Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 26

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Weekly Quiz Competition

Question 1
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Three consecutive whole numbers are such that if they be divided by $$5, 3$$ and $$4$$ respectively; the sum of their quotients is $$40$$. Find the numbers.

A
$$20, 35$$ and $$22$$

B
$$50, 51$$ and $$52$$

C
$$40, 11$$ and $$25$$

D
$$5, 5$$ and $$8$$

Solution

Let the three consecutive whole numbers be $$x,x+1,x+2$$
Now, according to question
$$\cfrac x5+\cfrac{x+1}3+\cfrac{x+2}4=40$$
$$\Rightarrow 12x+20x+20+15x+30=2400$$
$$\Rightarrow 47x=2350$$
$$\Rightarrow x=50$$
Hence, the three consecutive whole numbers are $$50,51$$ and $$52$$.
Question 2
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A student was asked to divide a number by $$6$$ and add $$12$$ to the quotient. He, however, first added $$12$$ to the number and then divided it by $$6$$, getting $$112$$ as the answer. The correct answer should have been.

A
$$122$$

B
$$118$$

C
$$114$$

D
$$124$$

Solution

Let the number be $$x$$.
Then as per the given condition, the student was asked to find out the result of $$\dfrac {x}{6}+12$$ ....(1)
But he write it as $$\dfrac {x+12}{6}$$ and got result as $$112$$.
Therefore, $$\dfrac {x+12}{6}=112$$
$$\Rightarrow x+12=112\times 6$$
$$\Rightarrow x=660$$
So, the number is $$660$$
So, the correct result would be $$\dfrac {x}{6}+12=\dfrac {660}{6}+12=122$$.
Hence, option A is correct.
Question 3
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The sum of the digits of a two digit number is $$ 6$$ and its ten's digit is twice its unit digit. Find the number.

A
$$58$$

B
$$20$$

C
$$42$$

D
$$76$$

Solution

Let the unit's digit be $$x$$, then the ten's digit = $$2x$$

Given sum of digits $$=6$$

$$\implies x+2x = 6\\\;\;\;\;\;\;\;\;\; 3x = 6\\ \;\;\;\;\;\;\;\;\; x=2$$

Ten's digit $$= 2x = 4$$

Required number =$$10(2x)+1(x)=10(4)+1(2)=42$$
Question 4
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Separate $$178$$ into two parts so that the first part is $$8$$ less than twice the second part.

A
$$126$$ and $$42$$

B
$$176$$ and $$102$$

C
$$126$$ and $$82$$

D
$$116$$ and $$62$$

Solution

Let the second part be $$x$$

First part is $$2x-8$$.

Therefore from question, we have
Sum of two parts is $$178$$

Therefore, $$x+2x-8=178$$

$$\Rightarrow 3x=178+8$$
$$\Rightarrow 3x=186$$

$$\Rightarrow x=62$$
First part is $$2x-8=2(62)-8=124-8=116$$

Therefore, the first and second parts are $$116$$ and $$62$$.
Question 5
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$$28$$ is $$12$$ less than $$4$$ times a number. Find the number.

A
$$10$$

B
$$12$$

C
$$14$$

D
$$8$$

Solution

Let the number be $$x$$
From question, we know
$$28$$ is $$12$$ less than $$4$$ times a number.
Therefore, $$28=4x-12$$
$$\therefore 4x=28+12$$
$$\therefore x=\dfrac {40}{4}=10$$
Therefore, the number is $$10$$.
Question 6
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The ten's digit of a two digit number is three times the unit digit. The sum of the number and the unit digit is $$32$$. Find the number.

A
$$31$$

B
$$12$$

C
$$19$$

D
$$39$$

Solution

Let the units digit be $$x$$.
The tens digit $$= 3x$$.
Original number $$= 10(3x) +x = 31x$$
Sum of number and unit digit $$= 32$$
$$31x + x = 32\\ \Rightarrow 32x = 32\\ \Rightarrow x = 1$$
Original number is $$ 31(x) = 31$$.
Question 7
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Twice a number is decreased by $$15$$, equals $$25$$. Find the number.

A
$$20$$

B
$$25$$

C
$$30$$

D
$$35$$

Solution

Let the number be $$x$$
According to given problem, we have
$$2x-15=25$$
$$2x=25+15$$
$$2x=40$$
$$\therefore x=20$$
Question 8
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AB is a segment. The point P is on the perpendicular bisector of segment AB such that length of AP exceeds length of AB by $$7 $$ cm. If the perimeter of $$\Delta ABP$$ is $$38$$ cm. Find the sides of $$\Delta ABP$$ .

A
$$5$$cm, $$12$$cm and $$19$$cm

B
$$8$$cm, $$15$$cm and $$15$$cm

C
$$2$$cm, $$17$$cm and $$ 11$$cm

D
$$6$$ cm, $$14$$cm and $$18$$cm

Solution

Let the length of the $$AB = x$$
$$\therefore AP = x+7 = BP$$ $$\because p $$ is on perpendicular bisector.
The perimeter of the $$\Delta ABP = AB+AP+BP = x+x+7+x+7 = 38 cm$$
On solving, it gives $$x = 8cm$$
$$\therefore$$ three sides are $$8, 8+7, 8+7 = 8cm, 15cm, 15 cm$$
Question 9
1 / -0

If Megha's age is increased by three times her age, the result is 60 years. Find her age.

A
$$11$$ years

B
$$20$$ years

C
$$15$$ years

D
$$19$$ years

Solution

Let Megha's age be $$x$$ years
Given that if Megha's age is increased by three times, the result is $$60$$ years.
$$\therefore x + 3x = 60$$
$$\Rightarrow 4x = 60$$
$$\therefore \ x = 15$$
Hence, Megha's age is $$ 15$$ years.
Question 10
1 / -0

The sum of two consecutive odd natural numbers is $$140$$. Find the bigger number.

A
$$85$$

B
$$71$$

C
$$91$$

D
$$69$$

Solution

Let the integer be $$x$$.
The two consecutive odd number $$= x, x+2$$

Sum $$= x+x+2 = 140$$
$$2x = 138\\ x = 69$$

The length of larger number $$= x+2 = 71$$.

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Linear Equations in One Variable Test - 26

Result

Linear Equations in One Variable Test - 26

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Question 1

$$20, 35$$ and $$22$$

$$50, 51$$ and $$52$$

$$40, 11$$ and $$25$$

$$5, 5$$ and $$8$$

Solution

Question 2

$$122$$

$$118$$

$$114$$

$$124$$

Solution

Question 3

$$58$$

$$20$$

$$42$$

$$76$$

Solution

Question 4

$$126$$ and $$42$$

$$176$$ and $$102$$

$$126$$ and $$82$$

$$116$$ and $$62$$

Solution

Question 5

$$10$$

$$12$$

$$14$$

$$8$$

Solution

Question 6

$$31$$

$$12$$

$$19$$

$$39$$

Solution

Question 7

$$20$$

$$25$$

$$30$$

$$35$$

Solution

Question 8

$$5$$cm, $$12$$cm and $$19$$cm

$$8$$cm, $$15$$cm and $$15$$cm

$$2$$cm, $$17$$cm and $$ 11$$cm

$$6$$ cm, $$14$$cm and $$18$$cm

Solution

Question 9

$$11$$ years

$$20$$ years

$$15$$ years

$$19$$ years

Solution

Question 10

$$85$$

$$71$$

$$91$$

$$69$$

Solution

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