Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 45

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Question 1
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The difference between $$\left(\displaystyle\frac{3}{5}\right)^{th}$$of $$\left(\displaystyle\frac{2}{3}\right)^{rd}$$ of a number and $$\left(\displaystyle\frac{2}{5}\right)^{th}$$ of $$\left(\displaystyle\frac{1}{4}\right)^{th}$$ of the same number is $$288$$. What is the number?

A
$$960$$

B
$$850$$

C
$$895$$

D
$$995$$

Solution

$$\Rightarrow$$ Let the number be $$x$$.
According to the given condition,
$$\dfrac{3}{5}\times \dfrac{2}{3}\times x-\dfrac{2}{5}\times \dfrac{1}{4}\times x=288$$

$$\Rightarrow$$ $$\dfrac{2}{5}x-\dfrac{2}{20}x=288$$

$$\Rightarrow$$ $$\dfrac{8x-2x}{20}=288$$

$$\Rightarrow$$ $$6x=5760$$

$$\therefore$$ $$x=\dfrac{5760}{6}=960$$

$$\therefore$$ The required number is $$960$$
Question 2
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There are four containers that are arranged in the ascending order of their heights. If the height of the smallest container given in the figure is expressed as $$\dfrac{7}{25} x = 10.5$$cm. Find the height of the largest container.

A
$$37.5$$cm

B
$$42.5$$cm

C
$$30.5$$cm

D
$$45.5$$cm

Solution

$$\Rightarrow$$ We can see $$x$$ is height of largest container.

$$\Rightarrow$$ $$\dfrac{7}{25}x=10.5$$

$$\Rightarrow$$ $$7x=262.5$$

$$\therefore$$ $$x=\dfrac{262.5}{7}=37.5\,cm$$

$$\therefore$$ Height of the largest container is $$37.5\,cm$$.
Question 3
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Two numbers are in the ratio $$\displaystyle1\frac{1}{2} : 2\frac{2}{3}$$. But if each number is increased by $$15$$. then the ratio becomes $$\displaystyle1\frac{2}{3} : 2 \frac{1}{2}$$. The numbers are:

A
$$27, 42$$

B
$$36, 48$$

C
$$27, 48$$

D
$$27, 36$$

Solution

The ratio of two numbers $$1\dfrac{1}{2}:2\dfrac{2}{3}=\dfrac{3}{2}:\dfrac{8}{3}$$ $$=9:16$$

Let the two numbers are $$9x$$ and $$16x$$.

According to the given condition,

$$\Rightarrow$$ $$\dfrac{9x+15}{16x+15}=1\dfrac{2}{3}:2\dfrac{1}{2}=\dfrac{5}{3}:\dfrac{5}{2}$$

$$\Rightarrow$$ $$\dfrac{9x+15}{16x+15}=\dfrac{2}{3}$$

$$\Rightarrow$$ $$3\times (9x+15)=2\times (16x+15)$$

$$\Rightarrow$$ $$27x+45=32x+30$$

$$\Rightarrow$$ $$5x=15$$

$$\therefore$$ $$x=3$$

$$\therefore$$ The numbers are $$9x=9\times 3=27$$ and $$16x=16\times 3=48$$.
Question 4
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If a boat goes $$7$$ km upstream in $$42$$ minutes and the speed of the stream is $$3$$ kmph, then the speed of boat in still water is:

A
$$4.2$$ km/hr

B
$$9$$ km/hr

C
$$13$$ km/hr

D
$$21$$ km/hr

Solution

As the boat goes 7 km upstream in 42 minutes,
Rate upstream $$= \cfrac{7}{42} \times 60 = 10 \; kmph$$
Speed of stream $$= 3 \; kmph \; \left\{ Given \right\}$$
Let speed of boat in still water be $$x \; km/hr$$.
As we know that,
speed upstream = speed of boat in still water - speed of stream
$$\therefore$$ speed upstream $$= \left( x - 3 \right) kmph$$
$$\Rightarrow \; \left( x - 3 \right) = 10$$
$$\Rightarrow \; x = 10 + 3 = 13 \; kmph$$
Hence, speed of boat in still water is 13 kmph.
Question 5
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Hamid has three boxes of different fruits. Box $$A$$ weights $$2\dfrac{1}{2}\ kg$$ more than Box $$B$$ and Box $$C$$ weights $$10\dfrac{1}{4}\ kg$$ more than Box $$B$$. The total weight of the boxes is $$48\dfrac{3}{4}$$. How many $$kgs$$ does Box A weights?

A
$$15\dfrac {1}{2}\ kg$$

B
$$14 \dfrac {1}{2}\ kg$$

C
$$13\dfrac {1}{2}\ kg$$

D
$$none\ of\ these$$

Solution

Let weight of $$B = x$$

$$\Rightarrow $$ Weight of $$A =x+2\dfrac12$$

$$= x + {\dfrac{5}{2}}$$

$$\Rightarrow $$ Weight of $$C =x+10\dfrac 14$$

$$= x +{\dfrac{{41}}{4}} $$

Total weight of bottles is $$48\dfrac 34=\dfrac {195}{4}$$

$$\Rightarrow x + x + x + {\dfrac{5}{2}} + {\dfrac{{41}}{4}} =\dfrac{195}{4}$$

$$\Rightarrow 3x+\dfrac{41+10}{4}=\dfrac{{195}}{4} $$

$$\Rightarrow 3x+\dfrac{51}{4}=\dfrac{{195}}{4} $$

$$\Rightarrow 3x=\dfrac{{195}}{4}- \dfrac{51}{4}$$

$$\Rightarrow 3x=\dfrac{{144}}{4}$$

$$\Rightarrow 3x=36$$

$$\Rightarrow x=12$$

$$\therefore $$ Weight of $$B=12\ kg$$

And weight of $$A = 12 + \dfrac{5}{2} = 14.5\ kg$$
Question 6
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A crate of mangoes contains one bruised mango for every $$30$$ mangoes in the crate. If $$3$$ out of every $$4$$ bruised mango are considered unsaleable, and there are $$12$$ unsaleable mangoes in the crate, how many mangoes are there in the crate?

A
$$480$$

B
$$500$$

C
$$420$$

D
$$520$$

Solution

Let $$x$$ be the total number of mangoes in the crate, Bruised mongo $$=$$ $$\cfrac{1}{{30}}x$$
unsaleable mangoes $$\cfrac{3}{4}\left( {\cfrac{1}{{30}}x} \right)$$
$$ \Rightarrow $$ $$\cfrac{1}{{40}}x\, = 12$$
$$ \Rightarrow $$ $$x = 12 \times 40$$
$$ \Rightarrow $$ $$ x= \,480$$
Question 7
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$$\dfrac{2x-3}{2}-\dfrac{(x+1)}{3}=\dfrac{3x-8}{4}$$

A
$$x=4$$

B
$$x=-2$$

C
$$x=2$$

D
$$x=-4$$

Solution

$$\cfrac { 2x-3 }{ 2 } -\cfrac { x+1 }{ 3 } =\cfrac { 3x-8 }{ 4 }$$

$$\Longrightarrow (6x-9-2x-2)\times 4=(3x-8)\times 6$$

$$\Longrightarrow16x-44=18x-48$$

$$ \Longrightarrow 2x=4$$

$$\Longrightarrow x=2$$
Question 8
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If $$t = x+2$$, find the value of x .If $$2t-7 +\dfrac{3(t-1)}{2}=3$$

A
$$\frac{23}{7}$$

B
$$\frac{3}{7}$$

C
$$\frac{9}{7}$$

D
$$\frac{37}{7}$$

Solution

$$(2t-7)+(\dfrac{3t-3}{2})=3\\(\dfrac{4t-14+3t-3}{2})=3\\7t-17=6\\\therefore 7t=6+17\\t=(\dfrac{23}{7})\\then x+2=(\dfrac{23}{7})\\\therefore x=(\dfrac{23}{7})-2\\=(\dfrac{23-14}{7})\\=(\dfrac{9}{7})$$
Question 9
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There is $$34000$$ rupees in a beg in the denomination of $$Rs 50,Rs 100,Rs 500$$ and $$Rs 1000$$ currencynotes. If the ratio of the number of these notes is $$2:3:4:1$$ then find the number of notes of $$500$$ denomination.

A
$$20$$

B
$$40$$

C
$$30$$

D
$$50$$

Solution

Let the number of notes be denoted by $$'x'$$
We have given ratio = $$2:3:4:1$$
$$\therefore$$ Acc to question:-
$$34,000=50(2x)+100(3x)+500(4x)+1000(x)$$
$$=100x+300x+2000x+1000x$$
$$34,000=3400 x$$
$$\dfrac{34,000}{3400}=x$$
$$x=10$$ notes
$$\therefore $$ No. of $$500$$ denominator notes $$=4\times 10$$
$$\Rightarrow 40$$
option $$B$$ is correct.
Question 10
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Find the two consecutive odd numbers whose sum is 76.

A
$$77$$ and $$39$$

B
$$37$$ and $$39$$

C
$$37$$ and $$99$$

D
$$56$$ and $$39$$

Solution

Let two consecutive odd numbers be $$x$$ and $$x+2$$
According to question
$$x+x+2=76$$
$$2x+2=76$$
$$2x=74$$
$$x=37$$
$$x+2=37+2=39$$
So, the numbers are $$37$$ and $$39$$

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Re-Attempt Weekly Quiz Competition

Linear Equations in One Variable Test - 45

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Linear Equations in One Variable Test - 45

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SHARING IS CARING

Question 1

$$960$$

$$850$$

$$895$$

$$995$$

Solution

Question 2

$$37.5$$cm

$$42.5$$cm

$$30.5$$cm

$$45.5$$cm

Solution

Question 3

$$27, 42$$

$$36, 48$$

$$27, 48$$

$$27, 36$$

Solution

Question 4

$$4.2$$ km/hr

$$9$$ km/hr

$$13$$ km/hr

$$21$$ km/hr

Solution

Question 5

$$15\dfrac {1}{2}\ kg$$

$$14 \dfrac {1}{2}\ kg$$

$$13\dfrac {1}{2}\ kg$$

$$none\ of\ these$$

Solution

Question 6

$$480$$

$$500$$

$$420$$

$$520$$

Solution

Question 7

$$x=4$$

$$x=-2$$

$$x=2$$

$$x=-4$$

Solution

Question 8

$$\frac{23}{7}$$

$$\frac{3}{7}$$

$$\frac{9}{7}$$

$$\frac{37}{7}$$

Solution

Question 9

$$20$$

$$40$$

$$30$$

$$50$$

Solution

Question 10

$$77$$ and $$39$$

$$37$$ and $$39$$

$$37$$ and $$99$$

$$56$$ and $$39$$

Solution

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