Maths-

General

Easy

Question

# Statement-I : If then A=

Statement-II : If then

Which of the above statements is true

- Only I is true
- Only II is true
- both I & II are true
- Neither I nor II true

Hint:

### In this question we will solve both the questions to know whether they are true or not. In both statements we will first solve the fractions in RHS. Then we will find the coefficients of , x and constant term after that we will compare both sides to get the value of A,B,C and p and q.

## The correct answer is: Only I is true

### Statement-I : If then A=

statement 1 is true.

Statement-II : If then

statement 2 is false.

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### If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:

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### If be that roots where , such that and then the number of integral solutions of λ is

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### The conic with length of latus rectum 6 and eccentricity is

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### Statement-I : If

Statement-II :If

Which of the above statements is true

Statement-I: If

Now, A=1 2A+B=3 A+B+C=1

A+B+C=1+1-1=-1

Statement 1 is not true.

Statement-II :If

Now, A=1, B=2, C=3

A+B-C=0

1+2-3=0

So, statement 2 is true.

Now, A=1 2A+B=3 A+B+C=1

A+B+C=1+1-1=-1

Statement 1 is not true.

Statement-II :If

Now, A=1, B=2, C=3

A+B-C=0

1+2-3=0

So, statement 2 is true.

### Statement-I : If

Statement-II :If

Which of the above statements is true

Maths-General

Statement-I: If

Now, A=1 2A+B=3 A+B+C=1

A+B+C=1+1-1=-1

Statement 1 is not true.

Statement-II :If

Now, A=1, B=2, C=3

A+B-C=0

1+2-3=0

So, statement 2 is true.

Now, A=1 2A+B=3 A+B+C=1

A+B+C=1+1-1=-1

Statement 1 is not true.

Statement-II :If

Now, A=1, B=2, C=3

A+B-C=0

1+2-3=0

So, statement 2 is true.

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### If x, y are rational number such that Then

### If x, y are rational number such that Then

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### A class contains 4 boys and g girls. Every Sunday five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of g is -

Number of groups having 4 boys and 1 girl

= (

and number of groups having 3 boys and 2 girls

= (

Thus, the number of dolls distributed

= g(1) + (2)[2g (g – 1)]

= 4g

We are given 4g

= (

^{4}C_{4}) (^{g}C_{1}) = gand number of groups having 3 boys and 2 girls

= (

^{4}C_{3}) (^{g}C_{2}) = 2g(g – 1)Thus, the number of dolls distributed

= g(1) + (2)[2g (g – 1)]

= 4g

^{2}– 3gWe are given 4g

^{2}– 3g = 85 g = 5.### A class contains 4 boys and g girls. Every Sunday five students, including at least three boys go for a picnic to Appu Ghar, a different group being sent every week. During, the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was 85, then value of g is -

maths-General

Number of groups having 4 boys and 1 girl

= (

and number of groups having 3 boys and 2 girls

= (

Thus, the number of dolls distributed

= g(1) + (2)[2g (g – 1)]

= 4g

We are given 4g

= (

^{4}C_{4}) (^{g}C_{1}) = gand number of groups having 3 boys and 2 girls

= (

^{4}C_{3}) (^{g}C_{2}) = 2g(g – 1)Thus, the number of dolls distributed

= g(1) + (2)[2g (g – 1)]

= 4g

^{2}– 3gWe are given 4g

^{2}– 3g = 85 g = 5.maths-

### There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls, is -

Let the number of yellow balls be x, that of black be 2x and that of green be y. Then

x + 2x + y = 20 or 3x + y = 20

y = 20 – 3x.

As 0 y 20, we get 0 20 – 3x 20

0 3x 20 or 0 x 6

The number of ways of selecting the balls is 7.

x + 2x + y = 20 or 3x + y = 20

y = 20 – 3x.

As 0 y 20, we get 0 20 – 3x 20

0 3x 20 or 0 x 6

The number of ways of selecting the balls is 7.

### There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls if the number of black balls to be selected is twice the number of yellow balls, is -

maths-General

Let the number of yellow balls be x, that of black be 2x and that of green be y. Then

x + 2x + y = 20 or 3x + y = 20

y = 20 – 3x.

As 0 y 20, we get 0 20 – 3x 20

0 3x 20 or 0 x 6

The number of ways of selecting the balls is 7.

x + 2x + y = 20 or 3x + y = 20

y = 20 – 3x.

As 0 y 20, we get 0 20 – 3x 20

0 3x 20 or 0 x 6

The number of ways of selecting the balls is 7.

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### If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a^{3} + b^{3} + c^{3} must be divisible by -

We have (a

= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)

Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,

6 | {(a

6 | (a

^{3}– a) + (b^{3}– b) + (c^{3}– c)= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)

Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,

6 | {(a

^{3}– a) + (b^{3}– b) + (c^{3}– c)}6 | (a

^{3}+ b^{3}+ c^{3}) as 6|(a + b + c).### If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a^{3} + b^{3} + c^{3} must be divisible by -

maths-General

We have (a

= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)

Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,

6 | {(a

6 | (a

^{3}– a) + (b^{3}– b) + (c^{3}– c)= (a – 1) (a) (a + 1) + (b – 1) (b) (b + 1) + (c – 1) (c) (c + 1)

Since (x – 1) x (x + 1) is divisible both 2 & 3, it is divisible by 6. Thus,

6 | {(a

^{3}– a) + (b^{3}– b) + (c^{3}– c)}6 | (a

^{3}+ b^{3}+ c^{3}) as 6|(a + b + c).maths-

### For x R, let [x] denote the greatest integer x, then value of++ +…+is -

For 0 r 66, 0 <

– < – 0

– – < – – –

= –1 for 0 r 66

Also, for 67 r 100, 1

–1 – –

– – 1 – – – –

= –2 for 67 r 100

Hence, = 67(–1) + 2(–34) = –135.

– < – 0

– – < – – –

= –1 for 0 r 66

Also, for 67 r 100, 1

–1 – –

– – 1 – – – –

= –2 for 67 r 100

Hence, = 67(–1) + 2(–34) = –135.

### For x R, let [x] denote the greatest integer x, then value of++ +…+is -

maths-General

For 0 r 66, 0 <

– < – 0

– – < – – –

= –1 for 0 r 66

Also, for 67 r 100, 1

–1 – –

– – 1 – – – –

= –2 for 67 r 100

Hence, = 67(–1) + 2(–34) = –135.

– < – 0

– – < – – –

= –1 for 0 r 66

Also, for 67 r 100, 1

–1 – –

– – 1 – – – –

= –2 for 67 r 100

Hence, = 67(–1) + 2(–34) = –135.

maths-

### The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 x 5, 12 y 18, z –1

Let t = z + 1

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

=

=

= 91 – 36 – 28 + 3 = 30.

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

^{25}in [(x + x^{2}+ x^{3}+ x^{4}+ x^{5}) (x^{12}+ x^{13}+…..+ x^{18}) (1 + x + x^{2}+ …..)]= Coefficient of x

^{12}in (1 + x + x^{2}+ x^{3}+ x^{4}) (1 + x + x^{2}+ x^{3}+ x^{4}+ x^{5}+ x^{6}) (1 + x + x^{2}+ …..)= Coefficient of x

^{12}in (1 – x)^{–1}= Coefficient of x

^{12}in (1 – x^{5}) (1 – x^{6}) (1 – x)^{–3}= Coefficient of x

^{12}in (1 – x^{5}– x^{6}+ x^{11}) (1 – x)^{–3}=

^{12+3–1}C_{3–1}–^{7+3–1}C_{3–1}–^{6+3–1}C_{3–1}+^{1+3–1}C_{3–1}=

^{14}C_{2}–^{9}C_{2}–^{8}C_{2}+^{3}C_{2}= 91 – 36 – 28 + 3 = 30.

### The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 x 5, 12 y 18, z –1

maths-General

Let t = z + 1

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

= Coefficient of x

=

=

= 91 – 36 – 28 + 3 = 30.

Equation reduces to x + y + t = 25

1 x 5, 12 y 18, t 0

Required number of ways

= Coefficient of x

^{25}in [(x + x^{2}+ x^{3}+ x^{4}+ x^{5}) (x^{12}+ x^{13}+…..+ x^{18}) (1 + x + x^{2}+ …..)]= Coefficient of x

^{12}in (1 + x + x^{2}+ x^{3}+ x^{4}) (1 + x + x^{2}+ x^{3}+ x^{4}+ x^{5}+ x^{6}) (1 + x + x^{2}+ …..)= Coefficient of x

^{12}in (1 – x)^{–1}= Coefficient of x

^{12}in (1 – x^{5}) (1 – x^{6}) (1 – x)^{–3}= Coefficient of x

^{12}in (1 – x^{5}– x^{6}+ x^{11}) (1 – x)^{–3}=

^{12+3–1}C_{3–1}–^{7+3–1}C_{3–1}–^{6+3–1}C_{3–1}+^{1+3–1}C_{3–1}=

^{14}C_{2}–^{9}C_{2}–^{8}C_{2}+^{3}C_{2}= 91 – 36 – 28 + 3 = 30.