Expansions Class 9 RS Aggarwal MCQs ICSE Foundation Maths Solutions

Expansions Class 9 RS Aggarwal MCQs ICSE Foundation Maths Solutions.  In this article you will learn how to solve Multiple Choice Questions on Expansion very easily. Hint and reason is also given in the favour of correct option. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.

Expansions Class 9 RS Aggarwal MCQs ICSE Foundation Maths Solutions

Expansions Class 9 RS Aggarwal MCQs ICSE Foundation Maths Solutions

Board ICSE
Subject Maths
Class 9th
Chapter-3 Expansions
Writer RS Aggrawal
Book Name Foundation
Topics Solution of MCQs
Academic Session 2024-2025

How to Solve MCQs Expansion

 Class 9 RS Aggarwal Multiple Choice Questions ICSE Foundation Maths Solutions

To solve MCQs on expansion one should know all the formula and practice exercise questions. After that you can easily solve MCQs on expansion.

Multiple Choice Questions : 

 Class 9 RS Aggarwal Multiple Choice Questions ICSE Foundation Maths Solutions

Page- 55

Que-1: If a+b = 12 and ab = 35, then (a-b)² =

(a) 2   (b) 4   (c) 16   (d) 20

Solution- (b) 4

Reason: a+b = 12
ab = 35
(a+b)² = 12² = 144
a² + 2ab + b² = 144
Now, we substitute the value of ab:
144 = a²+2⋅35+b²
a² + 70 + b² = 144
a²+b² = 144 – 70
a²+b² = 74
Now, we substitute a²+b² back into the expression for (a−b)²:
(a−b)² = a²−2ab+b²
(a−b)² = 74 − 2⋅35
(a−b)² = 74−70
(a−b)² = 4.

Que-2: If a+b =16 and a-b = 2, then ab =

(a) 48   (b) 56   (c) 63   (d) 65

Solution- (c) 63

Reason:  we know that ,
(a-b)^2 = (a+b)^2 – 4ab
now  , substitute the values
( 2)^2 = (16)^2 -4ab
4 =  256  – 4ab
4ab = 256  – 4
= 252
ab = 252/ 4
ab =  63.

Que-3: If p+1/p = 5/2, then p²+1/p² =

(a) 25/4   (b) 19/4   (c) 9/4   (d) 17/4

Solution- (d) 17/4

Reason: p+1/p = 5/2
First, square both sides of the given equation:
(p+1/p)² = (5/2)²
Expanding the left-hand side:
p²+2⋅p⋅1/p+1/p² = (5/2)²
Simplifying the left-hand side:
p²+2+1/p² = 25/4​
Subtract 2 from both sides to isolate p²+1/p²:
p²+1/p² = 25/4−2
p²+1/p² = (25-8)/4
p²+1/p² = 17/4.

Que-4: If x+1/x = 5, then x-1/x =

(a) ±√21  (b) ±√29   (c) ±3   (d) ±2

Solution- (a) ±√21

Reason: Let’s denote:
y = x+1/x
y = 5
First, square both sides to create a simpler form that can be used:
y² = (x+1/x)²
Substituting y = 5:
5² = x²+2+1/x²
25 = x²+2+1/x²
x²+1/x² = 25-2 = 23
Now, let z = x−1/x
Square both sides of z = x−1/x:
z² = (x−1/x)²
z² = x²−2+1/x²1​
z² = (x²+1/x²)−2
z² = 23−2 = 21
z = ±√21
x-1/x = ±√21.

Que-5: If y+1/y = 2, then y^5 + 1/y^5 =

(a) 5   (b) 3   (c) 2  (d) 1

Solution- (c) 2

Reason: y^5 + 1/y^5 = 2.

Que-6: If a²-1/a² = 7, then a^4 + 1/a^4 =

(a) 49   (b) 51   (c) 53   (d) 60

Solution- (b) 51

Reason: [a²-1/a²]²= (a²)²+(1/a²)²-2

(7)². = a⁴+1/a⁴-2

49+2. = a⁴+ 1/a⁴

51 = a⁴+ 1/a⁴.

Que-7: If x+y+z = 0, then x³+y³+z³ =

(a) xyz   (b) 3xyz   (c) 27xyz   (d) 0

Solution- (b) 3xyz

Reason: if x + y + z = 0
then x+y = -z

cubing both side

()x+y )^3 =(-z) ^3

x3 + y3 + 3xy (x+y) =- z3

x3+y3+3xy(-z) =- z3

x3 + y3 -3xyz = -z3

x³+y³+z³ = 3xyz
Que-8: If a/b + b/a = 1(a,b ≠ 0), then a³+b³ =

(a) 0   (b) 1   (c) 2   (d) 8

Solution- (a) 0

Reason: a/b+b/a = 1 , multiply both side by ab
a²+b² = ab
a²+b²−ab = 0−−−−−−−(1)
a³+b³ = (a+b)(a²+b²−ab) ,
put a²+b²−ab = 0 from eq.(1)
a³+b³ = (a+b)(0)
a³+b³ = 0.

Que-9: If p+1/p = x and p-1/p = y, then the relation between x and y is :

(a) x² = y²   (b) x²+y² = 4   (c) x²-y² = 4   (d) xy = 2

Solution- (c) x²-y² = 4

Reason:  p²-1/p² = y
p²+1/p²-2 = y²
p²+1/p² = y²+2  [+2 on both sides]
p²+1/p²+2 = y²+4
[p+1/p]² = y²+4        but p+1/p = x and [p+1/p]² = x² so
x² = y²+4
x²-y² = 4.

Que-10: If a = 1/(a-7), then a-(1/a) =
(a) 0   (b) 1   (c) 7   (d) 1/7

Solution- (c) 7

Reason:

a = 1/(a-7) 

a -7 = 1/a

a- 1/a  = 7

Que-11: 2.51 x 2.51 + 1.31 x 1.31 – 2.62 x 2.51 =

(a) 1.44   (b) 0.44   (c) 1   (d) 0

Solution- (a) 1.44

Reason:  let  = 2.51 =a and 1.31 = b

a2 + b2 – 2 x a x b

(a-b)2

(2.51-1.31) ^2 = 1.44

Que-12: If (x+1/x)² = 3, then x³+1/x³ =

(a) 9   (b) 3   (c) 2   (d) 0

Solution- (d) 0

Reason:  (x+1/x)² = 3
x + 1/x = √3
(x+1/x)³ = x³ + (1/x)³ + 3(x)(1/x) (x + 1/x)
√3³ = x³ + 1/x³ + 3(√3)
x³ + 1/x³ = 3√3 – 3√3
x³ + 1/x³ = 0.

Que-13: The value of ab if 3a+5b = 15 and 9a²+25b² = 75, is :

(a) 4   (b) 5   (c) 6   (d) 8

Solution- (b) 5

Reason: 3a +5b = 15
(3a+5b)²= (15)²
9a²+25b² + 30ab = 225
75 + 30ab = 225
30ab = 150
ab = 5.

Que-14: If l+m-n = 9 and l²+m²+n² = 31, then the value of mn+nl-lm is :

(a) -25   (b) 25   (c) -2   (d) -5

Solution- (a) -25

Reason:  l2 + m2 + n2 = 31
l + m – n = 9
On squaring both sides,
(l + m – n)2 = 81
⇒  l2 + m2 + n2 + 2 (lm – mn – nl) = 81
⇒  31 + 2(lm – mn – nl) = 81
⇒  2(lm – mn – nl) = 81 – 31 = 50
⇒  mn + nl – lm = -50/2 = -25.

Que-15: If a-b+c = 6 and a²+b²+c² = 38, then the value of ab+bc-ca is :

(a) 0   (b) 1   (c) -1    (d) not possible

Solution- (b) 1

Reason: a−b+c = 6
a²+b²+c² = 38
We need to find the value of ab+bc−ca
First, square the first equation a−b+c = 6:
(a−b+c)² = 6²
a²−2ab+b²−2bc+c²+2ac = 36
This simplifies to:
a²+b²+c²−2(ab+bc−ca) = 36
We are given:
a²+b²+c² = 38
Substitute this into the equation:
38−2(ab+bc−ca) = 36
Now, solve for ab+bc−ca:
38−36 = 2(ab+bc−ca)
2 = 2(ab+bc−ca)
ab+bc−ca = 1.

– : End of Expansions Class 9 RS Aggarwal MCQs ICSE Foundation Maths Solutions : —

Return to :- RS Aggarwal Solutions for ICSE Class-9 Mathematics

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