5)-5 \\ =40 \mathrm{yrs} \end{gathered}$$

Question 22:

After 12 years Manoj will be 3 times as old as he was 4 years ago. Find his present age.

Answer 22:

$$\begin{gathered} \begin{array}{l}\text{Let }\mathit{\text{x }}\text{be the present age of Manoj.}\end{array} \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \mathrm{we} \mathrm{have}: \\ \begin{array}{l}\Rightarrow x+\text{12}=\text{3}(x-\text{4})\\ \Rightarrow x+\text{12}=\text{3}x-\text{12}\\ \Rightarrow \text{2}x=\text{24}\\ \Rightarrow x=\text{12}\\ \therefore \text{Manoj's present age is 12 years}.\end{array} \end{gathered}$$

Question 23:

In an examination, a student requires 40% of the total marks to pass. If Rupa gets 185 marks and fails by 15 marks, find the total marks.

Answer 23:

$\begin{array}{l}\text{Let }\mathit{\text{x}}\text{ be}\text{the total marks.}\\ \text{According to the question, we have:}\\ \text{4}0\%\text{ of }\mathit{\text{x}}=\text{185}+\text{15}\\ \Rightarrow \frac{40x}{100}=200\\ \Rightarrow \text{4}0x=\text{2}00\times \text{1}00\\ \Rightarrow 40x=20000\\ \Rightarrow x=\text{5}00\\ \therefore \text{Total marks}=\text{5}00 \end{array}$

Question 24:

A number consists of two digits whose sum is 8. If 18 is added to the number its digits are reversed. Find the number.

Answer 24:

$\begin{array}{l}\text{Let }\mathit{\text{x}}\text{ be the digit in the units place.}\\ \text{Sum of the units and tens digits}=8\\ \text{Then, tens digit}=(8-x)\\ \therefore \mathrm{T}\text{he number is 10}(8-x)+x.\\ \text{Now, 10}(8-x)+x+18=10x+(8-x)\\ \Rightarrow \text{80}-10x+x+18=10x+8-x\\ \Rightarrow \text{98}-9x=9x+8\\ \Rightarrow \text{18}x=90\\ \Rightarrow x=5\\ \text{i.e., tens digit=}(8-5)\text{=3}\\ \therefore \mathrm{Required} \text{number=10}(8-5)\text{+5=10}\times \text{3+5=35 }\end{array}$

Question 25:

The total cost of 3 tables and 2 chairs is 1850. If a table costs 75 more than a chair, find the price of each.

Answer 25:

$\begin{array}{l}\text{Let Rs }\mathit{\text{x }}\text{be the}\text{cost of the chair.}\\ \text{Then, the cost of the table is Rs }(x+75).\\ \text{Now, 3}(x+75)+2x=1850\\ \Rightarrow \text{3}x+225+2x=1850\\ \Rightarrow \text{5}x=1625\\ \Rightarrow x=\frac{1625}{5}=325\\ \therefore \text{Cost of the chair = Rs 325; cost of the table = (325+75)=Rs 400}\\ \end{array}$

Question 26:

A man sold an article for 495 and gained 10% on it. Find the cost price of the article.

Answer 26:

$$\begin{gathered} \begin{array}{l}\text{Let the cost price of the article be Rs }x.\\ \text{According to the question, we have:} \\ \text{SP}=\mathrm{Rs} \text{495}\end{array} \\ \therefore \text{Gain \%=}\frac{\text{Gain}}{\mathrm{CP}}\text{×100} \\ \Rightarrow 10=\frac{\mathrm{Gain}}{x}\times 100 \\ \Rightarrow \mathrm{Gain}=\frac{10x}{100}=\mathrm{Rs} \frac{x}{10} \\ \begin{array}{l}\text{Now, CP}+\mathrm{Gain}=\text{SP}\\ \Rightarrow x+\frac{x}{10}=495\\ \Rightarrow \frac{x+10x}{10}=495\\ \Rightarrow 11x=495\times 10\\ \Rightarrow x=\frac{495\times 10}{11}\\ \Rightarrow x=\frac{4950}{11}\\ \Rightarrow x=450\\ \therefore \text{CP}=\text{Rs 45}0\end{array} \end{gathered}$$

Question 27:

The length of a rectangular field is twice its breadth. If the perimeter of the field is 150 metres, find its length and breadth.

Answer 27:

$\begin{array}{l}\text{Let the length and breadth of the rectangular field be }\mathit{\text{l }}\text{m and }\mathit{\text{b}}\text{ m, respectively.}\\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \mathrm{we} \mathrm{have}:\\ \text{2}(\mathit{\text{l}}+\mathit{\text{b}})=\text{15}0 ...\left(\text{i}\right)\\ \Rightarrow \mathit{\text{l}}+\mathit{\text{b}}=\text{75}\\ \text{Given that }\mathit{\text{l}}=\text{2}\mathit{\text{b ...}}\left(\text{ii}\right)\\ \text{Using }\left(\text{ii}\right)\text{ in }\left(\text{i}\right)\text{, we have:}\\ \text{2}\mathit{\text{b}}+\mathit{\text{b}}=\text{75}\\ \Rightarrow \text{3}\mathit{\text{b}}=\text{75}\\ \Rightarrow \mathit{\text{b}}=\text{25}\\ \therefore \mathit{\text{l}}=\text{5}0 \text{m and }\mathit{\text{b}}=\text{25 m} \end{array}$

Question 28:

Two equal sides of a triangle are each 5 metres less than twice the third side. If the perimeter of the triangle is 55 metres, find the lengths of its sides.

Answer 28:

$\begin{array}{l}\text{Let the length of third side be }x \mathrm{m}. \mathrm{Then}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \text{two equal sides will be }(2x-5) \mathrm{m}.\\ \therefore (2x-5)+(2x-5)+x=55\\ \Rightarrow \text{2}x-5+2x-5+x=55\\ \Rightarrow \text{5}x-10=55\\ \Rightarrow \text{5}x=65\\ \Rightarrow x=\frac{65}{5}=13\\ \therefore \text{Length of the third side=13 m}\\ \text{And length of the other two equal sides=}(2\times 13)-5=21 \mathrm{m}\\ \end{array}$

Question 29:

Two complementary angles differ by 8⁰. Find the angles.

Answer 29:

$\begin{array}{l}\text{Let the two complementary angles be }x°\text{ and }(90-x)°.\\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \mathrm{we} \mathrm{have}:\\ x-(90-x)=8 \\ \Rightarrow x-90+x=8\\ \Rightarrow \text{2}x=98\\ \Rightarrow x=49\\ \therefore \text{The measures of the complementary angles are 49}° \text{and }(\text{90}-49)°=41°.\end{array}$

Question 30:

Two supplementary angles differ by 44⁰. Find the angles.

Answer 30:

$\begin{array}{l}\text{Let the two supplementary angles be }x°\text{ and }(180-x)°.\\ \therefore x-(180-x)=44\\ \text{⇒}x-180+x={44}^0\\ \Rightarrow \text{2}x=224\\ \Rightarrow x=112\\ \therefore \text{The measures of the supplementary angles are 112}°\text{ and }(180-112)°, \mathrm{i}.\mathrm{e}., 68°.\end{array}$

Question 31:

In an isosceles triangle the base angles are equal and the vertex angle is twice of each base angle. Find the measures of the angles of the triangle.

Answer 31:

$\begin{array}{l}\text{Let the base angles of the isosceles triangle be }x° \mathrm{each}. \\ \text{Then, the measure the vertex angle will be (2x)°.}\\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \mathrm{we} \mathrm{have}:\\ x+x+2x=\text{18}0\left(\mathrm{S}\text{um of three sides of a triangle}\right)\\ \Rightarrow \text{ 4}x=\text{18}0\\ \Rightarrow x=\frac{180}{4}\\ \Rightarrow x=45\\ \\ \text{∴ Each base angle measures 45}°\text{ and the vertex angle measures (2}\times \text{45)}°, \mathrm{i}.\mathrm{e}., 90°.\\ \end{array}$

Question 32:

A man travelled $\frac{3}{5}$ of his journey by rail, $\frac{1}{4}$ by a taxi, $\frac{1}{8}$ by a bus and the remaining 2 km on foot. What is the length of his total journey?

Answer 32:

$\begin{array}{l}\text{Let the length of the total journey be }x\text{ km.}\\ \text{According to the question, we have:}\\ \frac{3}{5}x+\frac{1}{4}x+\frac{1}{8}x+2=x\\ \Rightarrow \frac{24x+10x+5x+80}{40}=x\\ \Rightarrow 39x+80=40x\\ \Rightarrow x=80\\ \therefore \text{The length of his total journey is 80 km.}\end{array}$

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Question 33:

A labourer is engaged for 20 days on the condition that he will receive 120 for each day he works and will be fined 10 for each day he is absent. If he receives 1880 in all, for how many days did he remain absent?

Answer 33:

$\begin{array}{l}\text{Let }\mathit{\text{x}}\text{ be the number of days of his absence. }\\ \therefore \text{Number of days of his presence = }(\text{2}0-x)\\ \text{Now,} (\text{2}0-x)\text{12}0-\text{1}0x=\text{188}0\\ \Rightarrow \text{24}00-\text{12}0x-\text{1}0x=\text{188}0\\ \Rightarrow \text{24}00-\text{188}0=\text{13}0x\\ \Rightarrow 130x=520\\ \Rightarrow \mathit{\text{x}}=\text{4}\\ \therefore \text{Number of days of his absence}=\text{4 }\end{array}$

Question 34:

Hari Babu left one-third of his property to his son, one-fourth to his daughter and the remainder to his wife. If his wife's share is 18000, what was the worth of his total property?

Answer 34:

$\begin{array}{l}\text{Let the worth of Hari Babu's property be Rs }x.\\ \text{According to the question, we have:}\\ \text{Son's share}=\frac{1}{4}x\\ \text{Daughter's share=}\frac{1}{3}x\\ \text{Wife's share}=\{x-(\frac{1}{4}x+\frac{1}{3}x\left)\right\}\\ \text{It is given that his wife's share }\mathrm{is} \mathrm{Rs} \text{18}000.\\ \text{i.e., }x-(\frac{1}{4}x+\frac{1}{3}x)=\text{18}000\\ \Rightarrow x-(\frac{1}{3}x+\frac{1}{4}x)=18000\\ \Rightarrow x-\frac{7x}{12}=18000\\ \Rightarrow \frac{5x}{12}=18000 \\ \Rightarrow \text{x}= \frac{\sout{1}\sout{8}\sout{0}\sout{0}{\sout{0}}^{3600}\times 12}{\sout{5}} \\ \Rightarrow x=\text{432}00\\ \therefore \text{Hari Babu}’\text{s total property is worth Rs 432}00.\\ \end{array}$

Question 35:

How much pure alcohol must be added to 400 mL of a 15% solution to make its strength 32%.

Answer 35:

$\begin{array}{l}\text{Let the volume of the pure alcohol be }x \mathrm{ml}.\\ \text{Initial concentration=}15\%\\ \text{So, initial amount of alcohol in the solution will be}=\frac{15}{100}\times \text{400= 60 ml}\\ \text{To make the strength of the solution 32\%, we will keep the amount of water constant a}\mathrm{nd add }x\text{ volume of pure alcohol}\text{.}\\ \mathrm{On adding pure alcohol}, the volume of \text{the solution increases to 400 + }x.\\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \mathrm{we} \mathrm{have}:\\ \frac{x+60}{400+x}=\frac{32}{100}\\ \text{⇒100}x+6000=12800+32x\\ \Rightarrow \text{100}x-32x=1\text{2800}-6000\\ \Rightarrow \text{68}x=6800\\ \Rightarrow x=100\\ \text{So, amount of pure alcohol to be added=100 ml }\end{array}$