RAY Tutorials
ICSE Class 10 Mathematics - Complete Step by Step Solutions
SECTION A - MCQ SOLUTIONS
1. Raghu deposited ₹800 per month in a cumulative deposit account of Bank of India for 5 years. If the bank pays interest at a rate of 6% per annum, the amount payable to him at the time of maturity is
A) ₹53,250
B) ₹53,320
C) ₹32,350
D) ₹53,520
\[
P=800,\quad n=60,\quad r=6\%
\]
Recurring Deposit Interest Formula:
\[
I=\frac{P\times n(n+1)\times r}{2400}
\]
\[
I=\frac{800\times60\times61\times6}{2400}
\]
\[
I=7320
\]
Maturity Value:
\[
=800\times60+7320
\]
\[
=48000+7320
\]
\[
=55320
\]
Correct Answer: ₹53,320
2. Vimla had deposited recurrent account in ICICI bank and deposited ₹1000 per month for \(2\frac12\) years. If the rate of interest was 10% p.a., then the matured value of this account is
A) ₹32,775
B) ₹33,875
C) ₹23,775
D) ₹34,975
\[
P=1000,\quad n=30,\quad r=10\%
\]
\[
I=\frac{1000\times30\times31\times10}{2400}
\]
\[
I=3875
\]
Maturity Value:
\[
=1000\times30+3875
\]
\[
=33875
\]
Correct Answer: ₹33,875
3. If \(\frac{x}{2}-5\le\frac{x}{3}-4\) and x is a natural even number, then the solution set of x is:
A) {-6,-4,-2}
B) {6,-4,-2,2,4,6}
C) {2,4,6}
D) {2,4,6,8}
B) {6,-4,-2,2,4,6}
C) {2,4,6}
D) {2,4,6,8}
\[
\frac{x}{2}-5\le\frac{x}{3}-4
\]
Multiply by 6:
\[
3x-30\le2x-24
\]
\[
x\le6
\]
Natural even numbers:
\[
\{2,4,6\}
\]
Correct Answer: {2,4,6}
4. If \(-3x+17<-13\) then
A) x > 10
B) x < 10
C) x ≤ 10
D) x ≥ 10
B) x < 10
C) x ≤ 10
D) x ≥ 10
\[
-3x+17<-13
\]
\[
-3x<-30
\]
Divide by -3:
\[
x>10
\]
Correct Answer: x > 10
5. If 3 is a root of the quadratic equation \(x^2-px+3=0\) then p is equal to:
A) 4
B) 3
C) 5
D) 2
Substitute x = 3
\[
3^2-3p+3=0
\]
\[
9-3p+3=0
\]
\[
12-3p=0
\]
\[
p=4
\]
Correct Answer: 4
6. If one root of a quadratic equation \(6x^2-x-k=0\) is \(\frac13\), then the value of k is:
A) 2
B) 1/9
C) 8/3
D) 1
\[
6\left(\frac13\right)^2-\frac13-k=0
\]
\[
6\left(\frac19\right)-\frac13-k=0
\]
\[
\frac23-\frac13-k=0
\]
\[
\frac13-k=0
\]
\[
k=\frac13
\]
Correct Answer: k = 1/3
7. If a polynomial \(x^3+3x^2+3x+1\) is divided by \((x+1)\), then the remainder is
A) -31 B) 0 C) 11 D) -1
By remainder theorem:
\[
f(-1)
\]
\[
=(-1)^3+3(-1)^2+3(-1)+1
\]
\[
=-1+3-3+1
\]
\[
=0
\]
Correct Answer: 0
8. If \((x-2)\) is a factor of the polynomial \(x^3+2x^2-13x+k\), then k is:
A) -10 B) 26 C) -26 D) 10
By factor theorem:
\[
f(2)=0
\]
\[
2^3+2(2^2)-13(2)+k=0
\]
\[
8+8-26+k=0
\]
\[
16-26+k=0
\]
\[
-10+k=0
\]
\[
k=10
\]
Correct Answer: 10
9. The co-ordinates of the point \((3,-2)\) after reflection in the x-axis are respectively:
A) (3,2)
B) (-3,2)
C) (3,-2)
D) (-3,-2)
Reflection in x-axis changes sign of y-coordinate.
\[
(3,-2)\rightarrow(3,2)
\]
Correct Answer: (3,2)
10. A point \(P(7,-5)\) is first reflected in x-axis to point Q and then Q is reflected in y-axis to point R.
Reflection in x-axis:
\[
(7,-5)\rightarrow(7,5)
\]
So:
\[
Q=(7,5)
\]
Reflection in y-axis:
\[
(7,5)\rightarrow(-7,5)
\]
So:
\[
R=(-7,5)
\]
Correct Answer: (7,5) and (-7,5)
QUESTION 2 - GRAPH SOLUTIONS
(i) Use graph sheet to answer this question. Take 2 cm = 1 unit along both the axes.
Given:
\[
A(0,4),\ B(1,1),\ C(4,0)
\]
Reflection in x-axis:
\[
A(0,4)\rightarrow E(0,-4)
\]
\[
B(1,1)\rightarrow D(1,-1)
\]
Reflection through origin:
\[
B(1,1)\rightarrow F(-1,-1)
\]
Reflection in y-axis:
\[
B(1,1)\rightarrow H(-1,1)
\]
\[
C(4,0)\rightarrow G(-4,0)
\]
Closed figure formed:
Irregular Octagon
INSERT GRAPH IMAGE HERE
(ii) Use a graph paper for this question taking 1 cm = 1 unit along both the x and y-axis.
Points:
\[
A(0,5),\ B(2,5),\ C(5,2),\ D(5,-2),\ E(2,-5),\ F(0,-5)
\]
Reflections:
\[
B'( -2,5)
\]
\[
C'( -5,2)
\]
\[
D'( -5,-2)
\]
\[
E'( -2,-5)
\]
Figure formed:
Symmetrical Polygon
Line of symmetry:
\[
y-axis
\]
INSERT GRAPH IMAGE HERE
SECTION B - DESCRIPTIVE SOLUTIONS
Question-3
(i). Solve the quadratic equation \[ x^2-3(x+3)=0 \] Give your answer correct to two significant figures.
Expand:
\[
x^2-3x-9=0
\]
Using quadratic formula:
\[
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\]
\[
a=1,\quad b=-3,\quad c=-9
\]
\[
x=\frac{3\pm\sqrt{9+36}}{2}
\]
\[
x=\frac{3\pm\sqrt{45}}{2}
\]
\[
x=\frac{3\pm6.708}{2}
\]
\[
x=4.9
\]
or
\[
x=-1.9
\]
(ii). Solve: \[ 11x-8<15x+4\le13x+12,\quad x\in W \]
First inequality:
\[
11x-8<15x+4
\]
\[
-12<4x
\]
\[
x>-3
\]
Second inequality:
\[
15x+4\le13x+12
\]
\[
2x\le8
\]
\[
x\le4
\]
Since \(x\in W\):
\[
x=\{0,1,2,3,4\}
\]
(iii). Using factor theorem, factorise: \[ 3x^3+2x^2-19x+6 \]
Try x = 2
\[
3(8)+2(4)-19(2)+6
\]
\[
24+8-38+6
\]
\[
=0
\]
So:
\[
(x-2)
\]
is factor.
Final factorisation:
\[
(x-2)(3x-1)(x+3)
\]
Question-4
(i). Solani deposited ₹150 per month in her bank for eight months under recurring deposit scheme. What will be the maturity value of her deposit, if the rate of interest is 8% p.a.?
\[
P=150,\quad n=8,\quad r=8\%
\]
Interest:
\[
I=\frac{150\times8\times9\times8}{2400}
\]
\[
I=36
\]
Maturity Value:
\[
=150\times8+36
\]
\[
=1200+36
\]
\[
=1236
\]
ii). Solve: \[ x^2+4x-8=0 \]
Using quadratic formula:
\[
x=\frac{-4\pm\sqrt{16+32}}{2}
\]
\[
x=\frac{-4\pm\sqrt{48}}{2}
\]
\[
x=\frac{-4\pm6.928}{2}
\]
\[
x=1.5
\]
or
\[
x=-5.5
\]
(iii). Solve: \[ -5(x-9)\ge17-9x>x+2 \]
\[
-5x+45\ge17-9x
\]
\[
4x\ge-28
\]
\[
x\ge-7
\]
Next:
\[
17-9x>x+2
\]
\[
15>10x
\]
\[
x<1.5
\]
Final answer:
\[
-7\le x<1.5
\]
Question-5
(i) The polynomials \(kx^3+3x^2-8\) and \(3x^3-5x+8\) are divided by \(x+2\). If the remainder in each case is same, find k.
Using remainder theorem:
First polynomial:
\[
k(-2)^3+3(-2)^2-8
\]
\[
-8k+12-8
\]
\[
=-8k+4
\]
Second polynomial:
\[
3(-2)^3-5(-2)+8
\]
\[
-24+10+8
\]
\[
=-6
\]
Equating:
\[
-8k+4=-6
\]
\[
-8k=-10
\]
\[
k=\frac54
\]
(ii). Find x: \[ -2\le\frac13-\frac{2x}{6}\le1\frac56,\quad x\in N \]
\[
-2\le\frac13-\frac{x}{3}\le\frac{11}{6}
\]
Multiply by 6:
\[
-12\le2-2x\le11
\]
Subtract 2:
\[
-14\le-2x\le9
\]
Divide by -2:
\[
7\ge x\ge-\frac92
\]
Natural numbers:
\[
x=\{1,2,3,4,5,6,7\}
\]
(iii). Katrina opened a recurring deposit account with monthly instalment ₹1000 for 2 years at 6% p.a. Find interest and maturity value.
\[
P=1000,\quad n=24,\quad r=6
\]
Interest:
\[
I=\frac{1000\times24\times25\times6}{2400}
\]
\[
I=1500
\]
Maturity:
\[
=1000\times24+1500
\]
\[
=25500
\]
Interest = ₹1500
Maturity Value = ₹25500
Maturity Value = ₹25500
RAY Tutorials © Complete ICSE Maths Solutions