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. = x^{5}.

Here “x” is called BASE and 5 is called Power or Index.

**Some Related Formulae.
**1. a

^{m}= a × a × a × …………… to m times.

2. a^{0} = 1 where a ≠ o; ∞

3. a^{-1} = \(\frac{1}{a}\)

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

- a
^{m}× a^{n}= a^{m+n} - a
^{m}× a^{n}× a^{r}× ………. = a^{m+n+r}…… - \(\frac{a^m}{a^n}\) = a
^{m-n} - \(\frac{a^m}{a^n}=\frac{1}{a^{n-m}}\)
- (a
^{m})^{n}= a^{mn} - a
^{mn}≠ a^{mn} - If a
^{m}= b^{m}Then a = b - If a
^{m}= a^{n}Then m = n - \(\sqrt[m]{a^n}\) = a
^{\(\frac{n}{m}\)} - √a = a
^{\(\frac{1}{2}\)} - 3√a = a
^{\(\frac{1}{3}\)} - If a
^{m}= k ⇒ a = k^{1/m} - If a
^{m}= k^{n}⇒ a = k^{n/m} - If a
^{1/m}= k ⇒ a = k^{m} - If a
^{1/m}= k^{n}⇒ a = k^{mn} - \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)
- (ab….)
^{m}= a^{m}.b^{m} - \(\sqrt[m]{a b \ldots \ldots \ldots}=\sqrt[m]{a} \cdot \sqrt[m]{b} \ldots \ldots .\)
- \(\sqrt{a b}\) = √a. √b.
- \(\left(\frac{a}{b}\right)^m=\left(\frac{b}{a}\right)^{-m}\)

5. If a^{b} = b^{a} 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 a^{x} < a^{y}

**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 (a^{1/8} + a^{-1/8})(a^{1/8} – a^{-1/8})

(a^{1/4} + a^{-1/4})(a^{1/2} + a^{-1/2}) is:

(a) a + \(\frac{1}{a}\)

(b) a – \(\frac{1}{a}\)

(c) a^{2} + \(\frac{1}{a^2}\)

(d) a^{2} – \(\frac{1}{a^2}\)

Answer:

[a^{1/8} + a^{-1/8}][a^{1/8} – a^{-1/8}][a^{1/4} + a^{-1/4}][a^{1/2} + a^{-1/2}]

[Use Formula (a + b)(a – b) = a^{2} – b^{2}]

= [(a^{1/8})^{2} – (a^{1/8})^{2}][a^{1/4} + a^{-1/4}][a^{1/2} + a^{-1/2}]

= (a^{1/4} – a^{-1/4})(a^{1/4} + a^{-1/4})(a^{1/2} + a^{-1/2})

= [(a^{1/4})^{2} – (a^{-1/4})^{2}][a^{1/2} + a^{-1/2}]

= (a^{1/2} – a^{-1/2})(a^{1/2} + a^{-1/2})

= (a^{1/2})^{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) x^{m}

(b) x^{-m}

(c) x^{n}

(d) x^{-n}

Answer:

\(\frac{x^{m+3 n} x^{4 m-9 n}}{x^{6 m-6 n}}\) = x^{m+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

Answer:

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 4^{x} = 5^{y} = 20^{z} then z is equal to:

(a) xy

(b) \(\frac{x+y}{x y}\)

(c) \(\frac{1}{x y}\)

(d) \(\frac{x y}{x+y}\)

Answer:

(d)

Let 4^{y} = 5^{y} = 2o^{z} =k

or 4 = k^{1/x}; 5 = k^{1/y}; 20 = k^{1/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) \(\frac{3}{9 \sqrt{3}}\)

Answer:

= (3^{-4})^{1/2}.3^{2} = 3^{-2}.3^{2} = 3^{-2+2} = 3^{0} = 1

(a) is correct.

Question 6.

If 2^{x} – 2^{x – 1} = 4, then the value of x^{x} is : [1 Mark, Feb. 2008, June 2010]

(a) 2

(b) 1

(c) 64

(d) 27

Answer:

2^{x} – 2^{x-1} = 4

or 2^{x-1} (2 – 1) = 4

or 2^{x-1} x1 = 22

or 2^{x-1} = 22

x – 1 = 2

😡 = 3

…x^{x} = 3^{3} =27

(d) is Correct

Tricks : Go by choices

For (d) 27 = 3^{3} = x^{x}

x = 3

Put x = 3 in 2^{x} – 2^{x-1} = 4

It satisfies it (d) is correct

Question 7.

If x = y^{a}, y = z^{b} and z = x^{c} then abc is : [1 Mark, June 2008]

(a) 2

(b) 1

(c) 3

(d) 4

Answer:

It x = y^{a}; y = Z^{b} & Z = x^{c}

Then abc = 1

Tricks : It is in cyclic order.

(b) is correct

Tricks : See Quicker BMLRS example

Question 8.

If x = 3^{1/3} + 3^{-1/3} then find value of 3x^{3} – 9x. [1 Mark, June 2009]

(a) 3

(b) 9

(c) 12

(d) 10

Answer:

Detail Method

It x = 3^{1/3} + 3^{-1/3} ……….. (I)

Cubing on both sides; we get x^{3} =(3^{1/3})^{3} + (3^{-1/3})^{3} +3.3^{1/3}.3^{-1/3}(3^{1/3} + 3^{-1/3})

= 3 + 3^{-1} + 3 × 1 × x

or x^{3} = 3 + \(\frac{1}{3}\) + 3x

or x^{3} – 3x = \(\frac{9+1}{3}\)

or 3x^{2} – 9x = 10

∴ (d) is correct

Tricks See Quicker BMLRS for CA-Found.

Question 9.

Find the value of: [1 – {1 – (1 – x^{2})^{-1}}^{-1}]^{-1/2} is

(a) 1/x

(b) x

(c) 1

(d) none of these

Answer:

∴ (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

Answer:

∴ (b) is correct

Tricks: Put n = 1

Question 11.

If 2^{x} × 3^{y} × 5^{z} = 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

Answer:

If 2^{x} × 3^{y} × 5^{z} = 360

2^{x} × 3^{y} × 5^{z} = 2^{3} × 3^{2} × 5^{1}

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= 2^{3} × 3^{2} × 5^{1} =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

Answer:

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

Answer:

(b) Tricks :

Put n = 0

Question 14.

Find the value of X, if x.(x)^{1/3} = (x^{1/3})^{z}. [1 Mark, Dec. 2012]

(a) 3

(b) 4

(c) 2

(d) 6

Answer:

(b) is correct

x.x^{\(\frac{1}{3}\)} = x^{\(\frac{x}{3}\)}

or x^{1+\(\frac{1}{3}\)} = x^{x/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) \(\frac{1}{a b c}\)

Answer:

(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

Answer:

(c) is correct

Tricks: Cyclic order

Question 17.

If p^{x} = q, q^{y} = r, r^{z} = p^{6}, then the value of xyz is ……….[1 Mark, June 2015]

(a) 0

(b) 1

(c) 3

(d) 6

Answer:

q^{y} = r ⇒ (p^{x})^{y} = r ⇒ r = p^{xy}

Now r^{z} = p^{6} ⇒ (p^{xy})^{z} = p^{6} ⇒ p^{xyz} = p^{6}

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) ∞

Answer:

(b) is correct

Tricks : Cyclic order

Question 19.

If 3^{x} = 5^{y} = (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

Answer:

See Short Cut Tricks Book “QUICKER BMLRS”

3^{x} = 5^{y} = (75)^{z} ………(1)

3^{1} × 5^{2} = 75^{1} ………(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) \(\frac{3}{5}\)

Answer:

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

Answer:

Option (a) is correct.

Question 22.

If u^{5x} = v^{5y} = w^{5z} and u^{2} = vw then xy + zx – 2yz

(a) 0

(b) 1

(c) 2

(d) None of these

Answer:

(a) u^{5x} = v^{5y} = w^{5z} ⇒ u^{x} = v^{y} = w^{z}

Tricks : See Quicker BMLRS Chapter : Indices

u^{2} = 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) x^{a+b+c}

(c) x^{(ap+bq+cr)}

(d) x^{abc} [1 Mark, June 2018]

Answer:

(b)

= x^{a}.x^{b}.x^{c}

= x^{a+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]

Answer:

(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) 3^{2m-2n}

(b) 3^{2n-2m}

(c) 1

(d) None

Answer:

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 2^{x} = 3^{y2} = 12^{z2} 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

Answer:

∵ 2^{x2} = 3^{y2} = 12^{z2} ………..(1) (Given)

Tricks:

Factorize 12 in terms of 2 & 3. We get

2^{2} × 3^{1} = 12^{1} ……….(2)

Always write as power of base of (2) ÷ Power on same base of 1 ; put “+”

Sign at the place of “×” Sign. So;

\(\frac{2}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}\)

So (c) is correct.

Details

(c) is correct.