Linear Equations in One Variable Test

Result

Linear Equations in One Variable Test - 30

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 consecutive odd numbers are such that when the smaller number is subtracted from three times the bigger number, the result is $56$ . The bigger number is:

A
$29$

B
$31$

C
$25$

D
$27$

Solution

Let the two consecutive odd numbers are $x$ and $x+2$ .
Two consecutive odd numbers are such that when the smaller number is subtracted from three times the bigger number, the result is $56$ .
Hence, equation is,
$3(x+2)-x=56$
$\therefore$ $2x=50$
$\therefore$ $x=25$
$\therefore$ Smaller number is $25$ and bigger number is $27$ .
Question 2
1 / -0

One number is 3 less than the two times of the other; if their sum is increased by 7 the result is 37. Find the numbers.

A
$9 , 11$

B
$11 , 13$

C
$11 , 19$

D
$9 , 13$

Solution

Let one number = x
other number = 2x-3
$\displaystyle (x+2x-3)+7=37$
3x = 33
x = 11
$\displaystyle \therefore$ The two number are 11 and 19
Question 3
1 / -0

The solution of the the equation
$5(3x+5)=2(7x-4)$ is

A
$x=17$

B
$x=33$

C
$x=-33$

D
$x=-17$

Solution

Given, $5(3x+5)=2(7x-4)$
$\therefore 15x+25=14x-8$
Putting $x$ terms on one side and constants on another side.
$\therefore 15x-14x=-8-25$
$\therefore x=-33$
Hence, option C is correct.
Question 4
1 / -0

If $\displaystyle \frac{7x+2}{6}-\frac{x-4}{3}=\frac{x+3}{3}$ , then $x=$

A
$\displaystyle -\frac{4}{3}$

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

C
$\displaystyle \frac{-1}{3}$

D
$\displaystyle 1\frac{1}{3}$

Solution

Given. $\displaystyle \frac{7x+2}{6}-\frac{x-4}{3}=\frac{x+3}{3}$ .
Multiply throughout by $3$ , we get
$\therefore$ $\displaystyle \frac{7x+2}{2}-(x-4)=x+3$
$\therefore$ $\displaystyle \frac{7x+2}{2}=2x-1$
On cross multiplying, we get
$3x=-4$
$\therefore$ $x=\dfrac {-4}{3}$
Question 5
1 / -0

The sum of 4 consecutive integers is 70. Then the greatest among them is:

A
19

B
23

C
17

D
16

Solution

Let the $4$ consecutive numbers be $x$ , $x+1$ , $x+2$ , $x+3$
Given: The sum of $4$ consecutive numbers is $70$ .
Therefore, $(x)+(x+1)+(x+2)+(x+3)=70$
$4x+6=70$
$4x=64$
$x=16$
Therefore,
$x+1=17, x+2=18, x+3=19$
Thus the greatest is $19$
Question 6
1 / -0

The total cost of three prizes is Rs. 2550. If the value of second prize is $\displaystyle \frac{3}{4}$ th of the first and the value of 3rd prize is $\displaystyle \frac{1}{2}$ of the second prize;
find the value of first prize.

A
Rs. $900$

B
Rs. $1500$

C
Rs. $1200$

D
Rs. $450$

Solution

Let First Prize be $x$
Then Second Prize $\displaystyle =\frac{3}{4}x$

Third Prize $\displaystyle =\frac{1}{2}\times \frac{3x}{4}=\frac{3x}{8}$

$\displaystyle x + \frac{3x}{4}+\frac{3x}{8}=2550$ (given)

L.C.M of $1,4,8$ is $8$ .
$\displaystyle \frac{x \times 8}{1 \times 8}+\frac{3x \times 2}{4 \times 2}+\frac{3x}{8} =2550$

$\displaystyle \frac{8x+6x+x}{8} =2550$

$\displaystyle \frac{17x}{8}=2550$
$\displaystyle x = \frac{2550 \times 8}{17}$
$x=Rs.1200$
$\therefore$ The value of first first prize is $Rs.1200$
Question 7
1 / -0

Solve for $x$ : $\displaystyle \frac{1}{4}(5x+4)=\frac{1}{3}(2x-1)$

A
$\displaystyle \frac{-16}{7}$

B
$\displaystyle \frac{10}{7}$

C
$\displaystyle \frac{8}{7}$

D
$\displaystyle \frac{-15}{7}$

Solution

Given, $\cfrac14(5x+4)=\cfrac13(2x-1)$
$\Rightarrow 3(5x+4)=4(2x-1)$ ( $\text{cross multiply}$ )
$\Rightarrow 15x+12=8x-4$
$\Rightarrow 15x-8x=-4-12$
$\Rightarrow 7x=-16$
$\Rightarrow x=-\cfrac{16}7$
Question 8
1 / -0

$\displaystyle \frac{1}{5}$ th of a flagpole is black and $\displaystyle \frac{1}{4}$ th is white and the remaining three metres is painted yellow then ffind the length of the flagpole

A
$\displaystyle 5\frac{5}{11}m$

B
$\displaystyle \frac{60}{11}cm$

C
$5 km$

D
None

Solution

Let the length of flagpole be $x$ m
Length of flagpole which is black $=\dfrac{1}{5}x$ m
Length of flagpole which is white $=\dfrac{1}{4}x$ m
Length of flagpole which is yellow $=3$ m
A.T.Q,
$\dfrac{1}{5}x+\dfrac{1}{4}x+3=x$
$\implies x-\dfrac{x}{5}-\dfrac{x}{4}=3$
$\implies \dfrac{20x-4x-5x}{20}=3$
$\implies 11x=3\times 20$
$\implies x=\dfrac{60}{11}$
$\implies x=5\dfrac{5}{11} m$
$\therefore$ The length of flagpole $=5\dfrac{5}{11}m$
Question 9
1 / -0

$\displaystyle \frac{1}{2}$ is subtracted from a number and the difference is multiplied by 4 , then 25 is added to the product and the sum is divided by 3 the result is equal to 10. Find the number

A
$\displaystyle \frac{3}{5}$

B
$\displaystyle \frac{7}{4}$

C
$\displaystyle \frac{6}{7}$

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

Solution

Let the number be $x$
Given $\displaystyle \left \{ \left ( x-\frac{1}{2} \right )\times4+25\right \}\div 3=10$
$\displaystyle \left \{ 4x-2+25 \right \}\times\frac{1}{3}=10$
$\displaystyle \left ( \frac{4x+23}{3} \right )=10$
$4x+23=30$
$\displaystyle 4x=7$
$\Rightarrow x=\frac{7}{4}$
Question 10
1 / -0

In a piggy bank, the number of $25$ paise coins are five times the number of $50$ paise coins. If there are $120$ coins find the amount in the bank ?

A
Rs. $25$

B
Rs. $10$

C
Rs. $35$

D
Rs. $40$

Solution

Let number of $50$ paise coins $= x$
then number of $25$ paise coins $= 5x$
Given total number of coins $= 120$
$\Rightarrow x + 5x = 120$
$\Rightarrow 6x = 120$
$\displaystyle\Rightarrow x=\frac{120}{6}=20$
$\therefore$ Number of $50$ paise coins $= 20$
Thus Amount $20 \times50 =$ Rs. $10$
And number of $25$ paise coins $= 5 \times 20=100$
Thus amount $= 25 \times 100 =$ Rs. $25$
Hence, total amount $25 + 10 =$ Rs. $35$ .

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 - 30

Result

Linear Equations in One Variable Test - 30

Score

-

Rank

-

SHARING IS CARING

Question 1

292929

313131

252525

272727

Solution

Question 2

9,119 , 119,11

11,1311 , 1311,13

11,1911 , 1911,19

9,139 , 139,13

Solution

Question 3

x=17x=17x=17

x=33x=33x=33

x=−33x=-33x=−33

x=−17x=-17x=−17

Solution

Question 4

−43\displaystyle -\frac{4}{3}−34​

13\displaystyle \frac{1}{3}31​

−13\displaystyle \frac{-1}{3}3−1​

113\displaystyle 1\frac{1}{3}131​

Solution

Question 5

19

23

17

16

Solution

Question 6

Rs. 900900900

Rs. 150015001500

Rs. 120012001200

Rs. 450450450

Solution

Question 7

−167\displaystyle \frac{-16}{7}7−16​

107\displaystyle \frac{10}{7}710​

87\displaystyle \frac{8}{7}78​

−157\displaystyle \frac{-15}{7}7−15​

Solution

Question 8

5511m\displaystyle 5\frac{5}{11}m5115​m

6011cm\displaystyle \frac{60}{11}cm1160​cm

5km5 km5km

None

Solution

Question 9

35\displaystyle \frac{3}{5}53​

74\displaystyle \frac{7}{4}47​

67\displaystyle \frac{6}{7}76​

23\displaystyle \frac{2}{3}32​

Solution

Question 10

Rs. 252525

Rs. 101010

Rs. 353535

Rs. 404040

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.

$29$

$31$

$25$

$27$

$9 , 11$

$11 , 13$

$11 , 19$

$9 , 13$

$x=17$

$x=33$

$x=-33$

$x=-17$

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

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

$\displaystyle \frac{-1}{3}$

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

Rs. $900$

Rs. $1500$

Rs. $1200$

Rs. $450$

$\displaystyle \frac{-16}{7}$

$\displaystyle \frac{10}{7}$

$\displaystyle \frac{8}{7}$

$\displaystyle \frac{-15}{7}$

$\displaystyle 5\frac{5}{11}m$

$\displaystyle \frac{60}{11}cm$

$5 km$

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

$\displaystyle \frac{7}{4}$

$\displaystyle \frac{6}{7}$

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

Rs. $25$

Rs. $10$

Rs. $35$

Rs. $40$