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MATHS 5 Marks Questions

Simple Equations · UP Board (UPMSP) STD 7 (UPMSP - English Medium). 8 questions.

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8 questions · timed · auto-graded

Question 15 Marks
Set up an equation on the basis of the statement given at the end and solve it to find the unknown quantity in the equation so obtained:
Anwar thinks of a number. If he takes away $7$ from $\frac{5}{2}$ of the number, the result is $23$
Answer
We have to frame the equation on the basis of given statement
And then solve the equation so framed.
Let the number be 'a'.
We know that,
$\frac{5}{2}$ of a is $\frac{5a}{2}$
Now, as per the statement,
We can write the equation as,
$\frac{5 a}{2}-7=23$
$\frac{5a}{2} = 23 + 7$
$\frac{5a}{2} = 30$
Multiplying $\frac{2}{5}$ to both the sides, we get
$\frac{5 a}{2} \times \frac{2}{5}=30 \times \frac{2}{5}$
Therefore, a $= 12$
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Question 25 Marks
Set up an equation on the basis of the statement given at the end and solve it to find the unknown quantity in the equation:
Ibenhal thinks of a number. If she adds $19$ to it and divides the sum by $5$, she will get $8.$
Answer
We have to frame the equation on the basis of given statement
And then solve the equation so framed.
Let the number be 'a'.
As per the statement,
We can write the equation as,
$\frac{a+19}{5}=8$
Multiplying 5 to both the sides, we get
$\frac{a+19}{5} \times 5=8 \times 5$
Therefore,
$a + 19 = 40$
$a = 40 – 19$
$a = 21$
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Question 35 Marks
Set up an equation on the basis of the statement given at the end and solve it to find the unknown quantity:
Munna subtracts thrice the number of notebooks he has from $50$, he finds the result to be $8.$
Answer
We have to frame the equation on the basis of given statement
And then solve the equation so framed.
Let the number be 'a'.
We know that,
Thrice of a is $3a$
Now, as per the statement,
We can write the equation as,
$50 – 3a = 8$
$-3a = 8 – 50$
$-3a = -42$
Dividing -3 on both the sides, we get
$\frac{-3 a}{-3}=\frac{-42}{-3}$
Therefore, $a = 14$
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Question 45 Marks
Set up an equation on the basis of the statement given at the end and solve the equation so obtained to find the unknown quantity.
when I subtracted $11$ from twice a number, the result was $15.$
Answer
We have to frame the equation on the basis of a given statement
And then solve the equation so framed.
Let the number be 'a'
We know that,
Twice of a is $2a$
Now as per the statement,
We can write the equation as,
$2a – 11 = 15$
$2a = 15 + 11$
$2a = 26$
Dividing $2$ into both the sides, we get
$\frac{2 a}{2}=\frac{26}{2}$
Therefore, $a = 13$
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Question 55 Marks
Set up an equation as per the given statement given at the end and solve it to find the unknown quantity in the equation.
If I take three-fourths of a number and add $3$ to it, I get $21.$
Answer
We have to frame the equation on the basis of given statement
And then solve the equation so framed.
Let the number be 'a'.
We know that,
Three fourth of a is $\frac{3 a}{4}$
Now, as per the statement,
We can write the equation as,
$\frac{3 a}{4}+3=21$
$\frac{3 a}{4} = 21 – 3$
$\frac{3 a}{4} = 18$
Now,
Multiplying $\frac{3}{4}$ to both the sides, we get
$\frac{3 a}{4} \times \frac{4}{3}=18 \times \frac{4}{3}$
Therefore, $a = 24$
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Question 65 Marks
Set up an equation on the basis of the statement given at the end and find the unknown quantity in the equation so obtained: One-fifth of a number minus $4$ gives $3$
Answer
We have to frame the equation on the basis of given statement
And then solve the equation so framed.
Let the number be $'a'.$
We know that, One fifth of a is $\frac{a}{5}$
Therefore, as per the statement,
We have,
$\frac{a}{5}-4=3$
$\frac{a}{5} = 3 + 4$
$\frac{a}{5} = 7$
Now,
Multiplying $5$ on both the sides, we get
$\frac{a}{5} \times 5=7 \times 5$
Therefore, $a = 35$
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Question 75 Marks
Set up an equation on the basis of the statement given at the end and solve it to find the unknown quantity.
Add $4$ to eight times a number you get $60.$
Answer
Here, we have to frame the equation on the basis of given statement
Let the number be $'a'$
Clearly, eight times of a is $8a$
According to the question, we have
$8a + 4 = 60$
$8a = 60 – 4$
$8a = 56$
Now,
Dividing by $8$ on both the sides, we get
$\frac{8 a}{8}=\frac{56}{8}$
Therefore, $a = 7.$
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Question 85 Marks
Complete the last column of the table.
S.no. Equation Value Say, whether the equation is satisfied. (yes/no)
$(i)$ $x + 3 = 0$ $x = 3$  
$(ii)$ $x + 3 = 0$ $x = 0$  
$(iii)$ $x + 3 = 0$ $x = -3$  
$(iv)$ $x - 7 = 1$ $x = 7$  
$(v)$ $x - 7 = 1$ $x = 8$  
$(vi)$ $5x = 25$ $x = 0$  
$(vii)$ $5x = 25$ $x = 5$  
$(viii)$ $5x = 25$ $x = -5$  
$(ix)$ $\frac{m}{3}=2$ $m = -6$  
$(x)$ $\frac{m}{3}=2$ $m = 0$  
$(xi)$ $\frac{m}{3}=2$ $m = 6$  
Answer
S.no. Given Equation Value at Check,$ whether the equation is satisfied or not. (Type yes/No)
$(i)$ $x + 3 = 0$ $x = 3$ Put $x = 3,$ in the equation.
$L.H.S. = 3 + 3 = 6 \ne RHS$
Thus, the answer is No. i.e. the equation is not satisfied.
$(ii)$ $x + 3 = 0$ $x = 0$ Put $x = 0,$ in the equation.
$L.H.S. = 0 + 3 = 3 \ne RHS$
Thus, the answer is No. i.e. the equation is not satisfied.
$(iii)$ $x + 3 = 0$ $x = -3$ Put $x = -3,$ in the equation.
$L.H.S. = -3 + 3 = 0 = RHS$
Thus, the answer is YES. i.e. the equation is satisfied.
$(iv)$ $x - 7 = 1$ $x = 7$ Put $x = 7,$ in the equation.
$L.H.S. = 7 - 7 = 0 \ne RHS$
Thus, the answer is No. i.e. the equation is not satisfied.
$(v)$ $x - 7 = 1$ $x = 8$ Put $x = 8,$ in the equation.
$L.H.S. = 8 - 7 = 1 = RHS$
Thus, the answer is YES i.e. the equation is satisfied.
$(vi)$ $5x = 25$ $x = 0$ Put $x = 0,$ in the equation.
$L.H.S. = 5(0) = 0 \ne RHS$
Thus, the answer is No. i.e. the equation is not satisfied.
$(vii)$ $5x = 25$ $x = 5$ Put $x = 5,$ in the equation.
$L.H.S. = 5(5) = 25 = RHS$
Thus, the answer is YES i.e. the equation is satisfied.
$(viii)$ $5x = 25$ $x = -5$ Put $x = -25,$ in the equation.
$L.H.S. = 5(-5) = 25 \ne RHS$
Thus, the answer is No. i.e. the equation is not satisfied.
$(ix)$ $\frac{m}{3}=2$ $m = -6$ Put $m = -6,$ in the equation.
$L.H.S. = \frac{-6}{3}=-2 \ne RHS$
Thus, the answer is No. i.e. the equation is not satisfied.
$(x)$ $\frac{m}{3}=2$ $m = 0$ Put $m = 0,$ in the equation.
$L.H.S. = \frac{0}{3}=0 \ne RHS$
Thus, the answer is No. i.e. the equation is not satisfied.
$(xi)$ $\frac{m}{3}=2$ $m = 6$ Put $m = 6,$ in the equation.
$L.H.S. = \frac{6}{3}=2 = RHS$
Thus, the answer is YES i.e. the equation is satisfied.
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