Linear Equations in One Variable Test

Result

Linear Equations in One Variable Test - 28

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

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.

A
$$8\displaystyle \frac{4}{29}$$ and $$14\displaystyle \frac{7}{29}$$

B
$$8\displaystyle \frac{3}{34}$$ and $$14\displaystyle \frac{7}{5}$$

C
$$8\displaystyle \frac{3}{34}$$ and $$14\displaystyle \frac{6}{25}$$

D
$$8\displaystyle \frac{4}{29}$$ and $$14\displaystyle \frac{2}{19}$$

Solution

Let the numbers be $$ 4x ; 7x $$

Given, $$ 3(7x) + 2(4x) = 59 $$
$$ => 21x + 8x = 59 $$
$$ 29x = 59 $$
$$ => x = \cfrac {59}{29} $$

Thus the numbers are $$ 4x = 4 \times \cfrac {59}{29} = \cfrac {236}{29} = 8 \cfrac {4}{29} $$
and $$ 7x = 7 \times \cfrac {59}{29} = \cfrac {413}{29} = 14 \cfrac {7}{29} $$
Question 2
1 / -0

In an examination, a student scores $$4$$ marks for every correct answer and loses $$1$$ mark for every wrong answer. If he attempts in all $$60$$ questions and secures $$130$$ marks, the number of questions he attempts correctly, is ___

A
$$35$$

B
$$38$$

C
$$40$$

D
$$42$$

Solution

Let the number of questions attempted correctly be $$x$$.
Then, number of incorrect ones $$= (60-x)$$.
$$\therefore 4x-1\left( 60-x \right) =130$$
$$\Rightarrow 5x=190$$
$$\Rightarrow x=38$$
So, number of questions student attempts correctly are $$38$$.
Question 3
1 / -0

Amit is now $$6$$ times as old as his son. Four years from now, the sum of their ages will be $$43$$ years. Determine Amit's present age:

A
$$30$$ years

B
$$32$$ years

C
$$34$$ years

D
$$28$$ years

Solution

Let present age of Amit's son be $$x$$ yrs and age of Amit be $$6x$$ yrs.
Four yrs from now, $$(6x+4)+(x+4)=43$$.....(Given condition)
$$\therefore 7x=35$$
$$\therefore x=5$$,
i.e., present age of Amit $$=6x=6\times 5=30$$yrs.
Hence, option $$A$$ is the correct answer.
Question 4
1 / -0

If $$\displaystyle \frac{1}{3}\left ( 7-4x \right )-\frac{1}{4}\left ( 11-5x \right )+1=x-\frac{1}{2}\left ( 3x-7 \right )$$, then $$x=$$

A
$$1$$

B
$$9$$

C
$$6$$

D
$$7$$

Solution

$$\cfrac { 1 }{ 3 } (7-4x) -\cfrac { 1 }{ 4 } (11-5x) + 1 =x - \cfrac { 1 }{ 2 } (3x-7)\\ LCM = 12\\ 4(7-4x) - 3(11-5x) +12 = 12x - 6(3x-7)\\ 28 - 16x -33 +15x +12 = 12x -18x +42\\ 7 -x = 42 -6x\\ 5x = 35\\ x = 7$$
Question 5
1 / -0

The ten's digit of a two digit number exceeds its unit's digit by $$4$$. If the ten's digit and the unit's digit are in the ratio $$3 : 1$$, find the number.

A
$$62$$

B
$$51$$

C
$$65$$

D
$$85$$

Solution

Let the units digit $$= x$$
Let the tens digit $$= x+4$$
Tens digit : One digit :: $$3:1$$
$$x+4 : x :: 3:1$$
$$\Rightarrow x+4 = 3x$$
$$\Rightarrow 2x = 4$$
$$\Rightarrow x = 2$$
Tens digit $$= 6$$
Therefore, the number is $$ 62$$.
Question 6
1 / -0

Solve the following equation:
$$\displaystyle \frac{4}{5}\, \left(x\, +\, \frac{5}{6} \right)\, +\, \frac{2}{3} \left(x\, -\, \frac{1}{4}\right)\, =\, {1}\frac{1}{9}$$

A
$$\displaystyle \frac{5}{19}$$

B
$$\displaystyle \frac{5}{16}$$

C
$$\displaystyle \frac{5}{14}$$

D
$$\displaystyle \frac{5}{12}$$

Solution

$$\displaystyle \frac{4}{5}\left ( x+\cfrac{5}{6} \right )+\cfrac{2}{3}\left ( x-\cfrac{1}{4} \right )=1\cfrac{1}{9}$$
$$=>\cfrac { 4x }{ 5 } +\cfrac { 20 }{ 30 } +\cfrac { 2x }{ 3 } -\cfrac { 2 }{ 12 } =\cfrac { 10 }{ 9 } $$
$$=>\cfrac { 4x }{ 5 } +\cfrac { 2 }{ 3 } +\cfrac { 2x }{ 3 } -\cfrac { 1 }{ 6 } -\cfrac { 10 }{ 9 } =0$$
$$=>\cfrac { 72x+60+60x-15-100 }{ 90 } =0$$
$$=>72x+60x=-60+15+100$$
$$=>132x=55$$
$$=>x=\cfrac{55}{132}$$
$$=>x=\cfrac{5}{12}$$
Question 7
1 / -0

A farmer divides his herd of $$x$$ cows among his $$4$$ sons so, that first son gets-one-half of the herd, the second son get, one-fourth, the third son gets one-fifth, and the fourth son gets $$7$$ cows, then the value of $$x$$ is :

A
$$100$$

B
$$140$$

C
$$160$$

D
$$180$$

Solution

Total number of cows $$= x$$
According to the question, we have
A $$= \displaystyle{\frac{x}{2}}$$, $$B = \displaystyle{\frac{x}{4}}$$, C $$= \displaystyle{\frac{x}{5}}$$, $$D = 7$$

$$\therefore$$$$\displaystyle{\frac{x}{2} + \frac{x}{4} + \frac{x}{5}} + 7 = x$$

$$\Rightarrow$$$$\displaystyle{\frac{10x + 5x + 4x}{20}} = x - 7$$

$$\Rightarrow19x = 20x - 140$$
$$\Rightarrow 140 = x$$
Hence, option 'B' is correct.
Question 8
1 / -0

The difference of the squares of two consecutive even natural numbers is 92. Taking x as the smaller of the two numbers, form an equation in x, and hence find the larger of the two numbers.

A
$$18$$

B
$$24$$

C
$$23$$

D
$$21$$

Solution

Given that,
The difference of squares of two consecutive even natural numbers is $$92$$.
The smaller number is $$x$$.

To find out,
$$(i)$$ The equation in $$x$$ representing the given information.
$$(ii)$$ The larger of the two numbers.

$$(i)$$ If the smaller number is $$x$$, then the larger number will be $$x+2$$

According to the question,
$$(x+2)^2-x^2=92$$

Hence, $$(x+2)^2-x^2=92$$ is the equation that represents the given information.

$$(ii)$$ Solving $$(x+2)^2-x^2=92$$

We know that, $$(a+b)^2=a^2+2ab+b^2$$

Applying the same identity, we get:
$$(x^2+4x+4)-x^2=92$$

$$\Rightarrow x^2+4x+4-x^2=92$$

$$\Rightarrow 4x+4=92$$

Dividing both sides by $$4$$, we get:
$$x+1=23$$

$$x=23-1$$

$$=22$$

Hence, $$x=22$$

$$\therefore\ x+2=22+2$$

$$=24$$

Hence, the larger of the two numbers is $$24$$.
Question 9
1 / -0

The difference between two numbers is $$8$$ and their ratio is $$1:5$$. Which of the following is the smaller number?

A
$$24$$

B
$$8$$

C
$$10$$

D
$$2$$

Solution

Let the numbers be $$x$$ and $$5x$$.
Given, $$5x-x=8$$
$$\Rightarrow 4x=8$$
$$\Rightarrow \displaystyle x=\frac{8}{4}=2$$
Hence, smaller number is $$2$$.
Question 10
1 / -0

The solution of the equation
$$\displaystyle \frac{3a-2}{3}+\frac{2a+3}{2}=a+\frac{7}{6}$$ is

A
$$a=2$$

B
$$\displaystyle a=\frac{1}{3}$$

C
$$a=-2$$

D
$$\displaystyle a=\frac{1}{2}$$

Solution

Given, $$\displaystyle \frac{3a-2}{3}+\frac{2a+3}{2}=a+\frac{7}{6}$$
L.C.M. of denominators is $$6$$.
Multiplying each term by $$6$$, we get
$$\displaystyle 6\times \left ( \frac{3a-2}{3} \right )+6\times \left(\frac{2a+3}{2}\right)=6\times a+6\times \frac{7}{6}$$
$$\Rightarrow 2(3a-2)+3(2a+3)=6a+7$$
$$\Rightarrow 6a-4+6a+9=6a+7$$
$$\Rightarrow 12a+5=6a+7$$
$$\Rightarrow 6a=2$$
$$\Rightarrow \displaystyle a=\frac{2}{6}=\frac{1}{3}$$
Hence, required solution is $$a=\cfrac{1}{3}$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Linear Equations in One Variable Test - 28

Result

Linear Equations in One Variable Test - 28

Score

-

Rank

-

SHARING IS CARING

Question 1

$$8\displaystyle \frac{4}{29}$$ and $$14\displaystyle \frac{7}{29}$$

$$8\displaystyle \frac{3}{34}$$ and $$14\displaystyle \frac{7}{5}$$

$$8\displaystyle \frac{3}{34}$$ and $$14\displaystyle \frac{6}{25}$$

$$8\displaystyle \frac{4}{29}$$ and $$14\displaystyle \frac{2}{19}$$

Solution

Question 2

$$35$$

$$38$$

$$40$$

$$42$$

Solution

Question 3

$$30$$ years

$$32$$ years

$$34$$ years

$$28$$ years

Solution

Question 4

$$1$$

$$9$$

$$6$$

$$7$$

Solution

Question 5

$$62$$

$$51$$

$$65$$

$$85$$

Solution

Question 6

$$\displaystyle \frac{5}{19}$$

$$\displaystyle \frac{5}{16}$$

$$\displaystyle \frac{5}{14}$$

$$\displaystyle \frac{5}{12}$$

Solution

Question 7

$$100$$

$$140$$

$$160$$

$$180$$

Solution

Question 8

$$18$$

$$24$$

$$23$$

$$21$$

Solution

Question 9

$$24$$

$$8$$

$$10$$

$$2$$

Solution

Question 10

$$a=2$$

$$\displaystyle a=\frac{1}{3}$$

$$a=-2$$

$$\displaystyle a=\frac{1}{2}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.