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[3 marks sum]

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14 questions · timed · auto-graded

Question 13 Marks
Find the product of $2a + 3b − 4c$ and $a − b − c$
Answer
$ (2a + 3b − 4c) \times (a − b − c)$
$= a \times (2a + 3b − 4c) − b \times (2a + 3b − 4c) − c \times (2a + 3b − 4c)$
$= 2a^2 + 3ab − 4ac − 2ab − 3b^2 + 4bc − 2ac − 3bc + 4c^2$
$= 2a^2 + 3ab − 2ab − 4ac − 2ac − 3b^2 + 4bc − 3dc + 4c^2$
$= 2a^2 + ab − 6ac − 3b^2 + bc + 4c^2$
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Question 23 Marks
Find the product of $5x − 6y − 7z$ and $2x + 3y z$
Answer
$(5x − 6y − 7z) \times (2x + 3y + z)$
$= 2x \times (5x − 6y − 7z) + 3y \times (5x − 6y − 7z) + z (5x − 6y − 7z)$
$= 10x^2 − 12xy − 14xz + 15xy − 18y^2 − 21yz + 5xz − 6yz − 7z^2$
$= 10x^2 − 12xy + 15xy − 14xz + 5xz − 18y^2 − 21yz − 6yz − 7z^2$
$= 10x^2 + 3xy − 9xz − 18y^2 − 21yz − 7z^2$
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Question 33 Marks
Find the product of $a + b − c$ and $2a − 3b$
Answer
$(a + b − c) \times (2a − 3b)$
$= 2a \times (a + b − c) − 3b \times (a + b − c)$
$= 2a^2 + 2ab − 2ac − 3ab − 3b^2 + 3bc$
$= 2a^2 − ab − 2ac − 3b^2 + 3bc$
$= 2a^2 − ab − 2ac + 3bc − 3b^2$
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Question 43 Marks
Multiply $: 2xyz + 3xy$ and $− 2y^2z$
Answer
$2xyz + 3xy \times − 2y^2z$
$= 2xyz \times − 2y^2z + 3xy \times − 2y^2z$
$= − 4x \times y^{2+1} \times z^{1+1} − 6 \times x \times y^{2+1} \times z$
$= − 4xy^3z^2 − 6xy^3z$
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Question 53 Marks
Subtract the sum of x + y and x – z from the sum of x – 2z and x + y + z
Answer
(x − 2z) + (x + y + z) − {(x + y) + (x − z)}
= x − 2z + x + y + z − {x + y + x − z}
= x − 2z + x + y + z − x − y − x + z
= x + x − x −x + y − y + z + z − 2z
= 0
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Question 63 Marks
Take $\frac{-3}{2} p+q-r$ from $\frac{1}{2} p-\frac{1}{3} q-\frac{3}{2} r \text {. }$
Answer
$\left(\frac{1}{2} p-\frac{1}{3} q-\frac{3}{2} r\right)-\left(-\frac{3}{2} p+q-r\right)$
$=\frac{1}{2} p-\frac{1}{3} q-\frac{3}{2} r+\frac{3}{2} p-q+r$
$=\frac{1}{2} p+\frac{3}{2} p-\frac{1}{3} q-q-\frac{3}{2} r+r$
$=\frac{3 p+9 p-2 q-6 q-9 r+6 r}{6} . . . . . .[$ Since $\text{L.C.M.} = 6]$
$=\frac{12 p}{6}-\frac{8 q}{6}-\frac{3 r}{6}$
$=2 p-\frac{4}{3} q-\frac{1}{2} r$
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Question 73 Marks
Find the sum of $3x + 4xy − y^2, xy − 4x + 2y^2$ and $3y^2 − xy + 6x$.
Answer
$(3x + 4xy − y^2) + (xy − 4x + 2y^2) + (3y^2 − xy + 6x)$
$= 3x + 4xy − y^2 + xy − 4x + 2y^2 + 3y^2 − xy + 6x$
$= 3x + 6x − 4x + 4xy + xy − xy + 2y^2 + 3y^2 − y^2$
$= 9x − 4x + 5xy − xy + 5y^2 − y^2$
$= 5x + 4xy + 4y^2$
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Question 83 Marks
Find the sum of $4x^2 + 7 − 3x, 4x − x^2 + 8$ and $− 10 + 5x − 2x^2.$
Answer
$(4x^2 + 7 − 3x) + (4x − x^2 + 8) + (− 10 + 5x − 2x^2)$
$= 4x^2 + 7 − 3x + 4x − x^2 + 8 − 10 + 5x − 2x^2$
$= 4x^2 − x^2− 2x^2 + 7 + 8 − 10 + 4x + 5x − 3x$
$= 4x^2 − 3x^2 + 15 − 10 + 9x − 3x$
$= x^2 + 5 + 6x$
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Question 93 Marks
Find the sum of $a^2 − ab + bc, 2ab + bc − 2a^2$ and $− 3bc + 3a^2 + ab.$
Answer
$(a^2 − ab + bc) + (2ab + bc − 2a^2) + (− 3bc + 3a^2 + ab)$
$= a^2 − ab + bc + 2ab + bc − 2a^2 − 3bc + 3a^2 + ab$
$= a^2 + 3a^2− 2a^2 + 2ab + ab − ab + bc + bc − 3bc$
$= 4a^2 − 2a^2 + 3ab − ab + 2bc − 3bc$
$= 2a^2 + 2ab − bc$
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Question 103 Marks
Find the sum of $x^2 − x + 1, − 5x^2 + 2x − 2$ and $3x^2 − 3x + 1$
Answer
$(x^2− x + 1) + (− 5x^2 + 2x − 2) + (3x^2 − 3x + 1)$
$= x^2− x + 1 − 5x^2 + 2x − 2 + 3x^2 − 3x + 1$
$= x^2 + 3x^2 − 5x^2 + 2x − x − 3x + 1 + 1 − 2$
$= 4x^2 − 5x^2 + 2x − 4x + 2 − 2$
$= − x^2− 2x$
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Question 113 Marks
Find the sum of $2x^2 + xy − y^2, − x^2 + 2xy + 3y^2$ and $3x^2 − 10xy + 4y^2.$
Answer
$(2x^2 + xy − y^2) + (− x^2 + 2xy + 3y^2) + (3x^2 − 10xy + 4y^2)$
$= 2x^2 + xy − y^2 − x^2 + 2xy + 3y^2+ 3x^2 − 10xy + 4y^2$
$= 2x^2 + 3x^2 − x^2 + xy + 2xy − 10xy + 3y^2 + 4y^2 − y^2$
$= 5x^2 − x^2 + 3xy − 10xy + 7y^2 − y^2$
$= 4x^2 − 7xy + 6y^2$
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Question 123 Marks
Find the sum of 3a + 4b + 7c, − 5a + 3b − 6c and 4a − 2b − 4c.
Answer
(3a + 4b + 7c) + (− 5a + 3b − 6c) + (4a − 2b − 4c)
= 3a + 4b + 7c − 5a + 3b − 6c + 4a − 2b − 4c
= 3a + 4a − 5a + 4b + 3b − 2b + 7c − 6c − 4c
= 7a − 5a + 7b − 2b + 7c − 10c
= 2a + 5b − 3c
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Question 133 Marks
Simplify: $6 p+\frac{2}{3} q-1 \frac{1}{2} p+\frac{1}{3} q+2 q$
Answer
$6 p+\frac{2}{3} q-1 \frac{1}{2} p+\frac{1}{3} q+2 q$
$=6 p-\frac{3}{2} p+\frac{2}{3} q+\frac{1}{3} q+2 q$
$=\left(\frac{12 p-3 p}{2}\right)+\left(\frac{2 q+q+6 q}{3}\right)$
$=\frac{9}{2} p+3 q$
$=4 \frac{1}{2} p+3 q$
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Question 143 Marks
Simplify : $22 m-12 \frac{1}{2} n-15 p+16 n$
Answer
$22 m-12 \frac{1}{2} n-15 p+16 n$
$=22 m-\frac{25}{2} n-15 p+16 n$
$=22 m+16 n-\frac{25}{2} n-15 p$
$=22 m+\frac{32 n-25 n}{2}-15 p$
$=22 m+\frac{7 n}{2}-15 p$
$=22 m+3 \frac{1}{2} n-15 p$
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[3 marks sum] - MATHS STD 6 Questions - Vidyadip