Question 14 Marks
Samidha and Shreya have pocket money $Rs.x$ and $Rs.y$ respectively at the beginning of a week. They both spend money throughout the week. At the end of the week, Samidha spends $Rs. 500$ and is left with as much money as Shreya had in the beginning of the week. Shreya spends $Rs. 500$ and is left with $\frac{3}{5}$ of what Samidha had in the beginning of the week. Find their pocket money.
AnswerPocket money of Samidha $= Rs. x$
Pocket money of Sherya $= Rs. y$
According to given information, we have
$ x-500=y \dots....(i)$
$\Rightarrow x-y=500 $
And, $y-500=\frac{3}{5} x$
$ \Rightarrow 5 y-2500=3 x$
$\Rightarrow 5(x-500)-2500=3 x$
$\Rightarrow 5 x-2500-2500=3 x$
$\Rightarrow 2 x=5000$
$\Rightarrow x=2500$
$\Rightarrow y$
$=2500-500 =2000 $
Thus, pocket money of Samidha is $Rs. 2500$ and that of Sherya is $Rs. 2000$.
View full question & answer→Question 24 Marks
Sunil and Kafeel both have some oranges. If Sunil gives $2$ oranges to Kafeel, then Kafeel will have thrice as many as Sunil. And if Kafeel gives $2$ oranges to Sunil, then they will have the same numbers of oranges. How many oranges does each have?
AnswerLet Sunil has number of oranges
and Kafeel has $y$ number of oranges.
In $1^{st}$ case $($if Sunil gives $2$ oranges to Kafeel$) :$
$3(x - 2) = y + 2$
$\Rightarrow 3x - 6 = y + 2$
$\Rightarrow 3x - y = 8 \dots....(i)$
In $2^{nd}$ case $($if Kafeel gives $2$ oranges to Sunil$) :$
$x + 2 = y - 2$
$\Rightarrow x - y = -4 \dots....(ii)$
Subtracting eqn. $(ii)$ from eqn. $(i),$ we get
$2x = 12$
$\Rightarrow x = 6$
$\Rightarrow 6 - y = -4$
$\Rightarrow y = 10$
Thus, Sunil has $6$ oranges and Kafeel has $10$ oranges.
View full question & answer→Question 34 Marks
$9$ pens and $5$ pencils cost $Rs.32$, and $7$ pens and $8$4 pencils cost $Rs.29$. Find the unit price for each pen and pencil.
AnswerLet the unit price for each pen$ = Rs. x$
and the unit price for each pencil $= Rs. y$
According to given information, we have
$9x + 5y = 32\dots ....(i)$
$7x + 8y = 29\dots ....(ii)$
Multiplying eqn. $(i)$ by $8$ and eqn.$ (ii)$ by $5,$ we get
$72x + 40y = 256 \dots....(iii)$
$35x + 40y = 145 \dots....(iv)$
Subtracting eqn. $(iv) $from eqn. $(iii),$ we get
$37x = 111$
$\Rightarrow x = 3$
$\Rightarrow 9(3) + 5y = 32$
$\Rightarrow 27 + 5y = 32$
$\Rightarrow 5y = 5$
$\Rightarrow y = 1$
Thus, the unit price for each pen is $Rs. 3$ and that for each pencil is $Rs.1.$
View full question & answer→Question 44 Marks
A solution containing $12\ \%\ $ alcohol is to be mixed with a solution containing $4\ \%\ $ alcohol to make $20$ gallons of solution containing $9\ \%\ $ alcohol. How much of each solution should be used?
AnswerLet $x$ gallons of $12\ \% \ $ alcohol and $y$ gallons of $4\ \% \ $ alcohol be mixed.
Then, we have
$ x+y=20 \dots....(i) $
And, $12\ \% \ $ of $x+4\ \% \ $ of $y=9\ \% \ $ of $20$
$ \Rightarrow \frac{12}{100} \times+\frac{4}{100} y$
$=\frac{9}{100} \times 20$
$\Rightarrow 12 x+4 y=180$
$\Rightarrow 3 x+y=45 \ldots .(i i) $
Subtracting eqn. $(i)$ from eqn. $(ii),$ we get
$ 2 x=25$
$\Rightarrow x=12.5$
$\Rightarrow 12.5+y=20$
$\Rightarrow y=7.5 $
Hence $,12.5$ gallons of $12\ \% \ $ alcohol and $7.5$ gallons of $4\ \% \ $ alcohol should be used.
View full question & answer→Question 54 Marks
An eraser costs $Rs. 1.50$ less than a sharpener. Also, the cost of $4$ erasers and $3$ sharpeners is $Rs.29.$ Taking $x$ and $y$ as the costs $($in $Rs.)$ of an eraser and a sharpener respectively, write two equations for the above statements and find the value of $x$ and $y.$
AnswerCost of an eraser $= Rs. x$
Cost of a sharpener $= Rs. y$
According to given information, we have
$x = y - 1.50$
$\Rightarrow x - y = -1.50 \dots....(i)$
And, $4x + 3y = 29\dots ....(ii)$
Multiplying eqn. $(i)$ by $3$, we get
$3x - 3y = -450 \dots....(iii)$
Adding eqns. $(ii)$ and $(iii),$ we get
$7x = 24.50$
$\Rightarrow x = 3.50$
$\Rightarrow 3.50 - y = -1.50$
$\Rightarrow y$
$= 3.50 + 1.50$
$= 5$
Thus, the cost of an eraser is $Rs.3.50$ and that of a sharpener is $Rs.5.$
View full question & answer→Question 64 Marks
In a triangle, the sum of two angles is equal to the third angle. If the difference between these two angles is $20^\circ ,$ determine all the angles.
AnswerLet the two angles of a triangle be $x$ and $y$ respectively.
Then, the $3^{rd}$ angle will be $180^\circ - (x + y).$
According to given information, we have
$x + y = 180^\circ - (x + y)$
$\Rightarrow 2(x + y) = 180^\circ$
$\Rightarrow x + y = 90^\circ \dots....(i)$
And,
$x + y = 20^\circ \dots....(ii)$
Adding eqns. $(i)$ and $(ii),$ we have
$2x = 110^\circ$
$\Rightarrow x = 55^\circ$
$\Rightarrow 55^\circ + y = 90^\circ$
$\Rightarrow y = 35^\circ$
$\Rightarrow 3^{rd}$ angle
$= 180^\circ - (55^\circ + 35^\circ )$
$= 180^\circ - 90^\circ$
$= 90^\circ$
Hence, the three angles of a triangle are $55^\circ , 35^\circ$ and $90^\circ .$
View full question & answer→Question 74 Marks
A father's age is three times the age of his child. After $12$ years, twice the age of father will be $36$ more than thrice the age of his child. Find his present age.$^*$ Question modified
AnswerLet the present age of father $= x$ years and that of his child $= y$ years
After $12$ years,
father's age $= (x + 12)$ years
child's age $= (y + 12)$ years
According to given information, we have
$x = 3y \dots....(i)$
Now, after $12$ years
$2(x + 12) = 3(y + 12) + 36$
$\Rightarrow 2x + 24 = 3y + 36 + 36$
$\Rightarrow 2x - 3y = 48 \dots....(ii)$
$\Rightarrow 2(3y) - 3y = 48$
$\Rightarrow 3y = 48$
$\Rightarrow y = 16$
$\Rightarrow x = 3 \times 16$
$= 48$
Thus, the present age of father is $48$ years.
$^*$ Question modified.
View full question & answer→Question 84 Marks
The present ages of Kapil and Dhruvi are in the ratio $2 : 3$. Six years later, the ratio will be $5 : 7$. Find their present ages.
AnswerLet the present age of Kapil $=x$ years and that of Dhruvi $= y$ years
After $6$ years,
Kapil's age $=(x+6)$ years
Dhruvi's age $=(y+6)$ years
According to given information, we have
$ \frac{x}{y}=\frac{2}{3}$
$\Rightarrow 3 x =2 y$
$\Rightarrow 3 x -2 y =0 \dots....(i) $
And,
$ \frac{x+6}{y+6}=\frac{5}{7}$
$\Rightarrow 7 x+42=5 y+30$
$\Rightarrow 7 x-5 y=-12 \ldots .(\text {ii) } $
Multiplying eqn. $(i)$ by $5$ and eqn.$ (ii)$ by $2 ,$ we get
$ 15 x-10 y=0\dots....(iii)$
$14 x-10 y=-24\dots ....(iv) $
Subtracting eqn. $(iv) $ from eqn. $(iii),$ we get
$ x=24$
$\Rightarrow 3(24)-2 y=0$
$\Rightarrow 72-2 y=0$
$\Rightarrow 2 y=72$
$\Rightarrow y=36 $
Thus, the present age og Kapil is $24$ years and that of Dhruvi is $36$ years.
View full question & answer→Question 94 Marks
The age of the father is seven times the age of the son. Ten years later, the age of the father will be thrice the age of the son. Find their present ages.
AnswerLet the present age of father $= x$ years and that of son $= y$ years
After $10$ years,
father's age$ = (x + 10)$ years
son's age$ = (y + 10)$ years
According to given information, we have
$x = 7y \dots....(i)$
And,
$(x + 10) = 3(y + 10)$
$\Rightarrow x + 10 = 3y + 30$
$\Rightarrow x - 3y = 20$
$\Rightarrow 7y - 3y = 20 \dots ...[$From $(i)]$
$\Rightarrow 4y = 20$
$\Rightarrow y = 5$
$\Rightarrow x = 7 x 5$
$= 35$
Thus, the present age of son is $5$ years and that of father is $35$ years.
View full question & answer→Question 104 Marks
The length of a rectangle is twice its width. If its perimeter is $30$ units, find its dimensions.
AnswerLet the length and breadth of a rectangle be $x$ units and $y$ units respectively.
According to given information, we have
$x = 2y$
$\Rightarrow x - 2y = 0 \dots....(i)$
ALso, perimeter of a rectangle $= 30$ units
$\Rightarrow 2(x + y) = 30$
$\Rightarrow x - y = 15 \dots....(ii)$
Subtracting eqn. $(ii)$ from eqn. $(i)$, we get
$-3y = -15$
$\Rightarrow y = 5$
Substituting the value of $y$ in eqn.$ (1)$, we get
$x - 2(5) = 0$
$\Rightarrow x - 10 = 0$
$\Rightarrow x = 10$
Thus, the length and breadth of a reactangle are $10$ units and $5$ units respectively.
View full question & answer→Question 114 Marks
Solve the following pairs of equations:$\frac{x+y}{x y}=2;\frac{x-y}{x y}=6$
Answer$ \frac{x+y}{x y}=2$
$\Rightarrow x + y =2 xy \dots....(i)$
$\frac{x-y}{x y}=6$
$\Rightarrow x-y=6 x y\dots ....(ii) $
Adding eqns. $(i)$ and $(ii),$ we get
$ 2 x=8 x y$
$\Rightarrow y=\frac{1}{4} $
Substituting the value of $y$ in eqn. $(i),$ we get
$ x+\frac{1}{4}=2 x \times \frac{1}{4}$
$\Rightarrow \frac{4 x+1}{4}=\frac{x}{2}$
$\Rightarrow 8 x +2=4 x$
$\Rightarrow 4 x =-2$
$\Rightarrow x =-\frac{1}{2} $
Thus, the solution set is $\left(-\frac{1}{2}, \frac{1}{4}\right)$.
View full question & answer→Question 124 Marks
Solve the following pairs of equations:$\frac{3}{5} x-\frac{2}{3} y+1=0;\frac{1}{3} y+\frac{2}{5} x=4$
Answer$ \frac{3}{5} x-\frac{2}{3} y+1=0$
$\Rightarrow 9 x-10 y+15=0$
$\Rightarrow 9 x-10 y=-15 \dots....(i)$
$\frac{1}{3} y+\frac{2}{5} x=4$
$\Rightarrow 5 y+6 x=60$
$\Rightarrow 6 x+5 y=60 \dots....(ii) $
Multiplying eqn. $(ii)$ by $2,$ we get
$ 12 x+10 y=120 \dots....(iii) $
Adding eqns. $(i)$ and $(iii),$ we get
$ 21 x =105$
$\Rightarrow x =5 $
Substituting the value of $x$ in eqn. $(ii),$ we get
$ 6(5)+5 y=60$
$\Rightarrow 30+5 y=60$
$\Rightarrow 5 y=30$
$\Rightarrow y=6 $
Thus, the solution set is $(5,6)$.
View full question & answer→Question 134 Marks
Solve the following simultaneous equations:$103a + 51b = 617;97a + 49b = 583$
AnswerThe given equations are
$103a + 51b = 617 \dots....(i)$
$97a + 49b = 583\dots ....(ii)$
Subtracting eqn. $(ii)$ from $(i)$. we get
$6a + 2b = 34$
$\Rightarrow 3a + b = 17 \dots....[$Dividing throughtout by $2] \dots....(iii)$
$200a + 100b = 1200$
$\Rightarrow 2a + b = 12\dots ...[$Dividing throughtout by $100] \dots....(iv)$
Subtracting eqn. $(iv)$ from eqn. $(iii)$, we get
$a = 5$
Substituting the value of an in eqn. $(iii),$ we get
$3(5) + b = 17$
$\Rightarrow 15 + b = 17$
$\Rightarrow b = 2$
Thus, the solution set is $(5,2).$
View full question & answer→Question 144 Marks
Solve the following simultaneous equations:$65x - 33y = 97;33x - 65y = 1$
AnswerThe given equations are
$65x - 33y = 97\dots ....(i)$
$33x - 65y = 1 \dots....(ii)$
Multiplying eqn. $(i)$ by $33$ and eqn.$ (ii)$ by $65$, we get
$2145x - 1089y = 3201 \dots....(iii)$
$2145x - 4225y = 65 \dots....(iv)$
Subtracting eqn. $(iv)$ from eq.$(iii),$ we get
$3136y = 3136$
$\Rightarrow y = 1$
Substituting the value of $y$ in eqn. $(ii),$ we get
$33x - 65(1) = 1$
$\Rightarrow 33x - 65 = 1$
$\Rightarrow 33x = 1 + 65$
$\Rightarrow 33x = 66$
$\Rightarrow x = 2$
Thus, the solution set is $(2, 1).$
View full question & answer→Question 154 Marks
Solve the following simultaneous equations:$41x + 53y = 135;53x + 41y = 147$
AnswerThe given equations are
$41x + 53y = 135 \dots....(i)$
$53x + 41y = 147 \dots....(ii)$
Multiplying eqn. $(i)$ by$ 53$ and eqn. $(ii)$ by $41,$ we get
$2173x + 2809y = 7155\dots ....(iii)$
$2173x + 1681y = 6027 \dots....(iv)$
Subtracting eqn. $(iv)$ from eq. $(iii)$, we get
$1128y = 1128$
$\Rightarrow y = 1$
Substituting the value of $y$ in eqn. $(i),$ we get
$41x + 53(1) = 135$
$\Rightarrow 41x + 53 = 135$
$\Rightarrow 41x = 135 - 53$
$\Rightarrow 41x = 82$
$\Rightarrow x = 2$
Thus, the solution set is $(2, 1).$
View full question & answer→Question 164 Marks
Solve the following simultaneous equations$:13a - 11b = 70,11a - 13b = 74$
AnswerThe given equation are
$13a - 11b = 70\dots ....(i)$
$11a - 13b = 74 \dots....(ii)$
Multiplying eqn. $(i)$ by $13$ and eqn. $(ii)$ by $4$, we get
$169a - 143b = 910 \dots....(iii)$
$121a - 143b = 814\dots ....(iv)$
Subtracting eqn. $(iv)$ from eq. $(iii),$ we get
$48a = 96$
$\Rightarrow a = 2$
Subtracting the value of $a$ in eqn. $(i),$ we get
$13(2) - 11b = 70$
$\Rightarrow 26 - 11b = 70$
$\Rightarrow -11b = 70 - 26$
$\Rightarrow -11b = 44$
$\Rightarrow b = -4$
Thus, the solution set is $(2, - 4).$
View full question & answer→Question 174 Marks
Solve the following simultaneous equations :$3(2u + v) = 7uv;3(u + 3v) = 11uv$
Answer$ 3(2 u+v)=7 u v$
$3(u+3 v)=11 u v $
Dividing by $uv,$ we get,
$ \frac{6}{v}+\frac{3}{u}=7 \ldots \ldots . . (1)$
$3(u+3 v)=11 u v$
$3 u+9 v=11 u v $
Dividing by $uv,$ we get
$ \frac{3}{v}+\frac{8}{u}=11 \dots...........(2) $
Multiplying $(1)$ by $3,$ we get,
$ \frac{18}{v}+\frac{9}{u}=21\dots ........(3) $
Subtracting $(2)$ from $(3),$ we get,
$ \frac{15}{v}=10$
$\Rightarrow v =\frac{15}{10}=\frac{3}{2}$
$\therefore \frac{3}{u} $
$ =7-6 \times \frac{2}{3}$
$=7-4$
$=3$
$\Rightarrow u=1 $
Thus, the solution set is $\left(1, \frac{3}{2}\right)$.
View full question & answer→Question 184 Marks
Solve the following simultaneous equations :$6x + 3y = 7xy,3x + 9y = 11xy$
Answer$ 6 x+3 y=7 x y$
$3 x+9 y=11 x y $
Dividing both sides of each equation by $x y,$ we get,
$ \frac{6}{y}+\frac{3}{x}=7$
$\frac{3}{y}+\frac{9}{x}=11 $
Multiplying $(2)$ by $2,$
$\frac{6}{y}+\frac{18}{x}=22$
Subtracting $(1)$ from $(3),$ we get,
$ \frac{15}{x}=15$
$\Rightarrow x=1$
$\therefore \frac{3}{y}+9=11$
$\Rightarrow \frac{3}{y}=11-9=2$
$\Rightarrow y=\frac{3}{2} $
Thus, the solution set is $\left(1, \frac{3}{2}\right)$.
View full question & answer→Question 194 Marks
Solve the following simultaneous equations by the substitution method:$5x + 4y - 23 = 0,x + 9 = 6y$
AnswerThe given equations are
$ 5 x+4 y-23=0\dots ....(i)$
$x+9=6 y\dots ....(ii) $
Now, consider equation
$ x+9=6 y$
$\Rightarrow x=6 y-9\dots ....(iii) $
Substituting the value of $x$ in eqn. $(i),$ we get
$ 5(6 y-9)+4 y-23=0$
$\Rightarrow 30 y-45+4 y-23=0$
$\Rightarrow 34 y-68=0$
$\Rightarrow 34 y=68$
$\Rightarrow y=\frac{68}{34}=2 $
Putting the value of $y$ in eqn. $(iii),$ we get
$ x=6(2)-9$
$=12-9$
$=3 $
Thus, the solution set is $(3,2)$.
View full question & answer→Question 204 Marks
Solve the following simultaneous equations by the substitution method:$x + 3y= 5,7x - 8y = 6$
AnswerThe given equations are
$x + 3y= 5 \dots....(i)$
$7x - 8y = 6\dots ....(ii)$
Now, consider equation
$x + 3y = 5$
$\Rightarrow x = 5 - 3y \dots....(iii)$
Substituting the value of $x$ in eqn. $(ii)$, we get
$7(5 - 3y) - 8y = 6$
$\Rightarrow 35 - 21y - 8y = 6$
$\Rightarrow 35 - 29y = 6$
$\Rightarrow -29y = -29$
$\Rightarrow y = 1$
Putting the value of $y$ in eqn.$ (ii),$ we get
$x = 5 - 3(1)$
$= 5 - 3 = 2$
Thus, the solution set is$ (2, 1).$
View full question & answer→