Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 34

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Question 1
1 / -0

Length of a rectangle is $8$ m less than twice its breadth. If the perimeter of the rectangle is $56$ m. Find its length and breadth.

A
Length $= 1$ 6m and breadth $= 12$ m

B
Length $= 13$ m and breadth $= 15$ m

C
Length $= 14$ m and breadth $= 17$ m

D
Length $= 18$ m and breadth $= 21$ m

Solution

Let the breadth of the rectangle be $x$ m. Then,
length of the rectangle $= (2x - 8)$ m
Given, perimeter $= 56$ m
$\Longrightarrow 2(2x-8+x)=56\\ \Longrightarrow 2(3x-8)=56\\ \Longrightarrow 6x-16=56\\ \Longrightarrow 6x=56+16\quad \quad \quad [Transposing\quad -16\quad to\quad RHS]\\ \Longrightarrow 6x=72\\ \Longrightarrow x=12$
$\text{Therefore, breadth of the rectangle}$ $= 12$ $\text{m and length of the rectangle}$ $= 2 \times 12 - 8 = 16$ $m$
Question 2
1 / -0

$70$ coins of $10$ paise and $50$ paise are mixed in a purse. If the total value of the money in the purse is Rs. $19$ , find the number of each type of coins in the purse.

A
Number of $10$ paise coins are $60$
Number of $50$ paise coins are $20$

B
Number of $10$ paise coins are $20$
Number of $50$ paise coins are $40$

C
Number of $10$ paise coins are $40$
Number of $50$ paise coins are $30$

D
Number of $10$ paise coins are $30$
Number of $50$ paise coins are $50$

Solution

Suppose the number of $10$ paise coins $= x$

$\therefore$ the number of $50$ paise coins $= 70 - x$

Now, value of $10$ paise coins $=$ Rs. $\displaystyle \frac{x}{10}$

and value of $50$ paise coins $=$ Rs. $\displaystyle \frac{70-x}{2}$

Total value of coins $=$ $\displaystyle \frac{x}{10}+ \frac{70-x}{2}$

$\therefore \displaystyle \frac{x}{10} + \frac{70-x}{2} = 19$

Multiplying both sides by $10$ , (LCM of $10,2 = 10$ ), we get

$x + 5 (70 - x) =19 \times 10$

$x + 350 - 5x = 190$

$-4x = -160$

$\displaystyle \frac{-4x }{-4} = \frac{-160}{-4}$

$\therefore x = 40$

$\therefore$ number of $10$ paise coins $= 40$

Number of $50$ paise coins $= 70 - 40 = 30$ .
Question 3
1 / -0

The digit in the units place of a $2$ digit number is $4$ times the digit in the tens place. The number obtained by reversing the digits exceeds the given number by $54$ . Find the given number.

A
$28$

B
$32$

C
$15$

D
$25$

Solution

Let one's digit be $4x$ . Then, ten's digit $= x$ .
Number $=10x+4x=14x$ ....(1)
Number obtained by reversing the digits $= 40x + x = 41x$
According to the given condition,
$41x - 14x = 54$
$27x =54$
$x=2$
Putting the value of $x$ in (1) , we get
Number $=14 \times 2 = 28$
Hence, $28$ is the required number.
Question 4
1 / -0

The difference between a number and two third of itself is $13$ . Find the number.

A
$31$

B
$35$

C
$39$

D
$33$

Solution

Let the number be $x$ .
According to the given condition,
$x-\cfrac { 2 }{ 3 } x=13$
$\Rightarrow \cfrac { 3x-2x }{ 3 } =13$ ....(By taking L.C.M.)
$\Rightarrow \cfrac { x }{ 3 } =13$
Multiply both sides by $3$
$3 \times \dfrac {x}{3}=13 \times 3$
$\Rightarrow x=39$
Hence, the number is $39$ .
Question 5
1 / -0

The distance between two towns is $300$ km. Car $A$ and Car $B$ starts simultaneously from these towns and move towards each other. The speed of Car $A$ is more than that of Car $B$ by $7$ km/h. If the distance between the cars after two hours is $34$ km. Find the speed of both cars.

A
Speed of car $B$ is $63$ km/h and speed of car $A$ is $70$ km/h

B
Speed of car $B$ is $70$ km/h and speed of car $A$ is $75$ km/h

C
Speed of car $B$ is $50$ km/h and speed of car $A$ is $65$ km/h

D
Speed of car $B$ is $72$ km/h and speed of car $A$ is $80$ km/h

Solution

Let the speed of car $B$ be $x$ km/h.
Then, speed of car $A$ $= (x + 7)$ km/h
Now, distance covered by car $B$ in $2$ hours $= 2x$ km and
Distance covered by car $A$ in $2$ hours $= 2(x + 7)$ km
Given, distance between the cars after two hours is $34$ km and distance between both the towns is $300$ km.
So, distance covered by both the cars in two hours moving in opposite direction $= 300 - 34 = 266$ km
$\Rightarrow 2(x+7)+2x=266$
$\Rightarrow 2x+14+2x=266$
$\Rightarrow 4x=252$
$\Rightarrow x=63$ km/h

Therefore, speed of car $B$ $= 63$ km/h and speed of car $A$ $= 63 + 7 = 70$ km/h.
Question 6
1 / -0

The difference of two numbers is $72$ and the quotient obtained by dividing the one by the other is $3$ . Find the numbers.

A
$60$ and $132$

B
$42$ and $114$

C
$36$ and $108$

D
$30$ and $102$

Solution

Let one number be $x$ .
According to given condition, second number $= 3x$
Given, difference of two numbers $= 72$
$\Rightarrow 3x-x=72\\ \Rightarrow 2x=72\\ \Rightarrow x=36$
So, one number $= 36$ and other number $= 3 \times 36 = 108$
Question 7
1 / -0

The sum of three consecutive odd numbers is $63$ . Find them.

A
$19, 21, 23$

B
$21,23,29$

C
$19,21,27$

D
$21,23,32$

Solution

Let the three consecutive odd numbers be $(2x+1), (2x+3)$ and $(2x+5)$ .
According to the given condition, we have
$(2x+1) + (2x+3) + (2x+5) = 63$
$6x + 9 = 63$
$6x = 54$ ...[By transposing $9$ to RHS]
$x = 9$
So, three consecutive odd numbers are $(2 \times 9 + 1), (2 \times 9 + 3)$ and $(2 \times 9 + 5)$ i.e., $19, 21$ and $23$ .
Question 8
1 / -0

Nineteen boys turn out for baseball. Of these 11 are wearing baseball shirts and 14 are wearing baseball pants. There are no boys without one or the other. The number of boys wearing full uniforms is ________.

A
3

B
5

C
6

D
8

Solution

Let the no. of boys wearing full uniform be x

∴ No. of boys wearing baseball shirts only =11−x

and No. of boys wearing baseball pants only =14−x

∴11−x+14−x+x=19

or, −x=19−11−14=−6

⟹x=6
Question 9
1 / -0

Sixteen years ago, Tanya's grandfather was $8$ times older to her. $8$ years from now he would be $3$ times for her age. Eight years ago. what was the ratio of Tanya's age to that of her grandfather?

A
$1:2$

B
$1:5$

C
$3:8$

D
None of these

Solution

Let, Tanya age $16$ years ago $= x$
Grandfather's age $16$ years ago $= 8x.$
$8$ years from now, $3(x+16+8)=8x+16+8$
$3x+48+24=8x+24$
$\Rightarrow 3x-8x=24-72$
$\Rightarrow -5x=-48$
$\Rightarrow x=\cfrac{48}{5}$
Also given, $8$ years ago, ratio was
$\cfrac{x+8}{8x+8}\Rightarrow \cfrac{\frac{48}{5}+8}{8\times \cfrac{48}{5}+8}$
$\Rightarrow\cfrac{ \cfrac{88}{5}}{\cfrac{424}{5}}$
$\Rightarrow \cfrac{88}{424}$
$\Rightarrow \cfrac{11}{53}$
The ratio of Tanya's age to that of her grandfather is $11:53$ .
Question 10
1 / -0

A labourer was engaged for $30$ days on the condition that he will be paid Rs. $35$ for each day he works and will be fined Rs. $10$ for each day he is absent. He earned Rs. $735$ in total. Find how many days did he work?

A
$21$

B
$22$

C
$25$

D
$23$

Solution

Let the number of days he worked be x. Then
$x \times 35 -10(30 -x) = 735$
or $35x -300 + 10x = 735$
or $45x=735+300$
or $x=\frac {1035}{45}=23$
$\therefore x=23 days$

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Linear Equations in One Variable Test - 34

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Question 1

Length =1= 1=16m and breadth =12= 12=12m

Length =13= 13=13m and breadth =15= 15=15m

Length =14= 14=14m and breadth =17= 17=17m

Length =18= 18=18m and breadth =21= 21=21m

Solution

Question 2

Number of 101010 paise coins are 606060 Number of 505050 paise coins are 202020

Number of 101010 paise coins are 202020Number of 505050 paise coins are 404040

Number of 101010 paise coins are 404040Number of 505050 paise coins are 303030

Number of 101010 paise coins are 303030Number of 505050 paise coins are 505050

Solution

Question 3

282828

323232

151515

252525

Solution

Question 4

313131

353535

393939

333333

Solution

Question 5

Speed of car BBB is 636363 km/h and speed of car AAA is 707070 km/h

Speed of car BBB is 707070 km/h and speed of car AAA is 757575 km/h

Speed of car BBB is 505050 km/h and speed of car AAA is 656565 km/h

Speed of car BBB is 727272 km/h and speed of car AAA is 808080 km/h

Solution

Question 6

606060 and 132132132

424242 and 114114114

363636 and 108108108

303030 and 102102102

Solution

Question 7

19,21,2319, 21, 2319,21,23

21,23,2921,23,2921,23,29

19,21,2719,21,2719,21,27

21,23,3221,23,3221,23,32

Solution

Question 8

3

5

6

8

Solution

Question 9

1:21:21:2

1:51:51:5

3:83:83:8

None of these

Solution

Question 10

212121

222222

252525

232323

Solution

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Length $= 1$ 6m and breadth $= 12$ m

Length $= 13$ m and breadth $= 15$ m

Length $= 14$ m and breadth $= 17$ m

Length $= 18$ m and breadth $= 21$ m

Number of $10$ paise coins are $60$
Number of $50$ paise coins are $20$

Number of $10$ paise coins are $20$
Number of $50$ paise coins are $40$

Number of $10$ paise coins are $40$
Number of $50$ paise coins are $30$

Number of $10$ paise coins are $30$
Number of $50$ paise coins are $50$

$28$

$32$

$15$

$25$

$31$

$35$

$39$

$33$

Speed of car $B$ is $63$ km/h and speed of car $A$ is $70$ km/h

Speed of car $B$ is $70$ km/h and speed of car $A$ is $75$ km/h

Speed of car $B$ is $50$ km/h and speed of car $A$ is $65$ km/h

Speed of car $B$ is $72$ km/h and speed of car $A$ is $80$ km/h

$60$ and $132$

$42$ and $114$

$36$ and $108$

$30$ and $102$

$19, 21, 23$

$21,23,29$

$19,21,27$

$21,23,32$

$1:2$

$1:5$

$3:8$

$21$

$22$

$25$

$23$