RS Aggarwal 2019,2020 solution class 7 chapter 7 Linear Equation In One Variable Test Paper 7

Test Paper 7

Page-118

Question 1:

Evaluate x³ + y³ + z³ −3xyz when x = −2, y = −1 and z = 3.

Answer 1:

$\begin{array}{l} \text{We have:}\\ x^3+y^3+z^3-3xyz\\ =(-2{)}^3+(-1{)}^3+(3{)}^3-3\times (-2)\times (-1)\times 3\\ =-8-1+27-18\\ =-27+27\\ =0\end{array}$

Question 2:

Write the coefficient of x in each of the following:

(i) −5xy
(ii) 2xy²z
(iii) $-\frac{3}{2}abc$

Answer 2:

$\begin{array}{l}\text{Coefficient of }\mathit{\text{x }}\text{in the given numbers are}\\ \text{(i}) -\text{5y} \\ \text{(ii}) {\text{2y}}^2\text{z} \\ \text{(iii}) -\frac{3}{2}ab\end{array}$

Question 3:

Subtract x² − 2xy + 5y² − 4 from 4xy − 5x² − y² + 6.

Answer 3:

$\begin{array}{l}\text{ We have:}\\ (4xy-5x^2-y^2+6)-(x^2-2xy+5y^2-4)\\ =4xy-5x^2-y^2+6-x^2+2xy-5y^2+4\\ =-6x^2-6y^2+6xy+10\\ =-2(3x^2+3y^2-3xy-5)\end{array}$

Question 4:

How much less is x² − 2xy + 3y² than 2x² − 3y² + xy?

Answer 4:

$$\begin{gathered} \begin{array}{l}\text{ We have:}\\ \text{(}2x^2-3y^2+xy)-(x^2-2xy+3y^2)\end{array} \\ =2x^2-3y^2+xy-x^2+2xy-3y^2) \\ \begin{array}{l}=2x^2-x^2-3y^2-3y^2+xy+2xy\\ =x^2-6y^2+3xy\end{array} \\ \therefore x^2-2xy+3y^2 \mathrm{is} \mathrm{less} \mathrm{than} 2x^2-3y^2+x by x^2-6y^2+3xy. \end{gathered}$$

Question 5:

Find the product $\left(\frac{3}{5}ab{c}^3\right)\times \left(\frac{-25}{12}{a}^3{b}^2\right)\times (-8b^{3}c).$

Answer 5:

$\begin{array}{l}\text{ We have: }\\ \frac{3}{5}ab{c}^3\times \frac{(-25)}{12}{a}^3{b}^2\times (-8b^{3}c)\\ =\frac{{\sout{3}}^1}{{\sout{5}}_1}ab{c}^3\times \frac{(-\sout{2}{\sout{5}}^5)}{\sout{1}{\sout{2}}_{{\sout{4}}_1}}{a}^3{b}^2\times (-{\sout{8}}^2{b}^{3}c)\\ =ab{c}^3\times (-5a^3{b}^2)\times (-2b^{3}c)\\ =10a^4{b}^6{c}^4\end{array}$

Question 6:

Simplify:

(3a + 4)(2a − 3) + (5a − 4)(a + 2)

Answer 6:

$\begin{array}{l}\text{ We have:}\\ (3a+4)(2a-3)+(5a-4)(a+2)\\ =\left\{3a\right(2a-3)+4(2a-3\left)\right\}+\left\{5a\right(a+2)-4(a+2\left)\right\}\\ =(6a^2-9a+8a-12)+(5a^2+10a-4a-8)\\ =(6a^2-a-12)+(5a^2+6a-8)\\ =(11a^2+5a-20)\\ \end{array}$

Question 7:

Solve: $\frac{3x}{10}+\frac{2x}{5}=\frac{7x}{25}+\frac{29}{25}.$

Answer 7:

$\begin{array}{l}\text{ We have:}\\ \frac{3x}{10}+\frac{2x}{5}=\frac{7x}{25}+\frac{29}{25}\\ \Rightarrow \frac{3x+4x}{10}=\frac{7x+29}{25}\\ \Rightarrow \frac{3x+4x}{10}=\frac{7x+29}{25}\\ \Rightarrow \frac{7x}{10}=\frac{7x+29}{25}\\ \Rightarrow \text{175}x=70x+290\\ \Rightarrow \text{105}x=290\\ \Rightarrow x=\frac{\sout{2}9{\sout{\sout{0}}}^{58}}{\sout{1}\sout{0}{\sout{5}}^{21}}\\ \Rightarrow x=\frac{58}{21}\\ \end{array}$

Question 8:

Solve: $0.5x+\frac{x}{3}=0.25x+7.$

Answer 8:

$\begin{array}{l}\text{We have:}\\ 0.5x+\frac{x}{3}=0.25x+7\\ \Rightarrow \frac{1.5x+x}{3}=0.25x+7\\ \Rightarrow 1.5x+x=3(0.25x+7)\\ \Rightarrow 2.5x=0.75x+21\\ \Rightarrow 2.5x-0.75x=21\\ \Rightarrow 1.75x=21\\ \Rightarrow x=\frac{21}{1.75}\\ \Rightarrow x=12\end{array}$

Question 9:

The sum of two consecutive odd numbers is 68. Find the numbers.

Answer 9:

$\begin{array}{l} \text{Let the consecutive odd numbers be }x\text{ and }(x+2).\\ x+(x+2)=68\\ \Rightarrow 2x+2=68\\ \Rightarrow 2x=68-2\\ \Rightarrow x=\frac{\sout{6}{\sout{6}}^{33}}{{\sout{2}}_1}\\ \Rightarrow x=33\\ \therefore \text{The required numbers are 33 and }(\text{33}+2), \mathrm{i}.\mathrm{e}., \text{35.}\end{array}$

Question 10:

Reenu's father is thrice as old as Reenu. After 12 years he will be just twice his daughter. Find their present ages.

Answer 10:

$\begin{array}{l}\text{Let Reenu's present age be }x.\\ \text{Then, her father's present age will be 3}x.\\ \text{Reenu's age after 12 years}=(x+12)\\ \text{Her father's age after 12 years}=(3x+12)\\ \text{Now, }(3x+12)=2(x+12)\\ \Rightarrow \text{3}x+12=2x+24\\ \Rightarrow x=12\\ \therefore \text{Reenu's present age }=12\text{ yrs}\\ \text{And her father's age}=(3\times 12)=36\text{ yrs}\end{array}$

Question 11:

Mark (✓) against the correct answer
If $2x+\frac{5}{3}=\frac{1}{4}x+4,$ then x = ?
(a) 3
(b) 4
(c) $\frac{3}{4}$
(d) $\frac{4}{3}$

Answer 11:

$$\begin{gathered} \left(\mathrm{d}\right) \frac{4}{3} \\ \begin{array}{l}2x+\frac{5}{3}=\frac{1}{4}x+4\\ \Rightarrow \text{2}x-\frac{1}{4}x=4-\frac{5}{3}\\ \Rightarrow \frac{8x-1x}{4}\text{=}\frac{12-5}{3}\\ \Rightarrow \frac{7x}{4}=\frac{7}{3}\\ \Rightarrow 21x=28\\ \Rightarrow x=\frac{\sout{2}{\sout{8}}^4}{\sout{2}{\sout{1}}_3}=\frac{4}{3}\\ \end{array} \end{gathered}$$

Question 12:

Mark (✓) against the correct answer
If $\frac{x}{2}-\frac{x}{3}=5$ then x = ?
(a) 8
(b) 16
(c) 24
(d) 30

Answer 12:

$\begin{array}{l}\left(\mathrm{d}\right) 30\\ \frac{x}{2}-\frac{x}{3}\text{=5}\\ \Rightarrow \frac{3x-2x}{6}=5\\ \Rightarrow x=30\\ \end{array}$

Question 13:

Mark (✓) against the correct answer
If $\frac{x-2}{3}=\frac{2x-1}{3}-1,$ then x = ?
(a) 2
(b) 4
(c) 6
(d) 8

Answer 13:

$$\begin{gathered} \begin{array}{l}\left(\mathrm{a}\right) 2\\ \frac{x-2}{3}\text{ =}\frac{2x-1}{3}-1\\ \Rightarrow \frac{x-2}{\cancel{3}}=\frac{2x-1-3}{\cancel{3}}\\ \Rightarrow x-2=2x-4\\ \Rightarrow \text{x-2x=-4+2}\\ \Rightarrow -x=-2\end{array} \\ \Rightarrow x=2 \end{gathered}$$

Question 14:

Mark (✓) against the correct answer
A number when multiplied by 4 is increased by 54. The number is

(a) 21
(b) 16
(c) 18
(d) 19

Answer 14:

$$\begin{gathered} \left(\mathrm{c}\right) 18 \\ \begin{array}{l}\text{Let the number be}x.\\ \text{According to the question, we have:}\\ \text{4}x=x+\text{54}\\ \Rightarrow \text{3}x=\text{54}\\ \Rightarrow x=\text{18}\end{array} \end{gathered}$$

Question 15:

Two complementary angles differ by 14°. The larger angle is

(a) 50°
(b) 52°
(c) 54°
(d) 56°

Answer 15:

$$\begin{gathered} \left(\mathrm{b}\right) 52° \\ \begin{array}{l}\text{Let the two complementary angles be }\mathrm{x}°\text{ and }(90-\mathrm{x})°.\\ \text{According to the question, we have: }\\ \mathrm{x}-(90-\mathrm{x})=14\\ \text{⇒2}\mathrm{x}=104\\ \Rightarrow \mathrm{x}=\text{52}\\ \Rightarrow (90-\mathrm{x})°=90°-52°=38°\\ \therefore \text{The larger angle is 52}°.\end{array} \end{gathered}$$

Question 16:

The length of a rectangle is twice its breadth and its perimeter is 96 m. The length of the rectangle is

(a) 28 m
(b) 30 m
(c) 32 m
(d) 36 m

Answer 16:

$$\begin{gathered} \left(\mathrm{c}\right) 32 \mathrm{m} \\ \begin{array}{l}\text{Let the length and breadth of the rectangle be l m and b m, respectively. }\\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \mathrm{we} \mathrm{have}: \\ \mathit{\text{l}}=\text{2}\mathit{\text{b ...}}\left(\text{i}\right)\\ \text{2}(\mathit{\text{l}}+\mathit{\text{b}})=\text{96 ...}\left(\text{ii}\right)\\ \text{Now, 2}(\text{2}\mathit{\text{b}}+\mathit{\text{b}})=\text{96}\\ \Rightarrow \text{6}\mathit{\text{b}}=96\\ \Rightarrow \mathit{\text{b}}=16\\ \therefore \text{Length}=16\times 2 \mathrm{m}=32 \text{m}\\ \end{array} \end{gathered}$$

Question 17:

The ages of A and B are in the ratio 4 : 3. After 6 years, their ages will be in the ratio 11 : 9. A's present age is

(a) 12 years
(b) 16 years
(c) 20 years
(d) 24 years

Answer 17:

$$\begin{gathered} \left(\mathrm{b}\right) 12 \mathrm{years} \\ \mathrm{Let} \mathrm{the} \mathrm{ages} \mathrm{of} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{be} x \mathrm{and} y \mathrm{years}, \mathrm{respectively}. \\ \begin{array}{l}\text{Now, }\frac{x}{y}=\frac{4}{3}\\ \Rightarrow 3x=4y\\ \Rightarrow x=\frac{4}{3}y\\ \text{After 6 years, we have: }\\ \frac{\text{x + 6 }}{y+6}=\frac{11}{9}\\ \Rightarrow \frac{\frac{4}{3}y+6}{y+6}=\frac{11}{9}\\ \Rightarrow \frac{4y+18}{\text{3(y}+6)}=\frac{11}{9}\\ \Rightarrow 36y+162=33y+198\\ \Rightarrow 3y=36\\ \Rightarrow y=12\\ \therefore x=\frac{4}{\sout{3}}\times \sout{1}{\sout{2}}^4=16\\ \text{Hence, A's present age is 16 years.}\end{array} \\ \begin{array}{}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{array} \end{gathered}$$

Question 18:

Fill in the blanks

(i) −2a²b is a ...... .
(ii) (a² − 2b²) is a ...... .
(iii) (a + 2b − 3c) is a ...... .
(iv) In −5ab, the coefficient of a is ...... .
(v) In x² + 2x − 5, the ...... term is −5.

Answer 18:

(i) −2a²b is a monomial.
(ii) (a² − 2b²) is a binomial.
(iii) (a + 2b − 3c) is a trinomial.
(iv) In −5ab, the coefficient of a is -5b.
(v) In x² + 2x − 5, the constant term is −5.

Question 19:

Write 'T' for true and 'F' for false

(i) In −x, the constant term is −1.
(ii) The coeffiecient of x in x² − 3x + 5 is 3.
(iii) (5x − 7) − (3x − 5) = 2x − 12.
(iv) (3x + 5y)(3x − 5y) = (9x² − 25y²).
(v) If a = 2 and b = $\frac{1}{2},$ then the value of ab (a² + b²) is $4\frac{1}{4}$.

Answer 19:

$$\begin{gathered} \left(i\right) \mathrm{F} \\ \mathrm{The} \mathrm{coefficient} \mathrm{of} x \mathrm{is} -1. \\ \left(ii\right) \mathrm{F} \\ \mathrm{The} \mathrm{coefficient} \mathrm{of} x \mathrm{is} -3. \\ \left(iii\right) \mathrm{F} \\ \mathrm{LHS}=(5x-7)-(3x-5) \\ =5x-7-3x+5 \\ =2x-2 \\ \left(iv\right) \mathrm{T} \\ \mathrm{LHS}=(3x+5y)(3x-5y) \\ =3x(3x-5y)+5y(3x-5y) \\ =9x^2-15xy+15xy-25y^2 \\ =9x^2-25y^2 \\ \left(v\right) \mathrm{T} \\ (a^2+b^2) \\ =\left[2^2+{\left(\frac{1}{2}\right)}^2\right] \\ =4+\frac{1}{4} \\ =4\frac{1}{4} \end{gathered}$$