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Question 15 Marks
The value of the expression $(10 y-20)$ depends on the value of $y$. Verify this by giving five different values to $y$ and finding for each $y$ the value of $(10 y-20)$. From the different values of $(10 y-20)$ you obtain, do you see a solution to $10 y-20=50$ ?
If there is no solution, try giving more values to $y$ and find whether the condition $10 y-20=50$ is met.
Answer
Let us find the value of the expression (10 y – 20) for different values of y.
Value of yValue of given expression ( 10 y-20 )
When y=-110 x (-1)-20=-10-20=-30
When y=010 x 0-20=0-20=-20
When y=110 x 1-20=10-20=-10
When y=210 x 2-20=20-20=0
When y=310 x 3-20=30-20=10
When y=410 x 4-20=40-20=20
When y=510 x 5-20=50-20=30
When y=610 x 6-20=60-20=40
When y=710 x 7-20=70-20=50
Thus, the condition 10y - 20 = 50, is true for y =7.
Hence, y =7 is the solution of the equation 10y — 20.
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Question 25 Marks
Complete the last column of the table.
S.No.EquationValueSay, whether the equation is satisfied (Yes/No)
(i)x+3=0x=3
(ii)x+3=0x=0
(iii)x+3=0x=-3
(iv)x-7=1x=7
(v)x-7=1x=8
(vi)5x=25x=0
(vii)5x=25x=5
(viii)5x=25x=-5
(ix)$\frac{m}{2}$=2m=-6
(x)$\frac{m}{2}$=2m=0
(xi)$\frac{m}{2}$=2m=6
Answer
S.No.EquationValueValue of LHSResult
(i)x+3=0x=3$\begin{aligned} x+3 & =3+3 =6 \neq R H S\end{aligned}$No
(ii)x+3=0x=0$\begin{aligned} x+3 & =0+3 =3 \neq \text { RHS }\end{aligned}$No
(iii)x+3=0x=-3$\begin{aligned} x+3 & =-3+3 =0=\text { RHS }\end{aligned}$Yes
(iv)x-7=1x=7$\begin{aligned} x-7 & =7-7 =0 \neq \text { RHS }\end{aligned}$No
(v)x-7=1x=8$\begin{aligned} x-7 & =8-7 =1=\text { RHS }\end{aligned}$Yes
(vi)5x=25x=0$5 x=5 \times 0=0 \neq R H S$No
(vil)5x=25x=5$\begin{array}{l}5 x=5 \times 5 =25=\text { RHS }\end{array}$Yes
(viii)5x=25x=-5$\begin{array}{l}5 x=5 \times(-5) =-25 \neq RHS \end{array}$No
(ix)$\frac{m}{3}$=2m=-6$m=-6 \frac{m}{3}=\frac{-6}{3}=-2 \neq$ RHSNo
(x)$\frac{m}{3}$=2m=0$m=0 \frac{m}{3}=\frac{0}{3}=0 \neq$ RHSNo
(x)$\frac{m}{3}$=2m=6$m=6 \frac{m}{3}=\frac{6}{3}=2= RHS$Yes
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5 Marks Questions - MATHS STD 7 Questions - Vidyadip