The number of roots between 0 and π of the equation 2sin2x + 1 = 3sinx equals
(A) 2 Correct Anwer : (D) 3 Solution: Now, 2sin2x + 1 = 3sinx 2sin2x – 3sinx + 1 = 0 (2sinx – 1)(sinx – 1) = 0 Option (4) is correct.
(B) 4
(C) 1
(D) 3
sinx = ½ or sinx = 1
x = π/6, π– π/6, x = π/2
For how many (not necessarily positive) integer values of n is the value of 4000 (2/5) n an integer ?
A) 3
B) 4
C) 6
D) 9
Philip lies a lot.
[1] He tells the truth on only one of the days of the week.
[2] The days of the week in order are:
Sunday, Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday,Monday,etc.
[3] One day he said: “I lie on Mondays and Tuesdays”.
[4] The next day he said: “Today is either Thursday, Saturday, or Sunday”.
[5] The next day he said: “I lie on Wednesdays and Fridays”.
On which day of the week does Philip tell the truth? Suppose his declaration in [3] is the only true one. Then, from his false declaration in [5], he made the declaration in [3] on a Wednesday or a Friday. Then, from [2], his false declaration in [4] was made on a Thursday or a Saturday. So his declaration in [5] is the only true one. Then, from his false declaration in [3], he made the declaration in [5] on a Monday or a Tuesday. Then, from [2], his false declaration in [4] was made on a Sunday or a Monday. His declaration in [4] would be true. So his declaration in [4] was made on a Monday. Then from [2], his true declaration in [5] was made on a Tuesday and Philip tells the truth on Tuesday.
From [1], Philip’s declarations in [3] and [5] cannot both be false; otherwise he tells the truth on more than one of the days of the week. From [1] and the fact that the declarations in [3] through [5] were made on consecutive days, Philip’s declarations in [3] and [5] cannot both be true; otherwise he tells the truth on more than one of the days of the week. So either Philip’s declaration in [3] is the only true one or his declaration in [5] is the only true one.
This situation is impossible.
Here is a list of words:
INN PEN PET PIE TEE TIE
[1] A wise man tells three knights, “I have told each of you a letter found in the three-letter word. I have told none of you the same letter.”
[2] He then tells them, “None of you can tell how many vowels the word has.”
[3] Then they are told, “At this point still none of you can tell how many vowels the word has.”
[4] Now knight no 3 says he knows what the word is.
What is the word?
From [1] and [2]. None of the three knights was told the letter N, since the words that contain ‘N’ have exactly one vowel. So the word is not INN or PEN. from [1] and [3]. None of the three knights was told Letter ‘I’, since after removing INN and PEN, whenever I is present word has exactly 2 vowels. Hence we are only left with PET and TEE from [4]. Letter ‘T’ and ‘E’ was not told to knight no. 4. Else he would not know the answer. So Knight 3 was given letter P. Hence the word is PET.
Chandni had twice as many blueberry jelly beans as cherry jelly beans. After eating 10 pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have?
1. 10
2. 20
3. 30
4. 40
Denote the number of blueberry and cherry jelly beans as b and c respectively. Then b = 2c and b – 10 = 3 (c – 10). Substituting, we have 2c – 10 = 3c – 30, so c = 20, b = 40.
The number 21! = 51090942171709440000 has over 60,000 positive integer divisors. One of them is chosen at random. What is the probability that it is odd ?
1. 1/21 21! = 218 × 39 × 54 × 73 × 11 × 13 × 17 × 19 Total factors = 19*10*5*4*2*2*2*2 since first multiple in above equation that is 19 is all the powers of 2, and out of 19 only 1 that is 20 is odd. In all other cases it is even, hence resultant factor will be even. The only thing that affects the parity of factors are the powers of 2. Therefore the answer is 1/ 19.
2. 1/18
3. 1/19
4. 1/2
A large bottle contains a collection of coins. When you examine the contents you are amazed to discover that there is an equal number of each type of denomination: 1 p, 2 p, 5 p, 10 p, 20 p, 50 p, Re 1, and Rs 2. Even more remarkable is that if you remove one coin there will be an exact amount in rupees. What is the smallest amount, in rupees, that could be in the bottle after removing that coin?
0.01 + 0.02 + 0.05 + 0.10 + 0.20 + 0.50 + 1 + 2 = 3.88 ( value in rupees) 1+2+5+10+20+50+100+200 = 388 (value is paise) => y = 388x-100I => y = 4(97x-25I) hence y should be a multiple of 4 Clearly y can not be 1, 2 , 5 , 10 , 50 For y = 20, minimum x will be 15 So, total amount = 388 × 15 = 5820, after removing 20 paise remaining value 5800 paise = 58 rupees
So, Total amount is 388x = 100I+y, where I is an integer and y is one of the coin in paise
For y = 100 and 200, minimum x will be 25
A three-dimensional rectangular box with dimensions X, Y and Z has faces whose surface areas are 24, 24, 48, 48, 72, and 72 square units. What is X + Y + Z?
Answer : 22
Solution :
XY = 24……(i)
YZ = 72….(ii)
XZ = 48….(iii)
(ii)/(i) = YZ/XY = 3 => Z/X = 3….(iv)
now (iv)*(iii) gives us Z2 = 144 => Z = 12
Following X = 4 and Y = 6.
X + Y + Z = 12 + 6 + 4 = 22.
