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Maths 3 Mark Question

Linear Equations in One Variable · Maharashtra Board STD 7 (MSBSHSE - Semi English Medium). 42 questions.

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Question 13 Marks
A sum of Rs. 500 is in the form of denominations of Rs. 5 and Rs. 10. If the total number of notes is 90. Find the number of notes of each type.
Answer
Total number of notes = 90
Let number of notes of Rs. 5 = x
Then number of notes of Rs. 10 = 90 - x
Then x × 5 + (90 - x) × 10 = 500
⇒ 5x + 900 - 10x = 500
⇒ -5x = 500 - 900 = -400
x = 80
Number of 5 rupees notes = 80
And ten rupees notes = 90 - 80 = 10
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Question 23 Marks
Solve the following equations. Check your result in case.
$\frac{\text{x}}{2}+\frac{\text{x}}{4}=\frac{1}{8}$
Answer
$\frac{\text{x}}{2}+\frac{\text{x}}{4}=\frac{1}{8}$
$\frac{2\text{x}+\text{x}}{4}=\frac{1}{8}$
$\Rightarrow\frac{3\text{x}}{4}=\frac{1}{8}$
$\Rightarrow\text{x}=\frac{1}{8}\times\frac{4}{3}=\frac{1}{6}$
$\therefore\text{x}=\frac{1}{6}$
Check:
$\text{L.H.S.}=\frac{\text{x}}{2}+\frac{\text{x}}{4}$
$=\frac{1}{6\times2}+\frac{1}{6\times4}=\frac{1}{12}+\frac{1}{24}$
$=\frac{2+1}{24}=\frac{3}{24}=\frac{1}{8}=\text{R.H.S.}$
Hence $\text{x}=\frac{1}{6}$
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Question 33 Marks
Solve the following equations. Check your result in case.
2(x - 2) + 3(4x - 1) = 0
Answer
2(x - 2) + 3(4x - 1) = 0
⇒ 2x - 4 + 12x - 3 = 0
⇒ 2x + 12x = 4 + 3 (By transposing)
⇒ 14x = 7
$\Rightarrow\text{x}=\frac{7}{14}=\frac{1}{2}$
Check: $\text{L.H.S}=2(\text{x}-2)+3(4\text{x}-1)$
$=2\Big(\frac{1}{2}-2\Big)+3\Big(4\times\frac{1}{2}-1\Big)$
$=2\times\Big(\frac{1-4}{2}\Big)+3(2-1)$
$=2\text{x}\Big(\frac{-3}{2}\Big)+3(1)$
$=-3+3=0=\text{R.H.S.}$
Hence $\text{x}=\frac{1}{2}$
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Question 43 Marks
The length of a rectangular field is twice its breadth. If the perimeter of the field is 150 metres, find its length and breadth.
Answer
Perimeter of field = 150m
Length + Breadth $=\frac{150}{2}=75\text{m}$
[Perimeter = 2(l + b)]
Let length = x Then breadth = 75 - x
Then x = 2(75 - x)
⇒ x = 150 - 2x
⇒ x + 2x = 150
⇒ 3x = 150
$\Rightarrow\text{x}=\frac{150}{3}=50$
Length = 50m
And breadth = 75 - 50 = 25m
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Question 53 Marks
Solve the following equations. Check your result in case.3 + 2x = 1 - x
Answer
3 + 2x = 1 – x
⇒ 2x + x = 1 – 3 (By transposing)
⇒ 3x = -2
$\Rightarrow\text{x}=\frac{-2}{3}$
Check:
$\text{L.H.S.}=3+2\text{x}=3+2\Big(\frac{-2}{3}\Big)$
$=\frac{3}{1}-\frac{4}{3}=\frac{9-4}{3}=\frac{2}{3}$
$\text{R.H.S.}=1-\text{x}=1-\big(\frac{-2}{3}\Big)=1+\frac{2}{3}$
$=\frac{3+2}{3}=\frac{5}{3}$
$\therefore\text{L.H.S = R.H.S}$
Hence $\text{x}=\frac{-2}{3}$
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Question 63 Marks
Solve the following equations. Check your result in case.
$\frac{1}{2}\text{x}-3=5+\frac{1}{3}\text{x}$
Answer
$\frac{1}{2}\text{x}-3=5+\frac{1}{3}\text{x}$
$\Rightarrow\frac{1}{2}\text{x}-\frac{1}{3}\text{x}=5+3$ (By transposing)
$\Rightarrow\frac{3\text{x}-2\text{x}}{6}=8$
$\Rightarrow\frac{\text{x}}{6}=8$
$\Rightarrow\text{x}=8\times6$
$\Rightarrow\text{x}=48$
$\therefore\text{x}=48$
Check:
$\text{L.H.S}=\frac{1}{2}\text{x}-3=\frac{1}{2}\times48-3$
$=24-3=21$
$\text{R.H.S.}=5+\frac{1}{3}\text{x}$
$=5+\frac{1}{3}\times48=5+16=21$
$\text{L.H.S.}=\text{R.H.S.}$
Hence $\text{x}=48$
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Question 73 Marks
Two supplementary angles differ by 44°. Find the angles.
Answer
Sum of two supplementary angles = 180°
Let first angle = x
Then second angle = 180° - x
x - (180° - x) = 44°
⇒ x - 180° + x = 44°
⇒ 2x = 44° + 180° = 224°
⇒ 2x = 224°
⇒ x = 112°
First angle = 112°
And second angle = 180° - 112° = 68°
Hence angles are 68°, 112°
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Question 83 Marks
Two complementary angles differ by 8°. Find the angles.
Answer
Sum of two complementary angles = 90°
Let first angle = x
Then second = 90° - x
x - (90° - x) = 8
⇒ x - 90° + x = 8
⇒ 2x = 8 + 90°
⇒ 2x = 98°
⇒ x = 49°
First angle = 49°
And second angle = 90° - 49° = 41°
Hence angles are 41°, 49°
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Question 93 Marks
Solve: $\frac{3\text{x}}{10}+\frac{2\text{x}}{5}=\frac{7\text{x}}{25}+\frac{29}{25}.$
Answer
We have:
$\frac{3\text{x}}{10}+\frac{2\text{x}}{5}=\frac{7\text{x}}{25}+\frac{29}{25}$
$\Rightarrow\frac{3\text{x}+4\text{x}}{10}=\frac{7\text{x}+29}{25}$
$\Rightarrow\frac{7\text{x}}{10}=\frac{7\text{x}+29}{25}$
$\Rightarrow175\text{x}=70\text{x}+290$
$\Rightarrow105\text{x}=290$
$\Rightarrow\text{x}=\frac{290}{105}$
$\Rightarrow\text{x}=\frac{58}{21}$
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Question 103 Marks
A labour is engaged for 20 days on the condition that he will receive Rs. 120 for each day he works and will be fined Rs. 10 for each day he is absent. If he receives Rs. 1880 in all, for how many days did he remain absent?
Answer
Let x be the number of days of his sbsence.
$\therefore$ Number of days of his presence = (20 - x)
Now, (20 - x) 120 - 10x = 1880
⇒ 2400 - 120x - 10x = 1880
⇒ 2400 - 1880 = 130x
⇒ 130x = 520
⇒ x = 4
$\therefore$ Number of days of his absence = 4
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Question 113 Marks
After 12 years Manoj will be 3 times as old as he was 4 years ago. Find his present age.
Answer
Let age of Manoj 4 years ago = x
Then his present age = x + 4
After 12 years his age will be = x + 4 + 12 = x + 16
x + 16 = 3(x)
x + 16 = 3x
⇒ 16 = 3x - x
⇒ 2x = 16
x = 8
His present age = 8 + 4 = 12 years
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Question 123 Marks
Solve the following equations. Check your result in case.
2.4(3 - x) - 0.6(2x - 3) = 0
Answer
2.4(3 - x) - 0.6(2x - 3) = 0
⇒ 7.2 - 2.4x - 1.2x + 1.8 = 0
⇒ -2.4x - 1.2x = -(7.2 + 1.8)
L.H.S. = 2.4(3 - x) - 0.6(2x - 3)
⇒ 2.4(3 - 2.5) - 0.6(2 × 2.5 - 3)
⇒ 2.4(0.5) - 0.6(5 - 3)
⇒ 1.2 - 0.6 × 2
= 1.2 - 1.2
R.H.S. = 0
Hence x = 2.5
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Question 133 Marks
Find three consecutive positive even integers whose sum is 90.
Answer
Let first positive even number = 2x
Second number = 2x + 2
Third number = 2x + 4
2x + 2x +2 + 2x + 4 = 90
⇒ 6x + 6 = 90
⇒ 6x = 90 - 6 = 84
x = 14
First even number = 2x = 2 × 14 = 28
Second number = 2x + 2 = 2 × 14 + 2 = 28 + 2 = 30
Third number = 30 + 2 = 32
Required numbers are 28, 30, 32
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Question 143 Marks
Solve the following equations. Check your result in case.
13(y - 4) - 3(y - 9) - 5(y + 4) = 0
Answer
13(y - 4) - 3(y - 9) - 5(y + 4) = 0
⇒ 13y - 52 - 3y + 27 - 5y - 20 = 0
⇒ 13y - 3y - 5y - 52 + 27 - 20 = 0
⇒ 13y - 8y - 72 + 27 = 0
⇒ 5y - 45 = 0
⇒ 5y = 45 (By transposing)
⇒ y = 9
Check:
L.H.S. = 13(y - 4) - 3(y - 9) - 5(y + 4)
= 13(9 - 4) - 3(9 - 9) - 5(9 + 4)
= 13 × 5 - 3 × 0 - 5 × 13
= 65 - 0 - 65 = 0 = R.H.S.
Hence y = 9
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Question 153 Marks
Two equal sides of a triangle are each 5 metres less than twice the third side. If the perimeter of the triangle is 55 metres, find the lengths of its sides.
Answer
Perimeter of an isosceles triangle = 55m
Let the third side of an isosceles triangle = x
Then each equal side = (2x - 5)m
According to the condition,
x + 2 (2x - 5) = 55
⇒ x + 4x - 10 = 55
⇒ 5x = 55 + 10
⇒ 5x = 65
⇒ x = 13
And 2x – 5 = 2 × 13 - 5 = 21m
Sides will be 13m, 21m, 21m
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Question 163 Marks
Solve the following equations. Check your result in case.
5(2x -3) - 3(3x - 7) = 5
Answer
5(2x - 3) - 3(3x - 7) = 5
⇒ 10x - 15 - 9x + 21 = 5
⇒ 10x - 9x - 15 + 21 = 5
⇒ 10x - 9x = 5 + 15 - 21 (By transposing)
⇒ x = 20 - 21 = -1
⇒ x = -1
Check:
L.H.S. = 5(2x - 3) - 3(3x - 7)
= 5[2 x (-1) -3] -3[3 (-1) -7] = 5[-2 - 3] - 3[-3 - 7]
= 5x(-5) -3x(-10)
= -25 + 30
= 5 = R.H.S.
Hence x = -1
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Question 173 Marks
Four added to twice a number yields $\frac{26}{5}$. Find the fraction.
Answer
Let the required fraction = x
then $2\text{x}+4=\frac{26}{5}$
$\Rightarrow2\text{x}=\frac{26}{5}-4$
$\Rightarrow2\text{x}=\frac{26-20}{5}=\frac{6}{5}$
$\therefore\text{x}=\frac{6}{5}\times\frac{1}{2}=\frac{3}{5}$
$\therefore$ Required fractiom $=\frac{3}{5}$
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Question 183 Marks
Solve the following equations. Check your result in case.
$\frac{2\text{m}+5}{3}=3\text{m}-10$
Answer
$\frac{2\text{m}+5}{3}=3\text{m}-10$
⇒ 2m + 5 = 3 (3m - 10) (By cross multiplication)
⇒ 2m + 5 = 9m - 30
⇒ 2m - 9m = -30 - 5
⇒ -7m = -35
⇒ m = 5
m = 5
Check:
$\text{L.H.S.}=\frac{2\text{m}+5}{3}=\frac{2\times5+5}{3}$
$=\frac{10+5}{3}=\frac{15}{3}=5$
R.H.S. = 3m - 10 = 3 × 5 - 10 = 15 - 10 = 5
L.H.S. = R.H.S.
Hence m = 5
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Question 193 Marks
Solve the following equations. Check your result in case.
$\frac{2\text{x}+5}{3\text{x}+4}=3$
Answer
$\frac{2\text{x}+5}{3\text{x}+4}=3$
(By cross multiplication)
$3(3\text{x}+4)=1(2\text{x}+5)$
$9\text{x}+12=2\text{x}+5$
$\Rightarrow9\text{x}-2\text{x}=5-12$
$\Rightarrow7\text{x}=-7$
$\Rightarrow\text{x}=\frac{-7}{7}=-1$
$\text{x}=-1$
Check:
$\text{L.H.S}.=\frac{2\text{x}+5}{3\text{x}+4}=\frac{2(-1)+5}{3(-1)+4}=\frac{-2\ +\ 5}{-3\ +\ 4}$
$=\frac{3}{1}=3=\text{R.H.S.}$
Hence $\text{x}=-1$
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Question 203 Marks
A number is $\frac{2}{5}$ times another number. If their sum is 70, find the numbers.
Answer
Let the second number = x
then first number $=\frac{2}{5}\text{x}$
their sum = 70
$\therefore\text{x}+\frac{2}{5}\text{x}=70$
$\Rightarrow\frac{5\text{x}+2\text{x}}{5}=70$
$\Rightarrow\frac{7\text{x}}{5}=70$
$\Rightarrow\text{x}=\frac{70\times5}{7}=50$
$\therefore$ Second number = 50
And first number $=\frac{2}{5}\times50=20$
$\therefore$ Numbers are 20, 50
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Question 213 Marks
Solve: $0.5\text{x}+\frac{\text{x}}{3}=0.25\text{x}+7$
Answer
We have:
$0.5\text{x}+\frac{\text{x}}{3}=0.25\text{x}+7$
$\Rightarrow\frac{1.5\text{x}+\text{x}}{3}=0.25\text{x}+7$
$\Rightarrow1.5\text{x}+\text{x}=3(0.25\text{x}+7)$
$\Rightarrow2.5\text{x}=0.75\text{x}+21$
$\Rightarrow2.5\text{x}-0.75\text{x}=21$
$\Rightarrow1.75\text{x}=21$
$\Rightarrow\text{x}=\frac{21}{1.75}$
$\Rightarrow\text{x}=12$
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Question 223 Marks
In an isosceles triangle the base angles are equal and the vertex angle is twice of each base angle. Find the measures of the angles of the triangle.
Answer
In an isosceles triangle
Let each equal base angles = x
Then vertex angle = 2x
According to the condition,
x + x + 2x = 180° (sum of angles of a triangle)
⇒ 4x = 180°
⇒ x = 45°
Then vertex angle = 2x = 2 x 45° = 90°
Angles of the triangle are 45°, 45° and 90°
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Question 233 Marks
Solve the following equations. Check your result in case.
$\frac{\text{x}+2}{\text{x}-2}=\frac{7}{3}$
Answer
$\frac{\text{x}+2}{\text{x}-2}=\frac{7}{3}$
By cross multiplication, We get
$7(\text{x}-2)=3(\text{x}+2)$
$7\text{x}-14=3\text{x}+6$
$\Rightarrow7\text{x}-3\text{x}=6+14$
$\Rightarrow4\text{x}=20$
$\Rightarrow\text{x}=\frac{20}{4}=5$
Check:
$\text{L.H.S.}=\frac{\text{x}+2}{\text{x}-2}$
$=\frac{5+2}{5-2}$
$=\frac{7}{3}=\text{R.H.S.}$
Hence $\text{x}=5$
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Question 243 Marks
Find the consecutive positive odd integers whose sum is 76.
Answer
Let first odd number = 2x + 1
Second odd number = 2x + 3
2x + 1 + 2x + 3 = 76
⇒ 4x + 4 = 76
⇒ 4x = 76 – 4 = 72
x = 18
First number = 2x + 1 = 2 × 18 + 1 = 36 + 1 = 37
Second number = 2x + 3 = 2 × 18 + 3 = 36 + 3 = 39
Numbers are 37, 39
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Question 253 Marks
A number when added to its half gives 72. Find the number.
Answer
Let the required number = x
And half of the number $=\frac{\text{x}}{2}$
Then, $\frac{\text{x}}{2}+\text{x}=72$
$\Rightarrow\frac{\text{x}+2\text{x}}{2}=72$
$\Rightarrow\frac{3\text{x}}{2}=72$
$\therefore\text{x}=\frac{72\times2}{3}=24\times2=48$
$\therefore$ Required number = 48
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Question 263 Marks
A man sold an article for Rs. 495 and gained 10% on it. Find the cost price of the article.
Answer
S.P of article = Rs. 495
gain = 10%
Let cost price = Rs. x
$\therefore\text{S.P.}=\frac{\text{x}\times(100+10)}{100}=\frac{110}{100}\text{x}$
$\therefore\frac{110}{100}\text{x}=495$
$\Rightarrow\text{x}=\frac{495\times100}{110}=45\times10=450$
$\therefore$ Cost price = Rs. 450
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Question 273 Marks
A man travelled $\frac{3}{5}$ of his journey by rail, $\frac{1}{4}$ by a taxi, $\frac{1}{8}$ by a bus and the remaining 2km on foot. What is the length of his total journey?
Answer
Let length of total journey = x km
According to the condition,
$\frac{3}{5}\text{x}+\frac{1}{4}\text{x}+\frac{1}{8}\text{x}+2=\text{x}$
LCM of 5, 4, 8 = 40
$\therefore\frac{3}{5}\text{x}+\frac{1}{4}\text{x}+\frac{1}{8}\text{x}+\frac{2}{1}=\frac{\text{x}}{1}$
$\frac{24\text{x}+10\text{x}+5\text{x}+80=40\text{x}}{40}$
⇒ 39x + 80 = 40x
⇒ 40x - 39x = 80
⇒ x = 80
Total journey = 80km
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Question 283 Marks
Sumitra has Rs. 34 in 50 paise and 25 paise coins. If the numbers of 25 paise coins is twice the number of 50 paise coins, how many coins of each kind does she have?
Answer
Amount of coins = Rs. 34
Let 50 paisa coins = x
then 25 paisa coins = 2x
Then $\frac{\text{x}\times50}{100}+\frac{2\text{x}\times25}{100}=34$
$\frac{\text{x}}{2}+\frac{\text{x}}{2}=34$
$\Rightarrow\frac{2\text{x}}{2}=34$
$\Rightarrow\text{x}=34$
Number of 50 paisa coins = 34
And number of 25 paisa coins = 2x = 2 × 34 = 68
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Question 293 Marks
Solve the following equations. Check your result in case.7 - 5x = 5 - 7x
Answer
7 - 5x = 5 - 7x
⇒ -5x + 7x = 5 - 7 (By transposing)
⇒ 2x = -2
x = -1
Check:
L.H.S. = 7 - 5x = 7 - 5(-1) = 7 + 5 = 12
R.H.S. = 5 - 7x = 5 - 7(-1) = 5 + 7 = 12
L.H.S. = R.H.S.
Hence x = -1
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Question 303 Marks
Two-thirds of a number is greater than one-third of the number by 3. Find the number
Answer
Let the required number = x
Two-third of the number $=\frac{2}{3}\text{x}$
And one third of the number $=\frac{1}{3}\text{x}$
Then $\frac{2}{3}\text{x}-\frac{1}{3}\text{x}=3$
$\Rightarrow\frac{2-1}{3}\text{x}=3$
$\Rightarrow\frac{1}{3}\text{x}=3$
$\Rightarrow\text{x}=\frac{3\times3}{1}=9$
$\therefore$ Required number = 9
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Question 313 Marks
Solve the following equations. Check your result in case.
0.5x - (0.8 - 0.2x) = 0.2 - 0.3x
Answer
0.5x - (0.8 - 0.2x) = 0.2 - 0.3x
⇒ 0.5x - 0.8 + 0.2x = 0.2 - 0.3x
⇒ 0.5x + 0.2x + 0.3x = 0.2 + 0.8
⇒ 1.0x = 1.0
⇒ x = 1
Check:
L.H.S. = 0.5x - (0.8 - 0.2x)
= 0.5 × 1 - (0.8 - 0.2 × 1)
= 0.5 - (0.8 - 0.2)
= 0.5 - 0.6
= -0.1
R.H.S. = 0.2 - 0.3x
= 0.2 - 0.3 × 1
= 0.2 - 0.3
= -0.1
L.H.S. = R.H.S.
Hence x = 1
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Question 323 Marks
A father is 30 years older than his son. In 12 years. The man will be three times as old as his son. Find their present ages.
Answer
Let present age of son = x years
Age of father = (x + 30) years
12 years after,
Father’s age = x + 30 + 12 = (x + 42) years
And son’s age = (x + 12) years
(x + 42) = 3(x + 12)
⇒ x + 42 = 3x + 36
⇒ 3x + 36 = x + 42
⇒ 3x - x = 42 - 36
⇒ 2x = 6
⇒ x = 3
Son’s age = 3 years
Father’s age = 3 + 30 = 33 years
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Question 333 Marks
The total cost of 3 tables and 2 chairs is Rs. 1850. If a table costs Rs. 75 more than a chair, find the price of each.
Answer
Cost of 3 tables and 2 chairs = 1850
Cost of table = Rs. 75 + cost of a chair
Let cost of chair = Rs. x,
Then Cost of table = Rs. 75 + x
According to the condition,
3(75 + x) + 2x = 1850
⇒ 225 + 3x + 2x = 1850
⇒ 225 + 5x = 1850
⇒ 5x = 1850 - 225 = 1625
x = 325
Cost of chair = Rs. 325
And cost of table = Rs. 325 + 75
= Rs. 400
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Question 343 Marks
Solve the following equations. Check your result in case.
3x - 5 = 0
Answer
$3\text{x}-5=0$
Adding 5 to both sides
$3\text{x}-5+5=0+5$
$\Rightarrow3\text{x}=5$
$\Rightarrow\text{x}=\frac{5}{3}$
Check:
$\text{L.H.S}=3\text{x}-5$
$=3\text{x}\frac{5}{3}-5$
$=5-5$
$=0$
$=\text{R.H.S}$
Hence $\text{ x}=\frac{5}{3}$
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Question 353 Marks
The fifth part of a number when increased by 5 equals its fourth part decreased by 5. Find the number.
Answer
Let the required number = x
Fifth part of the number $=\frac{\text{x}}{5}$
Fourth part of the number $=\frac{\text{x}}{4}$
$\therefore\frac{\text{x}}{5}+5=\frac{\text{x}}{4}-5$
$\Rightarrow\frac{\text{x}}{5}-\frac{\text{x}}{4}=-5-5$
$\Rightarrow\frac{4\text{x}-5\text{x}}{20}=-10$
$\Rightarrow\frac{-\text{x}}{20}=-10$
$\Rightarrow\frac{\text{x}}{20}=10$
$\Rightarrow\text{x}=10\times20=200$
$\therefore$ Required number = 200
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Question 363 Marks
Find two consecutive natural numbers whose sum is 63.
Answer
Let first natural number = x then
Next number = x + 1
x + x + 1 = 63
⇒ 2x = 63 - 1 = 62
x = 31
First number = 31
And second number = 31 + 1 = 32
Numbers are 31, 32
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Question 373 Marks
Raju is 19 years younger than his cousin. After 5 years, their ages will be in the ratio 2 : 3. Find their present ages.
Answer
Let present age of Raju’s cousin = x years
Then age of Raju = (x - 19) years
After 5 years,
Raju’s age = x - 19 + 5 = (x - 14) years
And his cousin age = x + 5
(x - 14) : (x + 5) = 2 : 3
$\Rightarrow\frac{\text{x}-14}{\text{x}+5}=\frac{2}{3}$
⇒ 3(x - 14) = 2 (x + 5) (By cross multiplication)
⇒ 3x - 42 = 2x + 10
⇒ 3x - 2x = 10 + 42
⇒ x = 52
Raju’s age = x - 19 = 52 - 19 = 33 years
And his cousin age = 52 years.
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Question 383 Marks
In an examination, a student requires 40% of the total marks to pass. If Rupa gets 185 marks and fails by 15 marks, find the total marks.
Answer
Let total marks = x
Pass marks = 40% of $\text{x}=\frac{40\text{x}}{100}=\frac{2}{5}\text{x}$
No. of marks got by Rupa = 185
No. of marks by which she failed = 15
Pass marks = 185 + 15 = 200
$\frac{2}{5}\text{x}=200$
$\Rightarrow\text{x}=\frac{200\times5}{2}\text{x}$
$\Rightarrow\text{x}=500$
Hence total marks = 500
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Question 393 Marks
Divide 184 into two parts such that one-third of one of part may exceed one-seventh of the other part by 8.
Hint. Let the two parts be x and (184 - x). Then $\frac{1}{3}\text{x}-\frac{1}{7}(184-\text{x})=8$
Answer
Sum of two numbers = 184
Let first number = x
Then second number = 184 - x
Then $\text{x}\times\frac{1}{3} = (184-\text{x})\times\frac{1}{7}+8$
$\frac{\text{x}}{3}=\frac{184-\text{x}}{7}+8$
$\frac{7\text{x }=\ 3(184-\text{x})+168}{21}$
(LCM of 3, 7 = 21)
$7\text{x}=552-3\text{x}+168$
$\Rightarrow7\text{x}+3\text{x}=552+168$
$\Rightarrow10\text{x}=720$
$\Rightarrow\text{x}=\frac{720}{10}=72$
First part = 72
Second part = 184 - 72 = 112
Hence parts are 72, 112
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Question 403 Marks
Solve the following equations. Check your result in case.
3x + 2(x + 2) = 20 - (2x - 5)
Answer
3x + 2(x + 2) = 20 - (2x - 5)
⇒ 3x + 2x + 4 = 20 - 2x + 5
⇒ 5x + 4 = 25 - 2x
⇒ 5x + 2x = 25 - 4 (By transposing)
⇒ 7x = 21
⇒ x = 3
Check:
L.H.S.= 3x + [2(x + 2)]
= 3 × 3 + 2(3 + 2)
= 9 + 2 × 5 = 9 + 10 = 19
R.H.S. = 20 - (2x - 5)
= 20 - (2 × 3 - 5)
= 20 - (6 - 5) = 20 - 1 = 19
L.H.S. = R.H.S.
Hence x = 3
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Question 413 Marks
Solve the following equations. Check your result in case.8x - 3 = 9 - 2x
Answer
$8\text{x}-3=9-2\text{x}$
$\Rightarrow8\text{x}+2\text{x}=9+3$ (By transposing)
$\Rightarrow10\text{x}=12$
$\Rightarrow\text{x}=\frac{12}{10}=\frac{6}{5}$
$\therefore\text{x}=\frac{6}{5}$
Check:
$\text{L.H.S}=8\text{x}-3=8\times\frac{6}{5}-3$
$=\frac{48}{5}-\frac{3}{1}=\frac{48-15}{5}=\frac{33}{5}$
$\text{R.H.S}=9-2\text{x}$
$=9-2\times\frac{6}{5}=\frac{9}{1}-\frac{12}{5}$
$=\frac{45-12}{5}=\frac{33}{5}$
$\therefore\text{L.H.S}=\text{R.H.S}$
Hence $\text{x}=\frac{6}{5}$
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Question 423 Marks
How much pure alcohol must be added to 400mL of a 15% solution to make its strength 32%?
Answer
Solution = 400ml
Quantity of alcohol = 15% of 400ml
$=\frac{400\times15}{100}=60\text{ml}$
Let pure alcohol added = x ml
Total solution = 400 + x
and total alcohol = (x + 60)
Now (400 + x) x 32% = x + 60
$\Rightarrow(400+\text{x})\times\frac{32}{100}=\text{x}+60$
⇒ 32(400 + x) = 100(x + 60)
⇒ 12800 + 32x = 100x + 6000
⇒ 12800 - 6000 = 100x - 32x
⇒ 6800 = 68x
⇒ x = 6800
Pure alcohol added = 100ml
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