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Paper - II
MAHESH TUTORIALS
S.S.C.
Batch : SB
Date :
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Time : 1 hr. 15 min.
Q.1. Solve the following :
4
1.2
0.57
0.03
– 0.23
(i)
Find the value of the following determinants :
(ii)
What is the equation of X - axis? Hence, find the point of intersection of
the graph of the equation x + y = 3 with the X - axis.
Q.2. Solve the following :
9
(i)
The length of the rectangle is greater than its breadth by 2 cm. The area
of the rectangle is 24 sq.cm, find its length and breadth.
(ii)
Solve the following simultaneous equations using Cramer’s rule :
y=
(iii)
5x – 10
; 4x + 5 = – y
2
The sum of a natural number and its reciprocal is
10
. Find the number.
3
Q.3. Solve the following : (Any Three)
(i)
The sum of the areas of two squares is 400sq.m. If the difference
between their perimeters is 16 m, find the sides of two square.
12
Paper - II
... 2 ...
(ii)
Solve the following simultaneous equations using graphical method :
4x = y – 5; y = 2x + 1
(iii)
The divisor and quotient of the number 6123 are same and the remainder
is half the divisor. Find the divisor.
(iv)
Durga’s mother gave some 10 rupee notes and some 5 rupee notes to
her, which amounts to Rs. 190. Durga said, ‘if the number of 10 rupee
notes and 5 rupee notes would have been interchanged, I would have
Rs. 185 in my hand.’ So how many notes of rupee 10 and rupee 5 were
given to Durga ?
Q.4. Solve the following : (Any One)
5
(i)
The cost of bananas is increased by Re. 1 per dozen, one can get 2 dozen
less for Rs. 840. Find the original cost of one dozen of banana.
(ii)
Students of a school were made to stand in rows for drill. If 3 students
less were standing in each row, 10 more rows were required and if 5
students more were standing in each row then the number of rows was
reduced by 10. Find the number of students participating in the drill.
Best Of Luck

Paper - II
S.S.C.
MAHESH TUTORIALS
Batch : SB
Test - III
Marks : 30
ALGEBRA – Chapter : 2, 3
Date :
MODEL ANSWER PAPER
Time : 1 hr. 15 min.
A.1. Solve the following :
1.2
0.03
(i)
0.57 – 0.23
= (1.2 × – 0.23) – (0.03 × 0.57)
= – 0.276 – 0.0171
= – 0.2931
(ii)
The equation of X-axis is y = 0
Let the point of intersection of graph x + y = 3 with X-axis be (h, 0)
 (h, 0) lies on the graph, it satisfies the equation
 Substituting x = h and y = 0 in the equation we get,
h + 0= 3

h = 3
1
1
1
1
 The line x + y = 3 intersects the X-axis at (3, 0).
A.2. Solve the following :
(i)
Let the breadth of the rectangle be ‘x’ cm.
 It’s length is (x + 2) cm.
As per the given condition.

Area of rectangle = Length × Breadth

24 = (x + 2) × x

24 = x2 + 2x

0 = x2 + 2x – 24
2

x + 2x – 24 = 0

x2 – 4x + 6x – 24 = 0

x (x – 4) + 6(x – 4) = 0

(x – 4) (x + 6) = 0

x – 4 = 0 or x + 6 = 0

x = 4 or x = -6
 The breadth of rectangle cannot be negative.
 x  -6
Hence x = 4 and x + 2 = 4 + 2 = 6
 The length of rectangle is 6 cm and its breadth is 4 cm.
1
1
1
Paper - II
... 2 ...
(ii)
5x – 10
2
2y
– 5x + 2y
4x + 5
4x + y
y=



D
=
Dx =
Dy =
=
=
=
=
–5 2
4
1
–10 2
–5 1
– 5 –10
4
–5
5x – 10
– 10
–y
–5
= (– 5 × 1) – (2 × 4)
1
= –5–8
= – 13
= (– 10 × 1) – (2 × – 5) = – 10 + 10 = 0
= (– 5 × – 5) – (– 10 × 4 ) = 25 + 40
= 65
1
By Cramer’s rule,
x
=
y
=
Dx
D
Dy
D
0
= –13 = 0
65
= –13 = – 5
 x = 0 and y = – 5 is the solution of given simultaneous equations.
(iii)









Let natural number be ‘x’
1
Its reciprocal is
x
From the given condition,
1
10
x+
=
x
3
Multiplying throughout by 3x,
3x2 + 3 = 10x
3x2 – 10x + 3 = 0
3x2 – 9x – x + 3 = 0
3x (x – 3) – 1 (x – 3) = 0
(3x – 1) (x – 3) = 0
3x – 1 = 0
or x – 3 = 0
3x = 1
or x = 3
1
x=
or x = 3
3
1
x 
because x is natural number
3
x=3
 The natural number is 3.
1
1
1
1
... 3 ...
A.3.
(i)
Solve the following : (Any Three)
Difference between the perimeters of two squares is 16 m
 The difference between the sides of the same squares is 4 m
Let the side of smaller square be x cm
 The side of bigger square is (x + 4) cm

Area of square = (side) 2
As per the given condition,
x2 + (x + 4)2 = 400
 x2 + x2 + 8x + 16 – 400 = 0

2x2 + 8x – 384 = 0
Dividing throughout by 2 we get,
x2 + 4x – 192 = 0
2

x – 12x + 16x – 192 = 0
 x (x – 12) + 16 (x – 12) = 0

(x – 12) (x + 16) = 0

x – 12 = 0 or x + 16 = 0

x = 12 or x = – 16
 The side of square cannot be negative
 x  – 16
Hence x = 12
 x + 4 = 12 + 4 = 16
 The side of smaller square is 12 m and bigger square is 16 m.
(ii)
Paper - II
1
1
1
1
4x = y – 5
 4x + 5 = y
 y = 4x + 5
x
0
–1
–2
y
5
1
–3
(x, y)
(0, 5) (–1, 1) (–2, –3)
y = 2x + 1
x
y
(x, y)
0
1
2
1
3
5
(0, 1) (1, 3) (2, 5)
1
Paper - II
... 4 ...
Y
Scale : 1 cm = 1 unit
on both the axes
6
(0, 5)
5
2
(2, 5)
4
(1, 3)
3
2
(–1, 1)
-5
-4
-3
-2
(0, 1)
0
-1
1
2
3
4
5
X
-1
-2
(–2, –3)
-3
-4
-5
y=
2x
+1
X
1
-6
-7
-8
Y
 x = – 2 and y = – 3 is the solution of given simultaneous equations.
1
... 5 ...
(iii)
Let divisor of 6123 be ‘x’
 Divisor = Quotient
[Given]
 Quotient = x
x
[Given]
Remainder =
2
We know,
1
Dividend = Divisor × Quotient + Remainder
x
6123 = x . x +
2
Multiplying throughout by 2,
2 (6123) = 2x2 + x

12246 = 2x2 + x
2

2x + x – 12246 = 0
 2x2 + 157x – 156x – 12246 = 0
1
 x (2x + 157) – 78 (2x + 157) = 0

(2x + 157) (x – 78) = 0
 2x + 157 = 0
or
(x – 78) = 0
 2x + 157 = 0
or
x – 78 = 0
– 157
 x=
or
x = 78
1
2
– 157
x=
cannot be acceptable because divisor cannot be negative.
2
 x = 78
 The divisor of 6123 is 78.
(iv)
Paper - II
Let the no. of Rs. 10 notes given to Durga be x and the no. of
Rs.5 notes given to her be y.
As per the first condition,
10x + 5y
=
190
.......(i)
As per the second condition,
5x + 10y
=
185
......(ii)
Adding (i) and (ii),
15x + 15y
=
375
Dividing throughout by 15 we get,
375
x+y
=
15

x+y
=
25
......(iii)
Subtracting (ii) from (i),
5x – 5y
=
5
Dividing throughout by 5 we get,
x–y
=
1
.......(iv)
Adding (iii) and (iv),
1
1
1
... 6 ...




x+y
=
x–y
=
2x
=
x
=
Substituting x = 13 in
13 + y
=
y
=
y
=
25
1
26
13
(iii),
25
25 – 13
12
 Durga had 13 notes of Rs. 10 rupee and 12 notes of Rs. 5.
A.4. Solve the following : (Any One)
(i)
Let the cost of banana per dozen be Rs. x.
Amount for which bananas are bought = Rs. 840
840
No. of dozens of bananas for Rs 840 is x
New cost of banana per dozen = Rs. (x + 1)
840
New No. of dozens of bananas for Rs 840 = x  1
As per the given condition,
840
840
–
2
x
x 1 =












1
1 
840  –
x  1
x
=
x  1 – x
840 
=

 x (x  1) 
 1 
840  2
 =
x  x 
840 =
2x2 + 2x – 840 =
Dividing throughout by 2,
2x2 + x – 420 =
2
x – 20x + 21x – 420 =
x (x – 20) + 21 (x – 20) =
x – 20 = 0 or
x = 20 or
The cost of bananas cannot
x  –21
Hence x = 20
Paper - II
1
1
1
0
2
1
2
2 (x2 + x)
0
0
0
0
x + 21 = 0
x = –21
be negative.
 The original cost of one dozen banana is Rs. 20.
1
1
1
... 7 ...
(ii)
Let the no. of students standing in each row be x and let no. of
rows be y.
 Total no. of students participating in the drill = xy
As per the first given condition,
(x – 3) (y + 10)
=
xy
 x (y + 10) – 3 (y + 10) =
xy

xy + 10x – 3y – 30
=
xy

10x – 3y
=
30
.......(i)
As per the second given condition,
(x + 5) (y – 10)
=
xy
 x (y – 10) + 5 (y – 10) =
xy

xy – 10x + 5y – 50
=
xy

– 10x + 5y
=
50
......(ii)
Adding (i) and (ii),
10x – 3y
=
30
– 10x + 5y
=
50
2y
=
80
80

y
=
2

y
=
40
Substituting y = 40 in (i),
10x – 3 (40)
=
30

10x – 120
=
30

10x
=
30 + 120

10x
=
150
150

x
=
10

x
=
15

xy = 15 × 40 = 600
 600 students were participating in the drill.

Paper - II
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1
1
1
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