# Indices – CA Foundation Maths Study Material

This Indices – CA Foundation Maths Study Material is designed strictly as per the latest syllabus and exam pattern.

## Indices – CA Foundation Maths Study Material

If a number x is multiplied 5 times written as.
x. x. x. x. x. = x5.
Here “x” is called BASE and 5 is called Power or Index.

Some Related Formulae.
1. am = a × a × a × …………… to m times.

2. a0 = 1 where a ≠ o; ∞

3. a-1 = $$\frac{1}{a}$$

4. a-m = $$\frac{1}{a^m}$$

• am × an = am+n
• am × an × ar × ………. = am+n+r ……
• $$\frac{a^m}{a^n}$$ = am-n
• $$\frac{a^m}{a^n}=\frac{1}{a^{n-m}}$$
• (am)n = amn
• amn ≠ amn
• If am = bm Then a = b
• If am = an Then m = n
• $$\sqrt[m]{a^n}$$ = a$$\frac{n}{m}$$
• √a = a$$\frac{1}{2}$$
• 3√a = a$$\frac{1}{3}$$
• If am = k ⇒ a = k1/m
• If am = kn ⇒ a = kn/m
• If a1/m = k ⇒ a = km
• If a1/m = kn ⇒ a = kmn
• $$\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}$$
• (ab….)m = am.bm
• $$\sqrt[m]{a b \ldots \ldots \ldots}=\sqrt[m]{a} \cdot \sqrt[m]{b} \ldots \ldots .$$
• $$\sqrt{a b}$$ = √a. √b.
• $$\left(\frac{a}{b}\right)^m=\left(\frac{b}{a}\right)^{-m}$$

5. If ab = ba Then
Either (i) a = b
or (ii) If a = 2
Then b = 4
or (Hi) If a = 4
Then b = 2

6. If a > 1 and x < y
Then ax < ay

Calculator Tricks

1. For a+m
Steps (i) Type a
(ii) Press x button.
(Hi) Continue pressing “=” button (Power -1 = m -1) times.

Example
4+5 = 1024
Type 4 Then Press x button Then “=” button 4 times; we get the result.

2. For a-m
Steps (i) Type “a”
(if) Press ÷ button.
(iii) Continue pressing “=” button (Power = m) times
Example:
2-4 = 0.0625
Type base 2 then press button then continue pressing = button 4 times. We get the required result.

3. For (1.05)60
1st Method:
Type base 1.05
Then press “×” button 60 – 1 = 59 times.
IInd Method:
(1.05)60 = [(1.05)12]5
1st work for 1.0512
Then “×” button and work for power 5.

Work As.:
Type 1.05 then press × button
Then press = button 12 – 1 = 11 times
Then press × button and continue
Pressing = button 5-1=4 times.
We get the required result.

Previous Year Exam Questions

Question 1.
Value of (a1/8 + a-1/8)(a1/8 – a-1/8)
(a1/4 + a-1/4)(a1/2 + a-1/2) is:
(a) a + $$\frac{1}{a}$$
(b) a – $$\frac{1}{a}$$
(c) a2 + $$\frac{1}{a^2}$$
(d) a2 – $$\frac{1}{a^2}$$
[a1/8 + a-1/8][a1/8 – a-1/8][a1/4 + a-1/4][a1/2 + a-1/2]
[Use Formula (a + b)(a – b) = a2 – b2]
= [(a1/8)2 – (a1/8)2][a1/4 + a-1/4][a1/2 + a-1/2]
= (a1/4 – a-1/4)(a1/4 + a-1/4)(a1/2 + a-1/2)
= [(a1/4)2 – (a-1/4)2][a1/2 + a-1/2]
= (a1/2 – a-1/2)(a1/2 + a-1/2)
= (a1/2)2 – (a-1/2)2 = a – a-1 = a – $$\frac{1}{a}$$
(b) is correct

Question 2.
Simplification of $$\frac{x^{m+3 n} x^{4 m-9 n}}{x^{6 m-6 n}}$$ is :
(a) xm
(b) x-m
(c) xn
(d) x-n
$$\frac{x^{m+3 n} x^{4 m-9 n}}{x^{6 m-6 n}}$$ = xm+3n+4m-9n-6m+6n = x-m = x-m
(b) is correct

Question 3.
On simplification $$\frac{1}{1+z^{a-b}+z^{a-c}}+\frac{1}{1+z^{b-c}+z^{b-a}}+\frac{1}{1+z^{c-a}+z^{c-b}}$$ reduces to : [1 Mark, Aug. 2007]
(a) $$\frac{1}{z^{2(a+b+c)}}$$
(b) $$\frac{1}{z^{(a+b+c)}}$$
(c) 1
(d) 0
Tricks
$$\frac{1}{1+Z^{a-b}+Z^{a-c}}+\frac{1}{1+Z^{b-c}+Z^{b-a}}+\frac{1}{1+Z^{c-a}+Z^{c-a}}$$
= 1 [it is in cyclic order]

Question 4.
If 4x = 5y = 20z then z is equal to:
(a) xy
(b) $$\frac{x+y}{x y}$$
(c) $$\frac{1}{x y}$$
(d) $$\frac{x y}{x+y}$$
(d)
Let 4y = 5y = 2oz =k
or 4 = k1/x; 5 = k1/y; 20 = k1/z
∴ 20 = 4 × 5

(d) is correct

Question 5.
$$\left(\frac{\sqrt{3}}{9}\right)^{5 / 2}\left(\frac{9}{3 \sqrt{3}}\right)^{7 / 2}$$ × 9 is equal to: [1 Mark, Nov. 2007]
(a) 1
(b) √3
(c) 3√3
(d) $$\frac{3}{9 \sqrt{3}}$$

= (3-4)1/2.32 = 3-2.32 = 3-2+2 = 30 = 1
(a) is correct.

Question 6.
If 2x – 2x – 1 = 4, then the value of xx is : [1 Mark, Feb. 2008, June 2010]
(a) 2
(b) 1
(c) 64
(d) 27
2x – 2x-1 = 4
or 2x-1 (2 – 1) = 4
or 2x-1 x1 = 22
or 2x-1 = 22
x – 1 = 2
😡 = 3
…xx = 33 =27
(d) is Correct
Tricks : Go by choices
For (d) 27 = 33 = xx
x = 3
Put x = 3 in 2x – 2x-1 = 4
It satisfies it (d) is correct

Question 7.
If x = ya, y = zb and z = xc then abc is : [1 Mark, June 2008]
(a) 2
(b) 1
(c) 3
(d) 4
It x = ya; y = Zb & Z = xc
Then abc = 1
Tricks : It is in cyclic order.
(b) is correct
Tricks : See Quicker BMLRS example

Question 8.
If x = 31/3 + 3-1/3 then find value of 3x3 – 9x. [1 Mark, June 2009]
(a) 3
(b) 9
(c) 12
(d) 10
Detail Method
It x = 31/3 + 3-1/3 ……….. (I)
Cubing on both sides; we get x3 =(31/3)3 + (3-1/3)3 +3.31/3.3-1/3(31/3 + 3-1/3)
= 3 + 3-1 + 3 × 1 × x
or x3 = 3 + $$\frac{1}{3}$$ + 3x
or x3 – 3x = $$\frac{9+1}{3}$$
or 3x2 – 9x = 10
∴ (d) is correct
Tricks See Quicker BMLRS for CA-Found.

Question 9.
Find the value of: [1 – {1 – (1 – x2)-1}-1]-1/2 is
(a) 1/x
(b) x
(c) 1
(d) none of these

∴ (b) is correct

Question 10.
$$\frac{2^n+2^{n-1}}{2^{n+1}-2^n}$$
(a) 1/2
(b) 3/2
(c) 2/3
(d) 1/3

∴ (b) is correct
Tricks: Put n = 1

Question 11.
If 2x × 3y × 5z = 360. Then what is the value of x, y, z. ? [1 Mark, Dec. 2009]
(a) 3, 2, 1
(b) 1,2,3
(c) 2, 3, 1
(d) 1, 3, 2
If 2x × 3y × 5z = 360
2x × 3y × 5z = 23 × 32 × 51
Comparing it; we get
x = 3; y = 2; z = 1;
∴ (a) is correct
Tricks : Go by choices.
For (a) x = 3; y = 2; Z = 1
LHS= 23 × 32 × 51 =360 = RHS.
(a) is correct

Question 12.
The recurring decimal 2.7777 ………. can be expressed as: [1 Mark, Dec. 2010]
(a) 24/9
(b) 22/9
(c) 26/9
(d) 25/9
Tricks : Go by choices.
By calculator
(a) $$\frac{24}{9}$$ = 2.666 …………………. ≠ 2.777
(b) $$\frac{22}{9}$$ = 2.444 ……….. ≠ 2.777
(c) $$\frac{26}{9}$$ = 2.888 …………… ≠ 2.77
(d) $$\frac{25}{9}$$ = 2.777
(d) is correct

Question 13.
The value of $$\frac{\left(3^{n+1}+3^n\right)}{\left(3^{n+3}-3^{n+1}\right)}$$ is equal to:
(a) 1/5
(b) 1/6
(c) 1/4
(d) 1/9
(b) Tricks :
Put n = 0

Question 14.
Find the value of X, if x.(x)1/3 = (x1/3)z. [1 Mark, Dec. 2012]
(a) 3
(b) 4
(c) 2
(d) 6
(b) is correct
x.x$$\frac{1}{3}$$ = x$$\frac{x}{3}$$
or x1+$$\frac{1}{3}$$ = xx/3
∴ 1 + $$\frac{1}{3}=\frac{x}{3}$$
∴ $$\frac{4}{3}=\frac{x}{3}$$
∴ x = 4
(b) is correct.

Question 15.
If $$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$$ = 0; then find the value of $$\left[\frac{a+b+c}{3}\right]^3$$ =
(a) 9abc
(b) $$\frac{1}{9 a b c}$$
(c) abc
(d) $$\frac{1}{a b c}$$
(c) is correct

Tricks:
Let a = -1; b= -1 and c =8, because $$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{-1}+\sqrt[3]{-1}+\sqrt[3]{8}$$
= -1 – 1 + 2 = 0(R.H.S) ∴ $$\left[\frac{a+b+c}{3}\right]^3=\left[\frac{-1-1+8}{3}\right]^3$$ = (2)3 = 8
= (-1).(-1).(8) = abc
∴ (c) is correct

Question 16.
The value of
$$\left(\frac{y^a}{y^b}\right)^{a^2+a b+b^2}\left(\frac{y^b}{y^c}\right)^{b^2+b c+c^2}\left(\frac{y^c}{y^a}\right)^{c^2+c a+a^2}$$ = . [1 Mark, June 2014]
(a) y
(b) -1
(c) 1
(d) None
(c) is correct
Tricks: Cyclic order

Question 17.
If px = q, qy = r, rz = p6, then the value of xyz is ……….[1 Mark, June 2015]
(a) 0
(b) 1
(c) 3
(d) 6
qy = r ⇒ (px)y = r ⇒ r = pxy
Now rz = p6 ⇒ (pxy)z = p6 ⇒ pxyz = p6
xyz = 6

Question 18.
The value of $$\frac{x^2-(y-z)^2}{(x+z)^2-y^2}+\frac{y^2-(x-z)^2}{(x+y)^2-z^2}+\frac{z^2-(x-y)^2}{(y+z)^2-x^2}$$. [1 Mark, June 2016]
(a) 0
(b) 1
(c) -1
(d) ∞
(b) is correct

Tricks : Cyclic order

Question 19.
If 3x = 5y = (75)z then
(a) $$\frac{1}{x}+\frac{2}{y}=\frac{1}{z}$$
(b) $$\frac{2}{x}+\frac{1}{y}=\frac{1}{z}$$
(c) $$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$$
(d) None
See Short Cut Tricks Book “QUICKER BMLRS”
3x = 5y = (75)z ………(1)
31 × 52 = 751 ………(2)

Tricks:
Power of (2) ÷ power of (1)
and put + sign at the place of “×”
We get $$\frac{1}{x}+\frac{2}{y}=\frac{1}{z}$$
So, (a) is correct

Question 20.
If abc = 2, then the value of $$\frac{1}{1+a+2 b^{-1}}+\frac{1}{1+\frac{b}{2}+c^{-1}}+\frac{1}{1+a^{-1}+c}$$ = [1 Mark, June 2016]
(a) 1
(b) 2
(c) $$\frac{1}{2}$$
(d) $$\frac{3}{5}$$
TRICKS
“Put a = 1, b = 2 & c = 1. So that abc = 2” in the given question. We get
$$\frac{1}{1+1+\frac{2}{2}}+\frac{1}{1+\frac{2}{2}+1^{-1}}+\frac{1}{1+1^{-1}+1}$$ = 1
Option (a) is correct.

Question 21.
If a = $$\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}$$, b = $$=\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}$$ then the value of $$\frac{1}{a^2}+\frac{1}{b^2}$$ is [1 Mark, June 2017]
(a) 486
(b) 484
(c) 482
(d) 500

Option (a) is correct.

Question 22.
If u5x = v5y = w5z and u2 = vw then xy + zx – 2yz
(a) 0
(b) 1
(c) 2
(d) None of these
(a) u5x = v5y = w5z ⇒ ux = vy = wz
Tricks : See Quicker BMLRS Chapter : Indices
u2 = vw;
∴ $$\frac{2}{x}=\frac{1}{y}+\frac{1}{z}=\frac{y+z}{y z}$$
or ; xy + zx = 2yz
or; xy + zx -2yz = 0

Question 23.

(a) x-(ap+bq+cr)
(b) xa+b+c
(c) x(ap+bq+cr)
(d) xabc [1 Mark, June 2018]
(b)

= xa.xb.xc
= xa+b+c
Option (b) is correct.

Question 24.
$$\frac{2^n+2^{n-1}}{2^{n+1}-2^n}$$
(a) $$\frac{1}{2}$$
(b) $$\frac{3}{2}$$
(c) $$\frac{2}{3}$$
(d) $$\frac{1}{3}$$ [1 Mark, May 2018]
(b)
Tricks:
Put minimum power = n – 1 = 0 or n = 1 in the question
∴ $$\frac{2^n+2^{n-1}}{2^{n+1}-2^n}=\frac{2^1+2^{1-1}}{2^{1+1}-2^1}=\frac{2+1}{4-2}=\frac{3}{2}$$

Question 25.
$$\frac{2^{m+1} \times 3^{2 m-n+3} \times 5^{n+m+4} \times 6^{2 n+m}}{6^{2 m+n} \times 10^{n+1} \times 15^{m+3}}$$ [1 Mark, Nov. 2018]
(a) 32m-2n
(b) 32n-2m
(c) 1
(d) None
Tricks
Put m = n = 0 in this equation. $$\frac{2^{m+1} \times 3^{2 m-n+3} \times 5^{n+m+4} \times 6^{2 n+m}}{6^{2 m+n} \times 10^{n+1} \times 15^{m+3}}$$ = 1

Question 26.
If 2x = 3y2 = 12z2 then [1 Mark, June 2019]
(a) $$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$$
(b) $$\frac{1}{x^2}+\frac{2}{y^2}=\frac{1}{z^2}$$
(c) $$\frac{1}{x^2}+\frac{2}{y^2}=\frac{1}{z^2}$$
(d) None
$$\frac{2}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$$