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Reducing simple equations to simpler form - class-IX

Description: reducing simple equations to simpler form
Number of Questions: 111
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Tags: linear equations in one variable linear equation in one variable equation formation of an equation and its solution maths equations
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Solve for $x$:

$\dfrac {9x}{7-6x}=15$

  1. $\dfrac{99}{105}$

  2. $-\dfrac{105}{99}$

  3. $\dfrac{105}{99}$

  4. $-\dfrac{99}{105}$


Correct Option: C
Explanation:

$\dfrac{9x}{7 - 6x} = 15$


$9x = 15(7 - 6x)  = 105 - 90x$

$9x + 90x = 105 $

$99x = 105$

$x = \dfrac{105}{99}$

If $17x+51y=85$, then $13x+39y=$

  1. $67$

  2. $61$

  3. $63$

  4. $65$


Correct Option: D
Explanation:

$17x+51y=85$

 dividing both sides by $17$

$\Rightarrow x+3y=5$ 
 on multiplying both sides by $13$
$\Rightarrow 13x+39y=65$
Therefore, $13x+39y=65$

Sabarmati express take 18 second to pass completely through a stations $162$m long and $15 second $ through another station $120m$ long. The length of the sabarmathi express os 

  1. $132m$

  2. $100m$

  3. $80m$

  4. $90m$


Correct Option: A

The manufacturer of a certain item can sell all he can produce at the selling price of $Rs. 60$ each. It costs him $Rs. 40$ in materials and labour to produce each item and he has overhead expenses of $Rs. 3000$ per week in order to operate the plant. The number of units he should produce and sell in order to make a profit of at least $Rs\,1000$ per week, is : 

  1. $200$

  2. $250$

  3. $300$

  4. $400$


Correct Option: A
Explanation:

Let the no of units sold be $x$ per week.

$60x$ = sales should include all the expenses and profit required to balance things out.
$60x = 1000+40x+3000$

$20x = 4000$
$x =200$

The value of $x$ for which $\cfrac{x-3}{4}--x< \cfrac{x-1}{2}-\cfrac{x-2}{3}$ and $2-x> 2x-8$

  1. $[-1,10/3]$

  2. $(1,10/3)$

  3. $R$

  4. none of these


Correct Option: A
Explanation:

$\dfrac{x-3}{4}-x<\dfrac{x-1}{2}-\dfrac{x-2}{3}$


$\dfrac{-3{x}-3}{4}<\dfrac{x+1}{6}$


$\implies x>-1$

$2-x>2{x}-8\implies x<\dfrac{10}{3}$

$\implies x\in (-1,10/3)$

If $\sqrt { 1+\dfrac { x }{ 289 }  } =1\dfrac { 1 }{ 17 }$ then $x=$

  1. $1$

  2. $13$

  3. $35$

  4. $15$


Correct Option: C
Explanation:

We have,

$\sqrt{1+\dfrac{x}{289}}=1\dfrac{1}{17}$

$\sqrt{\dfrac{289+x}{289}}=\dfrac{18}{17}$

 

On squaring both sides, we get

$ {{\left( \sqrt{\dfrac{289+x}{289}} \right)}^{2}}={{\left( \dfrac{18}{17} \right)}^{2}} $

$ \dfrac{289+x}{289}=\dfrac{324}{289} $

$ 289+x=324 $

$ x=324-289 $

$ x=35 $

 

Hence, this is the answer.

Solve for $x$:-
$\dfrac{{2x - 1}}{2}\,\,\, - \dfrac{{x + 3}}{3}\,\, = \dfrac{{x - 2}}{5}$

  1.  $x=\dfrac{5}{14}$

  2.  $x=\dfrac{3}{14}$

  3.  $x=\dfrac{33}{14}$

  4. None of these


Correct Option: C
Explanation:

$\dfrac{{2x - 1}}{2}\,\,\, - \dfrac{{x + 3}}{3}\,\, = \dfrac{{x - 2}}{5}\Rightarrow \dfrac{6x-3-2x-6}{6}=\dfrac{x-2}{5}\Rightarrow 20x-45=6x-12\Rightarrow 14x=33\Rightarrow x=\dfrac{33}{14}$

If $\sqrt{10+ \sqrt{25+ \sqrt{x+ \sqrt{154+ \sqrt{225}}}}} = 4$ find the value of $x$

  1. 110

  2. 108

  3. 100

  4. 114


Correct Option: B
Explanation:

We have,

$\sqrt {10+\sqrt {25+\sqrt {x+\sqrt{154+\sqrt{225}}}}}=4$
$\Rightarrow \sqrt {10+\sqrt {25+\sqrt {x+\sqrt{154+15}}}}=4$
$\Rightarrow \sqrt {10+\sqrt {25+\sqrt {x+\sqrt{169}}}}=4$
$\Rightarrow \sqrt {10+\sqrt {25+\sqrt {x+13}}}=4$
On squaring both sides, we get
$10+\sqrt {25+\sqrt {x+13}}=16$
$\sqrt {25+\sqrt {x+13}}=6$
On squaring both sides, we get
$25+\sqrt {x+13}=36$
$\Rightarrow \sqrt {x+13}=11$
On squaring both sides, we get
$x+13=121$
$x=121-13=108$
Hence, $x=108$

If $x(5\, -\, a)\, =\, 10\, -\, x^{2}$ and x = 2, find the value of 'a'.

  1. $1$

  2. $8$

  3. $4$

  4. $2$


Correct Option: D
Explanation:

$
x(5-a)\quad =\quad 10-{ x }^{ 2 }\ x\quad =\quad 2\ So\ 2(5-a)\quad =\quad 10\quad -{ 2 }^{ 2 }\ 10\quad -\quad 2a\quad =\quad 10-4\ -2a\quad =\quad -4\ a\quad =\quad 2\ 
$

Solve the following equation: 

$(x\, -\, 2)^{2}\, =\, (x\, +\, 1)\, (x\, -\, 1)$

  1. $1.5$

  2. $1.75$

  3. $2.25$

  4. $1.25$


Correct Option: D
Explanation:

Given, $(x-2)^2=(x+1)(x-1)$

$\Rightarrow x^2-4x+4=x^2-1$ .... Using $(a-b)(a+b)=a^2-b^2$
$\Rightarrow -4x+4=-1$
$\Rightarrow -4x=-5$
$\Rightarrow x=\dfrac {5}{4}=1.25$

Solve for $x$ : $\sqrt[3]{x}\,- 4\, =\, 0$

  1. $64$

  2. $36$

  3. $81$

  4. $49$


Correct Option: A
Explanation:

$\displaystyle \sqrt[3]{x}\,- 4\, =\, 0\, \Rightarrow\, \sqrt[3]{x}\, =\, 4$.
$\Rightarrow\, x\, =\, 4^{3}\, =\, 64$

Solve for $x$ : $5\, -\, \sqrt{x}\, =\, 0$

  1. $25$

  2. $54$

  3. $15$

  4. $20$


Correct Option: A
Explanation:
Given, $5 - \sqrt { x }  = 0$
$ \sqrt { x }  = 5$
Squaring on both sides, we get
$ x = { 5 }^{ 2 } = 25$

Simplify: 
$\displaystyle 6x-\left( -4y-8x \right) $

  1. $2x+y$

  2. $14x+4y$

  3. $12x+3y$

  4. $15x+5y$


Correct Option: B
Explanation:

On simplifying, we have

$\displaystyle 6x-\left( -4y-8x \right) =6x+4y+8x$
$=14x+4y$
Hence simplified form of the given expression is $14x+4y$.

Simplify: 
$\displaystyle x-\left[ y-{ x-\left( y-1 \right) -2x}  \right] $

  1. $2y+1$

  2. $-2y+1$

  3. $2x+y-1$

  4. $2x-y-1$


Correct Option: B
Explanation:

On simplified, we have

$\displaystyle x-\left[ y-{ x-\left( y-1 \right) -2x}  \right] $
$=x-\left[ y-{ x-y+1-2x}  \right] $
=$\displaystyle x-\left[ y-{ -x-y+1}  \right] =x-\left[ y+x+y-1 \right] $
=$\displaystyle x-\left[ 2y+x-1 \right] =x-2y-x+1=-2y+1$
Hence, simplified form of the given expression is $-2y+1$.

Number of variables in a simple linear equation

  1. Two

  2. One

  3. 0

  4. None


Correct Option: B
Explanation:

A linear equation can comprise of many variables. The most simple linear equation is in $one$ variable ,i.e, $ax+b=0$.

The value of $\displaystyle \sqrt{2\sqrt{2}\sqrt{2}}...\infty $ is 

  1. 0

  2. 1

  3. $\displaystyle 2\sqrt{2}$

  4. 2


Correct Option: D
Explanation:

Let  $ x =\sqrt{2\sqrt{2\sqrt{2.. }}}\infty$
square both side
$x^2=\sqrt{2\sqrt{2\sqrt{2...}}}\infty = 2x$
$x=2$



If $Rs.50$ is distributed among $150$ children giving $50p$ to each boy and $25p$ to each girl, then the number of boys is:

  1. $25$

  2. $40$

  3. $36$

  4. $50$


Correct Option: D
Explanation:
Let the number of boys $= x$
then number of girls $= 150-x$ 
According to the problem, the total money divided between girls and boys are:
$\cfrac { 50 }{ 100 } \times  (x)+\cfrac { 25 }{ 100 } (150x)=50$
Multiply equation by $100$, we get
$50x+(150x)25 = 5000$
$\Rightarrow 50x+375025x = 5000$
$\Rightarrow 25x = 1250$
$\Rightarrow x = 50$

If $\sqrt{x+1}=\sqrt{x-1}=1$, then x is equal to __________________.

  1. $\displaystyle\frac{5}{4}$

  2. $\displaystyle\frac{2}{3}$

  3. $\displaystyle\frac{4}{5}$

  4. $\displaystyle\frac{3}{5}$


Correct Option: A

The number of solution of the equation $\sqrt{x^{2}}=x-2$ is

  1. $0$

  2. $1$

  3. $2$

  4. $4$


Correct Option: B
Explanation:

Given equation:

$ \sqrt {x^2} = x-2$
$ =  |x-2|$ 
For $x>2$,  $ x=x-2$ No solution
For $x<2$,  $ 2x-2=0      \quad x=1$ 
One solution exists.

IF the lines $ \displaystyle y=m _{1}x+c $  and $  y=m _{2}x+c _{2}  $ are parallel , then 

  1. $ \displaystyle m _{1}=m _{2} $

  2. $ \displaystyle m _{1}=m _{2} =1 $

  3. $ \displaystyle m _{1}=m _{2} =-1 $

  4. $ \displaystyle m _{1}=m _{2} =0 $


Correct Option: A
Explanation:

Two lines are said to be parallel if the slopes of two line will be equal
$m _1=m _2$

If $\sqrt {x-1}-\sqrt {x + 1} + 1= 0$, then $4x$ equals

  1. $5$

  2. $4$

  3. $3$

  4. $2$


Correct Option: A
Explanation:

$\sqrt {x-1}-\sqrt {x+1}+1=0$
or $\sqrt {x-1}=\sqrt {x+1}=-1$
Squaring, we get
$x-1=x+1+1-2\sqrt {x+1}$
or $2\sqrt {x+1}=3$
or $4(x+1)=9$ or $4x=9-4=5$

Seamus has $3$ times as many marbles as Ronit, and Taj has $7$ times as many marbles as Ronit. If Seamus has $s$ marbles then, in terms of $s$, how many marbles do Seamus, Ronit and Taj have together?

  1. $\cfrac{3}{7}s$

  2. $\cfrac{7}{3}s$

  3. $\cfrac{11}{3}s$

  4. $7s$

  5. $11s$


Correct Option: C
Explanation:
The problem will be easier to solve if you can choose numbers that will give you all integers as you solve. both Seamus and Taj have a multiple of the number of marbles that Ronit has, so begin by picking for Ronit, not for Seamus.
If Ronit has $2$ marbles, then Seamus has $(3)(2)=6$ marbles and Taj has $(7)(2)=14$ marbles. Together, the three have $22$ marbles.
plug $s=6$ into the answer (remember that the problem asks about Seamus's starting number, not Ronit's and look for a match of $22$:
(A) $\cfrac{3}{7}s=$ not an integer
(B) $\cfrac{7}{3}s=\cfrac{7}{3}(6)=14$. Not a match
(C) $\cfrac{11}{3}s=\cfrac{11}{3}(6)=22$. Match!!
(D) $7s=42$. Not a match
(E) $11s=$ Too large
Alternately, you can use an algebraic approach. Begin by translating the first sentence into equations:
$s=3r$
$t=7r$
The question asks for the sum of the three:
$s+r+t=$?
The answer use only $s$, so figure out how to substitute to leave only $s$ in the equation
$r=\cfrac{s}{3}$
$t=7r=7(\cfrac{s}{3})$
Substitute those into the equation
$s+r+t$
$s+\cfrac{s}{3}+7(\cfrac{s}{3})$
$\cfrac{3s}{3}+\cfrac{s}{3}+\cfrac{7s}{3}$
$\cfrac{11s}{3}$
The correct answer is (C)

Solve the equation: $\dfrac{2z}{1-z}=6$

  1. $z = \dfrac{1}{4}$

  2. $z = \dfrac{3}{4}$

  3. $z = \dfrac{2}{4}$

  4. $z = \dfrac{3}{7}$


Correct Option: B
Explanation:

Given, $\dfrac{2z}{1-z}=6$
On multiplying both sides by $1 - z$, we get
$2z = 6(1 - z)$
$2z = 6 - 6z$
$8z = 6$
$z = \dfrac{3}{4}$

Solve the equation: $\dfrac{7x - 3}{3x}=2$

  1. $x=3$

  2. $x=2$

  3. $x=1$

  4. $x=-1$


Correct Option: A
Explanation:

$\dfrac{7x - 3}{3x}=2$
$7x - 3 = 6x$
On transposing $6x$ to the L.H.S and $3$ to the R.H.S we obtain
$7x - 6x = 3$
$x = 3$

Find the value of $ p$ in the linear equation: $4p + 2 = 6p + 10$

  1. $p=-4$

  2. $p=-3$

  3. $p=-2$

  4. $p=-1$


Correct Option: A
Explanation:

$4p + 2 = 6p + 10$

On transposing $4p+ 2$ to the R.H.S we obtain
$6p - 4p + 10 - 2 = 0$
$2p + 8 = 0$
$2p = -8$
$p = -4$

Solve the equation: $\dfrac{x+2}{2x}=1$

  1. $x=2$

  2. $x=3$

  3. $x=4$

  4. $x=5$


Correct Option: A
Explanation:

Given, $\dfrac{x+2}{2x}=1$
Multiplying both sides by $2x$, we obtain

$x + 2 = 2x$
$2x - x = 2$
$x = 2$

Reduce the following linear equation: $2x + 5 = 3$

  1. $x=1$

  2. $x=-1$

  3. $x=0$

  4. $x=2$


Correct Option: B
Explanation:

Given, $2x + 5 = 3$
On transposing $5$ to the R.H.S, we obtain 
$2x = 3 - 5$
$2x = -2$
$x = -1$

Calculate the value of $2x+y$, if $\dfrac{1}{2}x=5-\dfrac{1}{4}y$

  1. $20$

  2. $-20$

  3. $10$

  4. $-10$


Correct Option: A
Explanation:

Given, $\dfrac{1}{2}x=5-\dfrac{1}{4}y$

On multiplying both the sides by $4$, we get
$\Rightarrow 2x=20-y$

Adding $y$ on both the sides,
$\Rightarrow 2x+y=20$

Solve the following equation for the value of $x$: $6\sqrt [ 3 ]{ x } -24=6$.

  1. $25$

  2. $5$

  3. $100$

  4. $125$


Correct Option: D
Explanation:

$6\sqrt[3]{x}-24=6$

$\Rightarrow 6\sqrt[3]x=6+24=30$
$\Rightarrow \sqrt[3]x=\dfrac{30}{6}=5$
Taking cube on both sides, we get
$x=5^3=125$

On a car trip Sam drove  $m$  miles, Kara drove twice as many miles as Sam, and Darin drove  $20$  fewer miles than Kara. In terms of  $m$ , how many miles did Darin drive?

  1. $2m+20$

  2. $2m-20$

  3. $\frac{m}{2}+20$

  4. $\frac{m+20}{2}$


Correct Option: B
Explanation:

Given that

Number of miles driven by $Sam$ $=$ $m$
As $Kara$ drove twice as many miles as $Sam$,
Number of miles drove by $Kara$ $=$ $2m$
As $Darin$ drove 20 fewer miles than $Kara$,

Hence, Number of miles drove by $Darin$ $=$ Number of miles drove by $Kara$ $-$ $20$
$=$ $2m$ $-$ $20$ 
Therfore, $Darin$ drove $'2m$ $-$ $20'$ miles.

If $\dfrac{19}{5x+17} = \dfrac{19}{31}$, then find $x $.

  1. $0.4$

  2. $1.4$

  3. $2.8$

  4. $3.4$


Correct Option: C
Explanation:

Given $\dfrac{19}{5x+17} = \dfrac{19}{31}$

$\Rightarrow 19\times 31=19(5x+17)\ \Rightarrow 31=5x+17\ \Rightarrow 5x=14\ \Rightarrow x=\dfrac {14}{5}=2.8$

If $\cfrac{37}{4\sqrt{j}-19} = \cfrac{37}{17}$, then find the value of $j$.

  1. $64$

  2. $72.25$

  3. $81$

  4. $90.25$


Correct Option: C
Explanation:

Given, $\dfrac {{ 37 }}{{ (4\sqrt { j }  }-19)}={ 37 }/{ 17 }$

Since the numerators are equal, we can equate the denominators.
$\Rightarrow { (4\sqrt { j }  }-19)={ 17 }$
$\Rightarrow 4\sqrt { j } =36$
$\Rightarrow \sqrt { j } =9$
So, $j=81$
Hence, option C is correct.

If $\displaystyle \frac{a-b}{b}=\frac{3}{7}$, which of the following must also be true?

  1. $\displaystyle \frac{a}{b}=-\frac{4}{7}$

  2. $\displaystyle \frac{a}{b}=\frac{10}{7}$

  3. $\displaystyle \frac{a+b}{b}=\frac{10}{7}$

  4. $\displaystyle \frac{a-2b}{b}=-\frac{11}{7}$


Correct Option: B
Explanation:

Given: $\displaystyle \frac {a-b}{b}=\frac 37$

Separating denominators,

$\Rightarrow \displaystyle \frac ab -1=\frac 37$
$\Rightarrow \displaystyle \frac ab=\frac 37+1=\frac {10}{7}$
$\Rightarrow \dfrac {a}{b}=\dfrac {10}{7}$
Therefore, option B is correct.

If $( 2m) k=6$, then $mk = $

  1. $3$

  2. $4$

  3. $5$

  4. $6$


Correct Option: A
Explanation:

Given, $(2m)k$ $=$ $6$

$\Rightarrow 2$ $\times$ $m$ $\times$ $k$ $=$ $6$
$\Rightarrow m$ $\times$ $k$ $=$ $\dfrac {6}{2}$
$\Rightarrow mk$ $=$ $3$
Therefore, $mk$ $=$ $3$

If $a\neq 0$ and $\dfrac{5}{x}=\dfrac{5+a}{x+a}$, what is the value of $x$?

  1. $-5$

  2. $-1$

  3. $5$

  4. $2$


Correct Option: C
Explanation:

Given:

$a$ $\neq$ $0$ and $\dfrac {5}{x}$ $=$ $\dfrac {5 \space + \space a}{x \space + \space a}$
To find the value of $x$,
$\Rightarrow \dfrac {5}{x}$ $=$ $\dfrac {5 \space + \space a}{x \space + \space a}$
$\Rightarrow 5$ $\times$ $(x$ $+$ $a)$ $=$ $x$ $\times$ $(5$ $+$ $a)$
$\Rightarrow 5x$ $+$ $5a$ $=$ $5x$ $+$ $xa$
Get the co-efficients of $'x'$ on one-side,
$\Rightarrow 5x$ $+$ $xa$ $-$ $5x$ $=$ $5a$
$\Rightarrow xa$ $=$ $5a$
As $(a$ $\neq$ $0)$,    $x$ $=$ $5$
Therefore, the value of $'x'$ is $'5'$.

If one-third of a two digit number exceeds its one-fourth by $8$, then what is the sum of the digits of the number?

  1. $6$

  2. $13$

  3. $15$

  4. $17$


Correct Option: C
Explanation:

$(10x + y)\left (\dfrac {1}{3} - \dfrac {1}{4}\right ) = 8$
$\Rightarrow 10x + y = 96$
$\Rightarrow x = 9, y = 6$
$\therefore x + y = 15$

The total cost of three prizes is Rs. $2550$. If the value of second prize is $\left(\displaystyle\frac{3}{4}\right)^{th}$ of the first prize and the value of $3rd$ prize is $\displaystyle\frac{1}{2}$ of the second prize, then the value of the first prize is ___________.

  1. Rs. $1308$

  2. Rs. $1028$

  3. Rs. $1200$

  4. Rs. $1450$


Correct Option: C
Explanation:

let the first,second and third prizes be $x,y$ and $z$ respectively

$x+y+z=2550$.....(1)
$y=\dfrac{3}{4}x$
$z=\dfrac{1}{2}y= \dfrac{3}{8}x$
Putting values of $y$ and $z$ in equation (1)
$x+\dfrac{3}{4}x+\dfrac{3}{8}x=2550$

$\dfrac{8x+6x+3x}{8}=2550$
$17x=2550\times8$
$x=1200$

If $\sqrt {x-1}- \sqrt {x+1}+1 =0$, then $4x$ is equal to ____. 

  1. $4 \sqrt {-1}$

  2. $0$

  3. $5$

  4. $1 \dfrac {1}{4}$


Correct Option: C
Explanation:

We can write it as $\sqrt{x-1} + 1 = \sqrt{x+1}$

Squaring both sides we get,
$x-1 + 1 +2\sqrt{x-1} = x+1$
$\Rightarrow 2\sqrt{x-1} = 1$

Squaring both sides, we get
$4(x-1) = 1$ 
$\therefore 4x = 5$

Solve the following linear equations. If $\cfrac{x-5}{3} = \cfrac{x-3}{5}$, then $x  $is equal to

  1. $8$

  2. $6$

  3. $2$

  4. $3$


Correct Option: A
Explanation:
Given, $\cfrac { x - 5 }{ 3 }  =\cfrac { x-3 }{ 5 }$
Taking L.C.M., we get
$ \cfrac { 5x-25 }{ 15 }  = \cfrac { 3x - 9 }{ 15 } $
$5x - 25 = 3x -9$
Take $x$ terms on one sides and constants on another side, we get
$2x = 16$
$x = 8$

Solve the following linear equations. If $\cfrac{3t-2}{4}-\cfrac{2t+3}{3} = \cfrac{2}{3}-t$, then $t  $ is equal to

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: B
Explanation:
Given $\cfrac { 3t-2 }{ 4 } -\cfrac { (2t+3) }{ 3 }  = \cfrac { 2 }{ 3 }  - t$
L.C.M. of $4,3,3$ is $ 12$
$\Rightarrow  3(3t-2) - 4(2t+3) = 8 - 12t$
$\Rightarrow 9t - 6 -(8t +12) = 8 -12t$
$\Rightarrow t -18  = 8 -12t$
$\Rightarrow 13 t = 26$
$\Rightarrow t = 2$

Solve the following linear equations: $m-\cfrac{m-1}{2} = 1-\cfrac{m-2}{3}$

  1. $m = \cfrac{7}{5}$

  2. $m = \cfrac{2}{3}$

  3. $m = \cfrac{10}{3}$

  4. $m = \cfrac{3}{8}$


Correct Option: A
Explanation:
Given, $m - \cfrac { (m-1) }{ 2 }  = 1-\cfrac { (m-2) }{ 3 }$ 
L.C.M. of $2$ and $3$ is $6$
$6m -3(m-1) = 6 - 2(m-2)$
$ 6m - 3m +3 = 6 - 2m +4$
$5m = 7$
$ m = \cfrac { 7 }{ 5 }$

Which of the following is the solution of the equation $\displaystyle \frac{7y+4}{y+2}=\frac{-4}{3}$ ?

  1. $\displaystyle y = -\frac{4}{5}$

  2. $\displaystyle y = \frac{4}{5}$

  3. $\displaystyle y = -\frac{5}{4}$

  4. $\displaystyle y = \frac{5}{4}$


Correct Option: A
Explanation:

$\dfrac{7y+4}{y+2}=\dfrac{-4}{3}$
$21y+12=-4y-8$
$25y=-20$
$y=\dfrac{-20}{25} = \dfrac{-4}{5}$

Solve the following equations: $\cfrac{3y+4}{2-6y}=\cfrac{-2}{5}$

  1. $2$

  2. $-4$

  3. $4$

  4. $-8$


Correct Option: D
Explanation:

Given, $\dfrac{3y+4}{2-6y}=\dfrac{-2}{5}$

On cross multiplying, we get
$5(3y+4)=-2(2-6y)$
$\Rightarrow 15y+20=-4+12y$
$\Rightarrow 3y=-24$
$\Rightarrow y=-8$

Solve the following equations: $\cfrac{9x}{7-6x}=15$

  1. $x = \cfrac{25}{72}$

  2. $x = \cfrac{35}{33}$

  3. $x = \cfrac{45}{39}$

  4. $x = \cfrac{22}{45}$


Correct Option: B
Explanation:

Given, $\dfrac{9x}{7-6x}=15$
$9x=105-90x$

Add $90x$ on both the sides, we get
$9x+90x=105-90x+90x $
$99x=105$
$\therefore x=\dfrac{105}{99}=\dfrac{35}{33}$

Solve the following equations: $\cfrac{8x-3}{3x}=2$

  1. $x = \cfrac{5}{2}$

  2. $x = \cfrac{5}{6}$

  3. $x = \cfrac{9}{8}$

  4. $x = \cfrac{3}{2}$


Correct Option: D
Explanation:

Given, $\dfrac{8x-3}{3x}=2$

Multiply $3x$ on both the sides, we get
$8x-3=6x$
$8x-6x=3$
$2x=3$
$\therefore x=\dfrac{3}{2}$

Solve the following equations: $\cfrac{z}{z+15}=\cfrac{4}{9}$

  1. $2$

  2. $9$

  3. $12$

  4. $16$


Correct Option: C
Explanation:

Given, $\dfrac{z}{z+15}=\dfrac{4}{9}$

$9z=4(z+15)$
$9z=4z+60$
Subtract $4z$ from both the sides, we get
$9z-4z=4z+60-4z$
$5z=60$
$\therefore z=12$

In the expression $\cfrac { x+1 }{ x-1 } $ each $x$ is replaced by $\cfrac { x+1 }{ x-1 } $. The resulting expression, evaluated for $x=\cfrac { 1 }{ 2 } $ equals:

  1. $3$

  2. $-3$

  3. $1$

  4. $\dfrac12$


Correct Option: D
Explanation:

Expression is $\cfrac{x+1}{x-1}=y$

If $x$ is replaced by $\cfrac{x+1}{x-1}$
$\implies \cfrac{\cfrac{x+1}{x-1}+1}{\cfrac{x+1}{x-1}-1}$
$\implies \cfrac{x+1+x-1}{x+1-x+1}=\cfrac{2x}{2}$
The resultant expression is $x$.
When $x=\cfrac{1}{2}$
The value is $\cfrac{1}{2}$

A bag contains Rs. $90$ in coins. If coins of $50$ paise, $25$ paise, and $10$ paise are in the ratio $2 : 3: 5$, the number of $25$ paise coins in the bag is

  1. $80$

  2. $100$

  3. $120$

  4. $135$


Correct Option: C
Explanation:

Let  the number coins are $ 2x,3x$, and $5x$.

then 
rupees from 50 paise coins= $2x$ $\times$$ \dfrac{1}{2}=x$

rupees from 25 paise coins= $3x$ $\times$ $\dfrac{1}{4}=\dfrac{3}{4}x$

rupees from 10 paise coins=$5x$ $\times$ $\dfrac{1}{10}=\dfrac{x}{2}$
Now given $x+\dfrac{3}{4}x+\dfrac{x}{2}=90$

or, $4x+3x+2x=90$$\times$$4$
or, $9x=360$
or, $x=40$
so  number of 25 paise coins = $3x$=$3$x$40=120$

A candidate should score $45\%$ marks of the total marks to pass the examination. He gets $520$ marks and fails by $20$ marks. The total marks in the examination are

  1. $1000$

  2. $1100$

  3. $1200$

  4. $1400$


Correct Option: C
Explanation:

Given candidate should get $45\%$ of total marks to pass.

Let the total marks be $x$.
Given he got $520$ marks and fails by $20$ marks
Then to just pass
$540=\cfrac{45}{100}\times x$
$x=1200$
So, total marks $=1200$

Find the Solution  : $x - cy - bz = 0 $

                                 $ cx - y +az = 0 $
                                 $bx+ ay -z = 0 $

  1. $a^3 + b^3 + c^3 + 3abc = 1$

  2. $a^2 + b^2 + c^2 + 2abc = 1$

  3. $a^4 + b^4 + c^4 + 4abc = 1$

  4. $a^5 + b^5 + c^5 + 5abc = 1$


Correct Option: B
Explanation:

The given equations are 

$x-cy-bz = 0$
$cx - y + az = 0$
$bx + ay - z = 0$

Since x, y, z are not all zero, the system will have non-trivial solution if

$\triangle = \begin{vmatrix} 1 & -c & -b \ c & -1 & a \ b & a & -1\end{vmatrix} = 0$

or $1(1 - a^2) + c (-c -ab) -b (ac + b) = 0$

or $1 - a^2 - c^2 -abc - abc - b^2 = 0$

or $a^2 + b^2 + c^2 + 2abc = 1$

 Solve:$\dfrac{{7x - 2y}}{{xy}} = 5$ and $x=2y$

  1. $x=12/5 , y=6/5$

  2. $x=2/5 ,  y=1/5$

  3. $x=9 , y=9/2$

  4. $x=2 , y=1$


Correct Option: A
Explanation:
$\dfrac{7x-2y}{xy}=5$

$\Rightarrow 7x-2y=5xy$

$\Rightarrow \dfrac{7x}{xy}-\dfrac{2y}{xy}=5$

$\Rightarrow \dfrac{7}{y}-\dfrac{2}{x}=5$

Substitute $x=2y$ in the above equation, we get

$\dfrac{7}{y}-\dfrac{2}{2y}=5$

$\Rightarrow \dfrac{14-2}{2y}=5$

$\Rightarrow \dfrac{12}{2y}=5y$

$\Rightarrow 5y=6$

$\therefore y=\dfrac{6}{5}$

Put $y=\dfrac{6}{5}$ in $x=2y$ we get

$x=2\times \dfrac{6}{5}=\dfrac{12}{5}$

$\therefore x=\dfrac{12}{5}\,\,,y=\dfrac{6}{5}$

$\dfrac{2x-3}{2}-\dfrac{(x+1)}{3}=\dfrac{3x-8}{4}$

  1. $x=4$

  2. $x=-2$

  3. $x=2$

  4. $x=-4$


Correct Option: C
Explanation:

$\cfrac { 2x-3 }{ 2 } -\cfrac { x+1 }{ 3 } =\cfrac { 3x-8 }{ 4 } \ =>(6x-9-2x-2)\times 4=(3x-8)\times 6\ =>16x-44=18x-48\ =>2x=4\ =>x=2$

Solve: 

$3(x+1)=6$

  1. $x=-1$

  2. $x=\dfrac{1}{3}$

  3. $x=\dfrac{-1}{3}$

  4. $x={1}$


Correct Option: D
Explanation:
$3(x+1)=6$

$ \Rightarrow 3x+3=6$

$ \Rightarrow 3x=3$

$ \Rightarrow x=1$

The equation $x-\dfrac {8}{|x-3|}=3--\dfrac {8}{|x-3|}$ has

  1. Only one solution

  2. infinite solution

  3. no solution

  4. two solution


Correct Option: A
Explanation:

Consider the following equation.

$ x-\dfrac{8}{|x-3|}=3-\dfrac{8}{|x-3|} $

$ x-\dfrac{8}{\pm \left( x-3 \right)}=3-\dfrac{8}{\pm \left( x-3 \right)} $

$ x-\dfrac{8}{\left( x-3 \right)}=3-\dfrac{8}{\left( x-3 \right)} $

$ x=3 $

 $ x+\dfrac{8}{\left( x-3 \right)}=3+\dfrac{8}{\left( x-3 \right)} $

$ x=3 $

Hence, this is the correct answer.

If $t = x+2$, find the value of x .If $2t-7 +\dfrac{3(t-1)}{2}=3$

  1. $\frac{23}{7}$

  2. $\frac{3}{7}$

  3. $\frac{9}{7}$

  4. $\frac{37}{7}$


Correct Option: C
Explanation:

$(2t-7)+(\frac{3t-3}{2})=3\\(\frac{4t-14+3t-3}{2})=3\\7t-17=6\\\therefore 7t=6+17\\t=(\frac{23}{7})\\then x+2=(\frac{23}{7})\\\therefore x=(\frac{23}{7})-2\\=(\frac{23-14}{7})\\=(\frac{9}{7})$

Which equation is  non- linear 

  1. $x + 1 = 2x - 3$

  2. $9 - 5{x^2} = 4$

  3. $- 2x - 5 - \left( {x + 4} \right) = - 9$

  4. ${x} - \frac{1}{{15}} = 2$


Correct Option: B
Explanation:

Linear equations are those which have the degree of equation as $1$


Here every equation has the degree as $1$

Except $9-5x^2=4$

So its is not a linear equation 

A triangular number which is the sum of the square of two consecutive odd numbers is?

  1. $10$

  2. $15$

  3. $21$

  4. $28$


Correct Option: A
Explanation:
Let the two consecutive odd numbers be $n-2,n$
$(n-2)^2+n\\ n^2+4-4n+n^2\\2n^2+4-4n$
If we put $n=1\\ 2(1)-4+4=2\\n=2\\ 2(4)+4-4=4\\n=3\\2(9)+4-4(3)\\=18+4-12\\=10$
$\therefore $ Triangular no. is $10$
For any other value of $n$ the triangular no. is not in the required options
$\therefore$ we take $n=3$ and triangular no. is $10$

If $\displaystyle \frac{a}{3y}+\frac{3b}{x}=7$ and $\displaystyle a+1=2b+1=x=5,$ find the value of $'y'.$


  1. $\displaystyle \frac{10}{77}$

  2. $\displaystyle \frac{22}{69}$

  3. $\displaystyle \frac{20}{87}$

  4. $\displaystyle \frac{14}{93}$


Correct Option: C
Explanation:

Since, $ a+1=2b+1=x=5 $
$ => a = 5 - 1 = 4 $
$ 2b = 5 - 1 = 4 => b = 2 $

$ x = 5 $

Substituting these values in $ \frac { a }{ 3y } +\frac { 3b }{ x } =7 $, we get $ \frac { 4 }{ 3y } +\frac { 6 }{ 5 } =7 $
$ => \frac { 4 }{ 3y } = 7 - \frac { 6 }{ 5 } $
$ => \frac { 4 }{ 3y } =  \frac { 29 }{ 5 } $
$ => \frac { 3y }{ 4 } =  \frac { 5 }{ 29 } $
$ => y = \frac {20}{87} $

If $(2ax + 1) (3x + 1) = 6a (x + 1)$ and $x = 1$, find the value of $a$.

  1. $1$

  2. $4$

  3. $3$

  4. $2$


Correct Option: A
Explanation:

    $(2ax + 1) (3x + 1) = 6a (x + 1) $
$=>[2a(1)+1][3(1)+1]=6a[(1+1)]$
$=>(2a+1)(3+1)=6a(2)$
$=>(2a+1)(4)=12a$
$=>8a+4=12a$
$=>8a-12a=-4$
$=>-4a=-4$
$=>a=1$

Solve for $x$ : $\displaystyle \sqrt{\frac{x\, -\,2}{x\, +\, 1}}\, =\, \frac{1}{2}$

  1. 3

  2. 2

  3. 1

  4. 0


Correct Option: A
Explanation:
$\cfrac { \sqrt { x-2 }  }{ \sqrt { x+1 }  }  = \cfrac { 1 }{ 2 }$
Squaring on both sides we get
$\cfrac { x-2 }{ x+1 }  = \cfrac { 1 }{ 4 } \\ 4(x-2) = (x+1)\\ 4x - 8 = x +1\\ 3x = 9\\ x = 3$

Solve for $x$ : $\displaystyle \frac{4}{3\sqrt{x}}\, =\, \frac{1}{2}$

  1. ${5}\dfrac{6}{9}$

  2. ${7}\dfrac{1}{9}$

  3. $14$

  4. $27$


Correct Option: B
Explanation:
$\cfrac{4}{3\sqrt{x}}=\cfrac{1}{2}$
Squre both the sides we get
$\cfrac{16}{9x}=\cfrac{1}{4}$
$\Rightarrow 9x=16\times 4$
$\Rightarrow 9x=64 \Rightarrow x=\cfrac{64}{9}$
$\Rightarrow x=7\cfrac{1}{9}$

Solve:
$x\, +\, y\, =\, 7xy$
$2x\, -\, 3y\, =\, -xy$

  1. $x\, =\, \displaystyle \frac{1}{8}$ and $y\, =\, \displaystyle \frac{1}{4}$

  2. $x\, =\, \displaystyle \frac{1}{7}$ and $y\, =\, \displaystyle \frac{1}{4}$

  3. $x\, =\, \displaystyle \frac{1}{2}$ and $y\, =\, \displaystyle \frac{1}{4}$

  4. $x\, =\, \displaystyle \frac{1}{3}$ and $y\, =\, \displaystyle \frac{1}{4}$


Correct Option: D
Explanation:

Given equations are 

$x+y=7xy$ ....(1)
and $2x-3y=-xy$ ....(2)
Multiply equation (1) by $3$, we get
$3x+3y=21xy$ ....(3)
Add equations (2) and (3),
$5x=20xy$
$\Rightarrow y=\dfrac {1}{4}$
Put this value in equation (1), we get
$x+\dfrac {1}{4}=7x\times \dfrac {1}{4}$
$\Rightarrow x-\dfrac {7x}{4}=-\dfrac {1}{4}$
$\Rightarrow \dfrac {3x}{4}=\dfrac {1}{4}$
$\Rightarrow x=\dfrac {1}{3}$
Therefore, solution is $x=\dfrac {1}{3}, y=\dfrac {1}{4}$.

Find the value of $a$, if $x = 0.5$ is a solution of equation $ax^{2}\, +\, (a\, -\, 1)\,
x\, +\, 3\, =\, a$.

  1. $24$

  2. $15$

  3. $10$

  4. $8$


Correct Option: C
Explanation:

Given equation is $ax^2+(a-1)x+3=a$
The equation can be written as
$ax^2+(a-1)x+3-a=0$
Since $x=0.5$ is a solution of the equation, 
$a(0.5)^2 + (a-1)(0.5) + 3 - a = 0$
$\Rightarrow 0.25a + 0.5a - 0.5 + 3 - a = 0$
$\Rightarrow   (0.25 + 0.5 -1)a - 0.5+3 = 0$
$\Rightarrow  -0.25a + 2.5 = 0$
$\Rightarrow  0.25a = 2.5$
$\Rightarrow  a = \dfrac{2.5}{0.25}$
$\Rightarrow a = 10$

The solution of the equation $\displaystyle \frac{2x+4}{3x-1}=\frac{4}{3}$ is 

  1. 6

  2. 4

  3. $\displaystyle \frac{8}{3}$

  4. $\displaystyle \frac{3}{4}$


Correct Option: C
Explanation:

Given equation is $\displaystyle \frac{2x+4}{3x-1}=\frac{4}{3}$
Cross multiplying, we get
$3(2x+4)=4(3x-1)$
$\Rightarrow 6x+12=12x-4\Rightarrow 6x=16$
$\Rightarrow \displaystyle x=\frac{16}{6}=\frac{8}{3}$.
Hence, the solution is $x=\cfrac{8}{3}$.

Find the value of $y$ in the equation : 
$\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$

  1. $18$

  2. $9$

  3. $-38$

  4. $-32$


Correct Option: C
Explanation:

Given, $\displaystyle \frac{(2-3y)+4y}{9y-(8y+7)}=\frac{4}{5}$
$\Rightarrow \displaystyle \frac{2+y}{y-7}=\frac{4}{5}$

$\Rightarrow 5(2+y)=4(y-7) $ ....( cross multiplying )
$\Rightarrow 10+5y=4y-28$
$\Rightarrow y=-38$
Hence,. the solution is $x=-38$

A combination of locks requires 3 numbers to open. The second number is $\displaystyle 2d + 5$ greater than the first number. The third number is $\displaystyle 3d - 20$ less than the second number. The sum of the three numbers is $\displaystyle 10d + 9$. The first number is 

  1. $\displaystyle 5d-11$

  2. $\displaystyle 3d-7$

  3. $\displaystyle 2d+19$

  4. $\displaystyle 3d-11$


Correct Option: B
Explanation:

Let the first number be $x$
Then, second number = $x + 2d + 5$
Third number = $x + 2d + 5 -(3d -20)$ = $x -d + 25$
Sum of the three numbers = $ x + x+2d + 5 + x - d +25$ = $ 3x + d +30$

Thus, $ 3x + d +30$ = $10d + 9$
$3x = 10d + 9 - d - 30$ = $9d - 21$
$x = 3d - 7$

If $\displaystyle \sqrt{\left ( x-1 \right )\left ( y+2 \right )}=7$, $x$ and $y$ being positive whole numbers, then the values of $x$ and $y$ are, respectively

  1. $8,5$

  2. $15,12$

  3. $22,19$

  4. $6,8$


Correct Option: A
Explanation:

$ \sqrt{(x-1)(y+2)}=7\Rightarrow (x-1)(y+2)=7^{2}$
$ \Rightarrow (x-1)=7:and:(y+2)=7$
$ x=8$ and $\displaystyle y=5$

If $\displaystyle 4=\sqrt{x+\sqrt{x+\sqrt{x+....,}}}$ then the value of x will be 

  1. 20

  2. 16

  3. 12

  4. 8


Correct Option: C
Explanation:

$\displaystyle 4 = \sqrt{x+\sqrt{x+\sqrt{x+....}}}$
$\displaystyle \Rightarrow 4=\sqrt{x+4}$
$\displaystyle \Rightarrow 16={x+4}$
$\displaystyle \Rightarrow 16=x+4\Rightarrow x=12.$

$\displaystyle \sqrt{6+\sqrt{6+\sqrt{6+...}}}$ equals 

  1. $\displaystyle 6^{\frac{2}{3}}$

  2. 6

  3. $\displaystyle 6^{\frac{1}{3}}$

  4. 3


Correct Option: D
Explanation:

Let  $ x =\sqrt{6+\sqrt{6+\sqrt{6+...}}}$


sqare it on both sides


$x^2=\sqrt{6+\sqrt{6+\sqrt{6+...}}} = 6+x$



$\Rightarrow x^2=6+x $


$\Rightarrow x^2-x-6=0 $


$\Rightarrow x^2-3x+2x-6=0 $


$\Rightarrow x(x-3)+2(x-3)=0 $


$\Rightarrow (x-3)(x+2)=0 $


$either  x-3=0---> x=3 $


$ or  x+2=0 ----> x=-2 $


since result of sqrt of anything will be positive only, therefore,


Answer $x=3
$





If $\displaystyle \frac{x^2\, -\, (x\, +\, 1)(x\, +\, 2)}{5x\, +\, 1}\, =\, 6$, then $x$ is equal to

  1. $\displaystyle \frac{8}{33}$

  2. $\displaystyle \frac{8}{3}$

  3. $\displaystyle \frac{-8}{33}$

  4. $\displaystyle \frac{-6}{33}$


Correct Option: C
Explanation:

Given, $\displaystyle \frac{x^2\, -\, (x\, +\, 1)(x\, +\, 2)}{5x\, +\, 1}\, =\, 6$
$\Rightarrow x^2\, -\, (x^2\, +\, 3x\, +\, 2)\, =\, 6(5x\, +\, 1)$, .....(on cross multiplying )
$\Rightarrow x^2\, -\, x^2\, -\, 3x\, -\, 2\, =\, 30x\, +\, 6$
$\Rightarrow -3x - 2 = 30x + 6$
$\Rightarrow -3x - 30x = 6 + 2 \Rightarrow -33x = 8$
$\Rightarrow x\, =\, \displaystyle \frac{-8}{33}$
Hence, the solution is, $x=-\cfrac{8}{33}$.

After receiving two successive raises Hrash's salary became $\dfrac {15}{8}$ times of his initial salary. By how much percent was the salary raised the first time if the second raise was twice as much as high (in percent) as the first ?

  1. $15 \%$

  2. $20 \%$

  3. $25 \%$

  4. $30 \%$


Correct Option: C
Explanation:

Let initial salary was Rs. $100$
After two raise it become $\dfrac {15}{8}$ i.e. $\dfrac {(15 \times 100) }{ 8} =$ Rs. $187.5$
Raise $= 187.5 - 100 = 87.5$
Using formula,
[( first raise  + second raise) + (first raise * 2nd raise) / 100]  = 87.5
$x + 2x +\dfrac { (2x ^2)}{100} = 87.53$
$300x + 2x ^2 = 8750$
$x ^2 + 150x = 4375$
$x ^2 + 150x - 4375 = 0$
$x ^2 + 175x - 25x - 4375 = 0$
$x = -175, 25$ .....(Negative value is not possible)
So, $x= 25\%$
Second raise was $2x = 2 \times 25 = 50\%$.

Reduce the following linear equation: $6t - 1 = t - 11$

  1. $t=-1$

  2. $t=-2$

  3. $t=-3$

  4. $t=-4$


Correct Option: B
Explanation:

$6t - 1 = t - 11$
On transposing $t$ to the L.H.S.,  we obtain
$6t - t = -11 + 1$
$5t = -10$
$t = -2$

Solve the equation: $\dfrac{a+4}{6-3a}=\dfrac{1}{3}$

  1. $a=1$

  2. $a=-1$

  3. $a=2$

  4. $a=-2$


Correct Option: B
Explanation:

Given, $\dfrac{a+4}{6-3a}=\dfrac{1}{3}$


On multiplying $6 - 3a$ on both sides, we get

$a+4=\dfrac{6-3a}{3}$
$3a + 12 = 6 - 3a$
$6a = 6 - 12$
$6a = -6$
$a = -1$

Solve the linear equation: $5x - 12 = 2x + 18$

  1. $x=8$

  2. $x=9$

  3. $x=10$

  4. $x=11$


Correct Option: C
Explanation:

$5x - 12 = 2x + 18$
On transposing $2x$ to the L.H.S and $12$ to RHS, we obtain
$5x - 2x = 18 + 12$
$3x = 30$
$x = 10$

Reduce the linear equation: $x + 3-\dfrac{2x}{3}+\dfrac{x}{6}=0$

  1. $x=12$

  2. $x=10$

  3. $x=8$

  4. $x=-6$


Correct Option: D
Explanation:

Given, $x + 3-\dfrac{2x}{3}+\dfrac{x}{6}=0$
L.C.M of the denominator $3$ and $6$ is $6$.
Multiplying both the sides by $6$, we get
$6x + 18 - 4x + x = 0$
$7x - 4x + 18 = 0$
$3x = -18$
$x = -6$

Reduce the linear equation: $\dfrac{x}{2}+\dfrac{2x}{4}= 10$

  1. $x=6$

  2. $x=7$

  3. $x=8$

  4. $x=10$


Correct Option: D
Explanation:

Given, $\dfrac{x}{2}+\dfrac{2x}{4}= 10$
L.C.M of the denominator $2$ and $4$ is $4$.
Multiplying both the sides by $4$, we get
$2x + 2x = 40$
$4x = 40$
$x = 10$

A brand new car costs $ \$35,000$. For the first $50,000$ miles, it will depreciate approximately $\$0.15$ per mile driven. For every mile after that, it will depreciate by $\$0.10$ per mile driven until the car reaches its scrap value. Find the net worth of the car after it is driven $92,000$ miles.

  1. $\$11,300$

  2. $\$13,800$

  3. $\$17,000$

  4. $\$23,300$


Correct Option: D
Explanation:

Given the cost of car $=\$35000$

First $50000$ miles it will depreciated app $\$ 0.15$ per mile 
And After that it will depreciate by $\$0.10$ per miles 
Then  depreciate after $50000$ miles $=$  $50000\times $0.15=$7500$
And depreciate after $92000-50000=42000$ miles $= $ $42000\times $0.10=$4200$
Total  depreciate after $92000$ mole $=7500+4200=\$11700$
Then worth after it is driven $92000=35000-11700=\$23300$

If  $\sqrt{x+16} = x-4$, then the value of extraneous solution of the above equation is:

  1. $0$

  2. $4$

  3. $5$

  4. There are no extraneous solutions


Correct Option: A
Explanation:
Given, $\sqrt{x+16}=x-4$
Squaring both sides, we get
$x+16=(x-4)^{2}$
$\Rightarrow x+16=x^{2}-8x+16$
$\Rightarrow x^{2}-8x-x=16-16$
$\Rightarrow x^{2}-9x=0$
$\Rightarrow x(x-9)=0$
Then $x=0$ or $x=9$

A neighborhood recreation program serves a total $280$ children who are either $11$ years old or $12$ years old. The sum of the children's ages is $3,238$ years. How many $11$ year old children does the recreation program serve?

  1. $54$

  2. $122$

  3. $132$

  4. $158$

  5. $208$


Correct Option: B
Explanation:

Let the number of $11$ years children is $x$ and $12$ year children is $y$.

$\therefore  11x+12y=3238$.....(1)
$\Rightarrow x+y=280$
$\Rightarrow y=280-x$
Sustitute the value of $y$ in (1)
$\Rightarrow 11x+12(280-x)=3238$
$\Rightarrow 11x+3360-12x=3238$
$\Rightarrow 11x-12x=3238-3360$
$\Rightarrow x=122$
Hence, number of $11$ year children are $122$. 

If $3^{2x + 2} = 27^{2}$, find the value of $x$.

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: C
Explanation:

Given, $3^{2x+2}=27^{2}$ 

$3^{2x+2}=3^{6}$
Taking $\log$ on both sides give us 
$(2x+2)\log 3=6\log 3$
$\Rightarrow 2x+2=6$ 
$\Rightarrow x=2$

If $2^{x} + 2^{x + 2} = 40$, then the value of $x$ is

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: C
Explanation:

Given, ${ 2 }^{ x }+{ 2 }^{ x+2 }=40\ \Rightarrow { 2 }^{ x }+{ 2 }^{ 2 }{ 2 }^{ x }=40\ \Rightarrow { 2 }^{ x }+(4){ 2 }^{ x }=40\ \Rightarrow (5){ 2 }^{ x }=40\ \Rightarrow { 2 }^{ x }=8\ \Rightarrow { 2 }^{ x }={ 2 }^{ 3 }$

So, $ x=3$

When a number $x$ is subtracted from $36$ and the difference is divided by $x$, the result is $2$. Find the value of $x$.

  1. $2$

  2. $4$

  3. $6$

  4. $12$


Correct Option: D
Explanation:

According to the question:

$ \dfrac { 36-x }{ x } =2\ \Rightarrow 36-x=2x\ \Rightarrow 36=3x$
$ \Rightarrow x=12$

If $3^{n - 3} + 3^{2} = 18$, calculate the value of $n$.

  1. $1$

  2. $2$

  3. $3$

  4. $4$

  5. $5$


Correct Option: E
Explanation:

Given, ${ 3 }^{ n-3 }+{ 3 }^{ 2 }=18$

$ { 3 }^{ n-3 }+9=18$
${ 3 }^{ n-3 }=18-9=9$
$ { 3 }^{ n-3 }={ 3 }^{ 2 }$
So, $ n-3=2$
$n=5\ $

Let $a, b$ and $c$ be non-zero numbers such that $c$ is $24$ greater than $b$ and $b$ is $24$ greater than $a$. If $\dfrac {c}{a} = 3$, then find the value of $b$.

  1. $-48$

  2. $-24$

  3. $24$

  4. $48$


Correct Option: D
Explanation:

Let the value of $a$ be $x$

Thus, $b$ would be $24+x$
and $c$ is $24+24+x=48+x$.
Given, $\dfrac {c}{a}=3$
Therefore, $\dfrac {48+x}{x}=3$
$\Rightarrow 2x=48$
$\Rightarrow x=24$
Value of $b$ is $24+x=24+24=48$.

If $\sqrt[3]{8x+6} = -3$, calculate the value of $x$.

  1. $-4.125$

  2. $-2.625$

  3. $-1.875$

  4. $-1.125$


Correct Option: A
Explanation:

Given $\sqrt [ 3 ]{ 8x+6 } =-3$

Cubing on both sides, we get
$8x+6=(-3)^3$
$\Rightarrow 8x+6=-27\ \Rightarrow x=(-27-6)/8=-33/8\ \Rightarrow x=-4.125$

If $\sqrt[4]{\dfrac{x+1}{2}} = \dfrac{1}{2}$, then find $x $.

  1. $-0.969$

  2. $-0.875$

  3. $0$

  4. $0.875$


Correct Option: B
Explanation:
Given is $\sqrt [ 4 ]{ \dfrac { x+1 }{ 2 }  } = \dfrac { 1 }{ 2 } $
Now Raising the power to $4$ on both sides, we get
$\dfrac { x+1 }{ 2 } = { \left (\dfrac {1}{2}\right) }^{ 4 }\\ \Rightarrow \dfrac { x+1 }{ 2 } =\dfrac { 1 }{ 16 } \\ \Rightarrow 16x+16=2\\ \Rightarrow x=-0.875$

If $\dfrac{5}{x+3} = \dfrac{1}{x}+\dfrac{1}{2x}$, calculate the value of $x$.

  1. $\dfrac{3}{14}$

  2. $\dfrac{1}{3}$

  3. $\dfrac{6}{13}$

  4. $\dfrac{3}{4}$

  5. $\dfrac{9}{7}$


Correct Option: E
Explanation:

Given, $\dfrac { 5 }{ x+3 } =\dfrac { 1 }{ x } +\dfrac { 1 }{ 2x } $

Taking RHS:
$\dfrac { 1 }{ x } +\dfrac { 1 }{ 2x } $
LCM of these is $2x$
$\Rightarrow \dfrac { 2 }{ 2x } +\dfrac { 1 }{ 2x } \ \Rightarrow \dfrac { 3 }{ 2x } $
Now taking LHS:
$\dfrac { 5 }{ x+3 } $
It is given, LHS $=$ RHS
$\dfrac { 5 }{ x+3 } =\dfrac { 3 }{ 2x } $
$\Rightarrow 5\times 2x=3\times (x+3)\ \Rightarrow 10x=3x+9\ \Rightarrow x=\dfrac {9}{7}$

One of the requirements for becoming a court reporter is the ability to type  $225$  words per minute. Donald can currently type  $180$  words per minute, and believes that with practice he can increase his typing speed by  $5$  words per minute each month. Which of the following represents the number of words per minute that Donald believes he will be able to type  $m $ months from now?

  1. $5 + 180 m$

  2. $ 225 + 5 m$

  3. $ 180 + 5 m$

  4. $ 180 - 5 m$


Correct Option: C
Explanation:

With $ m $ months, Donald can type $5m $ more words per  minutes on top of $ 180 $ words.

So, total number of words Donald believes that he will be able to type $ m $ months from now $ = 180 + 5m $

If $\sqrt[3]{5j - 7} = -\cfrac{1}{2}$, calculate the value of $j$.

  1. $1.375$

  2. $2.118$

  3. $2.599$

  4. $5.125$


Correct Option: A
Explanation:

Given, $\sqrt [ 3 ]{ 5j-7 } =\dfrac { -1 }{ 2 } $

On cubing on both sides, we get
$5j-7=\dfrac { -1 }{ 8 } $
$\Rightarrow 5j=\dfrac { 55 }{ 8 } $
$\Rightarrow j=\dfrac { 11 }{ 8 } $
$\Rightarrow  j = 1.375$
Hence, option A is correct.

If $\dfrac {2}{3x + 12} = \dfrac {2}{3}$, then the value of $x + 4 $ is

  1. $\dfrac {1}{2}$

  2. $1$

  3. $\dfrac {3}{2}$

  4. $2$


Correct Option: B
Explanation:

Given $\dfrac { 2 }{ 3x+12 } =\dfrac { 2 }{ 3 } $

$\Rightarrow 3(2)=2(3x+12)$
$\Rightarrow 6=6x+24$
$\Rightarrow 6x=-18$
$\Rightarrow x=-3$
Therefore $x+4=-3+4=1$

If $\sqrt[5]{\cfrac{g-1}{4}} = \cfrac{1}{3}$, then find the value of $g$.

  1. $0.984$

  2. $0.996$

  3. $1.004$

  4. $1.016$


Correct Option: D
Explanation:

Given, $\sqrt [ 5 ]{ \dfrac { g-1 }{ 4 }  } =\dfrac { 1 }{ 3 } $

$ \Rightarrow  $ $\cfrac{g-1}{4} = \cfrac { 1 }{ { 3 }^{ 5 } } =\cfrac { 1 }{ 243 } $
$ \Rightarrow  $ $ g-1 = \cfrac{4}{243}$
$ \Rightarrow  $ $ g = 1 + \cfrac{4}{243} = 1.016$

The area of square $ABCD$ is three-fourths the area of parallelogram $EFGH$. The area of parallelogram $EFGH$ is one-third the area of trapezoid $IJKL$. If square $ABCD$ has an area of $125$ square feet, calculate the area of trapezoid $IJKL$, in square feet.

  1. $75$

  2. $225$

  3. $350$

  4. $500$


Correct Option: D
Explanation:

Given, area of square $ABCD$ is three fourth of area of parallelogram $EFGH$,

And the area of parallelogram $EFGH$ is one-third of the area of trapezoid $IJKL$ and area of square $ABCD$ is $125$.
Let the area of trapezoid $IJKL$ is $x$
Then  area of  parallelogram $EFGH =$ $\dfrac{1}{3}x$
And  area of square $ABCD=$ $\dfrac{3}{4}$ area of  parallelogram $EFGH=$ $\dfrac{3}{4}\times \dfrac{1}{3}x=\dfrac{1}{4}x$
But area of square $ABCD =125$
$\therefore \dfrac{1}{4}x=125$
$\Rightarrow x=500$
So, area of trapezoid $IJKL=500$.

Find the value of $x: \dfrac {1}{x} + \dfrac {4}{5x} = \dfrac {2}{x + 5}$

  1. $0.71$

  2. $3.57$

  3. $5.8$

  4. $45$


Correct Option: D
Explanation:
Given  $\dfrac { 2 }{ x+5 } =\dfrac { 1 }{ x } +\dfrac { 4 }{ 5x } $
Taking RHS:

$\dfrac { 1 }{ x } +\dfrac { 4 }{ 5x } $

LCM is $5x$
$\Rightarrow \dfrac { 5 }{ 5x } +\dfrac { 4 }{ 5x } \\ \Rightarrow \dfrac { 9 }{ 5x } $
Now taking LHS:
$\dfrac { 2 }{ x+5 } $
LHS $=$ RHS
$\dfrac { 2 }{ x+5 } =\dfrac { 9 }{ 5x } $
$\Rightarrow 5x\times 2=9(x+5)\\ \Rightarrow 10x=9x+45\\ \Rightarrow x=45$

Find the value of $\dfrac {4}{y} + 4$ given that $\dfrac {4}{y} + 4 = \dfrac {20}{y} + 20$

  1. $-1$

  2. $0$

  3. $1$

  4. $4$


Correct Option: B
Explanation:
The value of $\frac{20}{y}+20=5\left ( \frac{4}{y}+4 \right )$
Given $\frac{4}{y}+4=5\left ( \frac{4}{y}+4 \right )$
This when possible then value of $\frac{4}{y}+4=0$

Compute the approximate value of $x$: $\sqrt [3]{\dfrac {2x + 3}{5}} = \dfrac {2}{3}$

  1. $-0.76$

  2. $-0.69$

  3. $-0.67$

  4. $0.69$

  5. $0.76$


Correct Option: A
Explanation:
Given is $\sqrt [ 3 ]{ \dfrac { 2x+3 }{ 5 }  } = \dfrac { 2 }{ 3 } $
Now raising the power to $3$ on both sides, we get
$\dfrac { 2x+3 }{ 5 } = { (2/3) }^{ 3 }\\ \Rightarrow \dfrac { 2x+3 }{ 5 } =\dfrac { 8 }{ 27 } \\ \Rightarrow 54x+81=40\\ \Rightarrow x=-0.76$

If $\dfrac{3}{9}=\dfrac{3}{x+2}$, what is the value of $x$?

  1. $-\dfrac{5}{9}$

  2. $\dfrac{7}{3}$

  3. $3$

  4. $7$

  5. $\dfrac{25}{3}$


Correct Option: D
Explanation:

Given, $\dfrac {3}{9}=\dfrac {3}{x+2}$

On cross multiplying, we get
$\Rightarrow 3(x+2)=9(3)$
$\Rightarrow x=9-2$
$\Rightarrow x=7$

The square roots of Radhas and Krishs ages have a sum of $7$ and a difference of $1$. If Radha is older than Krish, how old is Radha?

  1. $13$

  2. $4$

  3. $9$

  4. $16$


Correct Option: D
Explanation:

Let the ages of Radha and krish be $x^{2}$ and $y^{2}$ respectively

Then square roots of their age will be $x$ and $y$ respectively.

Then according to the question,

$x+y=7$ ..(1)

$x-y=1$...(2)

Adding (1) and (2)

$2x=8$

$x=4$

From (1)

$4+y=7$

$y=3$

Therefore $x^{2}=16$ and $y^{2}=9$

As radha is older,hence age of radha is 16years.

What is the solution of $\displaystyle \frac{x-5}{2} - \frac{x-3}{5} = \frac{1}{2}$?

  1. $x = 5$

  2. $x = 7$

  3. $x = 8$

  4. $x = 9$


Correct Option: C
Explanation:

Given ,$\dfrac{x-5}{2}-\dfrac{x-3}{5}=\dfrac{1}{2}$

$\dfrac{5x-25-2x+6}{5\times 2}=\dfrac{1}{2}$

$3x-19=5$

$3x=24$

$x=8$

If $x=\displaystyle\frac{1}{\displaystyle 2-\frac{1}{\displaystyle 2-\frac{1}{2-x}}}, (x\neq 2)$, then the value of x is ________?

  1. $1$

  2. $3$

  3. $2$

  4. $5$


Correct Option: A
Explanation:

$x=\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}$


$\Rightarrow x=\dfrac{1}{2-\dfrac{2-x}{3-2x}}$


$\Rightarrow x=\dfrac{3-2x}{4-3x}$

On solving, we get $x=1$
So, Option (A)

Meera bought packs of trading cards that contain $10$ cards each. She gave away $7$ cards.
$x=$ Number of packs of trading cards
Which expression shows the number of cards left with Meera?

  1. $10x-7$

  2. $7x-8$

  3. $5-10x$

  4. $8-5x$


Correct Option: A
Explanation:

$\Rightarrow$  We have given, $x$ is number of packs of trading cards.

$\Rightarrow$  Pack of trading cards contain $10$ cards each.
$\therefore$    Total number of cards = $10x$
$\Rightarrow$  From total cards she gave away $7$ cards. 
$\therefore$    Number of cards left with Meera = $10x-7$

Solve for x : $\dfrac{(x + 2)(2x - 3) - 2x^2 + 6}{x - 5} = 2.$

  1. $5$

  2. $10$

  3. $15$

  4. $\frac{20}{3}$


Correct Option: B
Explanation:

$\dfrac{(x+2)(2x-3)-2x^2+6}{x-5}=2$


$\Rightarrow\dfrac{(2x^2+x-6)-2x^2+6}{x-5}=2$

$\Rightarrow\dfrac{x}{x-5}=2$

$\Rightarrow x=2x-10$

$\Rightarrow x=10$.  $[B]$

The present ages of a father and his son are in the ratio $7 : 3$ and the ratio of their ages will be $2 : 1$ after $10 $ years. Then, the present age of father (in years) is -

  1. $42$

  2. $56$

  3. $70$

  4. $77$


Correct Option: C
Explanation:

Let the present ages of father and son be $7x$ and $3x$.

After $10$ years, their ages will be $7x+10$ and $3x+10$.
According to the question, we have
$\dfrac {7x+10}{3x+10}=\dfrac {2}{1}$
$1(7x+10)=2(3x+10)$
$7x+10=6x+20$
$7x-6x=20-10$
$x=10$
Then present age of father is $7x$, i.e. $7\times 10=70$ years.

A number when added to its half gives $36$. Find the number.

  1. $24$

  2. $28$

  3. $20$

  4. $18$


Correct Option: A
Explanation:

Let the number be $x$

$\Rightarrow x+\cfrac { x }{ 2 } =36\ \Rightarrow \cfrac { 3x }{ 2 } =36\ \Rightarrow x=24$

$A$ has certain amount in his account. He gives half of this to his eldest son and one third of the remaining to his youngest son. The amount left with him now is

  1. $\cfrac { 1 }{ 3 } $ of the original

  2. $\cfrac { 2 }{ 5 } $ of the original

  3. $\cfrac { 3 }{ 4 } $ of the original

  4. $\cfrac { 1 }{ 6 } $ of the original


Correct Option: A
Explanation:

$x-\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 6 } x=\cfrac { 1 }{ 3 } x$

When an iron rod is cut into equal pieces of $30$ cm each, a piece of $4$ cm is left out. When cut into equal pieces of $29$ cm, a piece of $13$ cm is left out. The minimum length of rod is

  1. $270$ cm

  2. $272$ cm

  3. $274$ cm

  4. $280$ cm


Correct Option: C
Explanation:

Let the length of rod be x.

If rod cuts in 30 cm each, 4cm from left
$\implies x=(n _1)(30)+4-------(1)$
If cuts in 29 cm each 13 cm left
$x=(n _2)(29)+13------------(2)$
$\implies n _1(30)+4=(n _2)(29)+13$
$(n _1)(30)=n _2(29)+9---------(3)$
From (1) , for different values of n,
$x=34,64,94,124,154,184,214,244,274,304,.......$
Similarly from (2)
$x=42,71,100,129,158,187,216,245,274,303,.......$
From above it is clear that $274$ is minimum common value.So, minimum length must be $274$ cm.

It costs Rs. $10$ a kilometer to fly and Rs. $2$ a kilometer to drive. If one travels $200$ km covering $x$ km of the distance by flying and the rest by driving, then the cost of the trip is

  1. Rs. $2,000$

  2. Rs. $24,000$

  3. Rs.$ (8x + 400)$

  4. Rs. $(12x + 400)$


Correct Option: C
Explanation:

Let distance travelled by flying is $x$ km.
Hence, the distance  travelled  by  driving $= (200-x)$km.
The cost of flying $ = $ Rs. $ 10 $ per km
Hence, the cost of flying $x$ km $ = 10 \times x =10x$
The cost of driving $ = $ Rs. $ 2 $ per km
The cost of driving $(200-x)$ km $ = 2 \times (200-x) = 400 -2x$
Hence, the total cost $ =10x + 400-2x = $Rs.$400 + 8x$

The values of a so that the equation $\Vert x - 2\vert - 1\vert = a \vert x \vert$ does not contain any solution lying in the interval {2, 3} are

  1. $a \ \epsilon(-\infty \dfrac{1}{2})$

  2. $a \ \epsilon (1, \infty)$

  3. $a \ \epsilon (-\infty, 0) \cup (\dfrac{1}{2}, \infty )$

  4. None of these


Correct Option: A
Explanation:

The values of a so that the equation $\left| \left| x-2 \right| -1 \right| =a\left| x \right| $ does not contains any solution lying in the interval $\left{ 2,3 \right} $ are

$a\in (-\infty ,\cfrac { 1 }{ 2 } )\ \left| \left| x-2 \right| -1 \right| =a\left| x \right| \ =>\left| \left| 2-2 \right| -1 \right| =a\left| 2 \right| \ =>\left| -1 \right| =a\left| 2 \right| \ =>1=a\left| 2 \right| \ =>a=\cfrac { 1 }{ 2 } $
Obviously, it does not contains any solution lying in the interval $\left{ 2,3 \right} $ are $a\in (-\infty ,\cfrac { 1 }{ 2 } )$

The minimum value of $\displaystyle f(x)=|x-1|+|x-2|+|x-3|$ is equal to 

  1. $1$

  2. $2$

  3. $3$

  4. $0$


Correct Option: B
Explanation:

Solution:- (B) 2

The function $f$ is linear on each of the intervals $\left( - \infty, 1 \right], \left[ 1, 2\right], \left[2, 3\right] \text{ and } \left[ 3, \infty \right)$. Since a linear function on an interval always attains its minimum at one of the endpoints of the interval, and $f \left( x \right) = +\infty \text{ as } x = \pm \infty$, the function $f$ must attain its minimum at one of $x = 1, 2, 3$. Since $f(1)=3,  f \left( 2 \right) = 2 \text{ and } f \left( 3 \right) = 3$, the function $f$ attains a minimum of 2 at $x=2$.

A Gym sells two types of memberships. One packages costs $ $325$ for one year of membership with an unlimited number of visits. The second package has a $ $125$ enrolment fee, includes five free visits, and costs an additional $ $8$ per visit after first five. How many visits would a person need to use for each type of membership to cost the same amount over a one-year period?

  1. $20$

  2. $25$

  3. $30$

  4. $40$


Correct Option: C
Explanation:

Let the total number of visits be x. Hence equating the costs give us 

$ $325=$ 125+(x-5) $ 8$ $\rightarrow 200=8x-40$
$240=8x$ or $x=30$. Hence the person needs a total of 30 visits.

If $\cfrac{7}{m-\sqrt{3}} = \cfrac{\sqrt{3}}{m} + \cfrac{4}{2m}$, calculate the value of $m$.

  1. $-3.464$

  2. $-1.978$

  3. $-0.918$

  4. $1.978$


Correct Option: B
Explanation:

Given $\dfrac { 7 }{ m-\sqrt { 3 }  } =\dfrac { \sqrt { 3 }  }{ m } +\dfrac { 4 }{ 2m } =\dfrac { 4+2\sqrt { 3 }  }{ 2m } $
$\Rightarrow  14m=(4+2\sqrt { 3 } )m-\sqrt { 3 } (4+2\sqrt { 3 } )$
$ \Rightarrow (10-2\sqrt { 3 } )m=-4\sqrt { 3 } -6$
$\Rightarrow  m=\dfrac { -4\sqrt { 3 } -6 }{ 10-2\sqrt { 3 }  } =\dfrac { -12.928 }{ 6.536 } =-1.98$

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