8th CBSE Mathematics - Linear Equations in One Variable Five Mark Question Paper | QB365

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8th CBSE Mathematics - Linear Equations in One Variable Five Mark Question Paper

Published on: 28/09/2019

Linear Equations in One Variable

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Questions:

Generic 5

10 x 5 = 50

The digits of a two digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number, then we get 99. Find the original number.

In a rare coin collection, there is one gold coin for every three non-gold coins. If 10 more gold coins are added to the collection, the ratio of gold coins to non-gold coins becomes 1 : 2. Based on the information, find the total number of coins in the collection now?

Hamid has three boxes of different fruits. Box A weights 2\(\frac { 1 }{ 2 } \) kg more than box B and box C weights 10 \(\frac { 1 }{ 4 } \) kg more than box B. The total weights of the three boxes is 48\(\frac { 3 }{ 4 } \) How many kilograms does box A weigh?

Denominator of a number is 4 less than its numerator. If 6 is added to the numerator, it becomes thrice the denominator. Find the fraction.

The volume of water in a tank is twice of that in the other. If we draw out 25 L from the first and add it to the other, the volumes of the water in each tank will be the same.
(a) Find the volume of water in each tank.
(b) Also, check your answer.
(c) What do you understand from a scale?

Five years ago, a man was seven times as old as his son. Five years after, the father will be three times as old as his son. Find their present ages.

The difference between the squares of two consecutive numbers is 31. Find the numbers.

The length of a rectangle exceeds its breadth by 4 cm. If length and breadth are each increased by 3 cm then area of new rectangle will be 81 cm² more than that of the given rectangle. Find the length and breadth of the given rectangle.

Divide 34 into two parts in such a way that \(\left( \frac { 4 }{ 7 } \right) \) th of one part is equal to \(\left( \frac { 2 }{ 5 } \right) \) th of the other.

10.

I have a total of Rs. 300 in coins of denomination Rs. 1, Rs.2 and Rs. 5. The number of Rs. 2 coins is 3 times the number of Rs.5 coins. The total number of coins is 160.How many coins of each denomination are with me?

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Answers Key:

Generic 5

10 x 5 = 50

Let the digit at unit place be 'x'.
\(\therefore\) The digit at the tens place = (x + 5)
\(\therefore\) Original number = 10(x + 5) + x
With interchange of digits, the new number = 10x + (x + 5)
Now, according to the condition, we have
[original number] + [New number] = 99
or [10(x + 5) + x] + [10x + (x + 5)] = 99
or [10x + 50 + x] + [10x + x + 5] = 99
or 11x + 50 + 11x + 5 = 99
or 22x + 55 = 99
Transposing 55 to RHS, we have
22x = 99 - 55 = 44
Dividing both sides by 22, we have
x = 44 + 22 = 2
\(\therefore\) x = 2
i.e., Unit place digit = 2
\(\therefore\) Tens place digit = 2 + 5 = 7
Thus the original number = 72.

Let the number of gold coins initially be x
Then, the number of non-gold coins be 3x. When,
10 more gold coins added
Then, the total number of gold coins = (10 + x)
Then, according to the question , \(\frac { (10+x) }{ 3x } =\frac { 1 }{ 2 } \)
\(\Rightarrow\) 2 (10 + x) = 3x \(\Rightarrow\) 20 + 2x = 3x \(\Rightarrow\) x = 20
Then, total number of coins at last = 3x + 10 + x = 4x + 10 = 4 x 20 + 10 = 90

Let box B's weight be x kg.
Since, box A weights \(2\frac { 1 }{ 2 } \) kg more than box Band  box C weights \(10\frac { 1 }{ 4 } \) kg more than box B.
Weight of box A = \(\left( x+2\frac { 1 }{ 2 } \right) kg=\left( x+\frac { 5 }{ 2 } \right) kg\)
Weight of box B = \(\left( x+10\frac { 1 }{ 4 } \right) kg=\left( x+\frac { 41 }{ 4 } \right) kg\)
Total weight of all the boxes
\(\left( x+\frac { 5 }{ 2 } +x+x+\frac { 41 }{ 4 } \right) kg\)
According to the question,
Total weight = \(48\frac { 3 }{ 4 } kg=\frac { 195 }{ 4 } kg\)
\(\therefore\)  \(x+\frac { 5 }{ 2 } +x+x+\frac { 41 }{ 4 } =\frac { 195 }{ 4 } \)
\(\Rightarrow\) 4x + 10+ 4x + 4x + 41 = 195  [multiplying both sides by 4]
\(\Rightarrow\) 12x + 51 = 195 \(\Rightarrow\) 12x = 144 \(\Rightarrow\) x = 12
\(\therefore\) weight box
A = \(\left( 12+\frac { 5 }{ 2 } \right) kg=\frac { 29 }{ 2 } kg=14\frac { 1 }{ 2 } kg\)

Let the numerator of the number be x. Then, denominator of the number be (x - 4)
So, fraction \(\frac { x }{ x-4 } \)
According to the question, if 6 is added to numerator, it becomes thrice the denominator
\(\therefore\) \(\frac { x+6 }{ x-4 } =\frac { 3(x-4) }{ x-4 } \Rightarrow \frac { x+6 }{ x-4 } =3\)
\(\Rightarrow\) 3x -12 = x + 6 [by cross-multiplication]
\(\Rightarrow\) \(2x18\quad \Rightarrow x=9\)
Put x = 9 in Eq. (i), we get
Fraction = \(\frac { x }{ x-4 } =\frac { 9 }{ 9-4 } =\frac { 9 }{ 5 } \)

(a) Let volume of the water in second tank be x L³
Volume of the water in first tank = 2x L³
Now, as per the given condition, we have
2x - 25 = x + 25
\(\Rightarrow\) 2x - x = 25 + 25
\(\Rightarrow\) x = 50 L
\(\therefore\) Volume of water in second tank = 50 L³ tank
volume of the water in first tank is 2 x 50 =100L³
(b) Check On putting x = 50 in the obtained Eq. (i), we get
2 x 50 - 25 = 50 + 25
\(\Rightarrow\) 100 - 25 = 75 \(\Rightarrow\) 75 =75
\(\therefore\) LHS = RHS
So, the solution of the given equation is correct
(c) In a scale, the balance from both the sides is needed. In such a manner, a linear equation is also solved, where balance from the LHS and RHS is needed

10 yr and 40 yr

15,16

Length of rectangle = 14 cm and breadth of rectangle = 10 cm

First Part = 14, Second Part = 20

10.

Let the number of Rs. 5 coins be x.
Then, the number of Rs. 2 coins = 3x
The total number of coins is 160 .
The number of coins of Rs. 1= 160 - (x + 3x)
= (160 - 4x)
The amount that I have from Rs. 5 coins = 5 x x = 5x
The amount that I have from Rs. 2 coins = 2 x 3x = 6x
The amount that I have from Rs. 1 coins
= 1 x (160 - 4x) = 160 - 4x
According to the question,
Total amount = 300
\(\Rightarrow\) 5x + 6x + (160 - 4x) = 300
\(\Rightarrow\) 5x +6x +160 - 4x = 300
\(\Rightarrow\) 7x + 160 = 300 [transposing 160 to RHS]
\(\Rightarrow\) 7x = 300 -160 [transposing 160 to RHS]
\(\Rightarrow\) 7x = 140
\(\Rightarrow\) x = \(\frac { 140 }{ 7 } \) = 20 [dividing both sides by 7]
Number of Rs. 5 coins = x = 20
Number of Rs. 2 coins = 3x = 3 x 20 = 60
and number of Rs.1 coins = 160 - 4x = 160 - 4 x 20
= 160 - 80 = SO
Hence, I have SO,60 and 20 coins of denomination Rs.1, Rs. 2 and Rs 5, respectively.