Simple Equations(7th Grade > Mathematics ) Questions and answers for exam Preparation

7th Grade > Mathematics

SIMPLE EQUATIONS MCQs

Total Questions : 118 | Page 12 of 12 pages

Arjun has thrice as much money as Prateek, and Shubhendu has half as much as Arjun and Prateek put together. If Arjun has 120 rupees, then how much money do Prateek and Shubhendu have? [3 MARKS]

Discuss Question

Answer: Option A. ->
:

Framing the equation: 1 Mark
Steps: 1 Mark
Result: 1 Mark
Let Prateek have x rupees.
Then, Arjun will have = 3 times Prateek = 3x
Shubhendu has half as much as Prateek and Arjun put together
Or, Shubhendu has $\frac{x + 3 x}{2} = 2 x$
According to the question,
$3 x = 120$
Or, $x = 40$
Arjun has 120 rupees, Prateek has 40 rupees and Shubhendu has 80 rupees.

Question 112.

In a triangle ABC, angle A is half of B and angle C is thrice of A. Find the value of angle A, B and C? [4 MARKS]

Discuss Question

Answer: Option A. ->
:

Concept: 1 Mark
Framing of the equation: 1 Mark
Steps: 1 Mark
Result: 1 Mark
Let the value of angle B be $y^{\circ}$
Then, value of angle $A = \frac{y^{\circ}}{2}$
And, value of angle $C = \frac{3 y^{\circ}}{2}$
We know that sum of all angles in a triangle is $180^{\circ}$ .
$\Rightarrow y^{\circ} + \frac{y^{\circ}}{2} + \frac{3 y^{\circ}}{2} = 180^{\circ}$
$\Rightarrow 3 y^{\circ} = 180^{\circ}$
$\Rightarrow y^{\circ} = 60^{\circ} =$ Value of angle B
So, Value of angle $A = \frac{y^{\circ}}{2} = 30^{\circ}$
Value of angle $C = \frac{3 y^{\circ}}{2} = 90^{\circ}$ .

Question 113.

Check whether the following are satisfying the condition of a simple equation? [3 MARKS]
a) $6 x + \frac{4 x}{3} + 4$
b) $2 t + 34 = 42$
c) $2 x^{2} + 4 x + 2 = 0$

Discuss Question

Answer: Option A. ->
:

1 Mark each
a) No; It is an expression not equation because it does not have "=” sign in it.
b) Yes; It is an equation because "=” sign is present in it.
c) No; In a simple equation, the highest degree of the polynomial is one whereas here the highest power is 2.

Question 114.

63 cookies are distributed among four children Ram, Arjun, John, and Imran in such a way that Arjun gets twice the number of cookies than Ram. John gets half of the sum of cookies that Arjun and Ram got. Imran gets half of the number of cookies that John got. Find the number of cookies that each one received. [4 MARKS]

Discuss Question

Answer: Option A. ->
:

Framing the equation: 1 Mark
Steps: 2 Marks
Result: 1 Mark
Let the no. of cookies with Ram be y.
According to question, no. of cookies Arjun has = 2 times Ram = $2 y$
No. of cookies John has = Half of sum of Ram and Arjun. = $\frac{y + 2 y}{2}$
And, the number of cookies Imran has $= \frac{J o h n}{2}$
Since, total number of cookies = 63
$y + 2 y + (\frac{y + 2 y}{2}) + (\frac{(\frac{y + 2 y}{2})}{2}) = 63$
$\Rightarrow 3 y + \frac{3 y}{2} + \frac{3 y}{4} = 63$
$\Rightarrow \frac{12 y + 6 y + 3 y}{4} = 63$
$\Rightarrow 21 y = 63 \times 4$
$\Rightarrow y = 12$
So, No. of cookies Ram got $y = 12$
No. of cookies Arjun got $2 y = 24$
No. of cookies John got $\frac{y + 2 y}{2} = 18$
No. of cookies Imran got = 9

Question 115.

A carton has twenty chocolates worth rupees 100. It has Cadbury shots worth rupees 3 each and 5-star worth rupees 11 each. How many 5 star chocolates are there in the carton? [4 MARKS]

Discuss Question

Answer: Option A. ->
:

Framing the equation: 1 Mark
Steps: 2 Marks
Result: 1 Mark

Let the no. Of Cadbury shots $= y$
Then no. of 5-star chocolates $= 20 - y$
As given in the question, Cadbury shots worth rupees 3 and 5-star worth rupees 11.
Total cost of Cadbury shots is
$= N o . o f s h o t s \times c o s t o f e a c h = 3 \times y = 3 y$
Total cost of 5 star is
$= N o . o f 5 s t a r \times c o s t o f e a c h = (20 - y) \times 11 = 220 - 11 y$
Total worth of chocolates in carton $= 100$
$\Rightarrow 3 y + 220 - 11 y = 100$
$\Rightarrow 220 - 100 = 11 y - 3 y$
$\Rightarrow 120 = 8 y$
$\Rightarrow y = 15$
There are 15 Cadbury shots and 5 5-star.

Question 116.

Which of the following is a linear (or simple) equation in one variable? Give reasons. [4 MARKS]
(i) $x^{2} = 5$
(ii) $x + y = 4$
(iii) $x + 6 = 9$
(iv) $x^{2} + y^{2} = 9$

Discuss Question

Answer: Option A. ->
:
1 Mark each
(i)

x^{2} = 5

is not a linear equation since the highest exponent of the variable is 2.
(ii)

x + y = 4

has two variables in it. So, it is not a linear equation.
(iii)

x + 6 = 9

is a linear equation in one variable.
(iv)

x^{2} + y^{2} = 9

has two variables in it, and the highest exponent is 2. So, it is not a linear equation.

Question 117.

What is the value of the unknown in the following equations? [4 MARKS]
(i) $2 x + 4 = 10$
(ii) $11 x - 7 = 5 (x + 7)$
(iii) $7 m + \frac{19}{2} = 13$
(iv) $2 y + \frac{5}{2} = \frac{37}{2}$

Discuss Question

Answer: Option A. ->
:
Each part: 1 Mark each
(i)

2 x + 4 = 10

\Rightarrow 2 x = 10 - 4

\Rightarrow 2 x = 6

\Rightarrow x = \frac{6}{2} = 3

(ii)

11 x - 7 = 5 (x + 7)

\Rightarrow 11 x - 7 = 5 x + 35

\Rightarrow 11 x - 5 x = 35 + 7

\Rightarrow 6 x = 42

\Rightarrow x = \frac{42}{6} = 7

(iii)

7 m + \frac{19}{2} = 13

7 m = 13 - \frac{19}{2}

7 m = \frac{26 - 19}{2}

7 m = \frac{7}{2}

m = \frac{7}{2 \times 7}

m = \frac{1}{2}

(iv)

2 y + \frac{5}{2} = \frac{37}{2}

2 y = \frac{37}{2} - \frac{5}{2}

2 y = \frac{37 - 5}{2}

2 y = \frac{32}{2}

y = \frac{32}{4}

y = 8

Question 118.

Complete the last column of the table: [4 MARKS]
$\begin{matrix} S . N O & E q u a t i o n & V a l u e & S a y, w h e t h e r t h e e q u a t i o n i s s a t i s f i e d . (y e s / N o) (i) & x + 3 = 0 & x = 3 (i i) & x + 3 = 0 & x = 0 (i i i) & x + 3 = 0 & x = - 3 (i v) & x - 7 = 1 & x = 7 (v) & x - 7 = 1 & x = 8 (v i) & 5 x = 25 & x = 0 (v i i) & 5 x = 25 & x = 5 (v i i i) & 5 x = 25 & x = - 5 \end{matrix}$