4 Mark Question - Maths STD 7 Questions - Vidyadip

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4 Mark Question

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks

Solve the following equations. Check your result in case.
$2\text{x}-\frac{1}{3}=\frac{1}{5}-\text{x}$

Answer

$2\text{x}-\frac{1}{3}=\frac{1}{5}-\text{x}$
$\Rightarrow2\text{x}+\text{x}=\frac{1}{5}+\frac{1}{3}$ (By transposing)
$\Rightarrow3\text{x}=\frac{3+5}{15}$
$\Rightarrow3\text{x}=\frac{8}{15}$
$\Rightarrow\text{x}=\frac{8}{15}\times\frac{1}{3}=\frac{8}{45}$
$\text{x}=\frac{8}{45}$
Check:
$\text{L.H.S.}=2\text{x}-\frac{1}{3}$
$=2\times\frac{8}{45}-\frac{1}{3}=\frac{16}{45}-\frac{1}{3}$
$=\frac{16-15}{45}=\frac{1}{45}$
$\text{R.H.S}.=\frac{1}{5}-\text{x}=\frac{1}{5}-\frac{8}{45}$
$=\frac{9-8}{45}=\frac{1}{45}$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
Hence $\text{x}=\frac{8}{45}$

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Question 24 Marks

The ages of Sonal and Manoj are in the ratio 7 : 5 Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.

Answer

Ratio in present ages of Sonal and Manoj = 7 : 5
Let Sonal’s age = 7x
then Manoj’s age = 5x
10 years hence,
Sonal’s age will be = 7x + 10
And Manoj’s age = 5x + 10
$\therefore\frac{7\text{x}+10}{5\text{x}+10}=\frac{9}{7}$
(By cross multiplications)
$7(7\text{x}+10)=9(5\text{x}+10)$
$\Rightarrow49\text{x}+70=45\text{x}+90$
$\Rightarrow49\text{x}-45\text{x}=90-70$
$\Rightarrow4\text{x}=20$
$\Rightarrow\text{x}=\frac{20}{4}=5$
Sonal’s present age = 7x = 7 × 5 = 35 years
And Manoj’s age = 5x = 5 × 5 = 25 years

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Question 34 Marks

Solve the following equations. Check your result in case.
6(3x + 2) - 5(6x - 1) = 3(x - 8) - 5(7x - 6) + 9x

Answer

6(3x + 2) - 5(6x - 1) = 3(x - 8) - 5(7x - 6) + 9x
⇒ 18x + 12 - 30x + 5 = 3x - 24 - 35x + 30 + 9x
⇒ 18x - 30x + 12 + 5 = 3x - 35x + 9x - 24 + 30
⇒ -12x + 17 = -23x + 6
⇒ -12x + 23x = 6 - 17
⇒ 11x = -11
x = -1
Check:
L.H.S. = 6(3x + 2) - 5(6x - 1)
= 6[3x (-1) + 2] - 5[6x (-1) × -1]
= 6[-3 + 2] - 5[-6 - 1]
= 6x (-1) - 5 × (-7)
= -6 + 35 = 29
R.H.S. = 3(x - 8) - 5(7x - 6) + 9x
= 3[-1 - 8] - 5[7x(-1) - 6] + 9(-1)
= 3x(-9) - 5[-7 - 6] - 9
= -27 - 5(-13) - 9
= -27 + 65 - 9
= 65 - 36 = 29
L.H.S. = R.H.S.
Hence x = -1

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Question 44 Marks

Solve the following equations. Check your result in case.
$\frac{\text{y}-1}{3}-\frac{\text{y}-2}{4}=1$

Answer

$$$\frac{\text{y}-1}{3}-\frac{\text{y}-2}{4}=1$
$\frac{4(\text{y}-1)-3(\text{y}-2)}{12}=1$ (LCM of 3, 4 = 12)
$\Rightarrow\frac{4\text{y}-4-3\text{y}+6}{12}=1$
$\Rightarrow\frac{\text{y}+2}{12}=1$
$\Rightarrow\text{y}+2=12$
$\Rightarrow\text{y}=12-2=10$
$\Rightarrow\text{y}=10$
Check:
$\text{L.H.S.}=\frac{\text{y}-1}{3}-\frac{\text{y}-2}{4}$
$=\frac{10-1}{3}-\frac{10-2}{4}$
$=\frac{9}{3}-\frac{8}{4}=3-2$
$=1$
$=\text{R.H.S. = L.H.S.}$
Hence $= \text{y}=10$

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Question 54 Marks

Solve the following equations. Check your result in case.
$\frac{2}{7}(\text{x}-9)+\frac{\text{x}}{3}=3$

Answer

$\frac{2}{7}(\text{x}-9)+\frac{\text{x}}{3}=3$
$\Rightarrow\frac{6(\text{x}-9)+7\text{x}=63}{21}$ (LCM of 7, 3 = 21)
$\Rightarrow6\text{x}-54+7\text{x}=63$
$\Rightarrow6\text{x}+7\text{x}=63+54=117$
$\Rightarrow13\text{x}=117$
$\Rightarrow\text{x}=\frac{117}{13}=9$
Check:
$\text{L.H.S.}=\frac{2}{7}(\text{x}-9)+\frac{\text{x}}{3}$
$=\frac{2}{7}(9-9)+\frac{9}{3}$
$=\frac{2}{7}\times0+3$
$=0+3=3\text{ R.H.S.}$
Hence $\text{x}=9$

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Question 64 Marks

Solve the following equations. Check your result in case.
t - (2t + 5) - 5(1 - 2t) = 2(3 + 4t) - 3(t - 4)

Answer

t - (2t + 5) - 5(1 - 2t) = 2(3 + 4t) - 3(t – 4)
⇒ t - 2t - 5 - 5 + 10t = 6 + 8t - 3t + 12t
⇒ t - 2t + 10t - 10 = 8t - 3t + 18
⇒ 9t - 10 = 5t + 18
⇒ 9t - 5t = 18 + 10 (By transposing)
⇒ 4t = 28
⇒ t = 7
Check:
L.H.S. = t - [2t + 5] - 5[1 - 2t]
= 7 - [2 × 7 + 5] - 5[1 - 2 × 7]
= 7 - [14 + 5] - 5[1 - 14]
= 7 - 19 - 5(-13)
= 7 - 19 + 65
= 72 - 19 = 53
R.H.S. = 2[3 + 4t) - 3(t - 4)
= 2(3 + 4 × 7) - 3(7 – 4)
= 2(3 + 28) - 3(3)
= 2(31) - 9 = 62 - 9 = 53
L.H.S. = R.H.S.
Hence t = 7

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Question 74 Marks

Solve the following equations. Check your result in case.
$\frac{\text{y}+7}{3}=1+\frac{3\text{y}-2}{5}$

Answer

$\frac{\text{y}+7}{3}=1+\frac{3\text{y}-2}{5}$
$\frac{5(\text{y}+7)=15+3(3\text{y}-2)}{15}$ (LCM of 3, 5 = 15)
$5\text{y}+35=15+9\text{y}-6$
$5\text{y}-9\text{y}=15-6-25$ (By transposing)
$-4\text{y}=-26$
$\text{y}=\frac{-26}{-4}=\frac{13}{2}$
$\therefore\text{y}=\frac{13}{2}$
Check:
$\text{L.H.S.}=\frac{\text{y}-7}{3}=\frac{\frac{13}{2}+7}{3}$
$=\frac{13+14}{2\times3}$
$=\frac{27}{6}=\frac{9}{2}$
$\text{R.H.S.}=1+\frac{3\text{y}-2}{5}$
$=1+\frac{3\times\frac{13}{2}-2}{5}$
$=1+\frac{\frac{39}{2}-\frac{2}{1}}{5}$
$=1+\frac{39-4}{2\times5}$
$=1+\frac{35}{10}$
$=1+\frac{7}{2}=\frac{2+7}{2}=\frac{9}{2}$
$\therefore\text{L.H.S. = R.H.S.}$
Hence $\text{y}=\frac{13}{2}$

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Question 84 Marks

Solve the following equations. Check your result in case.
$\frac{2\text{x}-3}{5}+\frac{\text{x}+3}{4}=\frac{4\text{x}+1}{7}$

Answer

$\frac{2\text{x}-3}{5}+\frac{\text{x}+3}{4}=\frac{4\text{x}+1}{7}$
$\frac{28(2\text{x}-3)+35(\text{x}+3)=20(4\text{x}+1)}{140}$ (LCM of 5, 4, 7 = 140)
$28(2\text{x}-3)+35(\text{x}+3)=20(4\text{x}+1)$
$\Rightarrow56\text{x}-84+35\text{x}+105=80\text{x}+20$
$\Rightarrow56\text{x}+35\text{x}-80\text{x}=20+84-105$
$\Rightarrow91\text{x}-80\text{x}=140-105$
$\Rightarrow11\text{x}=-1$
$\Rightarrow\text{x}=\frac{-1}{11}$
$\therefore\text{x}=\frac{-1}{11}$
Check:
$\text{L.H.S.}=\frac{2\text{x}-3}{5}+\frac{\text{x}+3}{4}$
$=\frac{2\Big(-\frac{1}{11}\Big)-3}{5}+\frac{\frac{-1}{11}+3}{4}$
$=\frac{\frac{-2}{11}-3}{5}+\frac{\frac{-1}{11}+3}{4}$
$=\frac{-2-33}{11\times5}+\frac{-1+33}{11\times4}$
$=\frac{-35}{11\times5}+\frac{32}{11\times4}$
$=\frac{-7}{11}+\frac{8}{11}$
$=\frac{1}{11}$
$\text{R.H.S}.=\frac{4\text{x}+1}{7}$
$=\frac{4\Big(-\frac{1}{11}\Big)+1}{7}$
$=\frac{-\frac{4}{11}+1}{7}$
$=\frac{-4+11}{11\times7}$
$=\frac{7}{11\times7}$
$=\frac{1}{11}$
$\text{L.H.S. = R.H.S.}$

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Question 94 Marks

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

Answer

Five years ago,
Let Son’s age = x years
And father’s age = 7x years
Present age of son = (x + 5) years
And age of father = (7x + 5) years
5 years hence,
Father’s age = 7x + 5 + 5 = 7x + 10
And Son’s age = x + 5 + 5 = x + 10
(7x + 10) = 3(x + 10)
⇒ 7x + 10 = 3x + 30
⇒ 7x – 3x = 30 - 10
⇒ 4x = 20
⇒ x = 5
Father present age = 7x + 5 = 7 x 5 + 5 = 35 + 5 = 40 years
And son’s age = x + 5 = 5 + 5 = 10 years

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Question 104 Marks

Solve the following equations. Check your result in case.
$\frac{\text{x}+2}{6}-\Big(\frac{11-\text{x}}{3}-\frac{1}{4}\Big)=\frac{3\text{x}-4}{12}$

Answer

$\frac{\text{x}+2}{6}-\Big(\frac{11-\text{x}}{3}-\frac{1}{4}\Big)=\frac{3\text{x}-4}{12}$
$\frac{\text{x}+2}{6}-\Big(\frac{11-\text{x}}{3}+\frac{1}{4}\Big)=\frac{3\text{x}-4}{12}$
$\frac{2(\text{x}+2)-4(11-\text{x})+3=3\text{x}-4}{12}$
(LCM of 6, 3, 4, 12 = 12)
$\Rightarrow2\text{x}+4-44+4\text{x}+3=3\text{x}-4$
$\Rightarrow2\text{x}+4\text{x}-3\text{x}=-4-4+44-3$
$\Rightarrow6\text{x}-3\text{x}=44-11$
$\Rightarrow3\text{x}=33$
$\Rightarrow\text{x}=\frac{33}{3}=11$
$\therefore\text{x}=11$
check:
$\text{L.H.S.}=\frac{\text{x}+2}{6}-\Big[\frac{11\times\text{x}}{3}-\frac{1}{4}\Big]$
$=\frac{11+2}{6}-\Big[\frac{11-11}{3}-\frac{1}{4}\Big]$
$=\frac{13}{6}-\Big(0-\frac{1}{4}\Big)$
$=\frac{13}{6}=\frac{1}{4}$
$=\frac{26+3}{12}=\frac{29}{12}$
$\text{R.H.S.}=\frac{3\text{x}-4}{12}=\frac{3\times11-4}{12}$
$=\frac{33-4}{12}=\frac{29}{12}$
$\because\text{L.H.S. = R.H.S.}$
Hence $\text{x}=11$

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Question 114 Marks

Solve the following equations. Check your result in case.
$\frac{3}{4}(7\text{x}-1)-\Big(2\text{x}-\frac{1-\text{x}}{2}\Big)=\text{x}+\frac{3}{2}$

Answer

$\frac{3}{4}(7\text{x}-1)-\Big(2\text{x}-\frac{1-\text{x}}{2}\Big)=\text{x}+\frac{3}{2}$
$\Rightarrow\frac{3(7\text{x}-1)}{4}-\frac{4\text{x}-1+\text{x}}{2}=\frac{2\text{x}+3}{2}$
$\Rightarrow\frac{21\text{x}-3}{4}-\frac{5\text{x}-1}{2}=\frac{2\text{x}+3}{2}$
$=\frac{21\text{x}-3-10\text{x}+2=4\text{x}+6}{4}$ (LCM 4, 2 = 4)
$\Rightarrow21\text{x}-10\text{x}-4\text{x}=6+3-2$
(By transposing)
$\Rightarrow7\text{x}=7$
$\Rightarrow\text{x}=\frac{7}{7}=1$
$\therefore\text{x}=1$
Check:
$\frac{3}{4}(7\text{x}-1)-\Big[2\text{x}-\frac{1-\text{x}}{2}\Big]$
$=\frac{3}{4}(7\times1-1)-\Big[2\times1-\frac{1-1}{2}\Big]$
$=\frac{3\times6}{4}-(2-0)=\frac{18}{4}-2$
$=\frac{9}{2}-2=\frac{9-4}{2}=\frac{5}{2}$
$\text{R.H.S.}=\text{x}+\frac{3}{2}$
$=1+\frac{3}{2}$
$=\frac{2+3}{2}=\frac{5}{2}$
$\therefore\text{L.H.S. = R.H.S.}$
Hence $\text{x}=1$

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Question 124 Marks

Solve the following equations. Check your result in case.
$\frac{9\text{x}+7}{2}-\Big(\text{x}-\frac{\text{x}-2}{7}\Big)=36$

Answer

$\frac{9\text{x}+7}{2}-\Big(\text{x}-\frac{\text{x}-2}{7}\Big)=36$
$\Rightarrow\frac{9\text{x}+7}{2}-\frac{\text{x}}{1}+\frac{\text{x}-2}{7}=36$
$\Rightarrow\frac{7(9\text{x}+7)-14\text{x}+2(\text{x}-2)=36\times14}{14}$
(LCM of 2, 7 = 14)
$\Rightarrow63+49-14\text{x}+2\text{x}-4=504$
$\Rightarrow63\text{x}-14\text{x}+2\text{x}=504+4-49$
$\Rightarrow65\text{x}-14\text{x}=508-49$
$\Rightarrow51\text{x}=459$
$\Rightarrow\text{x}=\frac{459}{51}=9$
$\therefore\text{x}=9$
Check:
$\text{L.H.S.}=\frac{9\text{x}+7}{2}-\Big(\text{x}-\frac{\text{x}-2}{7}\Big)$
$=\frac{9\times9\times7}{2}-\Big(9-\frac{9-2}{7}\Big)$
$=\frac{81+7}{2}-\Big(9-\frac{7}{7}\Big)$
$=\frac{88}{2}-(9-1)$
$=44-8$
$\text{R.H.S.}=36$
$\text{R.H.S. = L.H.S.}$
Hence $\text{x}=9$

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Question 134 Marks

Solve the following equations. Check your result in case.
$\frac{\text{x}-2}{4}+\frac{1}{3}=\text{x}-\frac{2\text{x}-1}{3}$

Answer

$\frac{\text{x}-2}{4}+\frac{1}{3}=\frac{\text{x}}{1}-\frac{2\text{x}-1}{3}$
$\frac{3(\text{x}-2)+4=12\text{x}-4(2\text{x}-1)}{12}$ (LCM of 4, 3 = 12)
$3(\text{x}-2)+4=12\text{x}-4(2\text{x}-1)$
$\Rightarrow3\text{x}-6+4=12\text{x}-8\text{x}+4$
$\Rightarrow3\text{x}-12\text{x}+8\text{x}=4+6-4$
$\Rightarrow-12\text{x}+11\text{x}=10-4$
$\Rightarrow-\text{x}=6$
$\Rightarrow\text{x}=-6$
$\therefore\text{x}=-6$
Check:
$\text{L.H.S.}=\frac{\text{x}-2}{4}+\frac{1}{3}$
$=\frac{-6-2}{4}+\frac{1}{3}=\frac{-8}{4}+\frac{1}{3}$
$=-2+\frac{1}{3}$
$=\frac{-6+1}{3}=-\frac{5}{3}$
$\text{R.H.S.}=\text{x}-\frac{2\text{x}-1}{3}$
$=-6-\frac{2(-6)-1}{3}$
$=-6-\frac{-12-1}{3}=-6-\frac{(-13)}{3}$
$=-6+\frac{13}{3}=\frac{-18+13}{3}=-\frac{5}{3}$
$\therefore\text{ L.H.S.= R.H.S.}$
Hence $\text{x}=-6$

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Question 144 Marks

A number consists of two digits whose sum is 8. If 18 is added to the number its digits are reversed. Find the number.

Answer

Sum of digits = 8
Let units digit = x
Then tens digit = 8 - x
and number will be x + 10 (8 - x) ….(i)
By adding 18, the digits are reversed then
units digit = 8 - x
and tens digit = x
Number = (8 - x) = 10x
According to the condition,
(8 - x) + 10x = 18 + x + 10 (8 - x)
⇒ 8 - x + 10x = 18 + x + 80 - 10x
⇒ 10x - x - x + 10x = 18 + 80 - 8
⇒ 18x = 90
⇒ x = 5
Number is
x + 10(8 - x) = 5 + 10(8 - 5) = 5 + 10 × 3 = 35

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Question 154 Marks

Solve the following equations. Check your result in case.
$\frac{3\text{x}-1}{5}-\frac{\text{x}}{7}=3$

Answer

$\frac{21\text{x}-7-5\text{x}}{35}=\frac{3}{1}$ (LCM of 5, 7 = 35)
$\frac{16\text{x}-7}{35}=\frac{3}{1}$ (By cross multiplication)
$\Rightarrow16\text{x}-7=3\times35$
$\Rightarrow16\text{x}-7=105$
$\Rightarrow16\text{x}=105+7=112$
$\Rightarrow\text{x}=\frac{112}{6}=7$
$\Rightarrow\text{x}=7$
$\therefore\text{x}=7$
Check:
$\text{L.H.S.}=\frac{3\text{x}-1}{5}-\frac{\text{x}}{7}=\frac{3\times7-1}{5}=\frac{-7}{7}$
$=\frac{21-1}{5}-1=\frac{20}{5}-1=4-1=3=\text{R.H.S.}$
Hence $\text{x}=7$

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Question 164 Marks

Solve the following equations. Check your result in case.
0.18(5x - 4) = 0.5x + 0.8

Answer

$0.18(5\text{x} - 4) = 0.5\text{x} + 0.8$
$\frac{18}{100}(5\text{x}-4)=\frac{5}{10}\text{x}+\frac{8}{10}$
$\frac{18(5\text{x}-4)=50\text{x}+80}{100}$
$\Rightarrow90\text{x}-72=50\text{x}+80$
$\Rightarrow90\text{x}-50\text{x}=80+72$
$\Rightarrow40\text{x}=152$
$\Rightarrow\text{x}=\frac{152}{40}=\frac{38}{10}=3.8$
$\therefore\text{x}=3.8$
Check:
L.H.S. = 0.18(5x - 4)
= 0.18(5 × 3.8 - 4)
= 0.18(19.0 - 4)
= 0.18 × 15
= 2.70
= 2.7
R.H.S. = 0.5x + 0.8
= 0.5 × 3.8 + 0.8
= 1.90 + 0.8
= 1.9 + 0.8
= 2.7
$\because$ L.H.S. = R.H.S.
Hence x = 3.8

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Question 174 Marks

Solve the following equations. Check your result in case.
$2\text{x}-3=\frac{3}{10}(5\text{x}-12)$

Answer

$\frac{3\text{x}-3}{1}=\frac{3}{10}(5\text{x}-12)$
$\Rightarrow10(2\text{x}-3)=3(5\text{x}-12)$ (By cross multiplication)
$\Rightarrow20\text{x}-30=15\text{x}-36$
$\Rightarrow20\text{x}-15\text{x}=-36+30$
$\Rightarrow5\text{x}=-6$
$\Rightarrow\text{x}=\frac{-6}{5}$
$\therefore\text{x}=\frac{-6}{5}$
Check
$\text{L.H.S.}=2\text{x}-3=2\times\Big(\frac{-6}{5}\Big)-3$
$=\frac{-12}{5}-\frac{3}{1}=\frac{-12-15}{5}=\frac{-27}{5}$
$\text{R.H.S.}=\frac{3}{10}(5\text{x}-12)$
$=\frac{3}{10}\Big[5\text{x}\Big(\frac{-6}{5}\Big)-12\Big]$
$=\frac{3}{10}(-6-12)=\frac{3}{10}\times(-18)$
$\frac{3\times(-18)}{10}=\frac{3\times(-9)}{5}=\frac{-27}{5}$
$\therefore\text{L.H.S.=R.H.S.}$
Hence $\text{x}=\frac{-6}{5}$

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Question 184 Marks

Solve the following equations. Check your result in case.
$0.5\text{x}+\frac{\text{x}}{3}=0.25\text{x}+7$

Answer

$0.5\text{x}+\frac{\text{x}}{3}=0.25\text{x}+7$
$\Rightarrow\frac{5}{10}\text{x}+\frac{\text{x}}{3}=\frac{25}{100}\text{x}+7$
$\Rightarrow\frac{1}{2}\text{x}+\frac{\text{x}}{3}=\frac{\text{x}}{4}=7$
$\Rightarrow\frac{1}{2}\text{x}+\frac{1}{3}\text{x}-\frac{1}{4}\text{x}=7$
$\Rightarrow\frac{6\text{x}+4\text{x}-3\text{x}}{12}=7$
$\Rightarrow\frac{7\text{x}}{12}=7$
$\Rightarrow\text{x}=\frac{7\times12}{7}=12$
$\therefore\text{x}=12$
Check:
$\text{L.H.S.}=0.5\text{x}+\frac{\text{x}}{3}$
$=\frac{1}{2}\text{x}+\frac{\text{x}}{3}$
$=\frac{12}{2}+\frac{12}{3}$
$=6+4$
$=10$
$\text{R.H.S.}=0.25\text{x}+7$
$=\frac{1}{4}\times12+7$
$=3+7$
$=10$
$\therefore\text{L.H.S. = R.H.S.}$
Hence $\text{x}=12$

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Question 194 Marks

Solve the following equations. Check your result in case.
$\frac{2\text{x}-1}{3}-\frac{6\text{x}-2}{5}=\frac{1}{3}$

Answer

$\frac{2\text{x}-1}{3}-\frac{6\text{x}-2}{5}=\frac{1}{3}$
$=\frac{5(2\text{x}-1)-3(6\text{x}-2)}{5}=\frac{1}{3}$ (LCM of 3, 5 = 15)
$10\text{x}-5-18\text{x}+6=5$
$\Rightarrow-8\text{x}+1=5$
$\Rightarrow-8\text{x}=5-1=4$
$\therefore\text{x}=\frac{4}{-8}=\frac{-1}{2}$
Check:
$\text{L.H.S.}=\frac{2\text{x}-1}{3}-\frac{6\text{x}-2}{5}$
$=\frac{2\times\Big(-\frac{1}{2}\Big)-1}{3}-\frac{6\times\Big(-\frac{1}{2}\Big)-2}{5}$
$=\frac{-1-1}{3}-\frac{-3-3}{5}=\frac{-2}{3}-\frac{(-15)}{5}$
$=\frac{-2}{3}+1=\frac{1}{3}=\text{R.H.S.}$
Hence $\text{x}=\frac{-1}{2}$

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Question 204 Marks

Hari Babu left one-third of his property to his son, one-fourth to his daughter and the remainder to his wife. If his wife’s share is Rs. 18000, what was the worth of his total property?

Answer

Let the worth of hari babu's property be Rs x.
According to the question, We have
Son's share $=\frac{1}{4}\text{x}$
Daughter's share $=\frac{1}{3}\text{x}$
Wife's share $=\Big\{\text{x}-\Big(\frac{1}{4}\text{x}+\frac{1}{3}\text{x}\Big)\Big\}$
It is given that his wife's share is Rs. 18000.
i.e, $\text{x}-\Big(\frac{1}{4}\text{x}+\frac{1}{3}\text{x}\Big)=18000$
$\Rightarrow\text{x}-\Big(\frac{1}{3}\text{x}+\frac{1}{4}\text{x}\Big)=18000$
$\Rightarrow\text{x}-\frac{7\text{x}}{12}=18000$
$\Rightarrow\frac{5\text{x}}{12}=18000$
$\Rightarrow\text{x}=\frac{10000^{36000}\times12}{5}$
$\Rightarrow\text{x}=43200$
$\therefore$ Hari babu's total property is worth Rs. 43200.

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Question 214 Marks

Solve the following equations. Check your result in case.
$\frac{2}{3}\text{x}=\frac{3}{8}\text{x}+\frac{7}{12}$

Answer

$\frac{2}{3}\text{x}=\frac{3}{8}\text{x}+\frac{7}{12}$
$\Rightarrow\frac{2}{3}\text{x}-\frac{3}{8}\text{x}=\frac{7}{12}$ (By transposing)
$\Rightarrow\frac{16\text{x}-9\text{x}}{24}=\frac{7}{12}$ (LCM of 3, 8 = 24)
$\Rightarrow\frac{7\text{x}}{24}=\frac{7}{12}$
$\Rightarrow\text{x}=\frac{7}{12}\times\frac{24}{7}=2$
$\therefore\text{x}=2$
Check:
$\text{L.H.S.}=\frac{2}{3}\text{x}=\frac{2}{3}\times2=\frac{4}{3}$
$\text{R.H.S.}=\frac{3}{8}\text{x}+\frac{7}{12}$
$=\frac{3}{8}\times2+\frac{7}{12}=\frac{3}{4}+\frac{7}{12}$
$=\frac{9+7}{12}=\frac{16}{12}=\frac{4}{3}$
$\therefore\text{L.H.S.=R.H.S.}$
Hence $\text{x}=2$

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