Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 42

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Question 1
1 / -0

If $p$ is added to $-5$ , the result is $7$ . Hence $3p=$ ________

A
$6$

B
$18$

C
$36$

D
$4$

Solution

Given number is $p$
Adding $-5,$ we get number
$p-5$
But given that this is equal to $7$
$\therefore p-5=7$
Add $5$ on both sides, we get
$p-5+5=7+5$
$\Rightarrow p=12$
$\therefore 3p=3\times 12=36$
Question 2
1 / -0

Kishore has Rs. $p$ . Hari has Rs. $3$ more than Kishore and Ashok has Rs. $7$ less than Hari. Altogether the three boys have Rs. $20$ . How much money does Kishore have?

A
Rs. $30$

B
Rs. $10$

C
Rs. $7$

D
Rs. $3$

Solution

Given : Kishore have Rs $p$
And Hari have Rs $3$ more than Kishore
Then, Hari have $p+3$ Rs
Ashok have Rs $7$ less than Hari
Then, Ashok have $p+3-7$ Rs
But three boys all together have Rs $20$
$\therefore p+(p+3)+(p+3-7)=20$
$\Rightarrow p+p+3+p+3-7=20$
$\Rightarrow p+p+p=20-3-3+7$
$\Rightarrow 3p=21$
$\Rightarrow p=7$
Question 3
1 / -0

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?

A
$6$

B
$13$

C
$15$

D
$17$

Solution

Let the two digit number be $xy$ = $10x+y$ .
So, accoding to the problem.
$(\dfrac {1}{3}(10x + y)$ $-(10x + y)\dfrac {1}{4}) = 8$
$(10x + y)\left (\dfrac {1}{3} - \dfrac {1}{4}\right ) = 8$
$\Rightarrow 10x + y = 96$
$\Rightarrow x = 9, y = 6$
$\therefore x + y = 15$
Question 4
1 / -0

If $1$ is subtracted from the numerator of a fraction it becomes $\dfrac 13$ , and if $5$ added to the denominator, the fraction becomes $\dfrac 14$ . Which fraction shall result if $1$ is subtracted from the numerator and $5$ is added to the denominator?

A
$\dfrac 5{12}$

B
$\dfrac 7{12}$

C
$\dfrac 1{8}$

D
$\dfrac 23$

Solution

Let the fraction be $\dfrac {x}{y}$
$\dfrac {x - 1}{y} = \dfrac {1}{3}$ and $\dfrac {x}{y + 5} = \dfrac {1}{4}$
$\Rightarrow 3(x-1)=y\; and\; 4x=y+5$ ......eqn(1)
$\Rightarrow 3x-3=y\; and\; 4x=y+5$ .
$\Rightarrow 3x-3=4x-5$ .
$\Rightarrow x=-3+5$ .
$\Rightarrow x = 2$ and susbtitute in eqn (1) we get $y = 3$ . So, the fraction is $\dfrac {2}{3}$ .
New fraction $= \dfrac {2 - 1}{3 + 5} = \dfrac {1}{8}$ .
Question 5
1 / -0

If $14$ is taken away from one - fifth of a number, the result is $20$ . The equation expressing this statement is:

A
$\dfrac{x}{5}-14=20$

B
$x-\left (\dfrac{14}{5}\right)=\left (\dfrac{20}{5}\right)$

C
$x-14=\left (\dfrac{20}{5}\right)$

D
$x+4\left (\dfrac{14}{5}\right)=20$

Solution

Let the number be $x$ .
As per the question, we can write,
One- fifth of a number, i.e. $\dfrac {1}{5}x$ .
Now $14$ is taken away from this means $14$ is subtracted i.e., $\dfrac {x}{5}-14$ .
The result is equal to $20$ .
Thus all this can be written as ,
$\dfrac {x}{5}-14=20$
Question 6
1 / -0

If the sum of two numbers is $84$ and their difference is $30$ , the numbers are:

A
$-57$ and $27$

B
$57$ and $27$

C
$57$ and $-27$

D
$-57$ and $-27$

Solution

Given, the sum of two numbers is $84$ .
Let the smaller number be $x$ then, the larger number will be $84-x$ .

Given, their difference is $30$ .
$84-x-x=30$
$\Rightarrow 84-2x=30$
$\Rightarrow 84-30=2x$
$\Rightarrow 54=2x$
$\Rightarrow x=27$

Smaller number is $27$ .
Now, larger number is,
$84-x=84-27$
$=57$
Question 7
1 / -0

If five times a number increased by 8 is 83, then the number is:

A
$12$

B
$15$

C
$11$

D
$2$

Solution

Given that, five times a number increased by $8$ is $83$ .
To find out: The number.

Let the number be $x$ .
As per the question, we have
$5x+8=83$
$\Rightarrow 5x=83-8$
$\Rightarrow 5x=75$
$\Rightarrow x=15$

Hence, the number is $15$ .
Question 8
1 / -0

If $\cfrac{(x + 0.7x)}{2} = 0.85$ , then the value of $x$ is:

A
$2$

B
$1$

C
$-1$

D
$0$

Solution

Given, $\dfrac {x+0.7x}{2}=0.85$

Multiply $2$ on both the sides,

$\Rightarrow x+0.7x=1.7$

$\Rightarrow 1.7x=1.7$

Divide both sides by $1.7$ , we get

$\Rightarrow x=1$
Question 9
1 / -0

The numerator of a fraction is $6$ less than the denominator. If $3$ is added to the numerator, the fraction is equal to $\cfrac{2}{3}$ , find the original fraction

A
$\dfrac 39$

B
$\dfrac 69$

C
$\dfrac 37$

D
None of these

Solution

Let denominator $=x$

thus numerator $=x-6$

By condition $\dfrac{x-6+3}{x}=\dfrac{2}{3}$

$3(x-3)=2x$

$3x-9=2x$

$x=9$

So denominator $=x=9$

thus numerator $\Rightarrow x-6=9-6=3$

Original fraction $=\dfrac{3}{9}$
Question 10
1 / -0

Tank $A$ contains $5$ times as much water as Tank $B$ . How much water must be transfer from Tank $A$ to Tank $B$ so that each tank contains $45$ liters of water?

A
$30\ litres$

B
$60\ litres$

C
$75\ litres$

D
$45\ litres$

Solution

Let quantity of water in tank $B$ be $x$ litres.
Then quantity of water in the tank $A = 5x$ litres
Total water in both the tanks $= x + 5x$
$= 90\ liters$
$\Rightarrow 6x = 90 \Rightarrow x = 15$
$\therefore$ Water in tank $B = 15\ litres$
$\Rightarrow$ Water in tank $A = 15\times 5 = 75\ litres$
So, quantity of water to be transferred from tank $A$ to tank $B$ , so that each tank contains $45\ litres = 75 - 30 = 45\ litres$
Therefore, $30\ liters$ of water should be transferred.

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Linear Equations in One Variable Test - 42

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Linear Equations in One Variable Test - 42

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SHARING IS CARING

Question 1

666

181818

363636

444

Solution

Question 2

Rs.303030

Rs.101010

Rs.777

Rs.333

Solution

Question 3

666

131313

151515

171717

Solution

Question 4

512\dfrac 5{12}125​

712\dfrac 7{12}127​

18\dfrac 1{8}81​

23\dfrac 2332​

Solution

Question 5

x5−14=20\dfrac{x}{5}-14=205x​−14=20

x−(145)=(205)x-\left (\dfrac{14}{5}\right)=\left (\dfrac{20}{5}\right)x−(514​)=(520​)

x−14=(205)x-14=\left (\dfrac{20}{5}\right)x−14=(520​)

x+4(145)=20x+4\left (\dfrac{14}{5}\right)=20x+4(514​)=20

Solution

Question 6

−57-57−57 and 272727

575757 and 272727

575757 and −27-27−27

−57-57−57 and −27-27−27

Solution

Question 7

121212

151515

111111

222

Solution

Question 8

222

111

−1-1−1

000

Solution

Question 9

39\dfrac 3993​

69\dfrac 6996​

37\dfrac 3773​

None of these

Solution

Question 10

30 litres30\ litres30 litres

60 litres60\ litres60 litres

75 litres75\ litres75 litres

45 litres45\ litres45 litres

Solution

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Question Analysis

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$6$

$18$

$36$

$4$

Rs. $30$

Rs. $10$

Rs. $7$

Rs. $3$

$6$

$13$

$15$

$17$

$\dfrac 5{12}$

$\dfrac 7{12}$

$\dfrac 1{8}$

$\dfrac 23$

$\dfrac{x}{5}-14=20$

$x-\left (\dfrac{14}{5}\right)=\left (\dfrac{20}{5}\right)$

$x-14=\left (\dfrac{20}{5}\right)$

$x+4\left (\dfrac{14}{5}\right)=20$

$-57$ and $27$

$57$ and $27$

$57$ and $-27$

$-57$ and $-27$

$12$

$15$

$11$

$2$

$2$

$1$

$-1$

$0$

$\dfrac 39$

$\dfrac 69$

$\dfrac 37$

$30\ litres$

$60\ litres$

$75\ litres$

$45\ litres$