Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 31

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Question 1
1 / -0

Divide 224 into three parts so that the second will be twice the first and third will be twice the second

A
$$26, 52, 104$$

B
$$24, 48, 96$$

C
$$18, 36, 72$$

D
$$32, 64, 128$$

Solution

Let the first part be $$x$$
Then, second part $$= 2x$$
Third Part $$= 2 \times 2x = 4x$$

Given, $$x +2x + 4x = 224$$

$$\displaystyle \Rightarrow x=\frac{224}{7}=32$$

$$\displaystyle \therefore $$ The three parts are $$32, 2 \times 32, 4 \times 32$$ i.e. $$32,64,128$$.
Question 2
1 / -0

Ages of two friends are in the ratio $$2 : 1$$. If sum of their ages is $$51$$ then their ages are

A
$$34$$ yrs, $$20$$ yrs

B
$$34$$ yrs, $$17$$ yrs

C
$$20$$ yrs, $$10$$ yrs

D
$$30$$ yrs, $$15$$ yrs

Solution

Given ages in the ratio $$2 : 1$$
Thus let their ages be $$2x$$ and $$x$$
Given, the sum of their ages is $$51$$
Therefore, $$2x + x = 51$$
$$\Rightarrow 3x = 51$$
$$\Rightarrow x=17$$
Hence, required age's are $$34$$ yrs and $$17$$ yrs.
Question 3
1 / -0

Solve: $$\displaystyle 6x-1=2x+9$$

A
$$\displaystyle 1\frac{1}{2}$$

B
$$\displaystyle -2\frac{1}{2}$$

C
$$\displaystyle \frac{4}{10}$$

D
$$\displaystyle 2\frac{1}{2}$$

Solution

Given, $$6x - 1 = 2x + 9$$
Bringing all $$x$$ terms to one side and constants to another side.
$$ 6x - 2x = 9 + 1$$
$$\Rightarrow 4x = 10$$
$$\displaystyle\Rightarrow x =\frac{10}{4}=\frac{5}{2}=2\frac{1}{2}$$
Hence, solution is $$x=2\cfrac{1}{2}$$.
Question 4
1 / -0

In a two digit number the digit at the unit's place is four times the digit in the ten's place and the sum of the digits is equal to 10 What is the number?

A
14

B
44

C
82

D
28

Solution

Let the digit at ten's place $$= x$$
Then digit at unit's place $$= 4x$$
Sum of digit $$= 10$$
$$\displaystyle \Rightarrow x+4x=10\, \Rightarrow 5x=10\Rightarrow x=2$$
$$\displaystyle \therefore $$ The number $$= 10 \times x + 4x$$
$$= 10 \times 2 + 4 \times 2 = 20 + 8=28$$
Question 5
1 / -0

If the sum of one-half and one-fifth of a number exceeds one-third of that number by $$\displaystyle 7\frac{1}{3}$$ then the number is

A
15

B
18

C
20

D
30

Solution

Let the number be $$x$$
Given $$\displaystyle \left ( \frac{x}{2}+\frac{x}{5} \right )-\frac{x}{3}=7\frac{1}{3}=\frac{22}{3}$$
$$\displaystyle \Rightarrow \frac{15x+6x-10x}{30}=\frac{22}{3}$$
$$\displaystyle \Rightarrow \frac{11x}{30}=\frac{22}{3}\Rightarrow x=\frac{220}{11}=20$$
Question 6
1 / -0

A train starts with full of passengers. At the first station, the train drops one-third of the passengers and takes in 96 more. At the next station one-half of the passengers on board get down while 12 new passengers get on board If the passengers on board now are 240 the number of passengers in the beginning was

A
540

B
600

C
444

D
430

Solution

Let the full number of passengers be $$x$$
Number of passengers at the first station , after drops one-third of the passenger
$$\displaystyle =\left ( x-\frac{1}{3} x\right )+96=\frac{2x}{3}+96$$
Number of passengers at the second station , after one-half get down and 12 added
$$\displaystyle =\frac{1}{2}\left ( \frac{2x}{3}+96 \right )+12=\frac{x}{3}+48+12=\frac{x}{3}+60$$
Given $$\displaystyle \frac{x}{3}+60=240\Rightarrow \frac{x}{3}=240-60=180$$
$$\displaystyle \Rightarrow x=180\times3= 540$$
Question 7
1 / -0

Rs $$770$$ have to be divided among A, B and C such that A receives 2/9th of what B and C together receive. Then A's share is

A
Rs $$140$$

B
Rs $$154$$

C
Rs $$165$$

D
Rs $$170$$

Solution

Let, B and C together receives $$ Rs. x$$
$$\therefore $$ A's share $$=\dfrac{2}{9}x$$
A.T.Q,
$$x+\dfrac{2}{9}x=770$$
$$\implies \dfrac{9x+2x}{9}=770$$
$$\implies 11x=770 \times 9$$
$$\implies x=\dfrac{770 \times 9}{11}=630$$
A's Share $$ =\dfrac{2}{9}\times 630=140$$
Question 8
1 / -0

A sum of Rs 36.90 is made up of 180 coins which are either 10 paise coins or 25 paise coins Determine the number of each type of coins

A
$$126$$ of $$10$$ p coins and $$54$$ of $$25$$ p coins

B
$$54$$ of $$10$$ p coins and $$126$$ of $$25$$ p coins

C
$$90$$ of $$10$$ p coins and $$90$$ of $$25$$ p coins

D
$$54$$ of $$10$$ p coins and $$90$$ of $$25$$ p coins

Solution

Given : Total no of coins $$=180$$
Let, the no of $$10$$ paise coins $$x$$
No of $$25$$ paise coins $$=180-x$$
Value of $$10$$ paise coins $$=10x $$
Value of $$25$$ paise coins $$=25(180-x)$$
A.T.Q,
$$10x+25(180-x)=3690$$
$$\implies 10x+4500-25x=3690$$
$$\implies 4500-3690=25x-10x$$
$$\implies 15x=810$$
$$\implies x=\dfrac{810}{15}=54$$
$$\therefore$$ No of $$10 $$ paise coins $$=54$$
No of $$25$$ paise coins $$=180-54=126$$
Question 9
1 / -0

If the sum of four consecutive odd numbers is $$40$$, then the smallest number is

A
$$7$$

B
$$9$$

C
$$11$$

D
$$13$$

Solution

Let the first odd number be $$x$$.
Thus, the other consecutive numbers are $$x+2,x+4,x+6$$
Now, given that sum of all four numbers is 40,
$$x+ x+2+x+4+x+6=40$$
$$\Rightarrow 4x+12=40$$
$$\Rightarrow 4x=28$$
$$\Rightarrow x=7$$
Thus, the smallest number $$ 7$$.
Question 10
1 / -0

If $$50$$ is subtracted from two-third of a number, the result is equal to sum of $$40$$ and one-fourth of that number. What is the number?

A
$$256$$

B
$$216$$

C
$$186$$

D
$$144$$

Solution

Let the number be $$x$$.
According to the statement,
$$\displaystyle \frac{2x}{3}-50=\frac{x}{4}+40$$

$$\dfrac{2x}{3}-\dfrac{x}{4}=50+40$$

$$\dfrac{8x-3x}{12}=90$$

$$\dfrac{5x}{12}=90$$

$$x=\dfrac{90\times 12}{5}$$
$$=216$$

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Linear Equations in One Variable Test - 31

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Linear Equations in One Variable Test - 31

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Question 1

$$26, 52, 104$$

$$24, 48, 96$$

$$18, 36, 72$$

$$32, 64, 128$$

Solution

Question 2

$$34$$ yrs, $$20$$ yrs

$$34$$ yrs, $$17$$ yrs

$$20$$ yrs, $$10$$ yrs

$$30$$ yrs, $$15$$ yrs

Solution

Question 3

$$\displaystyle 1\frac{1}{2}$$

$$\displaystyle -2\frac{1}{2}$$

$$\displaystyle \frac{4}{10}$$

$$\displaystyle 2\frac{1}{2}$$

Solution

Question 4

14

44

82

28

Solution

Question 5

15

18

20

30

Solution

Question 6

540

600

444

430

Solution

Question 7

Rs $$140$$

Rs $$154$$

Rs $$165$$

Rs $$170$$

Solution

Question 8

$$126$$ of $$10$$ p coins and $$54$$ of $$25$$ p coins

$$54$$ of $$10$$ p coins and $$126$$ of $$25$$ p coins

$$90$$ of $$10$$ p coins and $$90$$ of $$25$$ p coins

$$54$$ of $$10$$ p coins and $$90$$ of $$25$$ p coins

Solution

Question 9

$$7$$

$$9$$

$$11$$

$$13$$

Solution

Question 10

$$256$$

$$216$$

$$186$$

$$144$$

Solution

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