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MATHS [4 marks sum]

Fundamental Concepts (Including Fundamental Operations) · ICSE STD 7 (ICSE - English Medium). 15 questions.

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15 questions · timed · auto-graded

Question 14 Marks
Simplify: $8 \mathrm{x}-[4 \mathrm{y}-\{4 \mathrm{x}+(2 \mathrm{x}-\overline{2 \mathrm{y}-2 \mathrm{x}})\}]$
Answer

$\begin{aligned} & 8 x-[4 y-\{4 x+(2 x-\overline{2 y-2 x})\}] \\ & =8 x-[4 y-\{4 x+(2 x-2 y+2 x)\}] \\ & =8 x-[4 y-\{4 x+(4 x-2 y)\}] \\ & =8 x-[4 y-\{4 x+4 x-2 y\}] \\ & =8 x-[4 y-\{4 x-4 x+2 y\}] \\ & =8 x-[-8 x+6 y] \\ & =8 x+8 x-6 y \\ & =16 x-6 y\end{aligned}$
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Question 24 Marks
Simplify: $7 \mathrm{a}[8 \mathrm{a}-(11 \mathrm{a}-(12 \mathrm{a}-\overline{6 \mathrm{a}-5 \mathrm{a}}))]$
Answer

$\begin{aligned} & 7 a[8 a-(11 a-(12 a-\overline{6 a-5 a}))] \\ & =7 a-[8 a-\{11 a-(12 a-6 a+5 a)\}] \\ & =7 a-[8 a-\{11 a-(17 a-6 a)\}] \\ & =7 a-[8 a-\{11 a-(11 a)\}] \\ & =7 a-[8 a-\{11 a-11 a\}] \\ & =7 a-8 a \\ & =-a\end{aligned}$
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Question 34 Marks
Simplify: $10-\{4 \mathrm{a}-(7-\overline{\mathrm{a}-5}-(5 \mathrm{a}-\overline{1+\mathrm{a}}))$
Answer

$\begin{aligned} & 10-\{4 a-(7-\overline{a-5}-(5 a-\overline{1+a})) \\ & =10-\{4 a-(7-a+5)-(5 a-1-a)\} \\ & =10-\{4 a-(12-a)-(4 a-1)\} \\ & =10-\{4 a-12+a-4 a+1\} \\ & =10-4 a+12-a+4 a-1 \\ & =10+12-1-4 a-a+4 a \\ & =21-a\end{aligned}$
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Question 44 Marks
Simplify: $\mathrm{p}^2-\left[\mathrm{x}^2-\left\{\mathrm{x}^2-\left(\mathrm{q}^2-\overline{\mathrm{x}^2-\mathrm{q}^2}\right)-2 \mathrm{y}^2\right\}\right]$
Answer

$\begin{aligned} & p^2-\left[x^2-\left\{x^2-\left(q^2-\overline{x^2-q^2}\right)-2 y^2\right\}\right] \\ & =p^2-\left[x^2-\left\{x^2-\left(q^2-x^2+q^2\right)-2 y^2\right\}\right] \\ & =p^2-\left[x^2-\left\{x^2-\left(2 q^2-x^2\right)-2 y^2\right\}\right] \\ & =p^2-\left[x^2-\left\{x^2-2 q^2+x^2-2 y^2\right\}\right] \\ & =p^2-x^2+2 x^2-2 q^2-2 y^2 \\ & =p^2+x^2-2 q^2-2 y^2\end{aligned}$
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Question 54 Marks
Simplify: $5 a-[6 a-\{9 a-(10 a-\overline{4 a-3 a})\}]$
Answer

$\begin{aligned} & 5 a-[6 a-\{9 a-(10 a-\overline{4 a-3 a})\}] \\ & =5 a-[6 a-\{9 a-(10 a-4 a+3 a)\}] \\ & =5 a-[6 a-\{9 a-10 a+4 a-3 a\}] \\ & =5 a-[6 a-9 a+10 a-4 a+3 a] \\ & =5 a-6 a+9 a-10 a+4 a-3 a \\ & =5 a+9 a+4 a-6 a-10 a-3 a \\ & =18 a-19 a \\ & =-a\end{aligned}$
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Question 64 Marks
The area of a rectangular field is 25x2 + 20xy + 3y2 square unit. If its length is 5x + 3y unit, find its breadth, Hence find its perimeter.
Answer
Area of a rectangle
$
=25 x^2+20 x y+3 y^2
$
Length $=(5 x+3 y)$ units
$
\begin{aligned}
& \therefore \text { Breadth }=\frac{\text { Area }}{\text { Length }} \\
& =\frac{25 x ^2-20 xy +3 y ^2}{5 x +3 y }
\end{aligned}
$
Image
Hence Breadth $=5 x+y$
Hence perimeter of rectangular field
$
\begin{aligned}
& =2(I+b) \\
& =2(5 x+3 y+5 x+y) \\
& =2(10 x+4 y) \\
& =20 x+8 y
\end{aligned}
$
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Question 74 Marks
The area of a rectangle is $6 x^2-4 x y-10 y^2$ square unit and its length is $2 x+2 y$ unit. Find its breadth.
Answer
Area of a rectangle
$
=6 x^2-4 x y-10 y^2 \text { sq.units }$
Length $=2 x+2 y$ units
$
\begin{aligned}
& \therefore \text { Breadth }=\frac{\text { Area }}{\text { Length }} \\
& =\frac{6 x ^2-4 xy -10 y ^2}{2 x +2 y }
\end{aligned}
$
Image
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Question 84 Marks
Evaluate: $(5 m-2 n)(5 m+2 n)\left(25 m^2+4 n^2\right)$
Answer
$(5m - 2n)(5m + 2n)(25m^2 + 4n^2)$
$= {5m (5m + 2n) - 2n(5m + 2n)} (25m^2 + 4n^2)$
$= (25m^2 + 10mn - 10mn - 4n^2)(25m^2 + 4n^2)$
$= (25m^2 - 4n^2)(25m^2 + 4n^2)$
$= 25m^2 (25m^2 + 4n^2) - 4n^2 (25m^2 + 4n^2)$
$= 625m^4 + 100m^2n^2 - 100m^2n^2 - 16n^4$
$= 625m^4 - 16n^4$
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Question 94 Marks
Evaluate: $(a+1)\left(a^2-a+1\right)$ and $(a-1)\left(a^2+a+1\right)$
Answer
$(a+1)\left(a^2-a+1\right)$ and $(a-1)\left(a^2+a+1\right)$
$= a (a^2 - a + 1) + 1 (a^2 - a + 1)$
$= a^3 - a^2 + a + a^2 - a + 1$
$= a^3 + 1$
$(a - 1)(a^2 + a + 1)$
$= a(a^2 + a + 1) - 1(a^2 + a + 1)$
$= a^3 + a^2 + a - a^2 - a - 1$
$= a^3 - 1$
Now, $(a + 1)(a^2 - a + 1) + (a - 1)(a^2 + a + 1)$
$= a^3 + 1 + a^3 - 1$
$= 2a^3$​​​​​​​
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Question 104 Marks
Evaluate: $(a + 5)(3a - 2)(5a + 1)$
Answer
$(a + 5)(3a - 2)(5a + 1)$
$= {a (3a - 2) + 5(3a - 2)} (5a + 1)$
$= (3a^2 - 2a + 15a - 10)(5a + 1)$
$= (3a^2 + 13a - 10)(5a + 1)$
$= 5a (3a^2 + 13a - 10) + 1 (3a^2 + 13a - 10)$
$= 15a^3 + 65a^2 - 50a + 3a^2 + 13a - 10$
$= 15a^3 + 68a^2 - 37a - 10$
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Question 114 Marks
Evaluate: $(3x - 2y)(4x + 3y) (8x - 5y)$
Answer
$(3x - 2y)(4x + 3y)(8x - 5y)$
$= 3x (4x + 3y) - 2y (4x + 3y)(8x - 5y)$
$= (12x^2 + 9xy - 8xy - 6y^2)(8x - 5y)$
$= (12x^2 + xy - 6y^2 )(8x - 5y)$
$= 8x (12x^2 + xy - 6y^2 ) - 5y (12x^2 + xy - 6y^2 )$
$= 96x^3 + 8x^2y - 48xy^2 - 60x^2y - 5xy^2 + 30y^3$
$= 96x^3 + 8x^2y - 60x^2y - 48xy^2- 5xy^2 + 30y^3$
$= 96x^3 - 52x^2y - 53xy^2 + 30y^3$
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Question 124 Marks
Evaluate: $(a^2 + b^2 + c^2 - ab - bc - ca)(a + b + c)$
Answer
$(a^2 + b^2 + c^2 - ab - bc - ca)(a + b + c)$
$= a (a^2 + b^2 + c^2 - ab - bc - ca) + b (a^2 + b^2 + c^2 - ab - bc - ca) + c (a^2 + b^2 + c^2 - ab - bc - ca)$
$= a^3 + ab^2 + ac^2 - a^2b - abc - ca^2 +a^2b + b^3 + bc^2 - ab^2 - b^2c - abc + a^2c + b^2c + c^3 - abc - bc^2 - c^2a$
$= a^3 + b^3 + c^3 -a^2b + a^2b - ca^2 + a^2c + bc^2 - bc^2 - ab^2 + ab^2 - abc - abc - abc + ac^2 - ac^2 + b^2c - b^2c$
$= a^3 + b^3 + c^3 - 3abc$
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Question 134 Marks
Simplify: 4(3x - 8) - 3(5x + 3) - 2(6x - 8)
Answer
4(3x - 8) - 3(5x + 3) - 2(6x - 8)
⇒ 12x - 32 - 15x - 9 - 12x + 16
⇒ 12x - 15x - 12x - 32 - 9 + 16
⇒ 12x - 27x - 41 + 16
⇒ - 15 x - 25
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Question 144 Marks
Evaluate:
$(i) \ (a + b)(a - b)$
$(ii) \ (a^2 + b^2)(a + b)(a - b)$; using the result of $(i).$
$(iii) \ (a^4 + b^4)(a^2 + b^2)(a + b)(a - b)$; using the result of $(ii).$
Answer
$ (i) \ (a + b)(a - b)$
$= a (a - b) + b(a - b)$
$= a^2 - ab + ab - b^2$
$= a^2 - b^2$
$(ii) \ (a^2 + b^2)(a + b)(a - b)$
$= (a^2 + b^2)(a^2 - b^2) ...{from(i)}$
$= a2 (a^2 - b^2) + b^2 (a^2 - b^2)$
$= a^4 - a^2b^2 + a^2b^2 - b^4$
$= a^8 - a^4b^4 + a^4b^4 - b^8$
$= a^4 - b^4$
$(iii) \ (a^4 + b^4)(a^2 + b^2)(a + b)(a - b)$
$= (a^4 + b^4) (a^4 - b^4) ....{from(ii)}$
$= a^4 (a^4 + b^4) + b^4 (a^4 + b^4)$
$= a^8 - a^4b^4 + a^4b^4 - b^8$
$= a^8 - b^8$
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Question 154 Marks
Subtract the sum of $3a^2 – 2a + 5$ and $a^2 – 5a – 7$ from the sum of $5a^2 -9a + 3$ and $2a – a^2 – 1 $
Answer
Sum of $3a^2 – 2a + 5$ and $a^2 – 5a – 7$
$= 3a^2 – 2a + 5 + a^2 – 5a – 7$
$= 3a^2 + a^2 – 2a – 5a + 5 – 7$
$= 4a^2 - 7a - 2$
and sum of $5a^2 -9a + 3$ and $2a – a^2 – 1$
$= 5a^2 - 9a + 3 + 2a – a^2 – 1 $
$= 5a^2 – a^2 - 9a + 2a + 3 - 1$
$= 4a^2 - 7a + 2$
Now $(4a^2 - 7a + 2) - (4a^2 - 7a - 2)$
$= 4a^2 - 7a + 2 - 4a^2 + 7a + 2$
$= 4a^2 - 4a^2 - 7a + 7a + 2 + 2$
$= 0 + 0 + 4 = 4$
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[4 marks sum] - MATHS STD 7 Questions - Vidyadip