Question 14 Marks
Mr. and Mrs. Ahuja weight $x \mathrm{~kg}$ and $y \mathrm{~kg}$ respectively. They both take a dieting course, at the end of which Mr. Ahuja loses $5 \mathrm{~kg}$ and weights as much as his wife weighed before the course. Mrs. Ahuja loses $4 \mathrm{~kg}$ and weighs $\frac{7}{8}$ th of what her husband weighed before the course. Form two equations in $x$ and $y$, find their weights before taking the dieting course.
AnswerWeight of Mr. Ahuja $=x \mathrm{~kg}$ and weight of Mrs. Ahuja $=y \mathrm{~kg}$.
After the dieting,
$x-5=y$
$x-y=5\ldots(1)$
and,
$y-4=\frac{7}{8} x$
$7 x-8 y=-32\ldots(2)$
Multiplying equation no. $(1)$ by $7$ , we get
$7 x-7 y=35\ldots .(3)$
Now subtracting equation $(2)$ from $(3),$
$7 x-7 y=35$
$7 x-8 y=-32$
$ y=67$
From$ (1)$
$x-67=5$
$ \Rightarrow x=72$
Thus, weight of Mr. Ahuja $=72 \mathrm{~kg}$
and that of Mrs. Anuja $=67 \mathrm{~kg}$.
View full question & answer→Question 24 Marks
Seven times a two digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is $3$ find the number.
AnswerLet the digit at ten’s place be $x$ And the digit at unit’s place be $y.$
Required number $= 10x + y$
When the digits are interchanged,
Reversed number $= 10y + x$
According to the question,
$7(10x + y) = 4(10y + x)$
$\therefore 70x + 7y = 40y + 4x$
$\therefore 66x = 33y$
$\therefore 2x - y = 0 ...(1)$
The difference between the digits is $3.$
$y - x = 3 ...(2)$
Adding equation $(1)$ and $(2)$
$y - x = 3$
$+ - y + 2x = 0$
$x = 3$
From $(1)$
$2(3) - y = 0$
$y = 6$
Thus, Required number $= 10(3) + 6 = 36$
View full question & answer→Question 34 Marks
The sum of a two digit number and the number obtained by reversing the order of the digits is $99.$ Find the number, if the digits differ by $3.$
AnswerLet the digits in the tens place be $x$ and the digit in the units place be $y.$
$\therefore $ Number $= 10x + y$
Number on reversing the digits $= 10y + x$
The difference between the digits $= x - y$ or $y - x$
Given $: ( 10x + y ) + ( 10y + x ) = 99$
$\Rightarrow 11x + 11y = 99$
$\Rightarrow x + y = 9 ...(1)x - y = 3 ...(2)$
or $y - x = 3 ...(3)$
On solving equations $(1)$ and $(2),$ We get
$2x = 12$
$\Rightarrow x = 6$
So, $y = 3$
On solving equation $(1)$ and $(3),$ We get
$2y = 12$
$ \Rightarrow y = 6$
So, $x = 3$
Number $= 10x + y = 10(6) + 3 = 63$
or Number $= 10x + y = 10(3) + 6 = 36$
$\therefore $ Required number $= 63$ or $36.$
View full question & answer→Question 44 Marks
$A$ and $B$ both the have some pencils. If $A$ gives $10$ pencils to $B,$ then $B$ will have twice as many as $A$. And if $B$ gives $10$ pencils to $A$, then they will have the same number of pencils. How many pencils does eachhave ?
AnswerLet the numberof pencils with $A = x$
and the number of pencils with $B = y.$
If $A$ gives $10$ pencils to $B,$
$y + 10 = 2( x - 10 )$
$2x - y = 30 ...(1)$
If $B$ gives to pencils to $A$
$y - 10 = x + 10$
$x - y = - 20 ...(2)$
Subtracting equation $(1)$ and $(2),$
$x - y = - 20$
$- 2x - y = 30 $
$- x = - 50$
$x = 50$
From $(1)$
$2(50) - y = 30$
$y = 70$
Thus, $A$ has $50$ pencils and $B$ has $70$ pencils.
View full question & answer→Question 54 Marks
In an examination, the ratio of passes to failures was $4 :1.$ Had $30$ less appeared and $20$ less passed, the ratio of passes to failures would have been $5 :1.$ Find the number of students who appeared for the examination.
AnswerLet the no. of pass candidates be $x$
and the no. of fail candidates be $y$.
According to the question,
$\frac{x}{y}=\frac{4}{1}$
$x-4 y=0\ldots(1)$
and
$\frac{x-20}{y-10}=\frac{5}{1}$
$x-5 y=-30\ldots(2)$
Subtracting equation $(2)$ from $(1),$
$\begin{gathered}x-4 y=0- x-5 y=-30y=30\end{gathered}$
From $(1)$
$x-4(30)=0$
$ x=120$
Total students appeared $=x+y=120+30=150$.
View full question & answer→Question 64 Marks
The sum of the numerator and the denominator of a fraction is equal to $7.$ Four times the numerator is $8$ less than $5$ times the denominator. Find the fraction.
AnswerLet the numerator and denominator of a fraction be $x$ and $y$ respectively.
let the fraction will be $\frac{x}{y}$.
According to the question,
$x+y=7 \dots...(1)$
$5 y-4 x=8 \dots...(2)$
Multiplying equation no. $(1)$ by $4$
$4 x+4 y=28\ldots(3)$
Adding equation $(3)$ and $(2)$
$4 x+4 y =28$
$+-4 x+5 y =8$
$9 y =36$
$y =4$
From $(1),$
$x+4=7$
$ x=3$
Required fraction $=\frac{3}{4}$.
View full question & answer→Question 74 Marks
A two digit number is obtained by multiplying the sum of the digits by $8.$ Also, it is obtained by multiplying the difference of the digits by $14$ and adding $2.$ Find the number.
AnswerLet the tens digit of the number be $x$ and the units digit be $y$.
So, the number is $10 x+y$.
According to the question,
$10 x+y=8(x+y)$
$ \Rightarrow 10 x+y=8 x+8 y$
$\Rightarrow 2 x=7 y\ldots .(1)$
and $10 x+y=14(x-y)+2$ or
$10 x+y=14(y-x)+2$
$\Rightarrow 4 x-15 y=-2 \ldots(2) \quad$ or
$24 x-13 y=2\ldots (3)$
Solving (1) and (2), We get
$y=2$ and $x=7$
Solving $(1)$ and $(3),$ We get
$y=\frac{2}{71}$
This is not possible, since $y$ is a digit and cannot be in fraction form.
So, the number is $72 .$
View full question & answer→Question 84 Marks
A Fraction becomes $\frac{1}{2}$ if $5$ subtracted from its numerator and $3$ is subtracted from its denominator. If the denominator of this fraction is $5$ more than its numerator. Find the fraction.
AnswerLet the numerator of the fraction be $x$ and denominator of the fraction be $y$.
Then, the fraction $=\frac{x}{y}$
According to given condition, we have
$\frac{x-5}{y-3}=\frac{1}{2}$
$ \Rightarrow 2 x-10=y-3$
$\Rightarrow 2 x-y=7\ldots(1)$
And,
$x+5=y$
$\Rightarrow x-y=-5\ldots(2)$
Subtracting $(2)$ from $(1),$ We get
$x=12$
$ \Rightarrow y=x+5$
$ \Rightarrow y=12+5=17$
Hence, the fraction is $\frac{12}{17}$.
View full question & answer→Question 94 Marks
If the numerator of a fraction is multiplied by $2$ and its denominator is increased by $1$ , it becomes $1 .$ However, if the numerator is increased bu $4$ and denominator is multiplied by $2$ , the fraction becomes $\frac{1}{2}$. Find the fraction.
AnswerLet the numerator of the fraction be $x$ and the denominator be $y$.
So, the Fraction is $\frac{x}{y}$
According to the question,
$\frac{2 x}{y+1}=1$
$ \Rightarrow 2 x=y+1$
$\Rightarrow 2 x-y=1 \dots...(1)$
and,
$\frac{x+4}{2 y}=\frac{1}{2}$
$ \Rightarrow 2 \mathrm{x}+8=2 \mathrm{y}$
$\Rightarrow 2 x-2 y=-8 \dots...(2)$
Solving equations $( 1 )$ and $( 2 ),$ We get
$y=9$
Putting the value of $y$ in $(1),$ we get
$2 x-9=1$
$ \Rightarrow 2 x=1+9$
$ \Rightarrow x=5$
So, the fraction is $\frac{5}{9}$.
View full question & answer→Question 104 Marks
Solve, using cross-multiplication $:4x - 3y - 11 = 0,6x + 7y - 5 = 0$
AnswerGiven equation are $4 x-3 y-11=0$ and
$6 x+7 y-5=0$
Comparing with $\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and
$\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$,
We have
$a_1=4, b_1=-3, c_1=-11$ and
$a_2=6, b_2=7, c_2=-5$
Now,
$x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow \mathrm{x}= \frac{-3 \times(-5)-7 \times(-11)}{4 \times 7-6 \times(-3)}$ and $y=\frac{-11 \times 6-(-5) \times 4}{4 \times 7-6 \times(-3)}$
$ \Rightarrow \mathrm{x}=\frac{15+77}{28+18}$ and $y=\frac{-66+20}{28+18}$
$ \Rightarrow \mathrm{x}=\frac{92}{46}$ and $y=-\frac{46}{46}$
$ \Rightarrow \mathrm{x}=2$ and $\mathrm{y}=-1$
View full question & answer→Question 114 Marks
Solve, using cross$-$multiplication $:8x + 5y = 9,3x + 2y = 4$
AnswerGiven equation are $8 x+5 y=9$ and
$3 x+2 y=4$
Comparing with $\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and
$\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$,
We have
$a_1=8, b_1=5, c_1=-9$ and $a_2=3, b_2=2, c_2=-4$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow \mathrm{x}=\frac{5 \times(-4)-2 \times(-9)}{8 \times 2-3 \times 5}$ and $y=\frac{-9 \times 3-(-4) \times 8}{8 \times 2-3 \times 5}$
$ \Rightarrow \mathrm{x}=\frac{-20+18}{16-15}$ and $y=\frac{-27+32}{16-15}$
$ \Rightarrow \mathrm{x}=-\frac{2}{1}$ and $y=\frac{5}{1}$
$ \Rightarrow \mathrm{x}=-2$ and $\mathrm{y}=5 .$
View full question & answer→Question 124 Marks
Solve, using cross-multiplication $:4x - 3y = 0,2x + 3y = 18$
AnswerGiven equation are $4 x-3 y=0$ and
$2 x+3 y=18$
Comparing with $\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and
$\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$,
We have
$a_1=4, b_1=-3, c_1=0$ and $a_2=2, b_2=3, c_2=-18$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow x= \frac{-3 \times(-18)-3 \times 0}{4 \times 3-2 \times(-3)}$ and $y=\frac{0 \times 2-(-18) \times 4}{4 \times 3-2 \times(-3)}$
$\Rightarrow x=\frac{54-0}{12+6} \quad$ and $y=\frac{0+72}{12+6}$
$\Rightarrow \mathrm{x}=\frac{54}{18}$ and $y=\frac{72}{18}$
$\Rightarrow x=3$ and $y=4$.
View full question & answer→Question 134 Marks
Solve, using cross-multiplication $:4x - y = 5,5y - 4x = 7$
AnswerGiven equation are $4 x-y=5$ and $5 y-4 x=7$
Comparing with $\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and $\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$,
We have
$a_1=41, b_1=-1, c_1=-5$ and $a_2=-4, b_2=5, c_2=-7$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow \mathrm{x}= \frac{-1 \times(-7)-5 \times(-5)}{4 \times 5-(-4) \times(-1)}$ and $y=\frac{(-5) \times(-4)-(-7) \times 4}{4 \times 5-(-4) \times(-1)}$
$\Rightarrow x=\frac{7+25}{20-4}$ and $y=\frac{20+28}{20-4}$
$ \Rightarrow \mathrm{x}=\frac{32}{16}$ and $y=\frac{48}{16}$
$ \Rightarrow \mathrm{x}=2$ and $\mathrm{y}=3$
View full question & answer→Question 144 Marks
Solve, using cross-multiplication$ :x - y + 2 = 0,7x + 9y = 130$
AnswerGiven equation are $x-y+2=0$ and $7 x+9 y=130$
Comparing with $\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and $\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$,
We have
$a_1=1, b_1=-1, c_1=2$ and $a_2=7, b_2=9, c_2=-130$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow x= \frac{-1 \times(-130)-9 \times 2}{1 \times 9-7 \times(-1)}$ and $y=\frac{2 \times 7-(-130) \times 1}{1 \times 9-7 \times(-1)}$
$\Rightarrow x=\frac{130-18}{9+7}$ and $y=\frac{14+130}{9+7}$
$\Rightarrow \mathrm{x}=\frac{112}{16} \quad$ and $y=\frac{144}{16}$
$\Rightarrow x=7$ and $y=9.$
View full question & answer→Question 154 Marks
Solve, using cross-multiplication $:5x + 4y + 14 = 0,3x = -10 - 4y$
AnswerGiven equation are $5 x+4 y+14=0$ and $3 x=-10-4 y$
Comparing with $\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and $\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$,
We have
$a_1=5, b_1=4, c_1=14$ and $a_2=3, b_2=4, c_2=10$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow \mathrm{x}=\frac{4 \times 10-4 \times 14}{5 \times 4-3 \times 4}$ and $y=\frac{14 \times 3-10 \times 5}{5 \times 4-3 \times 4}$
$ \Rightarrow \mathrm{x}=\frac{40-56}{20-12}$ and $y=\frac{42-50}{20-12}$
$ \Rightarrow \mathrm{x}=-\frac{16}{8}$ and $y=-\frac{8}{8}$
$ \Rightarrow \mathrm{x}=-2$ and $\mathrm{y}=-1$
View full question & answer→Question 164 Marks
Solve, using cross$-$multiplication $:6x + 7y - 11 = 0,5x + 2y = 13$
AnswerGiven equation are $6 x+7 y-11=0$ and $5 x+2 y=13$
Comparing with $\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and $\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$,
We have
$a_1=6, b_1=7, c_1=-11$ and $a_2=5, b_2=2, c_2=-13$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow x= \frac{7 \times(-13)-2 \times(-11)}{6 \times 2-5 \times 7}$ and $y=\frac{-11 \times 5-(-13) \times 6}{6 \times 2-5 \times 7}$
$ \Rightarrow x=\frac{-91+22}{12-35}$ and $y=\frac{-55+78}{12-35}$
$ \Rightarrow x=\frac{-69}{-23}$ and $y=\frac{23}{-23}$
$ \Rightarrow x=3$ and $y=-1$
View full question & answer→Question 174 Marks
Solve, using cross$-$multiplication $:4x + 3y = 17,3x - 4y + 6 = 0$
AnswerGiven equation are $4 x+3 y=17$ and $3 x-4 y+6=0$
Comparing with $\mathrm{a}_ 1 \mathrm{x}+\mathrm{b} _1 \mathrm{y}+\mathrm{c} _1=0$ and $\mathrm{a} _2 \mathrm{x}+\mathrm{b}_ 2 \mathrm{y}+\mathrm{c} _2=0$,
We have
$\mathrm{a} _1=4, \mathrm{~b} _1=3, \mathrm{c} _1=-17$ and $\mathrm{a} _2=3, \mathrm{~b} _2=-4, \mathrm{c} _2=6$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow \mathrm{x}= \frac{3 \times 6-(-4) \times(-17)}{4 \times(-4)-3 \times 3}$ and $y=\frac{-17 \times 3-6 \times 4}{4 \times(-4)-3 \times 3}$
$ \Rightarrow \mathrm{x}=\frac{18-68}{-16-9}$ and $y=\frac{-51-24}{-16-9}$
$ \Rightarrow x=\frac{-50}{-25}$ and $y=\frac{-75}{-25}$
$\Rightarrow x=2$ and $y=3$.
View full question & answer→Question 184 Marks
Solve, using cross$-$multiplication $:\sqrt2x - \sqrt3y = 0,\sqrt5x + \sqrt2y = 0$
AnswerGiven equation are $\sqrt{2 } x-\sqrt{3 } y=0$ and $\sqrt{ 5} x+\sqrt{2 } y=0$
Comparing with $\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and $\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$, We have
$a_1=\sqrt{2}, b_1=\sqrt{ 3} , c_1=0$ and $a_2=\sqrt{5}, b_2=\sqrt{2 } , c_2=0$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow \mathrm{x}=$
$\frac{(-\sqrt{3}) \times 0-\sqrt{2} \times 0}{\sqrt{2} \times \sqrt{2}-\sqrt{5} \times(-\sqrt{3})}$ and $y=\frac{0 \times \sqrt{5}-0 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}-\sqrt{5} \times(-\sqrt{3})}$
$\Rightarrow x=\frac{0}{2+\sqrt{15}}$ and $y=\frac{0}{2+\sqrt{15}}$
$\Rightarrow x=0$ and $y=0$.
View full question & answer→Question 194 Marks
Solve, using cross$-$multiplication $:4x + 6y = 15,3x - 4y = 7$
AnswerGiven equation are $4 x+6 y=15$ and $3 x-4 y=7$
Comparing with $\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and $\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$,
We have
$a_1=4, b_1=6, c_1=-15$ and $a_2=3, b_2=-4, c_2=-7$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow x= \frac{6 \times(-7)-(-4) \times(-15)}{4 \times(-4)-3 \times 6}$ and $y=\frac{-15 \times 3-(-7) \times 4}{4 \times(-4)-3 \times 6}$
$ \Rightarrow x=\frac{-42-60}{-16-18}$ and $y=\frac{-45+28}{-16-18}$
$ \Rightarrow x \frac{-102}{-34}$ and $y=\frac{-17}{-34}$
$ \Rightarrow x=3$ and $y=\frac{1}{2}$
View full question & answer→Question 204 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients $:3 - (x - 5) = y + 2,2 (x + y) = 4 - 3y$
Answer$ 3-(x-5)=y+2$
$ \therefore 3-x+5=y+2$
$ \therefore-x+8=y+2$
$\therefore x+y=6\ldots .(1)$
$ 2(x+y)=4-3 y$
$ \therefore 2 x+2 y=4-3 y$
$\therefore 2 x+5 y=4\dots(2)$
Multiplying equation no $(1)$ by $2 .$
$2 x+2 y=12\dots(3)$
Subtracting equation $(2)$ from $(3)$
$2 x+2 y=12$
$- 2 x+5 y=4$
$-3 y=8$
$y=-\frac{8}{3}$
From $(1)$
$x-\frac{8}{3}=6$
$ \Rightarrow x=\frac{26}{3}$
View full question & answer→Question 214 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$3 x-y=23, \frac{x}{3}+\frac{y}{4}=4$
Answer$3 x-y=23\ldots(1)$
$\frac{x}{3}+\frac{y}{4}=4$
$4 x+3 y=48\ldots(2)$
Multiplying equation no. $(1)$ by $3$
$9 x-3 y=69\ldots .(3)$
Adding equation $(3)$ and $(2)$
$9 x-3 y =69$
$+ 4 x+3 y =48$
$ 13 x =117$
$x =9$
From $(1)$
$3(9)-y=23$
$ \therefore 27-y=23$
$ \therefore y=27-23$
$ \therefore y=4$
View full question & answer→Question 224 Marks
The value of expression mx $-$ny is $3$ when $x = 5$ and $y = 6.$ And its value is $8$ when $x = 6$ and $y = 5.$ Find the values of $m$ and $n.$
AnswerThe value of expression $mx - ny$ is $3$ when $x = 5$ and $y = 6.$
$\Rightarrow 5m - 6n = 3 .....(1)$
The value of expression $mx - ny$ is $8$ when $x = 6$ and $y = 5.$
$\Rightarrow 6m - 5n = 8 ....(2)$
Multiply equation $(1)$ by $6$ and equation $(2)$ by $5,$ We get:
$30m - 36n = 18 ....(3)$
$30m - 25n = 40 .....(4)$
Subtracting equation $(4)$ from$ (3)$
$30m - 36n = 18$
$- 30m - 25n = 40$
$- 11n = - 22$
$n = 2$
Substituting $n = 2$ in equation $(1),$ we get
$5m - 6(2) = 3+$
$\Rightarrow 5m = 15$
$\Rightarrow m = 3$
$\therefore $ Solution is $m = 3 $ and $n = 2.$
View full question & answer→Question 234 Marks
If 2x + y = 23 and 4x - y = 19; find the values of x - 3y and 5y - 2x.
Answer2x + y = 23 ...(1)
4x - y = 19 ...(2)
Adding equation (1) and (2) we get,
2x + y = 23
+ 4x - y = 19
6x = 42
x = 7
From (1)
2x + y = 23
⇒ 2(7) + y = 23
⇒ 14 + y = 23
⇒ y = 23 - 14
y = 9
∴ x - 3y = 7 - 3(9) = -20
and 5y - 2x = 5(9) - 2(7) = 45 - 14 = 31.
View full question & answer→Question 244 Marks
Solve the following pair of linear (simultaneous) equation using method of elimination by substitution$:3x + 2y =11,2x - 3y + 10 = 0$
Answer$ 3 x+2 y=11$
$ \Rightarrow 3 x=11-2 y$
$\Rightarrow x=\frac{11-2 y}{3}$$\ldots(1)$
And,
$2 x-3 y+10=0$
$ \Rightarrow 2 x\left(\frac{11-2 y}{3}\right)-3 y+10=0$
$\Rightarrow \frac{22-4 y}{3}-3 y=-10$
$\Rightarrow \frac{22-4 y-9 y}{3}=-10$
$\Rightarrow 22-13 y=-30$
$\Rightarrow 13 y=52$
$ \Rightarrow y=4$
Substituting the value of $y$ in $(1),$ we have
$x=\frac{11-2(4)}{3}=\frac{11-8}{3}=\frac{3}{3}=1$
$\therefore$ Solution is $x=1$ and $y=4$.
View full question & answer→Question 254 Marks
A part of monthly expenses of a family is constants and the remainingvarywith the number of members in the family. For a family of $4$ person, the total monthly expenses are $Rs. 10,400$ whereas for a family of $7$ persons, the total monthly expenses are $Rs. 15,800$. Find the constant expenses per month and the monthly expenses of each member of a family.
AnswerLet $x$ be the constant expense per month of the family,
and $y$ be the expense per month for a single member of the family,For a family of $4$ people,
the total monthly expense is $Rs. 10,400.$
$x + 4y = 10,400 \dots...(1)$
For a family of $7$ people,
the total monthly expense is $Rs. 15,800.$
$x + 7y = 15,800 \dots...(2)$
Subtracting equation $(1)$ from equation $(2)$,We get :
$x + 7y = 15800$
$- x + 4y = 10400$
$- - - $
$3y = 5400$
$y = 1800$
Substituting $y = 1800$ in equation $(1)$, We get
$x + 4( 1800 ) = 10,400$
$\Rightarrow x = 3200.$
$\therefore $ The constant expense is $Rs. 3,200$ per month and the monthly expense of each member of a family is $Rs.1,800.$
View full question & answer→Question 264 Marks
The class $XI$ students of school wanted to give a farewell party to the outgoing students of class $XII$. They decided to purchase two kinds of sweets, one costing $Rs. 250$ per $kg$ and other costing $Rs. 350$ per $kg$. They estimated that $40 \ kg$ of sweets were needed. If the total budget for the sweets was $Rs. 11,800;$ find how much sweets of each kind were bought ?
AnswerAssume $x \ kg$ of the first kind costing $Rs. 250$ per $kg$ and $y \ kg$ of the second kind costing $Rs. 350$ per $kg$ sweets were bought.
It is estimated that $40 \ kg$ of sweets were needed.
$\Rightarrow x + y = 40\dots....(1)$
The total budget for the sweets was $Rs. 11,800.$
$\Rightarrow 250x + 350y = 11,800 \dots....(2)$
Multiply equation $(1)$ by $250$, We get :
$250x + 250y = 10,000\dots.....(3)$
Subtracting equation $(2)$ from $(3),$
$250x + 250y = 10,000$
$- 250x + 350y = 11,800$
$- - - $
$- 100y = - 1800$
$y = 18$
Substituting $y = 18$ in equation $(1)$, We get
$x + 18 = 40$
$\Rightarrow x = 22$
$\therefore 22 \ kgs$ of the first kind costing $Rs. 250$ per $kg$ and $18 \ kgs$ of the second kind costing $Rs. 350$ per $kg$ sweets were bought.
View full question & answer→Question 274 Marks
$90\%$ acid solution$ (90\%$ pure acid and $10\%$ water$)$ and $97\%$ acid solution are mixed to obtain $21$ litres of $95\%$ acid solution. How manylitresof each solution are mixed.
AnswerLet the quantity of $90\%$ acid solution be $x$ litres and The quantity of $97\%$ acid solution be ylitres
According to the question,
$x + y = 21 \dots...(1)$
and $90\%$ of $x + 97\%$ of $y = 95\%$ of $21$
$90x + 97y = 1995 \dots...(2)$
Multiplying equation no. $(1)$ by $90$, we get,
$90x + 90y = 1890 \dots....(3)$
Subtracting equation $(2)$ from $(3)$
$90x + 90y = 1890$
$- 90x + 97y = 1995$
$- - -$
$- 7y = - 105$
$y = 15$
From $(1)$
$x + 15 = 21$
$x = 6$
Hence, $90\%$ acid solution is $6$ litresand $97\%$ acid solution is $15$ litres.
View full question & answer→Question 284 Marks
The sum of digit of a two digit number is $11$. If the digit at ten's place is increased by $5$ and the digit at unit place is decreased by $5$, the digits of the number are found to be reversed. Find the original number.
AnswerLet $x$ be the number at the ten's place.
and y be the number at the unit's place.
So, the number is $10x + y$.
The sum of digit of a two digit number is $11.$
$\Rightarrow x + y = 11 \dots...(1)$
lf the digit at ten's place is ineased by $5$ and the digit at unit place is decreased by $5,$
the digits of the number are found to be reversed.
$\Rightarrow 10( x + 5 ) + ( y - 5 ) = 10y + x$
$\Rightarrow 9x - 9y = -45$
$\Rightarrow x - y = -5 \dots...(2)$
Subtracting equation $(1)$ from equation $(2)$, we get :
$x - y = - 5$
$- x + y = 11$
$- - -$
$- 2y = - 16$
$\Rightarrow y = 8$
Substituting $y = 8$ in equation $(1),$ we get
$x + 8 = 11$
$\Rightarrow x = 3$
$\therefore $ The number is $10x + y = 10(3) + 8 = 38.$
View full question & answer→Question 294 Marks
From Delhi station, if we buy $2$ tickets for station $A$ and $3$ tickets for station $B,$ the total cost is $Rs. 77$. But if we buy $3$ tickets for station $A$ and $5$ tickets for station $B$, the total cost is $Rs. 124$. What are the fares from Delhi to station $A$ and to station $B$ ?
AnswerLet, the fare of ticket for station $A$ be $Rs.$
$x$ and the fare of ticket for station $B$ be $Rs. y$
According, to the question
$2x + 3y = 77 \dots...(1)$ and
$3x + 5y = 124\dots ...(2)$
Multiplying equation no. $(1)$ by $3$ and $(2)$ by $2.$
$6x + 9y = 231 \dots...(3)$
$6x + 10y = 248 \dots....(4)$
Subtracting equation $(4)$ from $(3)$
$6x + 9y = 231$
$- 6x + 10y = 248$
$- - - $
$- y = - 17$
$y = 17$
From $(1)$
$2x + 3 (17) = 77$
$2x = 77 - 51$
$2x = 26$
$x = 13$
Thus, fare for station $A = Rs. 13$ and, fare for station $B = Rs. 17.$
View full question & answer→Question 304 Marks
Rohit says to Ajay, $“\ $Give mehundred, I shall then become twice as rich as you.$”\ $ Ajay replies, $“\ $if you give me ten, I shall be six times as rich as you.$”\ $ How much does each have originally ?
AnswerLet Rohit has $Rs. x$ and Ajay has $Rs. y$
When Ajay gives $Rs. 100$ to Rohit
$x + 100 = 2(y - 100)$
$x - 2y = -300 \dots...(1)$
WhenRohitgives $Rs. 10$ to Ajay
$6(x-10) = y + 10$
$6x - y = 70 \dots...(2)$
Multiplying equation no. $(2)$ By $2.$
$12x - 2y = 140 \dots...(3)$
Subtracting equation $(1)$ and $(3)$
$12x - 2y = 140$
$- x - 2y = - 300$
$- + + $
$11x = 440$
$x = 40$
From $(1)$
$40 - 2y = - 300$
$\Rightarrow - 2y = - 340$
$\Rightarrow y = 170$
Thus, Rohit has $Rs. 40$ and Ajay has $Rs. 170.$
View full question & answer→Question 314 Marks
It takes $12$ hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for $4$ hours and the pipe of smaller diameter is used for $9$ hours, only half of the pool is filled. How long would each pipe take to fill the swimming pool ?
AnswerLet the pipe with larger diameter and smaller diameter be pipes $A$ and $B$ respectively.
Also, let pipe $A$ work at a rate of $x\ hours/ unit$ and pipe $B$ work at a rate of $Y\ hours / unit.$
According to the question,
$x+y=\frac{1}{12}$
$\Rightarrow 12 x+12 y=1\ldots .(1)$
and
$4 x+9 y=\frac{1}{2}$
$\Rightarrow 8 x+18 y=1\ldots(2)$
Multiply $(1)$ by $2$ and $(2)$ by $3 $, We get
$24 x+24 y=2\ldots(3)$
$24 x+54 y=3\ldots(4)$
Subtracting equation $(4)$ from $(3),$
$24 x+24 y=2$
$- 24 x+54 y$
$-30 y=-1$
$y=\frac{1}{30}$
Putting $y=\frac{1}{30}$ in equation $(1)$
$12 x+12 y=1$
$ \therefore 12 x+12 x \frac{1}{30}=1$
$ \therefore 12 x+\frac{2}{5}=1$
$ \therefore 12 x=1-\frac{2}{5}$
$ \therefore x=\frac{3}{5} \times \frac{1}{12}$
$ \therefore x=\frac{1}{20}$
Hence, the pipe with larger diameter will take $20$ hours to fill the swimming pool and the pipe with smaller diameter will take $30$ hours to fill the swimming pool.
View full question & answer→Question 324 Marks
The area of a rectangle gets reduced by $9$ square units, if its length is reduced by $5$ units and breadth is increased by $3$ units. However, if the length of this rectangle increases by $3$ units and the breadth by $2$ units, the area increases by $67$ square units. Find the dimensions of the rectangle.
AnswerLet the length of the rectangle be $x$ units and the breadth of the rectangle be $y$ units.
We know that, area of a rectangle $=$ length $x$ breadth $=x y$ According to the question,
$ x y-9=(x-5)(y+3)$
$ \Rightarrow x y-9=x y+3 x-5 y-15$
$\Rightarrow 3 x-5 y=6\ldots...(1)$
$ x y+67=(x+3)(y+2)$
$ \Rightarrow x y+67=x y+2 x+3 y+6$
$\Rightarrow 2 x+3 y=61\ldots(2)$
Multiply $(1)$ by $2$ and $(2)$ by $3$ , we get
$6 x-10 y=12 \dots...(3)$
$6 x+9 y=183\ldots...(4)$
Subtracting equation $(4)$ from $(3),$
$6 x-10 y =12$
$- 6 x+9 y =183$
$- - -$
$-19 y =-171$
$y =9$
Putting $y=9$ in equation $(1)$
$3 x-5 y=6$
$ 3 x-5(9)=6$
$ 3 x=6+45$
$ x=\frac{51}{3}$
$ x=17$
Hence, the length of the rectangle is $17$ units and the breadth of the rectangle is $9$ units.
View full question & answer→Question 334 Marks
Two articles $A$ and $B$ are sold for $Rs. 1,167$ making $5\%$ profit on $A$ and$ 7\%$ profiton $A$ and $7\%$ profit on $B$. IF the two articles are sold for $Rs. 1,165,$ a profit of $7\%$ is made on $A$ and a profit of $5\%$ is made on $B$. Find the cost prices of each article.
AnswerLet the cost price of article $A=Rs.x$
and the cost price of articles $B=Rs.y$
According to the question,
$(x+5 \% \text { of } x)+(y+7 \% \text { of } y)=1167$
$ {\left[x+\frac{5}{100} x\right]+\left[y+\frac{7}{100} y\right]=1167}$
$ \frac{21 x}{20}+\frac{107 y}{100}=1167$
$105 x+107 y=116700\ldots..(1)$
and,
$[107 x] / 100+[105 y] / 100=1165$
$107 x+105 y=116,500\ldots..(2)$
Adding$ (1)$ and $(2)$
$105 x+107 y=116700$
$+ 107 x+105 y=116500$
$ 212 x+212 y=233200$
Dividing by $212 ,$
$x+y=1100\ldots..(3)$
subtracting $(2)$ from $(1)$
$105 x+107 y=116700$
$- 107 x+105 y=116500$
$-2 x+2 y=200$
Dividing by $2 ,$
$-x+y=100\ldots...(4)$
Adding equation $(4)$ and $(3)$
$-x+y =100$
$+ x+y =1100$
$2 y =1200$
$y =600$
from$ (3)$
$x+600=1100$
$ x=500$
Thus, cost price of article $A$ is $Rs. 500$ and that of article $B$ is $Rs. 600.$
View full question & answer→Question 344 Marks
The annual incomes of $A$ and $B$ are in the ratio $3 :4$ and their annual expenditure are in the ratio $5 : 7.$ If each $Rs. 5000;$ find their annual incomes.
AnswerLet $A\ 's$ annual in come $= Rs. x$
and $B\ 's$ annual income $= Rs. y$
According to the question,
$\frac{x}{y}=\frac{3}{4}$
$4 x-3 y=0\ldots...(1)$
and,
$\frac{x-5000}{y-5000}=\frac{5}{7}$
$7 x-5 y=10000\ldots...(2)$
Multiplying equation no. $(1)$ by $7$ and $(2)$ by $4.$
$28 x-21 y=0 \dots....(3)$
$28 x-20 y=40,000 \dots.....(4)$
Subtracting equation $(4)$ from $(3)$
$28 \mathrm{x}-21 \mathrm{y}=0$
$ -28 x-20 y=40,000$
$y=-40,000$
$ y=40,000$
From $(1)$
$4 x-3(40,000)=0$
$ x=30,000$
From $(1)$
$4 x-3(40,000)=0$
$ x=30,000$
Thus, $A\ 's$ income in $Rs. 30,000$ and $B\ 's$ income is $Rs. 40,000.$
View full question & answer→Question 354 Marks
Four years ago, a mother was four times as old as her daughter. Six years later, the mother will be two and a half times as old as her daughter at that time. Find the present ages of the mother and her daughter.
AnswerLet the present age of the mother be $x$ years.
and the present age of the daughter be $y$ year.
$4$ yrs ago age$=(x-4)$
mother's ag$e=4(y-4)=4 y-16$
According to the equation,
$x-4=4(y-4)$
$ \Rightarrow x-4=4 y-16$
$\Rightarrow x-4 y=-12 \dots...(1)$
And,
$6$ yrs later daughter's age $=x+6$
mother's age $=2 \frac{1}{2}(y+6)$
$x+6=2 \frac{1}{2}(y+6)$
$\Rightarrow \mathrm{x}+6=\frac{5}{2} \mathrm{y}+15$
$\Rightarrow \mathrm{x}-\frac{5}{2} \mathrm{y}=9$
Solving $(1)$ and $(2)$, We get $y=14$ and $x=44$
Hence, the present age of the mother is $44$ years and the present age of the daughter is $14$ years.
View full question & answer→Question 364 Marks
$A$ is $20$ years older than $B. 5$ years ago, $A$ was $3$ times as old as $B$. Find their present ages.
AnswerLet $A’s$ presentage be $x$ years
and $B’s$ present age be $y$ years
According to the question
$x = y + 20$
$x - y = 20\dots ...(1)$
Five years ago,
$x - 5 = 3(y - 5)$
$x - 5 = 3y - 15$
$x - 3y = - 10\dots ...(2)$
Subtracting $(1)$ and $(2),$
$x - 3y = - 10$
$- x - y = 20 $
$- + - $
$- 2y = - 30$
$y = 15$
$From (1)$
$x = 15 + 20$
$x = 35$
Thus, present ages of $A$ and $B$ are $35$ years and $15$ years.
View full question & answer→Question 374 Marks
Five years ago, $A's$ age was four times the age of $B$. Five years hence, $A’s$ age will be twice the age of $B$. Find their preset ages.
AnswerLet present age of $A = x$ years
And present age of $B = y$ years
According to the question,
Five years ago,
$x - 5 = 4(y - 5)$
$x - 4y = -15 \dots...(1$)
Five years later,
$x + 5 = 2(y + 5)$
$x + 5 = 2y + 10$
$x - 2y = 5 \dots....(2)$
Now subtracting $(1)$ from $(2)$
$x - 2y = 5$
$- x - 4y = - 15$
$- + + $
$2y = 20$
$y = 10$
From$ (1)$
$x - 4 (10) = -15$
$x = 25$
Present ages of $A$ and $B$ are $25$ years and $10$ years respectively.
View full question & answer→Question 384 Marks
Pooja and Ritu can do a piece of work in $17 \frac{1}{7}$ days. If one day work of Pooja be three fourth of one day work of Ritu' find in how many days each will do the work alone.
AnswerLet Pooja's $1$ day work $=\frac{1}{x}$
and Ritu's $1$ day work $=\frac{1}{y}$
According the question,
$\frac{1}{x}+\frac{1}{y}=\frac{7}{120}\ldots .(1)$
and
$\frac{1}{x}=\frac{3}{4} \cdot \frac{1}{y}$
$y=\frac{3}{4} x\ldots .(2)$
Using the value of $y$ from $(2)$ in $(1)$
$\frac{1}{x}+\frac{4}{3 x}=\frac{7}{120}$
$ \frac{1}{x}\left(\frac{7}{3}\right)=\frac{7}{120}$
$ x=40$
From $(2)$
$y=\frac{3}{4}(40)$
$ y=30$
Pooja will complete the work in $40$ days and Ritu will complete the work in $30$ days.
View full question & answer→Question 394 Marks
If the numerator of a fraction is increased by $2$ and denominator is decreased by $1 $, it becomes $\frac{2}{3}$. If the numerator is increased by $1$ and denominator is increased by $2 $, it becomes $\frac{1}{3}$. Find the fraction.
AnswerLet the numerator and denominator a fraction be $x$ and $y$ respectively.
According to the question,
$\frac{x+2}{y-1}=\frac{2}{3}$
$ 3 x+6=2 y-2$
$3 x-2 y=-8 \dots........(1)$
And,
$\frac{x+1}{y+2}=\frac{1}{3}$
$ 3 x+3=y+2$
$3 x-y=-1 \dots.......(2)$
Now,
Subtracting equation $(1)$ from $(2)$
$3 x-y=-1$
$- 3 x-2 y=-8$
$- + +$
$y=7$
From $(1),$
$3 x-2(7)=-8$
$ 3 x=-8+14$
$ x=2$
Required fraction $=\frac{2}{7}$
View full question & answer→Question 404 Marks
Two numbers are in the ratio $4:5$. If $30$ is subtracted from each of the numbers, the ratio becomes $1:2.$ Find the numbers.
AnswerLet the common multiple between the numbers be $x .$
So, the numbers are $x / y$
$\frac{x}{y}=\frac{4}{5}$
$\frac{x-30}{y-30}=\frac{1}{2}$
$2(x-30)=y-30$
$2 x-60=y-30$
$2 x-y=30$
From eq. $(i)\ \&\ (ii)$
$2\left(\frac{4}{5 y}\right)-y=30$
$\frac{8 y}{5}-y=30$
$\frac{8 y-5 y}{5}=30$
$3 y=150$
$y=\frac{150}{3}$
$y=50$
from eq. $(ii) $
$x=\frac{4}{5}(50)$
$x=4 \times 10$
$x=40$
View full question & answer→Question 414 Marks
The difference between two positive numbers $x$ and $y(x>y$ ) is $4$ and the difference between their reciprocals is $\frac{4}{21}$. Find the numbers.
AnswerTwo numbers are $x$ and $y$ respectively such that $x>y$.
Then,
$x-y=4\ldots...(i)$
$\Rightarrow x=4+y$
And,
$\frac{1}{y}-\frac{1}{x}=\frac{4}{21}$
$ \Rightarrow \frac{x-y}{x y}=\frac{4}{21}$
$\Rightarrow \frac{4}{x y}=\frac{4}{21} \dots......[$From$(1)]$
$\Rightarrow x y=21$
$ \Rightarrow(4+y) y=21$
$ \Rightarrow 4 y+y 2=21$
$ \Rightarrow y 2+4 y-21=0$
$ \Rightarrow y 2+7 y-3 y-21=0$
$ \Rightarrow y(y+7)-3(y+7)=0$
$ \Rightarrow(y-3)(y+7)=0$
$ \Rightarrow y=3$ and $y=-7$
We reject $y=-7$ since $y$ is positive.
$\Rightarrow y=3$
$ \Rightarrow x=4+y=4+3=7$
Thus, the two numbers are $7$ and $3$ respectively.
View full question & answer→Question 424 Marks
The sum of two numbers is $8$ and the sum of their reciprocals is $\frac{8}{15}$. Find the numbers.
AnswerLet the two numbers be $x$ and $y$ respectively.
Then,
$x+y=8\ldots (i)$
$\Rightarrow x=8-y$
And,
$\frac{1}{x}+\frac{1}{y}=\frac{8}{15}$
$ \Rightarrow \frac{y+x}{x y}=\frac{8}{15}$
$\Rightarrow \frac{8}{x y}=\frac{8}{15} \dots.....[$From$(1) ]$
$\Rightarrow x y=\frac{8 \times 15}{8}$
$\Rightarrow x y=15 \dots...(iii)$
$\Rightarrow(8-y) y=15$
$ \Rightarrow 8 y-y^2=15$
$ \Rightarrow y^2-8 y+15=0$
$ \Rightarrow y^2-3 y-5 y+15=0$
$ \Rightarrow y(y-3)-5(y-3)=0$
$ \Rightarrow(y-3)(y-5)=0$
$ \Rightarrow y=3 $ or $y=5$
$ \therefore x=8-y$
$ x=8-3=5$
$ x=8-5=3$
$ \Rightarrow x=5$ or $x=3$
Thus, the two numbers are $3$ and $5$ respectively.
View full question & answer→Question 434 Marks
The sum of two positive numbers $x$ and $y (x > y)$ is $50$ and the difference of their squares is $720$. Find the numbers.
AnswerTwo numbers are $x$ and $y$ such that $x>y$.
$x+y=50 \dots...(i)$
$ x^2-y^2=720$
$ \Rightarrow(x+y)(x-y)=720$
$ \Rightarrow 50(x-y)=720 \ldots...[$Using equation $(i)]$
$\Rightarrow \mathrm{x}-\mathrm{y}=\frac{720}{50}$
$\Rightarrow x-y=14.4 \dots....(ii)$
Adding $(i)$ and $(ii)$, we get
$\begin{gathered}x+y=50+x-y=14.4, 2 x=64.4\end{gathered}$
$x=\frac{64.4}{2}$
$ \Rightarrow x=32.2$
Substituting the value of $x$ in $(i),$ we have
$x+y=50$
$ 32.2+y=50$
$ \Rightarrow y=50-32.2$
$ \Rightarrow y=17.8$
Thus, the two numbers are $17.8$ and $32.2$ respectively.
View full question & answer→Question 444 Marks
When the greater of the two numbers increased by $1$ divides the sum of the numbers, the result is $\frac{3}{2}$. When the difference of these numbers is divided by the smaller, the result $\frac{1}{2}$. Find the numbers.
AnswerLet the two numbers be $a$ and $b$ respectively such that $b>a$. According to given condition,
$\frac{a+b}{b+1}=\frac{3}{2}$
$\Rightarrow 2 a+2 b=3 b+3$
$\Rightarrow 2 a-b=3\ldots . . .(1)$
Also,
$\frac{b-a}{a}=\frac{1}{2}$
$ \Rightarrow 2 \mathrm{~b}-2 \mathrm{a}=\mathrm{a}$
$\Rightarrow 2 b-3 a=0\ldots . . . .(2)$
Multiplying $(1)$ by $2$, We get
$4 a-2 b=6 \dots.......(3)$
Adding $(2)$ and $(3),$ We get
$\begin{gathered}4 a-2 b=6\pm3 a+2 b=0, a=6\end{gathered}$
Substituting $a=6$ in equation $(1)$, We get
$2(6)-b=3$
$ \Rightarrow 12-b=3$
$ \Rightarrow b=9$
Thus, the two numbers are $6$ and $9$ respectively.
View full question & answer→Question 454 Marks
Two numbers are in the ratio $4 :7$. If thrice the larger be added to twice the smaller, the sum is $59$. Find the numbers.
AnswerLet the smaller number be $x$ and the larger number be $y$.
According to the question,
$ \frac{x}{y}=\frac{4}{7}$
$ 7 x=4 y$
$7 x-4 y=0\ldots . . .(1)$
and, $3 y+2 x=59\ldots . . .(2)$
Multiplying equation no. $(1)$ by $3$ and $(2)$ by $4 .$
$21 x-12 y=0\ldots . . .(3)$
$8 x+12 y=236 \dots........(4)$
Adding equation $(3)$ and $(4),$
$\begin{gathered}21 x-12 y=0,+ 8 x+12 y=236, 29 x=236x=\frac{236}{29}\end{gathered}$
From$ (1)$
$7\left(\frac{236}{29}\right)=4 y$
$y=7\left(\frac{59}{29}\right)$
$ y=\left(\frac{413}{29}\right)$
Hence, the numbers are $\frac{236}{29}$ and $\frac{413}{29}$.
View full question & answer→Question 464 Marks
The ratio of two numbers is $\frac{2}{3}$. If $2$ is subtracted from the first and $8$ from the second, the ratio becomes the reciprocal of the oriainal ratio. Find the numbers.
AnswerLet the two numbers be $x$ and $y$.
According to the question,
$\frac{x}{y}=\frac{2}{3}$
$3 x-2 y=0\ldots . .(1)$
Also, $\frac{x-2}{y-8}=\frac{3}{2}$
$2 x-3 y=-20\ldots . . .(2)$
Multiplying equations no. $(1)$ by $2$ and $(2)$ by $3$
$6 x-4 y=0\ldots . .(3)$
$6 x-9 y=-60\ldots . . .(4)$
Subtracting equation $(4)$ from $(3)$
$6 x-4 y =0$
$- 6 x-9 y =-60$
$-++ $
$5 y =60$
$y =12$
From $(1),$ we get
$3 x-2(12)=0$
$ x=\frac{24}{3}$
$ x=8$
Thus, the numbers are $8$ and $12 .$
View full question & answer→Question 474 Marks
Four times a certain two digit number is seven times the number obtained on interchanging its digits. If the difference between the digits is $4;$ find the number.
AnswerLet $x$ be the number at the ten's place.
and $y$ be the number at the unit's place.
So, the number is $10x + y$.
Four times a certain two-digit number is seven times
the number obtained on interchanging its digits.
$\Rightarrow 4( 10x + y ) = 7( 10y + x )$
$\Rightarrow 40x + 4y = 70y + 7x$
$\Rightarrow 33x - 66y = 0$
$\Rightarrow x - 2y = 0 \dots....(1)$
If the difference between the digits is $4,$ then
$\Rightarrow x - y = 4 \dots...(2)$
Subtracting equation $(1)$ from equation $(2)$, we get :
$x - y = 4$
$- x - 2y = 0$
$- + - $
$y = 4$
Subtracting $y = 4$ in equation $(1),$ We get
$x - 2(4) = 0$
$\Rightarrow x = 8$
$\therefore $ The number is $10x + y = 10(8) + 4 = 84.$
View full question & answer→Question 484 Marks
A two$-$digit number is such that the ten’s digit exceeds twice the unit’s digit by $2$ and the number obtained by inter$-$changing the digits is $5$ more than thethesum of the digits. Find the two digit number.
AnswerLet the digit a unit’s place be $x$ and the digit at ten’s place be $y.$
Required no. $= 10y + x.$
According to the question,
$y - 2x = 2$
$-2x + y = 2 \dots...(1)$
and,
$(10x + y) -3 (y + x) = 5$
$7x - 2y = 5 \dots...(2)$
Multiplying equation no.$(1)$ by $2.$
$- 4x + 2y = 4 \dots...(3)$
Now adding$ (2)$ and $(3),$
$- 4x + 2y = 4$
$+ 7x - 2y = 5$
$3x = 9$
$x = 3$
From $(1),$ we get
$-2(3) + y = 2$
$y = 8$
Required number is $10(8) + 3 = 83.$
View full question & answer→Question 494 Marks
The ten’s digit of a two digit number is three times the unit digit. The sum of the number and the unit digit is $32$. Find the number.
AnswerLet the digit at unit’s place be $x$ and the digit at ten’s place be $y.$
Required no. $= 10y + x$
According to the question,
$y = 3x$
$\Rightarrow 3x - y = 0 \dots...(1)$
and,
$10y + x + x = 32$
$10y + 2x = 32\dots ...(2)$
Multiplying equation no. $(1)$ by $10.$
$30x - 10y = 0 \dots....(3)$
Now,
Adding equation $(3)$ and $(2)$
$30x - 10y = 0$
$+ 2x + 10y = 32$
$32x = 32$
$x = 1$
From $(1),$ we get
$y = 3(1) = 3$
Required no is $=10(3) + 1 = 31.$
View full question & answer→Question 504 Marks
The sum of the digits of a two digit number is $7.$ If the digits are reversed, the new number decreased by $2$, equals twice the original number. Find the number.
AnswerLet the digit at unit’s place be $x$ and the digit at ten’s place be $y.$
Required no. $= 10y + x$
If the digit’s are reversed
Reversed no. $= 10x + y$
According to the question,
$x + y = 7\dots ...(1)$
and,
$10x + y - 2 = 2(10y + x).$
$8x - 19y = 2\dots ...(2)$
Multiplying equation no. $(1)$ by $19.$
$19x + 19y = 133\dots ...(3)$
Now adding equation $(2)$ and $(3)$
$19x + 19y = 133$
$+ 8x - 19y = 2 $
$27x = 135$
$x = 5$
From $(1)$
$5 + y = 7$
$y = 2$
Required number is $= 10(2) + 5 = 25.$
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