MATHS M.C.Q. [1 Marks Each]

Let $5x$ be the smallest multiple of $5.$
Then, the three consecutive multiples shall be $5x, 5x + 5$ and $5x + 10$
According to the question,
$5x + 5x + 5 + 5x + 10 = 45$
$\therefore 5x + 5x + 5x = 45 - 15$
$\therefore 15x = 30$
$\therefore x = 2$
$\therefore$ The Smallest multiple $= 5x = 5 × 2 = 10$

A linear equation in one variable has only one solution. e.g. Solution of the linear equation $ax + b = 0$ is unique, i.e. $\text{x}=-\frac{\text{b}}{\text{a}}.$

$\frac{\text{4x}}{7}-12=0 \Rightarrow \frac{\text{4x}}{7}= 12$
$ \Rightarrow \text{4x}= 7 \times 12 = 84$
$\Rightarrow \text{x}=\frac{\text{84}}{4}= 21$

Taking, $3(x - 3) = 4(2x + 4)$
$⇒ 3x - 9 = 8x + 16$
$⇒ 3x - 8x = 16 + 9$
$⇒ -5x = 25$
$⇒ x = -5$
Therefore, the solution of the given equation is $-5.$

Let the required number be $x.$
$\text{x}+\frac{2}{5}=-\frac{3}{7}$
$\text{x}=\frac{3}{7}-\frac{3}{5}$
$\text{x} = -\frac{(15-14)}{35}$
$\text{x} = -\frac{15}{35}$

Given, $8x - 3 = 25 + 17x$
$8x - 17x = 25 + 3$
$-9x = 28$
$\text{x} = -\frac{28}{9}$
Hence, $x$ is a rational number.

Given, the two numbers are in the ratio $3 : 5.$
Then the numbers be $3x, 5x.$
According to the problem,
$3x + 5x = 96$
or, $8x = 96$
or, $x = 12.$
So the numbers are $36, 60.$

$2x - 3 = 7$
$2x = 7 + 3 = 10$
$\text{x} = \frac{10}{2} = 5$

We have two expressions such that,
$\Rightarrow 4x + 4 = 2x + 8$
$\Rightarrow 4x - 2x = 8 - 4$
$\Rightarrow 2x = 4$
$\Rightarrow x = 2$
Therefore, for $x = 2$ the given two expressions will become equal.

 Given,
Speed of the stream in still water $= 1\ km/ hr$
Let speed of the streamer $= x \ km/ hr$
Speed downstream $= (x + 1)\ km/ hr$
Speed upstream $= (x - 1)\ km/ hr$
According to the given condition,
$(x + 1) × 5 = (x - 1) × 6$
$5x + 5 = 6x - 6$
$x = 11$
Hence, Speed of streamer in still water is $11\ km/ hr$ and
Distance between two ports $= (11 + 1) × 5 = 60\ km/ hr.$

 Let the angles be $4x$ and $5x.$ Third angle $= 4x + 5x = 9x$
We have,
$4x + 5x + 9x = 180^\circ $
$x = 10^\circ $
$\therefore$ Third angle $= 9x = 9 \times 10^\circ = 90^\circ $

$2x + 3 = 2(x - 4)$
$⇒ 2x + 3 = 2x - 8$
$⇒ 3 = -8$ which is impossible.

 Current volume of alcohol $=\text{100ml}\times\frac{15}{100}=\text{60ml}.$
Required strength $=\frac{15}{100}\times400=\text{128ml}$
Required pure alcohol $=128-\text{60ml}=68$

Given, $\frac{1}{3}+\text{S}=\frac{2}{5}$
$\text{S}=\frac{2}{5}-\frac{1}{3}$
$\text{S}=\frac{6-5}{15}$
$\text{S}=\frac{1}{15}$

 Given, $\frac{5\text{x}}{3}-4=\frac{2\text{x}}{5}$
$\frac{5\text{x}}{3}-\frac{2\text{x}}{5}=4$
$\frac{25\text{x}-6\text{x}}{15}=4$
$19\text{x}=60$
$\frac{19\text{x}}{19}=\frac{60}{19}$
$\text{x}=\frac{60}{19}$
$\text{Now, }2\text{x}-7=2\times\frac{60}{19}-7$
$\frac{120-133}{19}=-\frac{13}{19}$
Hence, the numerical value of $2x - 7$ is $-\frac{13}{19}.$

$(x + 3) + x = 23$
$\Rightarrow 2x = 20$
$\Rightarrow x = 10$
$\Rightarrow x + 3 = 13$

$\frac{\text{7}}{\text{x}}= 3$
$ \Rightarrow \text{3x}= 7$
$ \Rightarrow \text{x}= \frac{7}{3}$

$x + 3 = 5$
$\Rightarrow x = 5 - 3$
$= 2.$

$X$ plus an odd number will always equal an even number if $x$ is odd.
Vice versa is true if $x$ is even.
Think about it: If $x = 1 (1$ is an odd number$),$ the next odd numbers would be $3, 5, 7, 9,$ etc.
Therefore, $x + 2$ would be the next odd number.

 Taking, $3x - 3 = 25 + 17x$
$⇒ 3x - 17x = 25 + 3$
$⇒ -14x = 28$
$\Rightarrow\text{x}=\frac{28}{(-14)}$
$⇒ x = -2$
$-2$ is an integer.

$(2\text{x} - 1) + (\text{x}-1) = \text{x} + 2$
$\Rightarrow\text{2y = 4 }\Rightarrow\text{x}= \frac{4}{2}= 2.$

Let the three consecutive integers be $x, x + 1$ and $x + 2.$
Equation $= \text{x} + \text{x} + 1 + \text{x} + 2 = 51$
$\Rightarrow3\text{x} + 3 = 51$
$\Rightarrow3\text{x} = 51 - 3$
$\Rightarrow3\text{x} = 48$
$\Rightarrow\text{x} = \frac{48}{3}= 16$
Middle integer $= \text{x}+1 = 16 + 1 = 17$

Let the age of the father is $x$
Given: $x = 3 × ($age of son$) = 3 × (15) = 45$ years.

 Let the equal side of the isosceles triangle be $x.$
Then, the perimeter of the triangle would be $(x + x + 6).$
$\therefore2\text{x} + 6 = 16 $
$\Rightarrow2\text{x} = 16 - 6$
$\Rightarrow2\text{x} = 10$
$\Rightarrow\text{x} = \frac{10}{2}= 5$
$\therefore$ Length of each equal side $= 5\ cm$

Speed of the man in still water $= 8\ km/ ph.$
Speed of the river $= 2\ km/ ph$
Downstream $= 8 + 2 = 10\ km/ ph$
Upstream $= 8 - 2 = 6\ km/ ph$
$\Rightarrow\frac{\text{x}}{10}+\frac{\text{x}}{6}=\frac{48}{60}$
$\Rightarrow\text{8x}=24$
$\Rightarrow\text{x}=\text{3km}$

Let the number be $x.$
$\therefore\frac{4}{5}\text{x}=\frac{3}{4}\text{x}+4$
$\Rightarrow\frac{4}{5}\text{x}=\frac{3\text{x}+16}{4}$
$\Rightarrow16\text{x} = 15\text{x} + 80$
$\Rightarrow16\text{x} - 15\text{x} = 80$
$\Rightarrow \text{x} = 80$

Given, Shilpa’s age three years ago $= x$
Then, Shilpa’s present age $= (x + 3)$
Arpita’s present age $3\ ×$ Shilpa’s present age $= 3 (x + 3)$

$\frac{\text{y}}{3}-7= 11$
$\Rightarrow \frac{\text{y}}{3}=7 + 11 = 18$
$\Rightarrow \text{y} = 3 \times 18 = 54$

Let the number be $x.$

 Let the number be $x.$

 $\frac{\text{5}}{12}\text{ x}= -27 \Rightarrow\text{x}=- \frac{\text{27}\times 2}{3}=-18$

 Let the number of boys in the class be $x.$
Then, the number of girls will be $(x - 8).$
The equation becomes:
$\Rightarrow\frac{\text{x}}{\text{x}-8}=\frac{7}{5}$
$\Rightarrow5\text{x} = 7\text{x} - 56$
$\Rightarrow5\text{x} -7\text{x} = -56$
$\Rightarrow -2\text{x} = -56$
$\Rightarrow \text{x} = \frac{-56} {-2} = 28$
Therefore, the number of boys is $28.$
Number of girls $= ( \text{x}- 8) = 28 -8 =20$
Total strength of the class $= 28 + 20 = 48$

$ x^\circ = (90^\circ - x^\circ ) + 10^\circ$
$\Rightarrow 2x^\circ = 100^\circ $
$\Rightarrow x^\circ = 50^\circ$
$\Rightarrow 90^\circ - x^\circ = 40^\circ .$

 Let the three consecutive multiplies of $7$ be $7x, (7x + 7), (7x + 14)$ where $x$ is a natural number.
According to question,
$7x + (7x + 7) + (7x + 14) = 357$
$21x + 21 = 357$
$21(x + 1) = 357$
$\frac{21(\text{x}+1)}{21}=\frac{357}{21}$
$x + 1 = 17$
$x = 17 - 1$
$x = 16$
Hence, the smallest multiple of $7$ is $7 × 16$ i.e., $112.$

Let the present ages of three persons be $x$ years, $y$ years & $z$ years, respectively.
According to the question, $x + y + z = 50$
After $5$ years,
The ages of the 3 persons will be is $(x + 5)$ years, $(y + 5)$ years and $(z + 5)$ years, respectively.
To find:$(x + 5) + (y + 5) + (z + 5) = x + y + z + 15$
$= 50 + 15$
$= 65$

 Consider $ax + b = O$ where $a$ and $b$ can take any value.
$x$ is the only variable with power one. Therefore, such equations are called linear equation in one variable.

Let the numbers be $x$ and $x + 15.$
$\therefore x + x + 15 = 95$
$\Rightarrow 2x + 15 = 95$
$\Rightarrow 2x = 95 - 15$
$\Rightarrow 2x = 80$
$\Rightarrow x = 40$
The smaller number is $40.$

$ X's$ one day's work $=\frac{1}{12}$
Let, $Y$ work for $x$ days
$\therefore$ $Y's$ one day's work $=\frac{1}{\text{x}}$​
One day work by $X$ and $Y$ together
$\frac{1}{12}+\frac{1}{\text{x}}=\frac{1}{8}$
$\frac{1}{\text{x}}=\frac{1}{8}-\frac{1}{12}=\frac{1}{24}$
$\therefore x = 24$ days
$\therefore Y$ can complete work alone in $24$ days.

 $\text{2x}-3=\text{x}+2$
$\Rightarrow2​​\text{x}-\text{x}=3+2$
$\Rightarrow​​\text{x}=5$

$\text{5x} -8 =7 \Rightarrow \text{5x = 8 + 7 = 15}$
$\Rightarrow \text{x}= \frac{15}{5}= 3.$

Let the greater number be $x$
Smaller number be $x - 1$
$\frac{\text{a}}{\text{q}}$
$x + x - 1 = 15$
$2x – 1 = 15$
$2x = 16$
$X = 8.$
Smaller number $= 8 - 1 = 7.$

Let the required number be $'x '.$
Then, $\frac{\text{a}}{9}=-2$
$⇒ a = -18$
The required number is $-18.$

$\frac{\text{y}}{5} = 10$
$\text{y} = 5\times10 = 50$

Let the number be $x$
$-\frac{7}{3}+\text{x} = \frac{3}{7}$
$\text{x}=\frac{3}{7}+\frac{7}{3} = \frac{(9+49)}{21} = \frac{58}{21}$

Given, $-\frac{4}{3}\text{y}=-\frac{3}{4}$
$\text{y}=-\frac{3}{4}\times-\frac{3}{4}$
$\text{y}=\Big(\frac{3}{4}\Big)^2$
Hence, the value of $y$ is $\Big(\frac{3}{4}\Big)^2.$

According to the question,
$\frac{\text{x}}{100}\times200=10$
$\therefore\frac{\text{x}}{100}\times200=10$
$\Rightarrow\text{2x}=10$
$\Rightarrow\text{x}=\frac{10}{2}$
$\therefore\text{x}=5$

$3x + x = 4$
when we transpose $-x$ to $LHS$ we get $+x$
$3x + x = 4$

 $13\text{x} - 14 = 9\text{x} + 10$
$\Rightarrow1\text{x}- 9\text{x} = 10 + 14$
$\Rightarrow\text{4x = 24 }\Rightarrow\text{x}= - \frac{24}{4}= 6$

$ 11x - 5 - x + 6 = 2x + 17$
$⇒ 8x = 16 $
$⇒ x = 2.$

Given that,
Two year ago, my age was $x$ years
Therefore,
Present age $=$ age two years ago $+\ 2$
Present age $= x + 2$
What was my age $5$ years ago$?$
Age $5$ years ago $=$ present age $- 5$
Age $5$ years ago $= x + 2 - 5$
Age $5$ years ago $= x - 3$
Therefore, my age $5$ years ago is $x - 3$

 Let $X$ be the number
$\frac{\text{X}}{8} = 6$
$\text{X} = 8 \times 6 = 48$

 Given that the equation is $7x + 15 = 50$
To find the root of the equation.
That is the root must satisfy the given equation.
Put $x = 5$ in given equation we get,
$7(5) + 15 = 50$
$35 + 15 = 50$
$50 = 50$
$\therefore 5 $ is a root.
$\therefore 5$ is the root of the given equation $7x + 15 = 50.$

 Let the first number be $x$ and the second number be $x + 15.$
According to the question,
$x + x + 15 = 95$
$2x = 80$
$x = 40$
Hence, the first number is $40$ and the second number is $55.$

$ 2x + 9 = 67 ⇒ x = 29.$

 Let the number be $x$
$-\frac{7}{3}+\text{x} =\frac{3}{7}$
$\text{x}=\frac{3}{7}+\frac{7}{3}= \frac{(9+49)}{21} = \frac{58}{21}$

 $\frac{\text{x}-1}{3}=\frac{\text{x}-2}{4}$
$\Rightarrow 4(x + 1) = 3(x - 1)$
$\Rightarrow 4x + 4 = 3x - 6$
$\Rightarrow 4x - 3x = -6 - 4$
$\Rightarrow x = -10$

 $\frac{2}{3}\text{y}=\frac{5}{12}$
$\Rightarrow\text{ y}=\frac{5}{12} \times\frac{3}{2}=\frac{5}{8}$

$ x^\circ + 2x^\circ = 180^\circ$
$\Rightarrow 3x^\circ = 180^\circ$
$\Rightarrow x^\circ = 60^\circ$
$\Rightarrow 2x^\circ = 120^\circ$

If $ax = b,$ then $\text{x}=\frac{\text{b}}{\text{a}}$
Since, $a$ and $b$ are positive integers. So, $\frac{\text{b}}{\text{a}}$ is also positive integer.
Hence, the solution of the given equation has to be always positive.

As given, digit in the units place $= a$
So, the digit in the tens place $= (4 + a)$
Using the values of the numbers,
$⇒ 10(4 + a) + a$
$⇒ 40 + 10a + a$
$⇒ 11a + 40$

 Given equation is
$ax + b = 0$
$ax = -b$
$\frac{\text{ax}}{\text{a}}=-\frac{\text{b}}{\text{a}}$
$\text{x}=-\frac{\text{b}}{\text{a}}$
Hence, the solution of the equation $ax + b = 0$ is $\text{x}=-\frac{\text{b}}{\text{a}}$

If any digit $q$ is appended to any number $x,$ it's value becomes $10x + q.$
So at first we have a number with tens' digit t and unit's digit $u.$
The number is $10t + u.$
After placing one more digit $1,$
it becomes $10(10t + u) + 1$
That is, $100t + 10u + 1.$

$x -$ Small number $= 30$
Small number $= x - 30$

$\frac{\text{x}}{2}+30=19.$
$\Rightarrow x + 60 = 38$
$\Rightarrow x = 38 - 60$
$\Rightarrow x = -22$

Let three consecutive even natural numbers be $x, x + 2,$ and $x + 4,$
As per the question,
$x - x + 2 + x + 4 = 54$
$⇒ 3x + 6 = 54$
$⇒ 3x = 54 - 6$
$⇒ 3x = 48$
$⇒ x = 16$
Therefore, the three consecutive even natural numbers are $16, 18$ and $20.$ The highest of these numbers is $20.$

 Let the number of males $7x$ and number of females $= 4x$
Given that,
$7x = 84 ⇒ x = 12$
Total number of members $7x + 4x = 11x = 11 × 12 = 132.$

 Let the two numbers be $2x$ and $5x$ since they are in the ratio of $2 : 5.$
The difference between $5x$ and $2x = 66$
$5x - 2x = 66$
$3x = 66$
$x = 22$
Hence, $2x = 2(22) = 44$ and $5x = 5(22) = 110.$

$ 2x - 30 = 60 - 3x $
$\Rightarrow 5x = 90$
$\Rightarrow x = 18.$

In $33(x + y) = 0, x$ and $y$ are two variables.

$2x + 10 = 76$
$2x = 76 - 10$
$2x = 66$
$\text{x} = \frac{66}{2}$
$x = 33$

$ ax - b = 0.$
$\Rightarrow\text{ax}=\text{b}$
$\Rightarrow\text{x}=\frac{\text{b}}{\text{a}}$
Therefore, $\text{x}=\frac{\text{b}}{\text{a}}$

 Let the three consecutive even number be $2x - 2, 2x, 2x + 2.$
We have,
$(2x - 2) + 2x + (2x + 2) = 234$
$⇒ 6x = 234$
$⇒ x = 39$
$\therefore$ Least even number is $2x - 2 = 2(39) - 2 = 76.$

 Let the original number be $x.$
According to the question we can write as $\Big(\frac{2}{3}\Big)\text{x}+\text{20x}$
On rearranging $\text{x}-\Big(\frac{2}{3}\Big)\text{x}=20$
Now taking the $L.C.M$ od $1$ and $3$ is $3$
$\frac{(\text{3x - 2x})}{ 3=20}$
$\frac{\text{x}}{3=20}$
Again by transposing $x = 60$
So the original number is $60$

 If $x$ is one number, then $x + 1$ would be the next consecutive number. Since the sum of the two consecutive numbers is $71,$ we can say,
$x + (x + 1) = 71$
$2x + 1 = 71$
$2x = 70$
$x = 35,$ so $x + 1 = 36$
$35 + 36 = 71$

 Perimeter of a rectangle $= 40\ cm$
Width $= 10\ cm$
Let the length be $x.$
Perimeter of rectangle $= 2($length $+$ width$)$
$40 = 2(x + 10)$
$\frac{40}{2} = \text{x} + 10$
$20 = x + 10$
$x = 20 - 10 = 10$
Thus, the length is also $10\ cm.$
Hence, we can say, that the given rectangle is basically a square, with all its sides equal.

 $\frac{\text{5x}}{3}= 30$
$\Rightarrow \text{5x}= 3 \times 30 = 90$
$ \Rightarrow \text{x}= \frac{90}{5}=18$

 Let the number be $x.$
$x + x = 24$
$2x = 24$
$\text{x} = \frac{24}{2}$
$x = 12$

 $3\text{x} + 8 = 14$
$\Rightarrow 3\text{x} = 14 - 8 = 6$
$\Rightarrow\text{x}= \frac{6}{3}= 2$

Let $CP = x$
$\text{Gain}=\frac{16}{100}\text{x}=\frac{\text{4x}}{25}$
$\text{CP}=\frac{\text{4x}}{25}+\text{x}=\frac{\text{29x}}{25}$
$\frac{\text{29x}}{25}=1885$
$\text{x}=\text{Rs. 1625}$

 Taking, $\text{m}-\frac{\text{m}-1}{3}=1-\frac{\text{m}-2}{2}$
$\Rightarrow\frac{\text{m}-\text{m}+1}{3}=\frac{2-\text{m}+2}{2}$
$\Rightarrow\frac{\text{2m}+1}{3}=\frac{4-\text{m}}{2}$
$\Rightarrow2{\text({2\text{m}}+1})=3(4-\text{m})$
$\Rightarrow{\text{4m}+2}=12-\text{3m}$
$\Rightarrow{\text{4m}+2}\text{3m}=12-2$
 $\Rightarrow\text{7m}=10$
$\Rightarrow\text{m}=\frac{10}{7}$

 $2\text{y} = 5(\text{7} - \text{y})$
$\Rightarrow2\text{y} = 35 - 5\text{y} $
$\Rightarrow2\text{y} + 5\text{y}= 35$
$\Rightarrow\text{7y = 35 }\Rightarrow\text{y}= - \frac{35}{7}= 5$

 Let $'a' a + 1,$ and $a + 2$ are the three consecutive integers.
According to the question,
$a + a + 1 + a + 2 = 180$
$\Rightarrow 3a + 3 = 180$
$\Rightarrow a + 1 = 160$
$\Rightarrow a = 60 - 1$
$\Rightarrow a = 59$
Therefore, the required integers are $59, 60$ and $61.$

 $3\text{y} + 4 = 5\text{y} - 4$
$\Rightarrow5\text{y} - 3\text{y} = 4 + 4$
$\Rightarrow\text{2y}=8 \Rightarrow\frac{8}{2}=4$

Let the number be $x.$

Before $10$ Years $x$
Today after $10$ Years $- x + 10$
After $10$ Years $x + 10 + 10 = x + 20$

 $3\text{x} + 4 = 13$
$\Rightarrow3\text{x}= 13 – 4 = 9$
$ \Rightarrow \text{x}= \frac{9}{3}=3$

Given, a number increased by $8\%$ of itself gives $135.$
Thus, $x + 8\%$ of $x = 135$
$\Rightarrow\frac{\text{108x}}{100}=135$
$\Rightarrow\text{x}=135\times\frac{100}{108}=125$
Thus, the number is $125.$

Let the number be $x.$

The number is $10x + y$
When digits are reversed, the number becomes $10y + x.$
We have,
$x + y = 8 (i)$
$10x + y + 18 = 10y + x$
$x - y + 2 = 0 (ii)$
Solving $(i)$ and $(ii),$ we get
$x = 3, y = 5$
$\therefore$ The number is $35.$

Adding $4$ times $x$ to $16$ is $45$ is $4x\ 16 = 45.$

$5\text{t}-3=3\text{t}-5$
$\Rightarrow5\text{t}-3\text{t}=3-5$
$\Rightarrow2\text{t}=-2$
$\Rightarrow\text{t}=\frac{-2}{2}=-1$

 Let the number be $x.$
Let $x$ be the common multiple of the ages of A$$ and $B$.Then. the ages of $A$ and $B$ would be $5x$ and $7x,$ respectively.  
$\therefore\frac{5\text{x}+4}{7\text{x}-4}=\frac{3}{4}$
$\Rightarrow4( 5\text{x} +4 ) = 3 ( 7\text{x} + 4 )$
$\Rightarrow20\text{x} + 16 = 21\text{x} + 12$
$\Rightarrow16 - 12 = 21\text{x} - 20\text{x}$
$\Rightarrow4 = \text{x}$
$\Rightarrow\text{x} = 4 $
$\therefore$ Age of $\text{B} = 7(\text{x}) = 7\times4 $
$= 28 \text{ years}$

$ x -$ smaller number $= 21$
$\Rightarrow $ smaller number $= x - 21$

 $-\frac{1}{3}= \text{2x}\Rightarrow \text{x}= - \frac{1}{3}$

 Let the number be $x.$

 Perimeter of rectangle $= 2($Length $+$ Breadth$)$
$20 = 2(6 + x)$
$6 + x =\frac{20}{2}$
$6 + x = 10$
$x = 10 - 6$
$x = 4\ cm$