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Roll No: [4] [4]
Page : [0] [1]
T1. P is the mid point of AB.
$\therefore AP=BP$
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\Rightarrow \angle APD = \angle BPE$
(i) Now, in $\Delta DAP$ and $\Delta EBP$ we have
$\angle PAD = \angle PBE$ ($\because \angle BAD = \angle ABE$)
$AP=BP$ (proved above).
$\angle DPA... | |
Roll No. : 4 4
Page : 0 2
(iii) prove that $AP$ bisects $\angle A$ as well as $\angle D$.
$\Delta ABD \cong \Delta ACD$ (By CPCT)
$\angle BAD = \angle CAD$ (By CPCT)
$\angle BAP = \angle CAP$
$\therefore AP$ bisects $\angle A$
$Ap$ bisects $\angle D,$
$\Delta BDP$ and $\Delta CDP$ (we have)
$DB = DC$ (Given)
$BP = CP$... | |
Roll No.: 5 5
Page : 0 1
T1
Given
p is the mid point
\angle BAD = \angle ABE (Given)
\angle EPA = \angle DPB (Given)
(i) Ap = AB
AD = BE
E = D
Then
\Delta DAP \cong \Delta EBP
\angle EPA = \angle DPB
(ii) \angle EPA + \angle EPB = \angle DPB + \angle EPD
\angle APD = \angle EPA + \angle EPD / \angle BPE = \angle DPB +... | |
Roll No.: 5 5
Page : 0 2
(ii) AB = AC
\angle BAD = \angle CAD
\angle BAD = \angle CAD (CPCT)
AP = AP (common)
\Delta ABP \cong \Delta ACP (SAS)
(iii) \Delta ABD \cong \Delta ACD
\angle BAD = \angle CAD (CPCT)
this means
\angle BAD = \angle CAD
\therefore AP \text{ bisects } \angle A
AP \text{ bisects } \angle AD
DB =... | |
Roll No.: 55
Page : 03
There fore
$\angle APB + \angle APC = 180^\circ$
$2 \angle APB = 180^\circ$
$\angle APB = \frac{180^\circ}{2}$
$\angle APB = 90^\circ$
AP is the perpendicular to BC
T3 | |
Roll No.: 3 8
Page : 0 1
I) AP = BP
$\angle DAP = \angle EBP$
$\angle DPA =$
1) Tap A fills = Both Taps fill + 4 hours = x
Tap B fills = Both taps fill + 9 hours = y
Both taps together open = z
z = 2x + 1
z + 9 = 2(z + 4) + 1 $\quad$ 2x + 1 $\quad$ z + 9 = 2(z + 4) + 1
z + 9 = 2z + 8 + 1
z + 9 - 1 = 2x
z + 8 = 2x
$\f... | |
Roll No.: 0 4
Page : 0 1
1) (a)
2) (d)
3) (a)
4) (c)
5) (c)
6)
7) $\angle BAC = 180 - (\angle ABC + \angle ACB)$
$= 180 - (69^\circ + 31^\circ)$
$= 180 - 100$
$= 80^\circ$
$\angle BAC = \angle BDC$ [Angles subtended by the same arc are equal]
8)
$\angle A + \angle B + \angle C = 180$ [WKT opposite angles are equal i... | |
Roll No: 3 8
Page : 0 2
2) $6 + \frac{2x}{3} + 8 = x = 14 + \frac{2x}{3} = 3x = 14 = 3x - 2x = 14 = x$
$x = 14$
$6 + \frac{2x}{3} = y = \frac{x}{3} + 4$
$y = \frac{14}{3} + 4$
$y = 5\dots + 4 \text{ (nearly)}$
$y = 9$
3) 15 apples + 10 oranges = $15x + 10y = 290$
12 apples + 18 oranges = $12x + 18y = 324$
$15... | |
Roll No.: 0 4
Page : 0 2
9) p(x-2) = 0 ; x=2
p(2) = g(2)
2(2)^3 + a(2)^2 + 3(2) - 5 = (2)^3 + a(2)^2 - 2(2) + a
16 + 4a + 6 - 5 = 8 + 4 - 4 + a
16 + 4a + 1 = 8 + a
4a - 1a = 8 - 17
3a = -9
a = -3
p(x) = 2x^3 + (-3)(x)^2 + 3(x) - 5
p(2) = 16 - 12 + 6 - 5 = 5
g(x) = x^3 + x^2 - 2x + (-3)
g(2) = 8 + 4 - 4 - 3 = 5
10)
... | |
Roll No.: 2 8
Page : 0 3
$\angle A + \angle B + \angle C = 180^\circ$
$90^\circ + \angle B + \angle B = 180^\circ$
$2\angle B = 180^\circ - 90^\circ$
$\angle B = \frac{90}{2}$
$\therefore \angle B = 45^\circ \quad \& \quad \angle C = 45^\circ$
$9. \quad 2x^3 + ax^2 + 3x - 5 \quad , \quad x^3 + x^2 - 2x + a$
$P(x) = 2... | |
Roll No.: 2 8
Page : 0 4
10.
i) $\Delta APB \cong \Delta AQB$
In $\Delta APB$
$AB = AB$ (common side) (S)
$\angle AQB = \angle APB (90^\circ)$ (A)
$\angle BAP = \angle QAB$ ($B$ is bisector) (A)
$\therefore \Delta APB \cong \Delta AQB$ by SAA congruency rule.
(ii) $BP = BQ$
From above solution by cpct
$BP = BQ$.... | |
Roll No.: 2 8
Page : 0 2
7. Given,
$\angle ABC = 69^\circ$; $\angle ACB = 31^\circ$
Find $\angle BDC = ?$
ABC is a triangle
Sum of angles in a triangle = 180^\circ
$\angle ACB + \angle ABC + \angle BAC = 180^\circ$
69 +
2 31 + 69 + $\angle BAC = 180^\circ$
$\angle BAC + 100 = 180^\circ$
$\angle BAC = 180 - 100$
= 80^... | |
Roll No. : 2 8
Page : 0 1
I
1. a
2. d
3. d a
4. d d
5. C
II
6. $a+b+c=5$ and $ab+bc+ca=10$, Given.
$a^3 + b^3 + c^3 - 3abc = -25$
WKT
$(a+b+c)^3 = a^3 + b^3 + c^3 - 3abc$
$= (a + b + c)^3 - 3abc$
$= (5)^3 - 3abc$
$= a^3 + b^3 + c^3 + 3ab + bc + ca - 3abc$
$= (a + b + c)^3 + 3(10) - 3abc$
$= (5)^3 + 30 - 3abc$
$= -25... | |
Roll No.: 1 4
Page : 0 4
10) Given
ACBD = quadrilateral
AC = AD
R.T.P :- $\Delta ABC \cong \Delta ABD$
construction :- Draw a line to
join CD
Proof :-
AC = AD (Given) (S)
AB = AB (Common) (S)
DC = DC (Common) (S)
$\therefore \Delta ABC \cong \Delta ABD$ by SSS congruency
rule
11)
$p(x) = x-1$
$2(1)^3 + 5(1)^2 - 5(1)... | |
Roll No.: 1 4
Page : 0 3
$p(x) = 2x^3 + (-3)(x)^2 + 3x - 5$
$p(2) = 16 - 12 + 6 - 5 = 5$
$g(x) = x^3 + x^2 - 2(x) + a$
$g(2) = 8 + 4 - 4 - 3 = 5$
Nill | |
Roll No.: 2 1
Page : 0 2
5Q
A) $3^{-7} \div 3^{-10} \times 3^{-5}$
=> $3^{-7-10} \times 3^{-5}$
=> $3^{-17} \times 3^{-5}$
=> $3^{-17+(-5)}$
=> $3^{-22}$
6Q
A) 10 9 5
60 cm x 5 cm x 30 cm
6 cm
=> 4500 450 cm^3
7Q
A) $6l^2 = 600 \text{ cm}^2$
=> $l^2 = \frac{600}{6}$
=> $l^2 = 100$
=> $l = \sqrt{100} \Rightarrow l = ... | |
Roll No.: 2 1
Page : 0 9
1Q)
A) $\frac{1}{2} \times d_1 \times d_2$
$\Rightarrow d_1 = 8 \text{ cm}$
$d_2 = 11 \text{ cm}$
$\Rightarrow \frac{1}{2} \times 8 \times 11$
$\Rightarrow 4 \times 11$
$\Rightarrow 44 \text{ cm}^2$
2Q)
A) $2a \times 2a \times 2a$
$\Rightarrow (2a)^3$
3Q)
A) $(-3)^{m+1} \times (-3)^5 = (-3)^... | |
Roll No.: 0 8
Page : 6 2
1 area of rhombus = $\frac{1}{2}(a \times b)$
$= \frac{1}{2}(8 \times 11)$
$= 4 \times 5.5 \quad \frac{1}{2}(88)$
$= 4.5 \text{ cm } 44 \text{ cm}$
2 $(2a)^3 = 8a^3$
3 $(-3)^{n+1} \times (-3)^5 = (-3)^7$
$(-3)^{n+1} \times (-3)^5 = (-3)^6$
$= (-3)^7 \div (-3)^6 = (-3)^1$
$= m = (-3)^{1+1}$
$... | |
Roll No.: 3 8
Page : 0 3
3) 5 notebooks + 3 pens = 190
Pens cost = y, note books cost = x
$5x + 3y = 190$ $\rightarrow$ (1)
3 notebooks + 2 pens = 118
$3x + 2y = 118$ $\rightarrow$ (2)
(1) - (2)
$5x + 3y = 190$
$(-) 3x + 2y = 118$
$2x + y = 72$ $\rightarrow$ (3)
(2) - (3)
$3x + 2y = 118$
... | |
Roll No.: 0 8
Page : 0 2
6 \frac{60^{10} \text{ cm} \times 8^9 \text{ cm} \times 30^5 \text{ cm}}{6}
= 10 \text{ cm} \times 4 \text{ cm} \times 5 \text{ cm} = 450 \text{ cm}
7 6a^2 = 600
a = \frac{600}{6} 100
a = 100 \text{ cm}^2
8 0.00007 = 7 \times 10^{-5}
9 3.06 \times 10^4 = 30600
10 5 \text{ books thickness} ... | |
Roll No.: 0 8
Page : 0 3
10 5 books thickness = 5 \times 2 = 100 \text{ nm}
5 paper sheets thickness in a book = 5 \times 0.016 = 0.080 \text{ nm}
25 paper sheets in 5 books = 0.080 \times 5 = 0.400
Total thickness of stack = 100 + 0.400 = 100.400
= 100.4 \times 10^{+3} | |
Roll No.: | 1 | 9 |
Page : | 0 | 1 |
1 The area of a rhombus is = 44cm
$\frac{1}{2} \times \overset{4}{8} \times 11$
$= 44 cm$
2 The volume of cube is
Edge is 2a
$(2a)^2$
3) M is = 1
M + 1 = 5 - 7
M = 5 - 7 - 1
M = 2 - 1 :
M = 1
4) The height of cuboid is 5
Edge is 2a
$h = (b + V) h = \frac{a}{b} = \frac{900^{10}}{... | |
Roll No.: 1 9
Page : 0 2
5 $3^{-7} \div 3^{-10} \times 3^{-5}$
$= \frac{3^{-7}}{3^{-10} \times 3^{-5}} = 3^{-2} = \frac{1}{9}$
6 $\frac{60 \times 54 \times 30}{6 \times 6 \times 6} \left(\frac{L \times b \times h}{L \times L \times L}\right)$
$= \frac{60 \times 54 \times 30}{6 \times 6 \times 6}$
$= 10 \times 9 \time... | |
Roll No. : | 1 | 0 |
Page : | 0 | 1 |
1. Area of rhombus = $d_1 \times d_2 \times \frac{1}{2} = 8 \times 11 \times \frac{1}{2}$
$= \frac{88}{2}$
$= 44 \text{cm}^2$
2. Volume of cube = $2a \times 2a \times 2a$
$= 8a^3$
3. $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$= (-3)^{m+6} = (-3)^7$
$= (-3)^m = (-3)^{7-6} = (-3)^1$
It ... | |
Roll No.: [1][0]
Page : [0][2]
7. T.S.A of cube $= 6l^2 = 600\text{cm}^2$
$= l^2 = 600 \div 6 = 100\text{cm}$
$= l = \sqrt{100} = 10$
The side of the cube $= 10\text{cm}$
8. $0.00007 = 7 \times 10^{-5}$
9. $3.06 \times 10^4 = 30600$
10. Total book thickness $= 5 \times 20\text{mm}$
$= 100\text{mm}$
Total paper shee... | |
Roll No.: 0 2
Page : 0 1
(1) Area of rhombus = $\frac{1}{2} \times d_1 \times d_2$
= $\frac{1}{2} \times 8 \times 11$
= 44cm
(2) volume of cube = $l^3$
edge of cube = $2a^3$
= 28a^3
= 2(2a)^3
(3) $(-3)^{1+1} \times (-3)^5 = (-3)^7$
$= (-3)^2 \times (-3)^5 = (-3)^7$
$a^{m+n} \times a^m$
(4) Given : Area = 80cm^2
vol... | |
Roll No. : 0 2
Page : 0 2
⑥ $2(lb + bh + hi)$
$2(60 \times 54 + 54 \times 30 + 30 \times 60)$
$2(3240 + 1670 + 1800)$
$2(6710)$
$13420$
$\Rightarrow l \times b = 13420$
$\Rightarrow l \times b = 13420$
$l = \frac{13420}{6}$
$l = 2236$
$\therefore$ Small Cubes with 6cm can be place in
the given Cuboid is $= 2236$
⑦ T.... | |
Roll No.: 0 3
Page : 0 1
1. A) area of rhombus = $\frac{1}{2}ab$
= $\frac{1}{2} \times 8 \times 11$
= 44 $\text{cm}^2$
2. A) volume of cube = s $\times$ s $\times$ s = $s^3$
= $(2a)^3$
= $8a^3$
3. A) $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$(-3)^{m+1} = \frac{(-3)^7}{(-3)^5}$
$(-3)^{m+1} = (-3)^2$
$\therefore$ we should... | |
Roll No.: 0 3
Page : 0 2
4. A) $h = \frac{1}{2}(a+b) \quad h = \frac{a}{b} \quad h = \frac{900}{180}$
$h = \frac{1}{2}(180+900) \quad h = \frac{900}{180} = 5$
$h = \frac{1}{2}(1080)$
$h = 540 \text{cm}$
5. A) $3^{-7} \div 3^{-10} \times 3^{-5}$
$= 3^3 \times 3^{-5}$
$= 3^{-2}$
$= \frac{1}{9}$
cubes
6. A) small $= \f... | |
Roll No.: 3 8
Page : 0 4
i) AP = BP
$\angle DAP = \angle EBP$
$\angle DPA = \angle EPB$
$\angle EPA + \angle EPD = \angle EPD + \angle DPB$
$\angle DPA = \angle EPD$
AS $\angle EPA = \angle DPB$ (Given)
$\angle DPE = \angle DPE$ (common)
$\angle DPA = \angle EPD$
$\Delta DAP \cong \Delta EPB$ (SAS ASA congruence rule)... | |
Roll No.: 03
Page : 03
7. A) Surface area of cube $= 6l^2$
$\Rightarrow 6l^2 = 600 \text{ cm}^2$
$\Rightarrow l^2 = \frac{600}{6}$
$\Rightarrow l^2 = 100$
$\Rightarrow l = \sqrt{100}$
$\Rightarrow l = 10$
8. A) 0.00007
$= \frac{7}{10^5} = 7 \times 10^{-5}$
9. A) $3.06 \times 10^4$
$= 30600$
10. A) thickness of 1 bo... | |
Roll No.: 0 4
Page : 0 1
8mmES
1) $\frac{1}{2} [d^1 \times d^2]$
$\Rightarrow \frac{1}{2} [8 \times 11]$
$\Rightarrow \frac{1}{2} \times \cancel{8}^4 \times 11$
$\Rightarrow 44 cm^2$
2) $2a^3$
3) $[-3]^{m+1} \times [-3]^5 = [-3]^7$
$\Rightarrow [-3]^{1+1} \times [-3]^5 = [-3]^7$
$\Rightarrow [-3]^2 \times [-3]^5 = ... | |
Roll No.: 0 4
Page : 0 1
8 MMES
$\Rightarrow 3^{-7-10} \times 3^{-5}$
$\Rightarrow 3^{+17} \times 3^{-5}$
$\Rightarrow 3^{+17+-5}$
$\Rightarrow 3^{22}$
6. $\frac{\overset{10}{60}\text{cm} \times \overset{9}{54}\text{cm} \times \overset{5}{30}\text{cm}}{6\text{cm}}$
450 cm are placed
in the given cuboid.
7. $\frac{6... | |
Roll No.: 0 3
Page : 0 3
(10) Thickness of 5 books in the stack =
20mm
Thickness of 5 paper in the stack =
0.016mm
Total Thickness of Stack in the
Standard form =
100 + 0.08
= 1.0008 \times 10^2
(11) d
(12) a
(13) d
(14) d
(15) c. | |
Roll No.: 0 6
Page : 0 1
(1) Area of rhombus = $\frac{1}{2} \times d, \times d_2$
=> $\frac{1}{2} \times 8^4 \times 11 = 44 \text{ cm}$
\framebox{Area of rhombus = 44 cm}
(2) Volume of cube = $l^3$
=> $2a \times 2a \times 2a$
\framebox{= 8a^3}
(3) $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$(-3)^{1+1} \times (-3)^5 = (-3)^... | |
Roll No.: 0 6
Page : 0 2
(6) Volume of cube = $l^3$
$\Rightarrow \frac{^{10}60cm \times ^{9}54cm \times ^{5}30cm}{_{1}6 \times _{1}6 \times _{1}6 cm}$
$\Rightarrow 10 \times 9 \times 5$
$= \boxed{450 cm}$
(7) T.S.A = $6l^2$
$\Rightarrow l^2 = \frac{600}{6} = 100$
$\Rightarrow \boxed{l^2 = 100}$
(8) 0.00007
$\frac{7}... | |
Roll No. : 0 9
Page : 0 1
1) $\left( \frac{1}{2} \times d_1 \times d_2 \right)$
$d_1 = 8 \text{cm}$
$d_2 = 11 \text{cm}$
$\frac{1}{2} \times 8 \times 11$
$= 44$
2) $(2a)^3$
$2a \times 2a \times 2a$
$= 8a^3$
3) $m+1+5 = 7$ $a^m \times a^n = a^{m+n}$
$m = 7 - 1 + 5$
$m = 7 - 6$
... | |
Roll No.: 0 9
Page : 0 2
5) $\frac{\cancel{60}^{10} \, \text{cm} \times \cancel{50}^{a} \, \text{cm} \times \cancel{30}^{5} \, \text{cm}}{\cancel{6}_{1} \, \text{cm} \times \cancel{5}_{1} \, \text{cm} \times \cancel{6}_{1} \, \text{cm}}$
$= 10 \times 9 \times 5 = 450 \, \text{cm}^3$
7) $6l^2 = 600$
$l^2 = \frac{600}{... | |
Roll No.: 1 1
Page : 0 1
1. Area of Rhombus = $\frac{1}{2} \times d_1 \times d_2$
$\Rightarrow \frac{1}{2} \times 8 \times 11$
$\Rightarrow 44 cm^2$
2. Given,
edge = $2a$
Volume of cube = $l^3$
$\Rightarrow (2a)^3$
$\Rightarrow 2a \times 2a \times 2a$
$\Rightarrow 8a^3$
3. $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$\Right... | |
Roll No.: 1 1
Page : 0 2
5. $3^{-7} \div 3^{-10} \times 3^{-5}$
Case 1
$\frac{3^{-7}}{3^{-10}} = 3^{+3}$
case 2
$3^{+3} \times 3^{-5} = 3^{-2} = \left(\frac{1}{3}\right)^2$
6. $\Rightarrow \frac{\text{volume of cuboid}}{\text{volume of cube}}$
$\Rightarrow \frac{\overset{10}{60} \times \overset{9}{54} \times \overset... | |
Roll No. : 4 5
Page : 0 1
I) Given
It is P is the mid point line segment
AB AP = BP
given
$\angle BAD = \angle ABE$
$\angle DAP = \angle EBP$
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\angle APD = \angle EPA + \angle EPD \quad \angle BPE = \angle DPB + \angle EPD$
$\therefore \angle APD = \angle BPE$
$AP = ... | |
Roll No.: [1] [1]
Page : [0] [3]
8. Standard form of 0.00007
= 7 \times 10^{-5}
9. 3.06 \times 10^4
= 3.06 \times 10000
= 30600
10. Thickness of 5 books = 5 \times 20
= 100 mm
Thickness of 5 paper sheets = 0.016 \times 5
= 0.08
Thickness of total books and = 100 + 0.08
paper = 100.08
Standard form = 10008 \times ... | |
Roll No.: 1 2
Page : 0 2
(6) Volume of Cubiod
Volume of cube
$= \frac{60 \times 54 \times 30}{6 \times 6 \times 6}$
$= 10 \times 9 \times 5$
$= 450 \text{ cubes}$
(7) Total Surface area of Cube $= 6l^2$
$600 = 6l^2$
$\frac{600}{6} = l^2$
$100 = l^2$
$\sqrt{100} = l$
$10 = l$
$\therefore L = 10 \text{ cm}$
(8) $0.000... | |
Roll No.: 1 2
Page : 0 1
(1) Area of Rhombus $= \frac{1}{2} \times d_1 \times d_2$
$= \frac{1}{\cancel{2}} \times \overset{4}{\cancel{8}} \times 11$
$= 44 \text{ cm}^2$
(2) Volume of Cube $= l^3$
$= (2a)^3$
$= 8a^3$
(3) $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$= (-3)^{m+1+5} = (-3)^7$
$= (-3)^{m+6} = (-3)^7$
$= (-3)^m =... | |
Roll No.: 1 2
Page : 0 3
(11) (d) = 20
(12) (a) = 6x^2 - 15x
(13) (a) = 154 \text{ cm}^2
(14) (d) = 1
(15) (c) = 4 | |
Roll No.: 1 3
Page : 0 1
1 Area of Rhombus = $\left( \frac{1}{2} \cdot d_1 \cdot d_2 \right) \text{ cm}^2$
$= \left( \frac{1}{2} \times 8 \times 11 \right) \text{ cm}^2$
$= (4 \times 11) \text{ cm}^2$
$= 44 \text{ cm}^2$
2 Volume of cube = $l^3$
$= (2a)^3$
$= 8a^3$
3 m = ?
$(-3)^{m+1} \times (-3)^5 = (-3)^7$
$\Right... | |
Roll No. : 1 3
Page : 0 2
5. $3^{-7} \div 3^{-10} \times 3^{-5}$
$= \frac{1}{3^7} \div \frac{1}{3^{10}} \times \frac{1}{3^5}$
$= \frac{1}{3^{7-10+5}}$
$= \frac{1}{3^2}$
$= \frac{1}{9}$
(or)
$= 3^{-2}$
6 Volume of cuboid = $60 \times 54 \times 30$
Volume of cube = $l^3 = 6 \times 6 \times 6$
no. of cubes can be placed... | |
Roll No.: 1 3
Page : 0 3
7 Surface Area $= 6l^2 = 600$
$\quad = l^2 = \frac{600}{6}$
$\quad = l^2 = 100$
$\quad = l = \sqrt{100}$
$\quad l = 10 \text{ cm}$
$\therefore$ Length of the side is $10 \text{ cm}$
8 $0.00007$
$\quad = \frac{7}{100000}$
$\quad = 7 \times 10^{-5}$
9 $3.06 \times 10^4$
$\quad = 3.06 \times 10... | |
Roll No.: 1 4
Page : 0 1
1. A. Area of rhombus = $\frac{1}{2} b_1 b_2 \text{ sq-units}$
= $\frac{1}{2} \cdot 8 \cdot 11$
= 44 cm
2. A. volume of cube = L X L X L
= 2a x 2a x 2a
= 8a^3 8a^2 + a
3. A. $(-3)^{m+1} \times (-3)^5 = (-3)^7$
= $(-3)^{m+5} = (-3)^7$
= m = 1
3
4. A. $\frac{180}{90} = \frac{3}{2}$
2 | |
Roll No.: 1 4
Page : 0 2
5. A. $3^{-7} \div 3^{-10} \times 3^{-5}$
$= 3^{-3} \times 3^{-5}$
$= 3^{+8}$
$= 6$
6. A. Area of cuboid = $L \times b \times h$
$= 60 \times 54 \times 30$
$= 1800 \times 54$
$= \frac{97200}{6}$
$= 151200$
7. A. cube Area = $L \times L \times L$
$= 600$
8. A. $7 \times 10^{-5}$
9. A. 30600... | |
Roll No.: 1 5
Page : 0 1
1) Area of rhombus = $A = \frac{1}{2} d_1 d_2$
$A = \frac{1}{2} 8\text{cm}, 11\text{cm}$
$A = 44\text{cm}$
2) Volume of cube = $l^3$
$2a \times 2a \times 2a$ or $= (2a)^3$
$= \boxed{8a^3}$
3) $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$(-3)^{4+1} \times (-3)^5 = (-3)^7$
4) $900 = \frac{1}{2} (180 ... | |
Roll No.: 4 5
Page : 0 2
ii) proof that $\Delta ABP \cong \Delta ACP$
Given
AB = AC
$\angle BAP = \angle CAP$
$\angle BAP = \angle CAP (CPCT)$
AP = AP (common side)
$\Delta ABP \cong \Delta ACP (SAS)$
iii) prove that AP bisects
$\Delta ABD \cong \Delta ACD$
$\angle BAD = \angle CAD (CPCT)$
this means
$\angle BAP = \a... | |
Roll No.: 1 5
Page : 0 2
6) Volume of cube = $l^3$
$10 \quad 9 \quad 5$
$60 \text{cm} \times 54 \text{cm} \times 30 \text{cm}$
$\cancel{6} \times \cancel{6} \times \cancel{6} \text{cm}$
$10 \times 9 \times 5$
$450 \text{cm}^3$
7) T.S.A = $6l^2$
$l^2 = \frac{600}{6}$
$l^2 = 100$
8) $0.00007$
$\frac{7}{10^5} = 7 \t... | |
Roll No.: 1 6
Page : 0 1
① $\frac{1}{2} \times d_1 \times d_2$
$\frac{1}{2} \times 8 \times 11$
$= 44 \text{cm}$
② $8a^3$
③ $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$(-3)^{7-5-1} = (-3^{m+1})$
$(-3)^1 = (-3^{m+1})$
$m = 1$
④ $\frac{h}{2} (a+b)$
$h = (a+b) \times 2$ $\frac{h}{2} (180+900)$
$= (180+900) \times... | |
Roll No.: 1 6
Page : 0 2
(6) 450
(7) 10
(8) 7 \times 10^{-5}
(9) 30600
(10) (5 \times 20) + (5 \times 0.016)
= 100 + 0.080
= 100.080
= 100 100 \times 8 \times 10^{-2}
(11) d
(12) A
(13) A
(14) D
(15) D | |
Roll No.: 1 7
Page : 0 1
Question - 1
$= \frac{1}{2} \times d_1 \times d_2$
$= \frac{1}{2} \times 8 \times 11$
$= 44$
Question - 2
$= l^3$
$= l = 2a$
$= (2a)^3$
$= 8a^3$
Question - 3
$= (-3)^{m + 1 + 5} = (-3)^7$
$= (-3)^{m + 6} = (-3)^7$
$= (-3)^m = (-3)^{7 - 6}$
$= (-3)^m = (-3)^1$
Question - 4
$= l \times b ... | |
Roll No.: | 1 | 7 |
Page : | 0 | 2 |
Question-5
$3^{-7} \div 3^{-10} \times 3^{-5}$
$= 3^{-7-10} \times 3^{-5}$
$= 3^{-17} \times 3^{-5}$
$= 3^{-22}$
Question-6
vol. of cuboid = l x b x h
vol. of cube = l x l x l
$= \frac{l \times b \times h}{l \times l \times l}$ $l = 60$ $l = 6$
... | |
Roll No.: 1 7
Page : 0 3
Question - 8
0.00007
= \frac{7}{100000}
= 7 \times 10^{-5}
Question - 9
3.06 \times 10^{4}
= 3.0600.0 \times 10000
= 30600
Question - 10
Thickness of one book = 20 mm
Thickness of five books = 100 mm
Thickness of one paper = 0.016
Thickness of five papers = 0.080
Total thickness = 100 + 0.08... | |
Roll No.: 2 0
Page : 0 1
1) Area of rhombus = $\frac{1}{2} d_1 d_2$
$= \frac{1}{\cancel{2}_1} \times \cancel{8}^4 \times 11$
$= 44 \text{ cm}$
2) Let 1 edge of cube be $2a^2$
If we double it we get $4a$
If we double it we get $16a$
The volume of cube is $16a$
3) $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$= (-3)^{m+1} = (-... | |
Roll No.: 2 0
Page : 0 2
5) $3^{-7} \div 3^{-10} \times 3^{-5}$
$= \frac{1}{3^7} \div \frac{1}{3^{10}} \times \frac{1}{3^5}$
$= \frac{1}{3^{-3}} \times \frac{1}{3^5}$
$= \frac{1}{3^2} = 3^{-2}$
6) Surface area of cuboid = $2(lb + lh + bh)$
$= 2(60 \times 54 + 60 \times 30 + 54 \times 30)$
$= 2(3240 + 1800 + 1620)$
$=... | |
Roll No.: | 2 | 0 |
Page : | 0 | 3 |
7) Surface area of cube = 600 \text{ cm}^2
8) 0.00007
= \frac{7}{10^5}
= 7 \times 10^{-5}
9) 3.06 \times 10^4
= 3.06 \times 10000
= 30600
10) Total thickness of 5 books = 20 \times 5 = 100 \text{ mm}
Total thickness of 5 papers heets = 0.016 \times 5 = 0.08
Total thickness of st... | |
Roll No.: 2 2
Page : 0 1
① Area of Rhombus = $\frac{1}{2} d_1 d_2$ sq.units
= $\frac{1}{2} (8 \times 11)$
= $\frac{1}{2} 88$
= $\frac{88}{2}$
= $44 \text{ cm}^2$
② Volume of cube = $s \times s \times s$
= $2a \times 2a \times 2a$
= $8a^3$
③ $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$m = 7 - 1 - 5$
$m = 1$
④ Volume of cub... | |
Roll No. : 4 5
Page : 0 3
DP = DP (common side)
SSS congurenc rule
ΔBDP ≅ ΔCDP
∠BDP = ∠CDP (CPCT)
∴ AP bisects ∠D
iv) AP bisects BC :
ΔACP ≅ ΔABP
BP = CP
This shows that P is mid point of BC
∴ AP bisects BC
AP perpendicular to BC
ΔABP ≅ ΔACP
∠APB = ∠APC (CPCT)
BC is a strign line
There fore
∠APB + ∠APC = 180°
2 ∠APB =... | |
Roll No.: $\boxed{2}\boxed{2}$
Page : $\boxed{0}\boxed{2}$
(6) $3^{-7} \div 3^{-10} \times 3^{-5}$
$= \frac{1}{3^7} \div \frac{1}{3^{10}} \times \frac{1}{3^5}$
$= \frac{1}{3^7} \times \frac{3^{10}}{1} \times \frac{1}{3^5}$
$= \frac{3^{10}}{3^7 \times 3^5}$
$= 3^{-2}$
$= \frac{1}{3^2}$
(6) Volume of cuboid $= l \times... | |
Roll No.: 2 2
Page : 0 3
(7) T.S.A of cube = $6l^2$
$6l^2 = 600$
$l^2 = \frac{600}{6}$
$l^2 = 100$
$l = \sqrt{100}$
$l = 10 \text{ cm}$
(8) $0.0002 \times (10^5)$
$2 \times 10^1$
(9) $3.06 \times 10^4$
$= 0.0306$
(10) 1 book's thickness = 20 mm
5 book's = $5 \times 20 = 100$
$100 = 1 \times 10^2$
1 papers thickness... | |
Roll No.: 2 4
Page : 0 1
8cm and 11cm
① $\frac{1}{2}(8 \times 11)$
$= \frac{1}{2}(8 \times 11)$
$= \frac{1}{2}(4 \times 11)$
$= 1(44)$
$= 44\text{cm}^2$
② $(8a)^3$
③ $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$m+1 = 5-7$
$m = 5-7-1$
$m = 2-1$
$m = 1$ | |
Roll No.: 2 4
Page : 6 2
(4)
l x b x h
height
volume of cuboid = base area of cuboid
height = 900 / 180
= 5 cm^2
(5)
3^{-7} \div 3^{-10} \times 3^{-5}
= 3^{-17} \times 3^{-5}
= 3^{-22}
(6)
volume of cuboid = l \times b \times h
= 60 \times 54 \times 30
= 97,200 \text{ cm}^2
cube = 6 cm
= 6^3
= 216 \text{ cm}^3
Ho... | |
Roll No.: 2 4
Page : 0 3
(8) $0.00007 = 7 \times 10^{-5}$
(9) $3.06 \times 10^4 = 30,600$
(10) 5 books of 20 mm thickness each
5 sheets of paper 0.016 mm each
The thickness of 5 books = $5 \times 20 = 100 \text{ mm}$
The thickness of 5 sheets = $0.016 \times 5 = 0.080$
the to thickness of the stack = $0.080 + 100... | |
Roll No.: 2 5
Page : 0 3
9 Usual form = 3.060000
= 30600
10 1 book thickness = 20 mm
1 sheet thickness = 0.016 mm
thickness of stack = 5 x 20 x 0.016
= 100 x 0.016
= 1.6 mm
11 (d)
12 (a)
13 (a)
14 (d)
15 (c) | |
Roll No.: 2 5
Page : 0 1
1 Area of rhombus
$= \frac{1}{2} \times d_1 \times d_2$
$= \frac{1}{2} \times 8 \times 11$
$= 44 \text{ cm}^2$
2 Volume of cube = $l^3$
$= 2a \times 2a \times 2a$
$= (2a)^3$
3 $(-3)^{m+1} \times (-3)^5 = (-3)^7$ $(a^m + a^n = a^{m+n})$
$m = (-3)^1 \times (-3)^5 \div (-3)^7$
$m = (-3)^6 \div ... | |
Roll No.: | 2 | 5 |
Page : | 0 | 2 |
5 $3^{-7} \div 3^{-10} \times 3^{-5}$
$= 3^{-7-10} \times 3^{-5}$
$= 3^{-17} \times 3^{-5}$
$= 3^{-22}$
$= \frac{1}{3^{22}}$
6. No. of Small cubes = $\frac{\overset{10}{60}\text{cm} \times \overset{9}{54}\text{cm} \times \overset{5}{30}\text{cm}}{\underset{1}{6}\text{cm} \times \u... | |
Roll No.: 2 6
Page : 0 1
(1) area of rhombus = $\frac{1}{2}(d_1d_2)$
= $\frac{1}{2} \times 8 \times 11$
= $44\text{ cm}$
(2) Volume of Cube = $(2a)^3$
= $8a$
(3) $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$(-3)^{m+5} = (-3)^7$
$(-3)^{m+6} = (-3)^{76}$
$(-3)^m = (-3)^1$
$m = 1$
(4) height of cuboid = $L \times b \times h$
... | |
Roll No.: 2 6
Page : 0 2
(5) $3^{-7} \div 3^{-10} \times 3^{-5}$
$\frac{1}{3^7} \div \frac{1}{3^{10}} \times \frac{1}{3^5}$
$= \frac{1}{3^{-3}} \times \frac{1}{3^5}$
$= \frac{1}{3^{-8}}$
(6) $\frac{10 \times 7 \times 5}{6 \times 6 \times 6} \quad$ Volume of Cuboid
$\quad \quad \quad \quad \quad \quad \quad \quad$ Sid... | |
Roll No.: 4 8
Page : 0 2
3) cost of 15 apples and 10 oranges = 290
cost of 12 apples and 18 oranges = 324
the cost of x, y =
x = apple
y = oranges
total x = 15+12 = 27, total y = 28
the vales are =
the cost of Apples = 574
the cost of orangs = 140 .
4) 5x and 3y = 190
3x and 2y = 118
x, y =
x =
1 Book = 30
1 Pen = 20... | |
Roll No.: 2 6
Page : 0 3
(10) Thickness of Books = $5 \times 20 = 100$
Thickness of paper sheets = $5 \times 0.016 = 0.08$
Total thickness = $100 + 0.08$
$= 100.08 = 1.0008 \times 10^2$
$= 1.0008 \times 10^2$
(11) (d) 20
(12) (a) $6x^2 - 15x$
(13) (a) $154 \text{ cm}^2$
(14) (b) $1+1+1$
(15) (c) 4 | |
Roll No.: 2 9
Page : 0 1
1) Area of Rhombus = $\frac{1}{2} d_1 d_2$
= $\frac{1}{2} \times 8 \times 11$
= $44 \text{ cm}^2$
2) Volume of cube = $s^3$
= $(2a)^3$
= $8a^3 \text{ cm}^3$
3) $(-3)^{m+1} \times (-3)^5 = (-3)^7$
$\Rightarrow (-3)^{m+1+5} = (-3)^7$
$\Rightarrow (-3)^{m+6} = (-3)^7$
$\Rightarrow 7 - 6 = m$
$\... | |
Roll No.: 2 9
Page : 0 2
4) Volume of cuboid = $l \times b \times h$
900 = 180 \times h
$\frac{900}{180} = h$
h = 5
5) $3^{-7} \div 3^{-10} \times 3^{-5}$
$\Rightarrow \frac{1}{3^7} \div \frac{1}{3^{10}} \times \frac{1}{3^5}$
$\Rightarrow \frac{1}{3^7} \times \frac{3^{10}}{1} \times \frac{1}{3^5}$
$\Rightarrow \frac{... | |
Roll No.: 2 9
Page : 0 3
6) Volume of Cuboid = $l \times b \times h$
= $60 \times 54 \times 30$
Volume of Cube = $s^3$
= $6 \times 6 \times 6$
No. of cubes in cuboid = $\frac{60 \times 54 \times 30}{6 \times 6 \times 6}$
= $10 \times 9 \times 5$
= $90 \times 5$
= $450$ cubes 4
7) T.S.A of cube = $6l^2$
$6l^2 = 6... | |
Roll No.: 2 9
Page : 0 4
8) $0.00007$
$\Rightarrow \frac{7}{100000}$
$\Rightarrow \frac{7}{10^5}$
$\Rightarrow 7 \times 10^{-5}$
9) $3.06 \times 10^4$
$\Rightarrow 30600$
10) Thickness of 1 book = $20\,\text{mm}$
Thickness of 5 books = $20 \times 5$
$= 100\,\text{mm}$
Thickness of 1 paper sheet = $0.016\,\text{mm}... | |
Roll No.: 3 0
Page : 0 1
1) $\frac{1}{2} [b_1 \times b_2]$
$\frac{1}{2} [8 \times 11]$
$\frac{1}{2} \times 8 \times 11$
$4 \times 11 = 44$
2) $2 \times a^2$
$\Rightarrow$
3) $(-3)^{m+1} \times (-3)^5 = -3^7$
$5 + 1 - 7 = m$
$6 - 7 = 1$
$m = 1$
4) $\frac{1}{2} \times (l + b)$
$\frac{180}{2} (l \times 90)$
$l = 10$
... | |
Roll No.:
Page :
7) $S \times 2.\text{th ickness} = 20 \times 5$
total thickness $= 100+0.08 \quad 5 \times 100 + 0.08$
$0 = 1.0008 \times 10^2$
$=$
8) $0.00007$
$70000 \quad 7 \times 10^7$
9) $3.06 \times 10^4$
$30.6000 \quad 30 \quad 600 \quad 30600$
10) $20 \text{ is to } 60$
11) d
12) a
13) a
14) d
15) c | |
Roll No.: 3 2
Page : 0 1
1. Area of rhombus = $\frac{1}{2} \times (a+b)$
$= \frac{1}{2} (8 \text{cm} + 11 \text{cm})$
$= \frac{1}{2} (19 \text{cm})$
$= \frac{1}{2} \times 19$
2. volume of a cube = $l^3$
$\Rightarrow 2a \times 2a \times 2a$
$\Rightarrow 8a^3$
3. $(-3)^{m+1} + (-3)^5 = (-3)^7$
$\dots... | |
Roll No.: 3 2
Page : 0 2
6. $\frac{\overset{10}{60}\text{cm} \times \overset{9}{54}\text{cm} \times \overset{5}{30}\text{cm}}{6 \times 6 \times 6}$
$\Rightarrow 10 \times 9 \times 5$
$= 90 \times 5$
$= 450 \text{cm}^3$
5. $\frac{3^{-7} \div 3^{-10} \times 3^{-5}}{3^{-3} \times 3^{-5}}$
$= 3^{-8}$
7. Total surface ar... | |
Roll No.: 3 2
Page : 0 3
11. d
12. 3x(2a-5)
6x^2 - 15x
option [a]
13. a
14. d
15. C | |
Roll No. : 4 8
Page : 0 1
1) Time taken by T-1 to fill tank = 4
Time taken by T-2 to fill tank = 9
the value of $y = 2x + 1$
$y = 2(4) + 1$
$y = 8 + 1$
$y = 9h$
$y = 2x + 1$
$9 = 2x + 1$
$9 = 3x$
$x = 9 - 3$
$= 6h$
the value of $x = 6$, $y = 9$
time taken by both taps = $9 - 6$
$= 3h$
2) Values of $x, y$
if Father gi... | |
Roll No.: 3 3
Page : 0 1
1. area of rhombus = $\frac{1}{2} \cdot d_1 \cdot d_2$
= $\frac{1}{2} \cdot 8 \cdot 11$
= $44 \text{ cm}^2$
2. volume of cube = $l^3$
= $(20)^3$
= $8000$
3. m = ?
$(-3)^{m+1} \times (-3)^5 = (-3)^7$ $\quad$ $(a^m \cdot a^n = a^{m+n})$
m = 1
4. area of a Cuboid = $180 \text{ cm}^2$
area ,, v... | |
Roll No.: 3 3
Page : 0 2
6) Cuboid of demensions $= 60 \text{cm} \times 54 \text{cm} \times 30 \text{cm}$
given cuboid $= 6$
$= \frac{60 \times 54 \times 30}{6}$
$= 450 \text{cm}^2$
7) area of a cube $= 600 \text{cm}^2$
$6l^2 = 600$
$l^2 = \frac{600}{6}$
$l = \sqrt{100}$
$l = 10$
8) : $0.0000\ 7$
$= 7 \times 10^6... | |
Roll No.: 3 1
Page : 0 1
1sol: Area of rhombus $\Rightarrow \frac{1}{2} \times d_1 \times d_2$
4cm
$\Rightarrow \frac{1}{2} \times 8cm \times 11cm$
$\Rightarrow 4cm \times 11cm$
$\Rightarrow 44cm^2$
2sol: Volume of cube $\Rightarrow l \times l \times l \Rightarrow l^3$
$\Rightarrow 2a \times 2a \times 2a$
$\Rightarro... | |
Roll No. : [3] [9]
Page : [0] [2]
6 sol : $\Rightarrow \frac{\overset{10}{60 \text{ cm}} \times \overset{9}{54 \text{ cm}} \times \overset{5}{30 \text{ cm}}}{6 \text{ cm} \times 6 \text{ cm} \times 6 \text{ cm}}$
$\Rightarrow 10 \text{ cm} \times 9 \text{ cm} \times 5 \text{ cm}$
$\Rightarrow 90 \text{ cm} \times 5 \t... | |
Roll No.: 3 1
Page : 6 3
11 . C
12 . A
13 . A
14 . D
15 . A
10010 (i) $\Rightarrow 0.016 \times 5$
$= 0.080 \text{ mm}$
(ii) $\Rightarrow 20 \times 5$
$\Rightarrow 1 \text{ mm}$
(iii) $\Rightarrow 0.080 \text{ mm} + 1 \text{ mm}$
$\Rightarrow 1.080 \text{ mm}$
(iv) $\Rightarrow 1.08 \times 10^1 \times 10^1$
$\Rightar... | |
Roll No.:
Page :
$\overset{3}{0.16} \quad \overset{3}{0.16}$
$\times 5 \quad \times 5$
$\overline{0.80} \quad \overline{0.80}$
$\quad \times 20$
$\quad 0000$
$- \overset{3}{0.160} <$
$\quad \overset{3}{0.1600}$
$\quad \times 5$
$\overline{\quad 0.8000 \text{ mm}}$ | |
Roll No.: 2 3
Page : 0 1
(1) Question 1
8cm and 11cm
Sol: = 8cm and 11cm
= l \times b
= \frac{8 \times 11}{2}
= \frac{88\text{cm}}{2}
= 44\text{cm}
Question 2
2a \times 2 \times a \times 2a = 8a
(3) Question 3
(-3)^{m+1} \times (-3)^5 = (-3)^7
= (-3)^{1+1} \times (-3)^5 = (-3)^7
= (-3)^2 \times (-3)^5 = (-3)^7
(4) ... | |
Roll No.: 3 3
Page : 0 2
Question 5
$3^{-7} \div 3^{-10} \times 3^{-5}$
$= 3^{-7 - (-10)}$
$= 3^{-17} \times 3^{-5}$
$= 3^{-17 + (-5)}$
$= 3^{-22}$
Question 6
$\frac{60 \times 50 \times 30}{6}$
So, 6 is is the smallest cube -> X
6 can be placed in all the given cuboide
Question 7
$\frac{600}{100} = 100 \text{ cm}^2$... | |
Roll No.: 2 1
Page : 0 3
10Q)
A) (5 \times 20) (5 \times 0.016)
=) 100 \times 0.0530
=) 0.0600
=) 600 \times 10^{-2}
II
11Q
A) d
12Q
A) a
13Q
A) a
14Q
A) d
15Q
A) d | |
(8) Given $HCF(306, 657) = 9$
Now , we need to find $LCM(306, 657)$
we know that
$HCF \times LCM = Product of two numbers$
$9 \times LCM(306, 657) = 306 \times 657$
$LCM(306, 657) = [(306) \times (657)] / 9$
$= 210042 / 9$
$= 33228$
$\therefore LCM(306, 657) = 33228$
(9) If any number ends with the digit 0 , it must b... | |
Roll No. : 4 8
Page : 0 3
T2 AB = AC
DB = DC
AD = AD
$\Delta ACD$ (SSS)
$\Delta ABD \cong \Delta ACD$
$\angle BAD = \angle CAD$ (CPCT)
AD = AD
(SSS) there fore $\Delta ABD \cong \Delta ACD$
Thuse, AD bisects both $\angle A$ and $\angle D$
2 $\angle ADB = 180^\circ$
$\angle ADB = \frac{180^\circ}{2} = 90^\circ$
AD $\p... | |
(4)
(1) The purpose of this activity is to study the nature, pasition and
relative size of the image formed by a convex lens
when an object is placed at different positions in front
of it.
(2) First, we take a convex lens and find its approximate
focal length. Using chalk, we draw five parallel straight
lines on a lon... | |
(3) Answer - (A) At the principal focus of the lens
(9) If any number ends with the digit odd , it must be divisible
by 11
That is , it must be divisible by both 2 and 7
Prime factorisation of $6^{n+1}$ is
$6^{n+1} = (2 \times 3)^{n+1} = 2^{n+1} + 3^{n+1}$
$\therefore 6^{n+1}$ is divisible by 5 and 6 cannot
end with d... | |
③ Answer - (b) At twice the focal length
When an object is placed at $2F_1$ of a convex lens, the
image formed is :
Position of Image : At $2F_2$
Size of Image : Same size as the object
Nature of Image : Real and Inverted.
⑧ Given $HCF(306, 657) = 9$
Now, we need to find $LCM(306, 657)$
We know that
$HCF \times LCM = ... | |
(17) A number of rays perpendicular to the principal axis are falling on a convex mirror
(18) The power of lens is defined as
$P = \frac{1}{v}$
(19) A number of rays parallel to the principal axis are falling on a concave mirror . After reflection from the mirror , all these parallel rays meet at point on the princip... |
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