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CBSE Questions for Class 8 Maths Linear Equations In One Variable Quiz 7 - MCQExams.com
CBSE
Class 8 Maths
Linear Equations In One Variable
Quiz 7
A daily wage worker was paid Rs. $$1700$$ during a period of $$30$$ days. During his period he was absent for $$4$$ days and was fined Rs. $$15$$ per day for absence. He was paid the full salary only for $$18$$ days as he came late on the other days. Those who came late were given only half the salary for that day. What was the total salary paid per month to a worker who came on time every day and was never absent?
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Rs. $$2400$$
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Rs. $$3000$$
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Rs. $$2700$$
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Rs. $$2250$$
Explanation
Let the salary of the worker per day be Rs. $$x$$.
Then, $$\displaystyle 18\times x+8\times \frac{x}{2}-4\times 15=1700 $$
$$\displaystyle \Rightarrow 18x+4x-60=1700$$
$$\Rightarrow 22x=1760$$
$$\displaystyle \rightarrow x=\frac{1760}{22}=80$$
$$\displaystyle \therefore $$ Total salary of a worker who came everyday on time for $$30$$ days $$= 30 \times$$ Rs. $$80 = $$ Rs. $$2400$$.
A grandfather is ten times older than his grand daughter. He is also $$54$$ years older than her. Find their present ages.
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The present age of granddaughter is $$6$$ years and that of grandfather is $$60$$ years.
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The present age of granddaughter is $$8$$ years and that of grandfather is $$80$$ years.
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The present age of granddaughter is $$7$$ years and that of grandfather is $$70$$ years.
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The present age of granddaughter is $$9$$ years and that of grandfather is $$90$$ years.
Explanation
Let the present age of granddaughter be $$x$$ years.
So, the present age of grandfather is $$10x$$ years.
According to the given condition, we have
$$10x-x=54$$
$$\Rightarrow 9x=54$$
$$ \Rightarrow x=6$$ .... On dividing $$9$$ from both the sides
Hence, the present age of granddaughter is $$6$$ years and that of grandfather is $$60$$ years.
If $$26 - (7m - 9) = -2 m$$, then $$m =$$
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$$\displaystyle \frac {9}{17}$$
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$$\displaystyle \frac {35}{9}$$
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$$7$$
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$$9$$
Explanation
Given, $$26 - (7m - 9) = -2 m$$
$$\Rightarrow 26-7m+9=-2m$$
$$\Rightarrow 35=5m$$
Divide $$5$$ on both the sides, we get
$$ \dfrac {35}{5}=\dfrac {5m}{5} $$
$$\therefore m=7$$
Solve $$\displaystyle \frac{x-2}{3} + \frac{3x-5}{5} = \frac{5x-7}{6} - \frac{1}{10}$$
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$$x=4$$
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$$x=8$$
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$$x=5$$
0%
$$x=12$$
Explanation
Given, $$ \cfrac { x-2 }{ 3 } +\cfrac { 3x-5 }{ 5 } =\cfrac { 5x-7 }{ 6 } -\cfrac { 1 }{ 10 }$$
Multiplying both sides of the given equation by $$30$$
....(LCM of $$3, 5, 6$$ and $$10$$), we get
$$10(x-2)+6(3x-5)=5(5x-7)-3\times 1$$
$$10x-20+18x-30=25x-35-3$$
$$28x-50=25x-38$$
Transposing the terms, we get
$$28x-25x=50-38$$
$$3x = 12$$
Dividing both sides by $$3$$, we get
$$x = 4$$
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 are their present ages?
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Rahul's age $$= 10$$ years and Haroon's age $$=17$$ years
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Rahul's age $$= 25$$ years and Haroon's age $$= 42$$ years
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Rahul's age $$= 28$$ years and Haroon's age $$= 50$$ years
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Rahul's age $$= 20$$ years and Haroon's age $$= 28$$ years
Explanation
Let the common ratio between Rahul's age and Haroon's age be $$x$$.
Therefore, age of Rahul and Haroon will be $$5x$$ and $$7x$$ years respectively.
After $$4$$ years, their ages will be $$(5x+4)$$ and $$(7x+4)$$ rexpectively.
According to the given condition, we have
$$5x+4+7x+4=56$$
$$\Rightarrow 12x+8=56$$
$$\Rightarrow x=4$$
Therefore, present age of rahul is $$5x=5\times 4=20$$ years and present age of Haroon is $$7x=7\times 4=28$$ years.
3 years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is same today. The present age of the baby is
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3 years
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2 years
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$$\displaystyle 1\frac{1}{2}$$ years
0%
1 years
Explanation
Total age of the family three years ago = 17 x 5 = 85 years.
Let the present age of the child be x years.
Present total age of the family = 85 + 5 x 3 + x = (100 + x) years.
Given $$\displaystyle \frac{100+x}{6}=17 $$
$$\displaystyle \Rightarrow 100+x=102\Rightarrow x=2 $$ years
Solve the linear equation:
$$3 (t - 3) = 5 (2t +1)$$
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$$t = -2$$
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$$t = 1$$
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$$t = 6$$
0%
$$t = -9$$
Explanation
We have, $$3(t-3)=5(2t+1)$$
$$\Rightarrow 3t-9=10t + 5$$
$$\Rightarrow 3t-10t=5+9$$ [Transposing 10t to LHS and -9 to RHS]
$$\Rightarrow-7t = 14$$
$$\Rightarrow \displaystyle \frac{7t}{7} = - \frac{14}{7}$$
$$\Rightarrow t = - 2$$
The organisers of an essay competition decide that a winner in the competition gets a prize of Rs. $$100$$ and a participant who does not win gets a prize of Rs. $$25$$. The total prize money distributed is Rs. $$3000$$. Find the number of winners, if the total number of participants is $$63$$.
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$$14$$
0%
$$16$$
0%
$$19$$
0%
$$12$$
Explanation
Let x be the number of the winners. So, the number of participants who does not win $$(63-x)$$.
Prize money given to $$x$$ winners = Rs. $$(100 \times x) =$$ Rs. $$100x$$
and prize money given to $$(63 - x)$$ participants = Rs. $$25 (63-x)$$
Given, total prize money distributed = Rs. $$3000$$
$$\therefore 100 x + 25(63-x) = 3000$$
$$\Rightarrow 4x + (63-x) = 120$$ ...[On dividing both sides by 25]
$$\Rightarrow 3x + 63 =120$$
$$\Rightarrow 3x = 120 -63$$ ...[Transposing 63 to RHS]
$$\Rightarrow 3x - 57 \Rightarrow x = \displaystyle \frac{57}{3} = 19$$
$$\therefore$$ The number of winners is $$19$$.
If a scooterist drives at the rate of $$25\text{ km/h}$$ he reaches his destination $$7\text{ min}$$ late and if he drives at the rate of $$30\text{ km/h}$$ he reaches his destination $$5\text{ min}$$ earlier. How far is his destination?
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$$20\text{ km}$$
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$$25\text{ km}$$
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$$30\text{ km}$$
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$$32\text{ km}$$
Explanation
Let the distance be $$x\text{ km}$$ far.
From the given information we can conclude that the difference between the time taken at the given two speeds is $$12 \text{ min}.$$
Now, by using the relation between distance, time, and speed,
$$\begin{aligned}{}\frac{x}{{25}} - \frac{x}{{30}}& = \frac{{12}}{{60}}\\\frac{{6x - 5x}}{{150}}& = \frac{1}{5}\\x &= \frac{{150}}{5}\\& = 30\text{ km}\end{aligned}$$
Hence, the desired destination is $$30 \text{ km}$$ far.
A grand father is ten times older than his grand daughter. He is also $$54$$ years older than her. Find their present ages.
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The present age of granddaughter is $$5$$ years and that of grandfather is $$50$$ years.
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The present age of granddaughter is $$4$$ years and that of grandfather is $$40$$ years.
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The present age of granddaughter is $$7$$ years and that of grandfather is $$70$$ years.
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The present age of granddaughter is $$6$$ years and that of grandfather is $$60$$ years.
Explanation
Let the present age of granddaughter be $$x$$ years.
So, the present age of grandfather is $$10x$$ years.
According to the given condition, we have
$$10x-x=54$$
$$\Rightarrow 9x=54$$
$$ \Rightarrow x=6$$ ....[On dividing both sides by $$9$$]
Hence, the present age of granddaughter is $$6$$ years and that of grandfather is $$60$$ years.
The age of a girl in months is equal to the age of her grandmother in years. If the difference between their ages is $$66$$ years, find their ages.
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The age of girl is $$5$$ years and that of grandmother is $$60$$ years
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The age of girl is $$6$$ years and that of grandmother is $$72$$ years
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The age of girl is $$7$$ years and that of grandmother is $$84$$ years
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The age of girl is $$8$$ years and that of grandmother is $$96$$ years
Explanation
Let the age of grandmother is $$12x$$ years.
Then, the age of girl $$=$$ $$\cfrac { 12x }{ 12 } =x$$ years
According to the given condition,
$$12x - x = 66$$
$$\Rightarrow 11x=66$$
$$\Rightarrow x=6$$
Therefore, the age of girl $$= 6$$ years and that of grandmother $$= 12 \times 6 = 72$$ years.
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.
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Length $$= 1$$6m and breadth $$= 12$$m
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Length $$= 13$$m and breadth $$= 15$$m
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Length $$= 14$$m and breadth $$= 17$$m
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Length $$= 18$$m and breadth $$= 21$$m
Explanation
Let the breadth of the rectangle be $$x$$ m. Then,
length of the rectangle $$= (2x - 8)$$ m
Given, perimeter $$= 56$$ m
$$\Longrightarrow 2(2x-8+x)=56\\ \Longrightarrow 2(3x-8)=56\\ \Longrightarrow 6x-16=56\\ \Longrightarrow 6x=56+16\quad \quad \quad [Transposing\quad -16\quad to\quad RHS]\\ \Longrightarrow 6x=72\\ \Longrightarrow x=12$$
$$\text{Therefore, breadth of the rectangle}$$ $$= 12$$ $$\text{m and length of the rectangle}$$ $$= 2 \times 12 - 8 = 16$$ $$m$$
$$70$$ 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.
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Number of $$10$$ paise coins are $$60$$
Number of $$50$$ paise coins are $$20$$
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Number of $$10$$ paise coins are $$20$$
Number of $$50$$ paise coins are $$40$$
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Number of $$10$$ paise coins are $$40$$
Number of $$50$$ paise coins are $$30$$
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Number of $$10$$ paise coins are $$30$$
Number of $$50$$ paise coins are $$50$$
Explanation
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 $$= 70 - 40 = 30$$.
When $$19$$ is subtracted from the product of $$p$$ and $$4$$, the result is $$17$$.
Then, the value of $$p$$ is
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$$\displaystyle \dfrac { -1 }{ 9 } $$
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$$\displaystyle \dfrac { 1 }{ 9 } $$
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$$\displaystyle -9$$
0%
$$\displaystyle 9$$
Explanation
According to the question, $$\displaystyle 4p-19=17$$
Adding $$19$$ to both sides we get,
$$\Longrightarrow \displaystyle 4p=17+19$$
$$\Longrightarrow\displaystyle 4p=36$$
$$\Longrightarrow \displaystyle p=\frac { 36 }{ 4 } $$ ...
(Divide by $$4$$ on both sides)
$$\therefore \displaystyle p=9$$
So, option D is correct.
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?
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$$88$$
0%
$$26$$
0%
$$62$$
0%
$$68$$
Explanation
Let the digit in units place be $$x$$.
Then, digit in ten's place $$= 3x$$
So, original number $$= 10(3x) + x = 31x$$
On interchanging the digits, new number $$= 10x + 3x = 13x$$
According to the given condition,
$$ {31x + 13x = 88} $$
$$\Rightarrow 44x=88\\ \Rightarrow x=2$$
So, the original number is $$31x$$ or $$31 \times 2 = 62$$.
On the other hand, if we consider the digit in ten's place as $$x$$, then the digit in the unit's place will be $$3x$$.
So, original number $$= 10(x) + 3x = 13x$$
On interchanging the digits, new number $$= 10(3x) + x = 31x$$
According to the given condition,
$$ {13x + 31x = 88} $$
$$\Rightarrow 44x=88\\ \Rightarrow x=2$$
So, the original number is $$13x$$ or $$13 \times 2 = 26$$.
Here, both answers are correct, as $$26 + 62 = 88$$.
The difference between a number and two third of itself is $$13$$. Find the number.
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$$31$$
0%
$$35$$
0%
$$39$$
0%
$$33$$
Explanation
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 $$39$$.
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.
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Speed of car $$B$$ is $$63$$ km/h and speed of car $$A$$ is $$70$$ km/h
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Speed of car $$B$$ is $$70$$ km/h and speed of car $$A$$ is $$75$$ km/h
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Speed of car $$B$$ is $$50$$ km/h and speed of car $$A$$ is $$65$$ km/h
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Speed of car $$B$$ is $$72$$ km/h and speed of car $$A$$ is $$80$$ km/h
Explanation
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.
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.
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$$28$$
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$$32$$
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$$15$$
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$$25$$
Explanation
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.
The difference of two numbers is $$72$$ and the quotient obtained by dividing the one by the other is $$3$$. Find the numbers.
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$$60$$ and $$132$$
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$$42$$ and $$114$$
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$$36$$ and $$108$$
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$$30$$ and $$102$$
Explanation
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 $$= 3 \times 36 = 108$$
The sum of three consecutive odd numbers is $$63$$. Find them.
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$$19, 21, 23$$
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$$21,23,29$$
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$$19,21,27$$
0%
$$21,23,32$$
Explanation
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 $$23$$.
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 ________.
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3
0%
5
0%
6
0%
8
Explanation
Let the no. of boys wearing full uniform
be x
∴
∴
No. of boys wearing baseball shirts only
=
11
−
x
=11−x
and No. of boys wearing baseball pants only
=
14
−
x
=14−x
∴
11
−
x
+
14
−
x
+
x
=
19
∴11−x+14−x+x=19
or,
−
x
=
19
−
11
−
14
=
−
6
⟹
x
=
6
Sixteen years ago, Tanya's grandfather was $$8$$ times older to her.
$$8$$ years from now h
e would be $$3$$ times for her age. Eight years ago. what was the ratio of Tanya's age to that of her grandfather?
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$$1:2$$
0%
$$1:5$$
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$$3:8$$
0%
None of these
Explanation
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 $$11:53$$.
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?
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0%
$$21$$
0%
$$22$$
0%
$$25$$
0%
$$23$$
Explanation
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 $$x=\frac {1035}{45}=23$$
$$\therefore x=23 days$$
A piece of string is $$80$$ cm long. It is cut into three pieces. The longest piece is $$3$$ times as long as the middle sized piece and the shortest piece is $$46$$ cm shorter than the longest piece. Find the length of the shortest piece.
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0%
$$14$$ cm
0%
$$10$$ cm
0%
$$8$$ cm
0%
$$18$$ cm
Explanation
Let the length of middle piece $$=$$ $$x$$
Then length of longest piece $$=$$ $$3x$$
Then length of shortest piece $$=$$ $$3x - 46$$
Thus, $$x + 3x + 3x - 46 = 80$$
$$\Rightarrow 7x = 126$$
$$\Rightarrow x = 18$$
Hence, shortest piece $$=$$ $$18 \times 3 - 46 = 54 - 46 = 8$$ cm
The sum of one half, one third and one fourth of a number exceed the number itself by $$12$$. The number is
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0%
$$72$$
0%
$$144$$
0%
$$180$$
0%
$$244$$
Explanation
Let the number be $$x$$
$$\Rightarrow\;\begin{bmatrix}\displaystyle\frac{x}{2}+\displaystyle\frac{x}{3}+\displaystyle\frac{x}{4}\end{bmatrix}-x=12$$
$$\Rightarrow\;\displaystyle\frac{6x+4x+3x-12x}{12}=12$$
$$\Rightarrow\;x=144$$
The perimeter of an isosceles triangle is $$\displaystyle 3\frac { 9 }{ 4 } $$ cm. The base of an isosceles triangle is $$\dfrac{3}{2}$$ cm. What is the length of either of the remaining equal sides?
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0%
$$\displaystyle \frac { 15 }{ 4 } $$ cm
0%
$$\displaystyle \frac { 15 }{ 8 } $$ cm
0%
$$\displaystyle \frac { 8 }{ 15 } $$ cm
0%
$$\displaystyle \frac { 4 }{ 15 } $$ cm
Explanation
Perimeter = $$\displaystyle 3\frac { 9 }{ 4 } =\frac { 21 }{ 4 } $$ cm
Property of an isosceles triangle = two sides are equal
Perimeter of triangle = Sum of three sides
$$\therefore \displaystyle \frac { 21 }{ 4 } =a+a+\frac { 3 }{ 2 } $$
$$\Longrightarrow\displaystyle 2a=\frac { 21 }{ 4 } -\frac { 3 }{ 2 } $$
$$\Longrightarrow 2a=\dfrac { 21-6 }{ 4 } $$
$$\therefore a= \dfrac { 15 }{ 8 } $$
The sum of three consecutive multiples of $$9$$ is $$135$$, then the multiples are
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0%
$$36, 45, 54$$
0%
$$45, 54, 63$$
0%
$$27, 36, 45$$
0%
None of the above.
Explanation
If $$p$$ is a multiple of $$9$$, then next two multiples are $$\displaystyle p+9$$ and $$p+18$$
According to the question:
$$\Longrightarrow \displaystyle p+(p+9)+(p+18)=135$$
$$\Longrightarrow\displaystyle 3p+27=135$$
$$\Longrightarrow \displaystyle 3p=135-27$$
$$\Longrightarrow \displaystyle 3p=108$$
$$\Longrightarrow \displaystyle p=36$$
Thus,
$$p=36$$
$$p+9=45$$
$$p+18=54$$
Hence the required multiples are $$36,45,54$$
In a division, a student took $$63$$ as divisor instead of $$36$$. His answer was $$24$$. The correct answer is -
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0%
$$42$$
0%
$$32$$
0%
$$48$$
0%
$$28$$
0%
$$38$$
Explanation
From the given condition we have,
$$\dfrac {x}{63} = 24$$
$$x = 24 \times 63$$
So, correct answer would be,
$$\dfrac {24 \times 63}{36} = 42$$
In a contest there were $$10$$ problems. Each correct answer is eligible for $$5$$ points while each incorrect answer is penalized with a deduction of $$2$$ points. Venkat answered all the questions but got only $$29$$ points. The number of correct answers of Venkat was
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0%
$$5$$
0%
$$2$$
0%
$$7$$
0%
$$4$$
Explanation
Let the correct answer of Venkat be x. Then
$$5x -2(10 -x) =29$$
or $$5x -20 + 2x = 29$$
or $$7x=49$$
or $$x=7$$
When a number is divided by $$4$$, the result is $$-9$$. The number is
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0%
$$36$$
0%
$$-36$$
0%
$$\dfrac{1}{36}$$
0%
$$\dfrac{-1}{36}$$
Explanation
Let the required number be $$x$$
According to the question, $$\displaystyle \frac { x }{ 4 } =-9\\ $$
$$\Longrightarrow\displaystyle x=-9\times 4$$ ...(Multiply both sides by $$4$$)
$$\Longrightarrow\displaystyle x=-36$$
So, option B is correct.
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