Question 12 Marks
A sum of $Rs. 500$ is in the form of denominations of $Rs. 5$ and $Rs. 10$. If the total number of notes is $90.$ Find the number of notes of each type.
AnswerTotal number of notes $= 90$
Let number of notes of $Rs. 5 = x$
Then number of notes of $Rs. 10 = 90 - x$
Then $x \times 5 + (90 - x) \times 10 = 500$
$\Rightarrow 5x + 900 - 10x = 500$
$\Rightarrow -5x = 500 - 900 = -400$
$x = 80$
Number of $5$ rupees notes $= 80$ And ten rupees notes $= 90 - 80 = 10$
View full question & answer→Question 22 Marks
Solve the following equations. Check your result in case.
$\frac{\text{x}}{2}+\frac{\text{x}}{4}=\frac{1}{8}$
Answer $\frac{\text{x}}{2}+\frac{\text{x}}{4}=\frac{1}{8}$
$\frac{2\text{x}+\text{x}}{4}=\frac{1}{8}$
$\Rightarrow\frac{3\text{x}}{4}=\frac{1}{8}$
$\Rightarrow\text{x}=\frac{1}{8}\times\frac{4}{3}=\frac{1}{6}$
$\therefore\text{x}=\frac{1}{6}$
Check:
$\text{L.H.S.}=\frac{\text{x}}{2}+\frac{\text{x}}{4}$
$=\frac{1}{6\times2}+\frac{1}{6\times4}=\frac{1}{12}+\frac{1}{24}$
$=\frac{2+1}{24}=\frac{3}{24}=\frac{1}{8}=\text{R.H.S.}$
Hence $\text{x}=\frac{1}{6}$
View full question & answer→Question 32 Marks
Solve the following equations. Check your result in case. $2(x - 2) + 3(4x - 1) = 0$
Answer$2(x - 2) + 3(4x - 1) = 0$
$\Rightarrow 2x - 4 + 12x - 3 = 0$
$\Rightarrow 2x + 12x = 4 + 3$ (By transposing) $\Rightarrow 14x = 7$
$\Rightarrow\text{x}=\frac{7}{14}=\frac{1}{2}$ Check: $\text{L.H.S}=2(\text{x}-2)+3(4\text{x}-1)$
$=2\Big(\frac{1}{2}-2\Big)+3\Big(4\times\frac{1}{2}-1\Big)$
$=2\times\Big(\frac{1-4}{2}\Big)+3(2-1)$
$=2\text{x}\Big(\frac{-3}{2}\Big)+3(1)$
$=-3+3=0=\text{R.H.S.}$
Hence $\text{x}=\frac{1}{2}$
View full question & answer→Question 42 Marks
The length of a rectangular field is twice its breadth. If the perimeter of the field is $150$ metres, find its length and breadth.
AnswerPerimeter of field $= 150m$
Length $+$ Breadth $=\frac{150}{2}=75\text{m} [$Perimeter $= 2(l + b)]$
Let length $= x$ Then breadth $= 75 - x$
Then $x = 2(75 - x)$
$\Rightarrow x = 150 - 2x$
$\Rightarrow x + 2x = 150$
$\Rightarrow 3x = 150$
$\Rightarrow\text{x}=\frac{150}{3}=50$
Length $= 50m$ And breadth $= 75 - 50 = 25m$
View full question & answer→Question 52 Marks
Solve the following equations. Check your result in case.$3 + 2x = 1 - x$
Answer$3 + 2x = 1 – x$
$\Rightarrow 2x + x = 1 – 3$ (By transposing)
$\Rightarrow 3x = -2$
$\Rightarrow\text{x}=\frac{-2}{3}$
Check: $\text{L.H.S.}=3+2\text{x}=3+2\Big(\frac{-2}{3}\Big)$
$=\frac{3}{1}-\frac{4}{3}=\frac{9-4}{3}=\frac{2}{3}$
$\text{R.H.S.}=1-\text{x}=1-\big(\frac{-2}{3}\Big)=1+\frac{2}{3}$
$=\frac{3+2}{3}=\frac{5}{3}$
$\therefore\text{L.H.S = R.H.S}$ Hence $\text{x}=\frac{-2}{3}$
View full question & answer→Question 62 Marks
Solve the following equations. Check your result in case.
$\frac{1}{2}\text{x}-3=5+\frac{1}{3}\text{x}$
Answer $\frac{1}{2}\text{x}-3=5+\frac{1}{3}\text{x}$
$\Rightarrow\frac{1}{2}\text{x}-\frac{1}{3}\text{x}=5+3$ (By transposing)
$\Rightarrow\frac{3\text{x}-2\text{x}}{6}=8$
$\Rightarrow\frac{\text{x}}{6}=8$
$\Rightarrow\text{x}=8\times6$
$\Rightarrow\text{x}=48$
$\therefore\text{x}=48$
Check:
$\text{L.H.S}=\frac{1}{2}\text{x}-3=\frac{1}{2}\times48-3$
$=24-3=21$
$\text{R.H.S.}=5+\frac{1}{3}\text{x}$
$=5+\frac{1}{3}\times48=5+16=21$
$\text{L.H.S.}=\text{R.H.S.}$
Hence $\text{x}=48$
View full question & answer→Question 72 Marks
Two supplementary angles differ by $44^\circ .$ Find the angles.
AnswerSum of two supplementary angles $= 180^\circ $
Let first angle $= x$
Then second angle $= 180^\circ - x$
$x - (180^\circ - x) = 44^\circ $
$\Rightarrow x - 180^\circ + x = 44^\circ $
$\Rightarrow 2x = 44^\circ + 180^\circ = 224^\circ $
$\Rightarrow 2x = 224^\circ $
$\Rightarrow x = 112^\circ $
First angle $= 112^\circ $
And second angle $= 180^\circ - 112^\circ = 68^\circ $
Hence angles are $68^\circ , 112^\circ $
View full question & answer→Question 82 Marks
Two complementary angles differ by $8^\circ .$ Find the angles.
AnswerSum of two complementary angles $= 90^\circ $
Let first angle $= x$
Then second $= 90^\circ - x $
$x - (90^\circ - x) = 8 $
$\Rightarrow x - 90^\circ + x = 8 $
$\Rightarrow 2x = 8 + 90^\circ $
$\Rightarrow 2x = 98^\circ $
$ \Rightarrow x = 49^\circ $
First angle $= 49^\circ $ And second angle $= 90^\circ - 49^\circ = 41^\circ $
Hence angles are $41^\circ , 49^\circ $
View full question & answer→Question 92 Marks
Solve: $\frac{3\text{x}}{10}+\frac{2\text{x}}{5}=\frac{7\text{x}}{25}+\frac{29}{25}.$
AnswerWe have: $\frac{3\text{x}}{10}+\frac{2\text{x}}{5}=\frac{7\text{x}}{25}+\frac{29}{25}$ $\Rightarrow\frac{3\text{x}+4\text{x}}{10}=\frac{7\text{x}+29}{25}$ $\Rightarrow\frac{7\text{x}}{10}=\frac{7\text{x}+29}{25}$ $\Rightarrow175\text{x}=70\text{x}+290$ $\Rightarrow105\text{x}=290$ $\Rightarrow\text{x}=\frac{290}{105}$ $\Rightarrow\text{x}=\frac{58}{21}$
View full question & answer→Question 102 Marks
A labour is engaged for $20$ days on the condition that he will receive $Rs. 120$ for each day he works and will be fined $Rs. 10$ for each day he is absent. If he receives $Rs. 1880$ in all, for how many days did he remain absent$?$
AnswerLet $x$ be the number of days of his sbsence.
$\therefore$ Number of days of his presence $= (20 - x)$
$Now, (20 - x) 120 - 10x = 1880$
$\Rightarrow 2400 - 120x - 10x = 1880$
$\Rightarrow 2400 - 1880 = 130x$
$\Rightarrow 130x = 520$
$\Rightarrow x = 4$
$\therefore$ Number of days of his absence $= 4$
View full question & answer→Question 112 Marks
After $12$ years Manoj will be $3$ times as old as he was $4$ years ago. Find his present age.
AnswerLet age of Manoj $4$ years ago $= x$
Then his present age $= x + 4$
After $12$ years his age will be $= x + 4 + 12$
$= x + 16$
$x + 16 = 3(x)$
$x + 16 = 3x$
$\Rightarrow 16 = 3x - x$
$\Rightarrow 2x = 16$
$x = 8$ His present age $= 8 + 4 = 12$ years
View full question & answer→Question 122 Marks
Solve the following equations. Check your result in case. $2.4(3 - x) - 0.6(2x - 3) = 0$
Answer$2.4(3 - x) - 0.6(2x - 3) = 0$
$\Rightarrow 7.2 - 2.4x - 1.2x + 1.8 = 0$
$\Rightarrow -2.4x - 1.2x = -(7.2 + 1.8)$
$L.H.S. = 2.4(3 - x) - 0.6(2x - 3)$
$\Rightarrow 2.4(3 - 2.5) - 0.6(2 \times 2.5 - 3)$
$\Rightarrow 2.4(0.5) - 0.6(5 - 3)$
$\Rightarrow 1.2 - 0.6 \times 2 = 1.2 - 1.2$
$R.H.S. = 0$ Hence $x = 2.5$
View full question & answer→Question 132 Marks
Find three consecutive positive even integers whose sum is $90.$
AnswerLet first positive even number $= 2x$ Second number $= 2x + 2$
Third number $= 2x + 4 2x + 2x +2 + 2x + 4 = 90$
$\Rightarrow 6x + 6 = 90$
$\Rightarrow 6x = 90 - 6 = 84$
$x = 14$
First even number $= 2x = 2 \times 14 = 28$
Second number $= 2x + 2 = 2 \times 14 + 2 = 28 + 2 = 30$
Third number $= 30 + 2 = 32$
Required numbers are $28, 30, 32$
View full question & answer→Question 142 Marks
Solve the following equations. Check your result in case. $13(y - 4) - 3(y - 9) - 5(y + 4) = 0$
Answer$13(y - 4) - 3(y - 9) - 5(y + 4) = 0$
$\Rightarrow 13y - 52 - 3y + 27 - 5y - 20 = 0$
$\Rightarrow 13y - 3y - 5y - 52 + 27 - 20 = 0$
$\Rightarrow 13y - 8y - 72 + 27 = 0$
$\Rightarrow 5y - 45 = 0$
$\Rightarrow 5y = 45$ (By transposing)
$\Rightarrow y = 9$
Check: $L.H.S. = 13(y - 4) - 3(y - 9) - 5(y + 4)$
$= 13(9 - 4) - 3(9 - 9) - 5(9 + 4)$
$= 13 \times 5 - 3 \times 0 - 5 \times 13$
$= 65 - 0 - 65 = 0$
$= R.H.S.$ Hence $y = 9$
View full question & answer→Question 152 Marks
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.
AnswerPerimeter of an isosceles triangle $= 55m$
Let the third side of an isosceles triangle $= x$
Then each equal side $= (2x - 5)m$
According to the condition, $x + 2 (2x - 5) = 55$
$\Rightarrow x + 4x - 10 = 55$
$\Rightarrow 5x = 55 + 10$
$\Rightarrow 5x = 65$
$\Rightarrow x = 13$ And $2x – 5 = 2 \times 13 - 5 = 21m$ Sides will be $13m, 21m, 21m$
View full question & answer→Question 162 Marks
Solve the following equations. Check your result in case. $5(2x -3) - 3(3x - 7) = 5$
Answer$5(2x - 3) - 3(3x - 7) = 5$
$\Rightarrow 10x - 15 - 9x + 21 = 5$
$\Rightarrow 10x - 9x - 15 + 21 = 5$
$\Rightarrow 10x - 9x = 5 + 15 - 21$ (By transposing)
$\Rightarrow x = 20 - 21 = -1$
$\Rightarrow x = -1$ Check:
$L.H.S. = 5(2x - 3) - 3(3x - 7) = 5[2 x (-1) -3]$
$-3[3 (-1) -7] = 5[-2 - 3] - 3[-3 - 7] = 5x(-5) -3x(-10) = -25 + 30 = 5 = R.H.S.$ Hence $x = -1$
View full question & answer→Question 172 Marks
Four added to twice a number yields $\frac{26}{5}$. Find the fraction.
AnswerLet the required fraction $= x$
then $2\text{x}+4=\frac{26}{5}$
$\Rightarrow2\text{x}=\frac{26}{5}-4$
$\Rightarrow2\text{x}=\frac{26-20}{5}=\frac{6}{5}$
$\therefore\text{x}=\frac{6}{5}\times\frac{1}{2}=\frac{3}{5}$
$\therefore$ Required fractiom $=\frac{3}{5}$
View full question & answer→Question 182 Marks
Solve the following equations. Check your result in case. $\frac{2\text{m}+5}{3}=3\text{m}-10$
Answer$\frac{2\text{m}+5}{3}=3\text{m}-10$
$\Rightarrow 2m + 5 = 3 (3m - 10)$ (By cross multiplication)
$\Rightarrow 2m + 5 = 9m - 30$
$\Rightarrow 2m - 9m = -30 - 5$
$\Rightarrow -7m = -35$
$\Rightarrow m = 5$
$m = 5$
Check: $\text{L.H.S.}=\frac{2\text{m}+5}{3}=\frac{2\times5+5}{3}$
$=\frac{10+5}{3}=\frac{15}{3}=5$
$R.H.S. = 3m - 10$
$= 3 \times 5 - 10$
$= 15 - 10 = 5$
$L.H.S. = R.H.S.$
Hence $m = 5$
View full question & answer→Question 192 Marks
Solve the following equations. Check your result in case.
$\frac{2\text{x}+5}{3\text{x}+4}=3$
Answer $\frac{2\text{x}+5}{3\text{x}+4}=3$
(By cross multiplication)
$3(3\text{x}+4)=1(2\text{x}+5)$
$9\text{x}+12=2\text{x}+5$
$\Rightarrow9\text{x}-2\text{x}=5-12$
$\Rightarrow7\text{x}=-7$
$\Rightarrow\text{x}=\frac{-7}{7}=-1$
$\text{x}=-1$
Check:
$\text{L.H.S}.=\frac{2\text{x}+5}{3\text{x}+4}=\frac{2(-1)+5}{3(-1)+4}=\frac{-2\ +\ 5}{-3\ +\ 4}$
$=\frac{3}{1}=3=\text{R.H.S.}$
Hence $\text{x}=-1$
View full question & answer→Question 202 Marks
A number is $\frac{2}{5}$ times another number. If their sum is $70,$ find the numbers.
AnswerLet the second number $= x$
then first number $=\frac{2}{5}\text{x}$ their sum $= 70$
$\therefore\text{x}+\frac{2}{5}\text{x}=70$
$\Rightarrow\frac{5\text{x}+2\text{x}}{5}=70$
$\Rightarrow\frac{7\text{x}}{5}=70$
$\Rightarrow\text{x}=\frac{70\times5}{7}=50$
$\therefore$ Second number $= 50$ And first number $=\frac{2}{5}\times50=20$
$\therefore$ Numbers are $20, 50$
View full question & answer→Question 212 Marks
Solve: $0.5\text{x}+\frac{\text{x}}{3}=0.25\text{x}+7$
AnswerWe have: $0.5\text{x}+\frac{\text{x}}{3}=0.25\text{x}+7$ $\Rightarrow\frac{1.5\text{x}+\text{x}}{3}=0.25\text{x}+7$ $\Rightarrow1.5\text{x}+\text{x}=3(0.25\text{x}+7)$ $\Rightarrow2.5\text{x}=0.75\text{x}+21$ $\Rightarrow2.5\text{x}-0.75\text{x}=21$ $\Rightarrow1.75\text{x}=21$ $\Rightarrow\text{x}=\frac{21}{1.75}$ $\Rightarrow\text{x}=12$
View full question & answer→Question 222 Marks
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.
AnswerIn an isosceles triangle
Let each equal base angles $= x$
Then vertex angle $= 2x$
According to the condition,
$x + x + 2x = 180^\circ ($sum of angles of a triangle$)$
$\Rightarrow 4x = 180^\circ $
$\Rightarrow x = 45^\circ $
Then vertex angle $= 2x = 2 \times 45^\circ = 90^\circ $
Angles of the triangle are $45^\circ , 45^\circ $ and $90^\circ $
View full question & answer→Question 232 Marks
Solve the following equations. Check your result in case. $\frac{\text{x}+2}{\text{x}-2}=\frac{7}{3}$
Answer$\frac{\text{x}+2}{\text{x}-2}=\frac{7}{3}$ By cross multiplication, We get $7(\text{x}-2)=3(\text{x}+2)$ $7\text{x}-14=3\text{x}+6$ $\Rightarrow7\text{x}-3\text{x}=6+14$ $\Rightarrow4\text{x}=20$ $\Rightarrow\text{x}=\frac{20}{4}=5$ Check: $\text{L.H.S.}=\frac{\text{x}+2}{\text{x}-2}$ $=\frac{5+2}{5-2}$ $=\frac{7}{3}=\text{R.H.S.}$ Hence $\text{x}=5$
View full question & answer→Question 242 Marks
Find the consecutive positive odd integers whose sum is $76.$
AnswerLet first odd number $= 2x + 1$
Second odd number $= 2x + 3 $
$2x + 1 + 2x + 3 = 76$
$ \Rightarrow 4x + 4 = 76 $
$\Rightarrow 4x = 76 – 4 = 72 $
$x = 18$
First number $= 2x + 1 $
$= 2 \times 18 + 1 $
$= 36 + 1 $
$= 37$
Second number $= 2x + 3 $
$= 2 \times 18 + 3 $
$= 36 + 3 = 39$
Numbers are $37, 39$
View full question & answer→Question 252 Marks
A number when added to its half gives $72.$ Find the number.
AnswerLet the required number $= x$ And half of the number $=\frac{\text{x}}{2}$
Then, $\frac{\text{x}}{2}+\text{x}=72$
$\Rightarrow\frac{\text{x}+2\text{x}}{2}=72$
$\Rightarrow\frac{3\text{x}}{2}=72$
$\therefore\text{x}=\frac{72\times2}{3}=24\times2=48$
$\therefore$ Required number $= 48$
View full question & answer→Question 262 Marks
A man sold an article for $Rs. 495$ and gained $10\%$ on it. Find the cost price of the article.
Answer$S.P$ of article $= Rs. 495$ gain $= 10\%$
Let cost price $= Rs. x$
$\therefore\text{S.P.}=\frac{\text{x}\times(100+10)}{100}=\frac{110}{100}\text{x}$
$\therefore\frac{110}{100}\text{x}=495$
$\Rightarrow\text{x}=\frac{495\times100}{110}=45\times10=450$
$\therefore$ Cost price $= Rs. 450$
View full question & answer→Question 272 Marks
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$?$
AnswerLet length of total journey $= x\ km$
According to the condition,
$\frac{3}{5}\text{x}+\frac{1}{4}\text{x}+\frac{1}{8}\text{x}+2=\text{x}$
$LCM$ of $5, 4, 8 = 40$
$\therefore\frac{3}{5}\text{x}+\frac{1}{4}\text{x}+\frac{1}{8}\text{x}+\frac{2}{1}=\frac{\text{x}}{1}$
$\frac{24\text{x}+10\text{x}+5\text{x}+80=40\text{x}}{40}$
$\Rightarrow 39x + 80 = 40x$
$\Rightarrow 40x - 39x = 80$
$\Rightarrow x = 80$
Total journey $= 80\ km$
View full question & answer→Question 282 Marks
Sumitra has $Rs. 34$ in $50$ paise and $25$ paise coins. If the numbers of $25$ paise coins is twice the number of $50$ paise coins, how many coins of each kind does she have$?$
AnswerAmount of coins $= Rs. 34$
Let $50$ paisa coins $= x$
then $25$ paisa coins $= 2x$
Then $\frac{\text{x}\times50}{100}+\frac{2\text{x}\times25}{100}=34$
$\frac{\text{x}}{2}+\frac{\text{x}}{2}=34$
$\Rightarrow\frac{2\text{x}}{2}=34$
$\Rightarrow\text{x}=34$
Number of $50$ paisa coins $= 34$ And number of $25$ paisa coins $= 2x = 2 \times 34 = 68$
View full question & answer→Question 292 Marks
Solve the following equations. Check your result in case. $7 - 5x = 5 - 7x$
Answer$7 - 5x = 5 - 7x$
$\Rightarrow -5x + 7x = 5 - 7 ($By transposing$)$
$\Rightarrow 2x = -2 x = -1$ Check: $L.H.S. = 7 - 5x = 7 - 5(-1)$
$= 7 + 5 = 12$
$R.H.S. = 5 - 7x$
$= 5 - 7(-1) = 5 + 7 = 12$
$L.H.S. = R.H.S.$
Hence $x = -1$
View full question & answer→Question 302 Marks
Two-thirds of a number is greater than one-third of the number by $3.$ Find the number
AnswerLet the required number $= x$
Two-third of the number $=\frac{2}{3}\text{x}$ And one third of the number $=\frac{1}{3}\text{x}$
Then $\frac{2}{3}\text{x}-\frac{1}{3}\text{x}=3$
$\Rightarrow\frac{2-1}{3}\text{x}=3$
$\Rightarrow\frac{1}{3}\text{x}=3$
$\Rightarrow\text{x}=\frac{3\times3}{1}=9$
$\therefore$ Required number $= 9$
View full question & answer→Question 312 Marks
Solve the following equations. Check your result in case.
$0.5x - (0.8 - 0.2x) = 0.2 - 0.3x$
Answer$0.5x - (0.8 - 0.2x) = 0.2 - 0.3x$
$\Rightarrow 0.5x - 0.8 + 0.2x = 0.2 - 0.3x$
$\Rightarrow 0.5x + 0.2x + 0.3x = 0.2 + 0.8$
$\Rightarrow 1.0x = 1.0$
$\Rightarrow x = 1$
Check:
$L.H.S. = 0.5x - (0.8 - 0.2x)$
$= 0.5 \times 1 - (0.8 - 0.2 \times 1)$
$= 0.5 - (0.8 - 0.2)$
$= 0.5 - 0.6$
$= -0.1$
$R.H.S. = 0.2 - 0.3x$
$= 0.2 - 0.3 \times 1$
$= 0.2 - 0.3$
$= -0.1$
$L.H.S. = R.H.S.$
Hence $x = 1$
View full question & answer→Question 322 Marks
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.
AnswerLet present age of son $= x$ years
Age of father $= (x + 30)$ years
$12$ years after,
Father’s age $= x + 30 + 12 = (x + 42)$ years
And son’s age $= (x + 12)$ years
$(x + 42) = 3(x + 12)$
$\Rightarrow x + 42 = 3x + 36$
$\Rightarrow 3x + 36 = x + 42$
$\Rightarrow 3x - x = 42 - 36$
$\Rightarrow 2x = 6$
$\Rightarrow x = 3$
Son’s age $= 3$ years
Father’s age $= 3 + 30 = 33$ years
View full question & answer→Question 332 Marks
The total cost of $3$ tables and $2$ chairs is $Rs. 1850.$ If a table costs $Rs. 75$ more than a chair, find the price of each.
AnswerCost of $3$ tables and $2$ chairs $= 1850$
Cost of table $= Rs. 75 +$ cost of a chair
Let cost of chair $= Rs. x,$
Then Cost of table $= Rs. 75 + x$
According to the condition,
$3(75 + x) + 2x = 1850$
$⇒ 225 + 3x + 2x = 1850$
$⇒ 225 + 5x = 1850$
$⇒ 5x = 1850 - 225 = 1625$
$x = 325$
Cost of chair $= Rs. 325$
And cost of table $= Rs. 325 + 75$
$= Rs. 400$
View full question & answer→Question 342 Marks
Solve the following equations. Check your result in case. $3x - 5 = 0$
Answer$3\text{x}-5=0$ Adding $5$ to both sides $3\text{x}-5+5=0+5$
$\Rightarrow3\text{x}=5$
$\Rightarrow\text{x}=\frac{5}{3}$ Check: $\text{L.H.S}=3\text{x}-5$
$=3\text{x}\frac{5}{3}-5$
$=5-5$
$=0$
$=\text{R.H.S}$ Hence $\text{ x}=\frac{5}{3}$
View full question & answer→Question 352 Marks
The fifth part of a number when increased by $5$ equals its fourth part decreased by $5.$ Find the number.
AnswerLet the required number $= x$
Fifth part of the number $=\frac{\text{x}}{5}$
Fourth part of the number $=\frac{\text{x}}{4}$
$\therefore\frac{\text{x}}{5}+5=\frac{\text{x}}{4}-5$
$\Rightarrow\frac{\text{x}}{5}-\frac{\text{x}}{4}=-5-5$
$\Rightarrow\frac{4\text{x}-5\text{x}}{20}=-10$
$\Rightarrow\frac{-\text{x}}{20}=-10$
$\Rightarrow\frac{\text{x}}{20}=10$
$\Rightarrow\text{x}=10\times20=200$
$\therefore$ Required number $= 200$
View full question & answer→Question 362 Marks
Find two consecutive natural numbers whose sum is $63.$
AnswerLet first natural number $= x$
then Next number $= x + 1$
$x + x + 1 = 63$
$\Rightarrow 2x = 63 - 1 = 62$
$x = 31$
First number $= 31$ And second number $= 31 + 1 = 32$
Numbers are $31, 32$
View full question & answer→Question 372 Marks
Raju is $19$ years younger than his cousin. After $5$ years, their ages will be in the ratio $2 : 3.$ Find their present ages.
AnswerLet present age of Raju’s cousin $= x$ years
Then age of Raju $= (x - 19)$ years After $5$ years,
Raju’s age $= x - 19 + 5 = (x - 14)$ years And his cousin age $= x + 5 (x - 14) : (x + 5) = 2 : 3$
$\Rightarrow\frac{\text{x}-14}{\text{x}+5}=\frac{2}{3}$
$\Rightarrow 3(x - 14) = 2 (x + 5) ($By cross multiplication$)$
$\Rightarrow 3x - 42 = 2x + 10$
$\Rightarrow 3x - 2x = 10 + 42$
$\Rightarrow x = 52$
Raju’s age $= x - 19 = 52 - 19 = 33$
years And his cousin age $= 52$ years.
View full question & answer→Question 382 Marks
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.
AnswerLet total marks $= x$
Pass marks $= 40\%$ of $\text{x}=\frac{40\text{x}}{100}=\frac{2}{5}\text{x}$
No. of marks got by Rupa $= 185$
No. of marks by which she failed $= 15$
Pass marks $= 185 + 15 = 200$
$\frac{2}{5}\text{x}=200$
$\Rightarrow\text{x}=\frac{200\times5}{2}\text{x}$
$\Rightarrow\text{x}=500$
Hence total marks $= 500$
View full question & answer→Question 392 Marks
Divide $184$ into two parts such that one-third of one of part may exceed one-seventh of the other part by $8.$ Hint. Let the two parts be $x$ and $(184 - x).$ Then $\frac{1}{3}\text{x}-\frac{1}{7}(184-\text{x})=8$
AnswerSum of two numbers $= 184$
Let first number $= x$
Then second number $= 184 - x$
Then $\text{x}\times\frac{1}{3} = (184-\text{x})\times\frac{1}{7}+8$
$\frac{\text{x}}{3}=\frac{184-\text{x}}{7}+8$
$\frac{7\text{x }=\ 3(184-\text{x})+168}{21}$
$(LCM$ of $3, 7 = 21)$
$7\text{x}=552-3\text{x}+168$
$\Rightarrow7\text{x}+3\text{x}=552+168$
$\Rightarrow10\text{x}=720$
$\Rightarrow\text{x}=\frac{720}{10}=72$
First part $= 72$
Second part $= 184 - 72 = 112$
Hence parts are $72, 112$
View full question & answer→Question 402 Marks
Solve the following equations. Check your result in case. $3x + 2(x + 2) = 20 - (2x - 5)$
Answer$3x + 2(x + 2) = 20 - (2x - 5)$
$\Rightarrow 3x + 2x + 4 = 20 - 2x + 5$
$\Rightarrow 5x + 4 = 25 - 2x$
$\Rightarrow 5x + 2x = 25 - 4 ($By transposing$)$
$\Rightarrow 7x = 21$
$\Rightarrow x = 3$
Check: $L.H.S.= 3x + [2(x + 2)]$
$= 3 \times 3 + 2(3 + 2)$
$= 9 + 2 \times 5 = 9 + 10$
$= 19$
$R.H.S. = 20 - (2x - 5)$
$= 20 - (2 \times 3 - 5)$
$= 20 - (6 - 5) = 20 - 1$
$= 19$
$L.H.S. = R.H.S.$
Hence $x = 3$
View full question & answer→Question 412 Marks
Solve the following equations. Check your result in case.$8x - 3 = 9 - 2x$
Answer$8\text{x}-3=9-2\text{x}$
$\Rightarrow8\text{x}+2\text{x}=9+3$ (By transposing)
$\Rightarrow10\text{x}=12$
$\Rightarrow\text{x}=\frac{12}{10}=\frac{6}{5}$
$\therefore\text{x}=\frac{6}{5}$
Check:
$\text{L.H.S}=8\text{x}-3=8\times\frac{6}{5}-3$
$=\frac{48}{5}-\frac{3}{1}=\frac{48-15}{5}=\frac{33}{5}$
$\text{R.H.S}=9-2\text{x}$
$=9-2\times\frac{6}{5}=\frac{9}{1}-\frac{12}{5}$
$=\frac{45-12}{5}=\frac{33}{5}$
$\therefore\text{L.H.S}=\text{R.H.S}$
Hence $\text{x}=\frac{6}{5}$
View full question & answer→Question 422 Marks
How much pure alcohol must be added to $400\ mL$ of a $15\%$ solution to make its strength $32\%?$
AnswerSolution $= 400\ ml$
Quantity of alcohol $= 15\%$ of $400\ ml $
$=\frac{400\times15}{100}=60\text{ml}$
Let pure alcohol added $= x\ ml$
Total solution $= 400 + x$ and total alcohol $= (x + 60)$
Now $(400 + x) x 32\% = x + 60$
$\Rightarrow(400+\text{x})\times\frac{32}{100}=\text{x}+60$
$\Rightarrow 32(400 + x) = 100(x + 60)$
$\Rightarrow 12800 + 32x = 100x + 6000$
$\Rightarrow 12800 - 6000 = 100x - 32x$
$\Rightarrow 6800 = 68x$
$\Rightarrow x = 6800$ Pure alcohol added $= 100\ ml$
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