MathematicsQ51–53 of 80 questions
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51
MathematicsEasyMCQWBJEE · 2012
An urn contains $8$ red and $5$ white balls. Three balls are drawn at random. Then,the probability that balls of both colours are drawn is
A
$\frac{40}{143}$
B
$\frac{70}{143}$
C
$\frac{3}{13}$
D
$\frac{10}{13}$
Solution
(D) Total number of ways to select $3$ balls from $13$ balls is ${}^{13}C_{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$.
The event that balls of both colours are drawn means we select either ($2$ red and $1$ white) or ($1$ red and $2$ white) balls.
Number of ways to select $2$ red and $1$ white ball $= {}^{8}C_{2} \times {}^{5}C_{1} = 28 \times 5 = 140$.
Number of ways to select $1$ red and $2$ white balls $= {}^{8}C_{1} \times {}^{5}C_{2} = 8 \times 10 = 80$.
Total favorable outcomes $= 140 + 80 = 220$.
Required probability $= \frac{220}{286} = \frac{10}{13}$.
The event that balls of both colours are drawn means we select either ($2$ red and $1$ white) or ($1$ red and $2$ white) balls.
Number of ways to select $2$ red and $1$ white ball $= {}^{8}C_{2} \times {}^{5}C_{1} = 28 \times 5 = 140$.
Number of ways to select $1$ red and $2$ white balls $= {}^{8}C_{1} \times {}^{5}C_{2} = 8 \times 10 = 80$.
Total favorable outcomes $= 140 + 80 = 220$.
Required probability $= \frac{220}{286} = \frac{10}{13}$.
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52
MathematicsEasyMCQWBJEE · 2012
The remainder obtained when $1! + 2! + 95!$ is divided by $15$ is
A
$14$
B
$3$
C
$1$
D
$0$
Solution
(B) We need to find the remainder of $(1! + 2! + 95!) \pmod{15}$.
First,calculate the factorials:
$1! = 1$
$2! = 2$
$3! = 6$
$4! = 24$
$5! = 120$
Since $5! = 120$ and $120 = 15 \times 8$,$5!$ is divisible by $15$.
Consequently,all factorials $n!$ for $n \geq 5$ are divisible by $15$.
Thus,$95! \equiv 0 \pmod{15}$.
The expression becomes $1! + 2! + 95! \equiv 1 + 2 + 0 \pmod{15} = 3 \pmod{15}$.
Therefore,the remainder is $3$.
First,calculate the factorials:
$1! = 1$
$2! = 2$
$3! = 6$
$4! = 24$
$5! = 120$
Since $5! = 120$ and $120 = 15 \times 8$,$5!$ is divisible by $15$.
Consequently,all factorials $n!$ for $n \geq 5$ are divisible by $15$.
Thus,$95! \equiv 0 \pmod{15}$.
The expression becomes $1! + 2! + 95! \equiv 1 + 2 + 0 \pmod{15} = 3 \pmod{15}$.
Therefore,the remainder is $3$.
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53
MathematicsEasyMCQWBJEE · 2012
Let $f(x)=ax^{2}+bx+c$ and $g(x)=px^{2}+qx+r$ such that $f(1)=g(1)$,$f(2)=g(2)$ and $f(3)-g(3)=2$. Then,$f(4)-g(4)$ is:
A
$4$
B
$5$
C
$6$
D
$7$
Solution
(C) Let $h(x) = f(x) - g(x) = (a-p)x^2 + (b-q)x + (c-r)$.
Since $f(1) = g(1)$,we have $h(1) = 0$.
Since $f(2) = g(2)$,we have $h(2) = 0$.
Since $h(x)$ is a quadratic polynomial with roots $1$ and $2$,we can write $h(x) = k(x-1)(x-2)$ for some constant $k$.
We are given $f(3) - g(3) = 2$,which means $h(3) = 2$.
Substituting $x=3$ into $h(x) = k(x-1)(x-2)$,we get $h(3) = k(3-1)(3-2) = k(2)(1) = 2k$.
Since $h(3) = 2$,we have $2k = 2$,which implies $k = 1$.
Thus,$h(x) = 1(x-1)(x-2) = (x-1)(x-2)$.
We need to find $f(4) - g(4)$,which is $h(4)$.
$h(4) = (4-1)(4-2) = (3)(2) = 6$.
Since $f(1) = g(1)$,we have $h(1) = 0$.
Since $f(2) = g(2)$,we have $h(2) = 0$.
Since $h(x)$ is a quadratic polynomial with roots $1$ and $2$,we can write $h(x) = k(x-1)(x-2)$ for some constant $k$.
We are given $f(3) - g(3) = 2$,which means $h(3) = 2$.
Substituting $x=3$ into $h(x) = k(x-1)(x-2)$,we get $h(3) = k(3-1)(3-2) = k(2)(1) = 2k$.
Since $h(3) = 2$,we have $2k = 2$,which implies $k = 1$.
Thus,$h(x) = 1(x-1)(x-2) = (x-1)(x-2)$.
We need to find $f(4) - g(4)$,which is $h(4)$.
$h(4) = (4-1)(4-2) = (3)(2) = 6$.
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