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Given below are two statements:
Statement I : A coin is tossed three times. The probability of getting exactly two heads is 3/8
Statement II: In tossing of 10 coins, the probability of getting exactly 5 heads is 63/256
In the light of the above statements, choose the correct answer from the options given below.
Statement I:
Probability of getting head = P(H) = $$\frac{1}{2}$$
Probability of getting tail = P(T) = $$\frac{1}{2}$$
3 coins are tossed and probability of getting 2 heads = $$3_{C_2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)$$ = $$\frac{3}{8}$$
Statement I is true.
Statement II:
10 coins are tossed and probability of getting exactly 5 heads = $$10_{C_5}\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^5$$ = $$\frac{63}{256}$$
Statement II is true
Answer is option A.
Which of the following statements regarding quadratic equations are true?
(A) Solution set for the equation $$6x^{2}-5x=4$$ is $$\left\{-\frac{1}{2},\frac{4}{3}\right\}$$
(B) The nature of the roots of the equation $$9x^{2}+6x+1=0$$ is equal, real and rational.
(C) If one root of $$4x^{2}-3x+K=0$$ is 3 times the other, then $$K=\frac{27}{256}$$
We have :
$$6x^2-5x-4\ =0\ $$
=$$6x^2-8x+3x-4\ =0\ $$
(2x+1)(3x-4)=0
so solution set is {-1/2 ,4/3}
$$9x^2+6x+1\ =0$$
D = 36-36 =0
So we can say roots are real and equal
$$4x^2-3x+k$$=0
let roots be p and 3p
we get 4p =3/4
p =3/16
Now k/4 = 3p^2
we get k = 12p^2
we get k = 27/64
So A and B are true and C is false
Given below are two statements
Statement I: In the sequence of numbers 30, 90, 182, 306, 462,P........ . the term P is 650.
Statement II : There are 8 digits in $$9^{8}$$ when it is expressed in decimal form.
In light of the above statements, choose the correct answer from the options given below
We have :
30, 90, 182, 306, 462,P.
Now 90-30 =60 (1)
182-90 = 92 (2)
306-182 = 124 (3)
462 -306 = 156
If you observe , the first difference of the terms form an arithmetic progression
So we can say the next term will be
462 +156+32 = 462 +188 = 650
So P is 650
So statement 1 is true .
Taking statement 2 :
we get 9^8
Now number of digits = [log9^8 +1]
we get [8log9+1]= 8
so number of digits in 9^8 is 8
Given below are two statements
Statement I : $$(543)_6$$ is equivalent to $$(317)_8$$
Statement II : The last 4 bits in the binary representation of a multiple of 16 is 1000.
In light of the above statements, choose the correct answer from the options given below
543 in base 6 expressed in decimal form =5(36)+4(6)+3 =207
317 in base 8 expressed in decimal form = 3(64)+8+7 =207
So statement 1 is correct
Now 16 in base 2 = 10000
32 in base 2 = 100000
Therefore statement 2 is false
Hence C is the correct answer
Given below are two statements
Statement I : The set of numbers(5,6, 7, p, 6, 7, 8, q) has an arithmetic mean of 6 and mode (most frequently occurring number) of 7. Then $$p\times q=16$$.
Statement II: Let p and q be two positive integers such that $$p+q+p\times q=94$$. Then $$p+q=20$$.
In light of the above statements, choose the correct answer from the options given below
Statement 1:
It is given that the mode of the set is 7. Thus, either p or q(or both) will be 7.
Now, p*q will be a multiple of 7. But it is given that p*q = 16. Thus, this statement is false.
Statement 2:
It is given that p + q + p*q = 94
If p + q = 20, then p*q = 74.
Thus, either (p,q) will be (2,37), (37,2), (1,74) or (74,1).
Since none of the values will give a sum of 20, this statement is also false.
Thus, the correct option is B.
Given below are two statements:
Statement I: $$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=2$$
Statement II: If $$a+b+c=0$$, then $$(a^{3}+b^{3}+c^{3})\div abc=3$$
In light of the above statements, choose the correct answer from the options given below
From statement 1 we get
Adding taking LCM
$$\frac{\left(\sqrt{\ 7}+\sqrt{\ 5}\right)^2+\left(\sqrt{\ 7}-\sqrt{\ 5}\right)^2}{7-5}$$
= 12
So statement 1 is false
Now statement 2 is an identity
so 2 is true .
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