Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 25

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Weekly Quiz Competition

Question 1
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Baichungs father is $$26$$ years younger than Baichungs grandfather and $$29$$ years older than Baichung. The sum of the ages of all the three is $$135$$ years. What is the age of each one of them?

A
Baichungs age: 14 years; Baichungs fathers age: $$28$$ years; Baichungs grandfathers age $$= 42$$ years

B
Baichungs age: $$17$$ years; Baichungs fathers age: $$46$$ years; Baichungs grandfathers age $$= 72$$ years

C
Baichungs age: $$16$$ years; Baichungs fathers age: $$38$$ years; Baichungs grandfathers age $$= 52$$ years

D
Baichungs age: $$25$$ years; Baichungs fathers age: $$58$$ years; Baichungs grandfathers age $$= 82$$ years

Solution

Let the father's age be $$x$$.
Grandfather's age $$= x+26$$
Baichung age $$= x -29$$
Sum of ages $$= 135$$
$$x+26 + x + x -29 = 135$$
$$3x = 135 +3 = 138$$
$$x = 46$$
So, grandfather's age $$= 46 + 26 = 72$$ yrs
Father's age $$= 46$$ yrs
Baichungs age $$= 46 - 29 = 17$$ yrs
Question 2
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The present ages of a father and his son are in the ratio $$7 : 3$$ and the ratio of their ages will be $$2 : 1$$ after $$10 $$ years. Then, the present age of father (in years) is

A
$$42$$

B
$$56$$

C
$$70$$

D
$$77$$

Solution

Let the present ages of father and son be $$7x$$ and $$3x$$.
After $$10$$ years, their ages will be $$7x+10$$ and $$3x+10$$.
According to the question, we have
$$\dfrac {7x+10}{3x+10}=\dfrac {2}{1}$$
$$1(7x+10)=2(3x+10)$$
$$7x+10=6x+20$$
$$7x-6x=20-10$$
$$x=10$$
Then present age of father is $$7x$$, i.e. $$7\times 10=70$$ years.
Question 3
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It costs Rs. $$10$$/kilometer to fly and Rs. $$2$$/kilometer to drive. If one travels $$200$$ km covering $$x$$ km of the distance by flying and the rest by driving, then the cost of the trip is

A
Rs. $$2,000$$

B
Rs. $$24,000$$

C
Rs.$$ (8x + 400)$$

D
Rs. $$(12x + 400)$$

Solution

Let distance travelled by flying is $$x$$ km.
Hence, the distance travelled by driving $$= (200-x)$$km.
The cost of flying $$ = $$ Rs. $$ 10 $$ per km
Hence, the cost of flying $$x$$ km $$ = 10 \times x =10x$$
The cost of driving $$ = $$ Rs. $$ 2 $$ per km
The cost of driving $$(200-x)$$ km $$ = 2 \times (200-x) = 400 -2x$$
Hence, the total cost $$ =10x + 400-2x = $$Rs.$$8x+400$$
Question 4
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If you subtract $$\cfrac{1}{2}$$ from a number and multiply the result by $$\cfrac{1}{2}$$, you get $$\cfrac{1}{8}$$. The number obtained is

A
$$\dfrac 39$$

B
$$\dfrac 34$$

C
$$\dfrac {3}{13}$$

D
$$\dfrac 32$$

Solution

Let the number be $$x$$
We need to subtract $$\dfrac {1}{2}$$ from $$x$$ and multiply the resultant by $$\dfrac {1}{2}$$,
So, we get $$\left (x-\cfrac { 1 }{ 2 } \right)\cfrac { 1 }{ 2 } $$
The resultant is equal to $$\dfrac {1}{8}$$
$$\left (x-\dfrac {1}{2}\right)\dfrac {1}{2}=\dfrac {1}{8}$$
$$(2x-1)\cfrac { 1 }{ 4 } = \cfrac { 1 }{ 8 }$$
$$4x - 2 = 1$$
$$4x = 3$$
$$x = \cfrac { 3 }{ 4 }$$
Question 5
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Lakshmi is a cashier in a bank. She has currency notes of denominations Rs. $$100$$, Rs. $$50$$ and Rs. $$10$$, respectively. The ratio of the number of these notes is $$2:3:5$$. The total cash with Lakshmi is Rs. $$4,00,000$$. How many notes of each denomination does she have?

A
Rs. $$100$$ $$\rightarrow 2000$$ notes; Rs. $$50 \rightarrow 3000$$ notes; Rs. $$10 \rightarrow 5000$$ notes

B
Rs. $$100 \rightarrow 1500$$ notes; Rs. $$50 \rightarrow 2000$$ notes; Rs. $$10\rightarrow 3000$$ notes

C
Rs. $$100 \rightarrow 1000$$ notes; Rs. $$50 \rightarrow 2500$$ notes; Rs. $$10 \rightarrow 4000$$ notes

D
Rs. $$100 \rightarrow 3000$$ notes; Rs. $$50 \rightarrow 2000$$ notes; Rs. $$10 \rightarrow 4000$$ notes

Solution

Let the number of $$100, 50$$ and $$10$$ Rs notes be $$2x, 3x, 5x$$.
Total cash $$=$$ Rs. $$4,00,000$$

$$\therefore 2x(100) + 3x(50) + 5x(10) =4,00,000$$

$$200x +150x +50x = 4,00,000$$

$$400x = 4,00,000$$

$$\therefore x = 1,000$$

Noumber of Rs. $$100$$ notes $$= 2 \times 1000=2000$$
Number of Rs. $$50$$ notes $$=3 \times 1000= 3000$$
Number of Rs. $$10$$ notes $$=5\times 1000= 5000$$
Question 6
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One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get $$88$$. What is the original number?

A
$$26$$

B
$$12$$

C
$$43$$

D
$$16$$

Solution

Let one of the digits be $$x$$.
Then the other digit is $$3x$$.
Original number $$= 10x +3x = 13x$$
Interchanged number $$= 10(3x) +x = 31x$$
$$\Rightarrow 13x+31x = 88$$
$$\Rightarrow 44x = 88$$
$$\Rightarrow x = 2$$
The other digit $$= 3x= 3\times 2 = 6$$
Therefore the original number is $$ 26$$.
Question 7
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The sum of three non-zero prime numbers is $$100$$. One of them exceeds the other by $$36$$. Then the largest number is

A
$$73$$

B
$$91$$

C
$$67$$

D
$$57$$

Solution

The sum of three prime numbers can only be even if one of the prime numbers is even, the only prime number that is even is $$2$$.
So, the sum of the other two prime numbers must be $$98$$.
$$x+x+36=98$$
$$\Rightarrow 2x=62$$
$$\Rightarrow x=31$$
Numbers are $$2, 31$$ and $$67$$.
The highest among them is $$67$$.
Question 8
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The ages of Rahul and Haroon are in the ratio $$5:7$$. Four years later the sum of their ages will be $$56$$ years. What is the present age of rahul?

A
$$14$$

B
$$18$$

C
$$20$$

D
$$11$$

Solution

Let the ages of Rahul and Haroon be $$5x$$ and $$7x$$.
Four years later, their ages will be $$5x+4$$ and $$7x+4$$.
Given, the sum of their ages is $$56$$.
$$\therefore 5x+4+7x+4 = 56$$
$$12x = 56 -8$$
$$12 x = 48$$
$$\therefore x = 4$$
Therefore, age of Rahul will be $$5x=5\times 4=20$$ and that of Haroon will be $$7x=7\times 4=28$$.
Thus, Rahul is $$20$$ years and Haroon is $$28$$ years old.
Question 9
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Three consecutive integers are such that when they are taken in increasing order and multiplied by $$2, 3$$ and $$4$$ respectively, they add up to $$74$$. Find these numbers.

A
$$4, 5, 6$$

B
$$5, 6, 7$$

C
$$8, 9, 10$$

D
$$7, 8, 9$$

Solution

Let the consecutive integers be $$x, x+1, x+2$$
Given that they are taken in increasing order and multiplied by $$2,3,4$$.
$$\therefore (x)2 + 3(x+1) + 4(x+2) = 74$$
$$2x+3x+3 + 4x + 8 = 74$$
$$9x = 74 -11 = 63$$
$$\therefore x = 7$$
Then, the numbers would be $$x=7$$,
$$x+1=7+1=8$$,
$$x+2=7+2=9$$.
So, the integers are $$7, 8, 9$$.
Question 10
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The sum of three consecutive odd numbers is $$57$$. Find the numbers.

A
$$17, 19$$ and $$21$$

B
$$19, 15$$ and $$25$$

C
$$13, 17$$ and $$26$$

D
$$21, 13$$ and $$29$$

Solution

Let the $$3$$ consecutive odd numbers be $$x, x+2, x+4$$
$$ x+x+2 +x+4 = 57\\ 3x = 51\\ x = 17$$
So the numbers are $$17, 19$$ and $$21.$$

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Re-Attempt Weekly Quiz Competition

Linear Equations in One Variable Test - 25

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Linear Equations in One Variable Test - 25

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Question 1

Baichungs age: 14 years; Baichungs fathers age: $$28$$ years; Baichungs grandfathers age $$= 42$$ years

Baichungs age: $$17$$ years; Baichungs fathers age: $$46$$ years; Baichungs grandfathers age $$= 72$$ years

Baichungs age: $$16$$ years; Baichungs fathers age: $$38$$ years; Baichungs grandfathers age $$= 52$$ years

Baichungs age: $$25$$ years; Baichungs fathers age: $$58$$ years; Baichungs grandfathers age $$= 82$$ years

Solution

Question 2

$$42$$

$$56$$

$$70$$

$$77$$

Solution

Question 3

Rs. $$2,000$$

Rs. $$24,000$$

Rs.$$ (8x + 400)$$

Rs. $$(12x + 400)$$

Solution

Question 4

$$\dfrac 39$$

$$\dfrac 34$$

$$\dfrac {3}{13}$$

$$\dfrac 32$$

Solution

Question 5

Rs. $$100$$ $$\rightarrow 2000$$ notes; Rs. $$50 \rightarrow 3000$$ notes; Rs. $$10 \rightarrow 5000$$ notes

Rs. $$100 \rightarrow 1500$$ notes; Rs. $$50 \rightarrow 2000$$ notes; Rs. $$10\rightarrow 3000$$ notes

Rs. $$100 \rightarrow 1000$$ notes; Rs. $$50 \rightarrow 2500$$ notes; Rs. $$10 \rightarrow 4000$$ notes

Rs. $$100 \rightarrow 3000$$ notes; Rs. $$50 \rightarrow 2000$$ notes; Rs. $$10 \rightarrow 4000$$ notes

Solution

Question 6

$$26$$

$$12$$

$$43$$

$$16$$

Solution

Question 7

$$73$$

$$91$$

$$67$$

$$57$$

Solution

Question 8

$$14$$

$$18$$

$$20$$

$$11$$

Solution

Question 9

$$4, 5, 6$$

$$5, 6, 7$$

$$8, 9, 10$$

$$7, 8, 9$$

Solution

Question 10

$$17, 19$$ and $$21$$

$$19, 15$$ and $$25$$

$$13, 17$$ and $$26$$

$$21, 13$$ and $$29$$

Solution

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