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:
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
If be that roots where , such that and then the number of integral solutions of λ is
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with roots 
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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.
Maths-
If x, y are rational number such that 
Then
If x, y are rational number such that Then
Maths-General
maths-
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
= (4C4) (gC1) = g
and number of groups having 3 boys and 2 girls
= (4C3) (gC2) = 2g(g – 1)
Thus, the number of dolls distributed
= g(1) + (2)[2g (g – 1)]
= 4g2 – 3g
We are given 4g2 – 3g = 85
g = 5.
= (4C4) (gC1) = g
and number of groups having 3 boys and 2 girls
= (4C3) (gC2) = 2g(g – 1)
Thus, the number of dolls distributed
= g(1) + (2)[2g (g – 1)]
= 4g2 – 3g
We are given 4g2 – 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
= (4C4) (gC1) = g
and number of groups having 3 boys and 2 girls
= (4C3) (gC2) = 2g(g – 1)
Thus, the number of dolls distributed
= g(1) + (2)[2g (g – 1)]
= 4g2 – 3g
We are given 4g2 – 3g = 85 g = 5.
= (4C4) (gC1) = g
and number of groups having 3 boys and 2 girls
= (4C3) (gC2) = 2g(g – 1)
Thus, the number of dolls distributed
= g(1) + (2)[2g (g – 1)]
= 4g2 – 3g
We are given 4g2 – 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 6The 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 6The number of ways of selecting the balls is 7.maths-
If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a3 + b3 + c3 must be divisible by -
We have (a3 – a) + (b3 – b) + (c3 – 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 | {(a3 – a) + (b3 – b) + (c3 – c)}
6 | (a3 + b3 + c3) as 6|(a + b + 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 | {(a3 – a) + (b3 – b) + (c3 – c)}
6 | (a3 + b3 + c3) as 6|(a + b + c).If a, b, c are the three natural numbers such that a + b + c is divisible by 6, then a3 + b3 + c3 must be divisible by -
maths-General
We have (a3 – a) + (b3 – b) + (c3 – 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 | {(a3 – a) + (b3 – b) + (c3 – c)}
6 | (a3 + b3 + c3) as 6|(a + b + 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 | {(a3 – a) + (b3 – b) + (c3 – c)}
6 | (a3 + b3 + c3) 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.
r 66, 0 < – < – 0– – < – – – = –1 for 0 r 66Also, for 67
r 100, 1–1 – – – – 1 – – – – = –2 for 67 r 100Hence,
= 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.
r 66, 0 < – < – 0– – < – – – = –1 for 0 r 66Also, for 67
r 100, 1–1 – – – – 1 – – – – = –2 for 67 r 100Hence,
= 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 x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
= Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
= Coefficient of x12 in
(1 – x)–1
= Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
= Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
= 12+3–1C3–1 – 7+3–1C3–1 – 6+3–1C3–1 + 1+3–1C3–1
= 14C2 – 9C2 – 8C2 + 3C2
= 91 – 36 – 28 + 3 = 30.
Equation reduces to x + y + t = 25
1
x 5, 12 y 18, t 0Required number of ways
= Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
= Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
= Coefficient of x12 in
(1 – x)–1= Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
= Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
= 12+3–1C3–1 – 7+3–1C3–1 – 6+3–1C3–1 + 1+3–1C3–1
= 14C2 – 9C2 – 8C2 + 3C2
= 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 x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
= Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
= Coefficient of x12 in (1 – x)–1
= Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
= Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
= 12+3–1C3–1 – 7+3–1C3–1 – 6+3–1C3–1 + 1+3–1C3–1
= 14C2 – 9C2 – 8C2 + 3C2
= 91 – 36 – 28 + 3 = 30.
Equation reduces to x + y + t = 25
1
x 5, 12 y 18, t 0Required number of ways
= Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
= Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
= Coefficient of x12 in
(1 – x)–1= Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
= Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
= 12+3–1C3–1 – 7+3–1C3–1 – 6+3–1C3–1 + 1+3–1C3–1
= 14C2 – 9C2 – 8C2 + 3C2
= 91 – 36 – 28 + 3 = 30.
