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Answer is (iii) 2⁴¹

(i) 0 

(ii) Odd

6² x 6^m = 6⁵

6^2+m = 6⁵ 

[Since a^m x aⁿ = a^m+n]

Equating the powers, we get

2 + m = 5

m = 5 – 2

= 3

In 124¹²⁸, the unit digit of base 124 is 4 and the power is 128 (even power). 

Therefore, unit digit of 124¹²⁸ is 4. 

Also in 126¹²⁴, the unit digit of base 126 is 6 and the power is 124 (even power). 

Therefore, unit digit of 126¹²⁴ is 6.

Product of the unit digits = 6 x 6 = 36 

∴ Unit digit of the 124¹²⁸ x 126¹²⁴ is 6.

To draw a chord we need two points on a circle. 

∴ Number chords through 21 points on a circle 

= ²¹C₂ = \(\frac{21\times20}{2\times1}\) = 210

(C) (x – 5) years

Current age of amulya = x

Then, 5 years ago her age was = (x – 5) years

(A) 6x

6 × x can be represented as 6x.

(C) x – 1 = 0

An expression with a variable, constants and the sign of equality (=) is called an equation.

(B) 12

From the question it is given that, value of x is 2.

Now substitute the value of x in x + 10

= 2 + 10

= 12

(B) x – 2 = 0

Transforming – 2 from left hand side to right hand side it becomes 2.

Then, x = 2

(A) x + y = y + x

Let us assume a and b are the two integers,

Then, commutative law of addition = a + b = b + a

Where, a = x, b = y

Therefore, commutative law of addition = x + y = y + x

(A) x + y = y + x 

The correct option is (B) x – 2 = 0.

(B) (x ÷ 6) metres

We know that, perimeter of hexagon = number of sides × length of each sides

Given, the perimeter of a regular hexagon is x metres

Then, the length of each of its sides = (x/6) metres

= (x ÷ 6) metres

(B) (x ÷ 6) metres 

It can be observed that the expressions in alternatives (c) and (d) are formed by using numbers only.

Many expressions can be formed by using the 3 numbers 5, 7 and 8 Some of these are as follows 5 × (8 – 7)

5 × (8 + 7)

(8 + 5) × 7

(8 – 5) × 7

(7 + 5) × 8

(7 – 5) × 8

True

We have, 2a – 1 = 5

⇒ 2a – 1 + 1 = 5 + 1 [Adding 1 to both sides]

⇒ 2a = 6

⇒ 2a/2 = 6/2 [Dividing both sides by 2]

⇒ a = 3, which is the solution of the given equation.

False.

In the equation 7k – 7 = 7, the variable is k.

True

In an equation, the LHS is equal to the RHS.

False.

An expression with a variable, constants and the sign of equality (=) is called an equation.

A hexagon has six vertices. By joining any three vertices of a hexagon we get a triangle. 

∴ Number of triangles formed by joining the vertices of a hexagon 

= ⁶C₃ = \(\frac{6\times5\times4}{3\times2\times1}\) = 20

In this problem first, we have to select consonants and vowels. 

Then we arrange a five-letter word using 3 consonants and 2 vowels.

Therefore here both combination and permutation involved. 

The number of ways of selecting 3 consonants from 7 is ⁷C₃. 

The number of ways of selecting 2 vowels from 4 is ⁴C₂. 

The number of ways selecting 3 consonants from 7 and 2 vowels from 4 is ⁷C₃ x ⁴C₂. 

Now with every selection number of ways of arranging 5 letter word

= 5! x ⁷C₃ x ⁴C₂

= 120 x \(\frac{7\times6\times5}{3\times2\times1}\) x \(\frac{4\times3}{2\times1}\)

= 25200

(c) 5

nP₂ = 20 

n(n – 1) = 20 

n(n – 1) = 5 x 4 

∴ n = 5

(d) 5

5C₄ = 5C₁ = 5

The letters of the word CHAT in alphabetical order are A, C, H, T. To arrive the word CHAT, first, we have to go through the word that begins with A. If A is fixed as the first letter remaining three letters C, H, T can be arranged among themselves in 3! ways. Next, we select C as the first letter and start arranging the remaining letters in alphabetical order. Now C and A is fixed remaining two letters can be arranged in 2! ways. Next, we move on H with C, A, and H is fixed the letter T can be arranged in 1! ways.

∴ Rank of the word CHAT = 3! + 2! + 1! = 6 + 2 + 1 = 9

Note: The rank of a given word is basically finding out the position of the word when possible words have been formed using all the letters of the given word exactly once and arranged in alphabetical order as in the case of dictionary. The possible arrangement of the word CHAT are (i) ACHT, (ii) ACTH, (iii) AHCT, (iv) AHTC, (v) ATCH, (vi) ATHC, (vii) CAHT, (viii) CATH, (ix) CHAT. So the rank of the word occurs in the ninth position. 

∴ The rank of the word CHAT is 9.

Given that ⁿP_r = 1680, ⁿC_r = 70 

We know that ⁿC_r = \(\frac{nP_r}{r!}\)

70 = \(\frac{1680}{r!}\)

r! = \(\frac{1680}{70}\) = 24 

r! = 4 x 3 x 2 x 1 = 4! 

∴ r = 4

Answer is (b) n + 1

2[x³ + x² – x – 1] = 2[x²(x + 1) - 1 (x + 1)]

= 2(x + 1) (x² – 1)

= 2(x + 1) (x + 1) (x – 1)

= 2(x + 1)² (x – 1)

3[x³ + 3x² – x – 3] = 3[x²(x + 3) - 1 (x + 3)]

= 3[(x + 3)(x² – 1)]

= 3(x + 3)(x + 1) (x – 1)

L.C.M. = 6(x + 1)² (x – 1) (x + 3)

The number of students dropping out this year is 30 less than the number of students dropped out last year.

Sharad reduced the consumption of tea per day by 5 cups after having some health problem.

Last year Jay planted 10 more plants than twice the plants planted by his friend John.

The minimum temperature on a day in Delhi was 10°C less than the maximum temperature.

Age of Kartik’s father is seven times the age of Kartik.

Let the daughter’s age be m years. 

5 times of m is 5m. 

2 more than 5m is the expression 5m + 2. 

The age of Megha (in years) is (5m + 2).

The required equation is 2t + 3 = 3, which has solution t = 0.

Anagha is at step p. 

Sushant is 10 steps ahead of Anagha. That is, he is at the step p + 10.

Faizal is 6 steps behind Anagha. That is, he is at step p – 6. 

8 times of p = 8p 

3 less than 8p = 8p – 3 

So, the total number of steps = 8 p – 3

The required equation is x + 1 = 0, its solution is x = -1, which is not a whole number.

Amount left with Leela is Rs. 10,000 more than the amount she contributed towards the Prime Minister’s Relief Fund.

A pen costs 5 times the cost of a pencil.

Anagha is at step p. 

Sushant is 10 steps ahead of Anagha. That is, he is at the step p + 10.

Faizal is 6 steps behind Anagha. 

That is, he is at step p – 6. 

8 times of p = 8p 

3 less than 8p = 8p – 3 

So, the total number of steps = 8 p – 3

Cost of a pen is 6 times the cost of a pencil.

Algebric expression : 

(i) 3x + 5 

(iv) 2x + 1

(v) (x/2) + 5 – x 

(vi) 3x 

Arithmetic expression : 

(ii) 5 × 4 +7 

(ii) 3 + 4 × 3 + 5

(a) 9

\(\frac{kx}{(x+4)(2x-1)}\) = \(\frac{4}{x+4}+\frac{1}{2x-1}\)

kx = 8x – 4 + x + 4 

kx = 9x 

k = 9

(b) 16

(√2)⁸ = \((2^{\frac{1}{2}})^8\) = 2⁴ = 16

Product of (zeroes) rots = c / b

= 2a / 1 = α. 2 / α

or, 2a = 2

:.    a = 1