Rationale: Total subjects = 4 +3 +3 = 10
Max subjects by 1 teacher = 2
So, min of teachers required = 10/2 = 5
The answer: (B)
Question: a school has 4 maths 3 physics and 3 chemistry teachers each teacher can teach 2 subjects max what is he minimum number of teachers required
Options:
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Reasoning: 7a/30 tells us that a is a factor of 30, which has 2, 3, and 5 as its prime factors. Each of them is distinct. Moreover, all multiples of 30 less than 200 can be derived by multiplying these prime factors alone. Thus, number of different prime factors is 3.
Ans: (B)
Question: In a is a positive integer smaller than 200 and (14a)/60 is an integer, then a has how many different positive prime factors?
Options:
(A) 2
(B) 3
(C) 5
(D) 6
(E) 8
Rationale: Each of the numbers except 28, is a multiple of 3.
The answer: (D)
Question: Select Odd number:
6, 9, 15, 21, 24, 28, 30
Options:
(A) 20
(B) 21
(C) 24
(D) 28
(E) 30
Reasoning: 44 + 28 + 10 = 82%
100 82 = 18%
850 * 18/100 = 153
Ans: (C)
Question: In a school of 850 boys, 44% of Muslims, 28% Hindus, 10% Sikhs and the remaining of other communities. How many belonged to the other communities?
Options:
(A) 125
(B) 627
(C) 153
(D) 721
(E) 159
Rationale: (x + 1)(2x - 2)(3x + 3) = ax3+bx2+cx+dax3+bx2+cx+d
6(x + 1)(x - 1)(x + 1) = ax3+bx2+cx+dax3+bx2+cx+d
6(x2x2 - 1)(x + 1) = ax3+bx2+cx+dax3+bx2+cx+d
6(x3+x2x1)6(x3+x2x1) = ax3+bx2+cx+dax3+bx2+cx+d
6x3+6x26x66x3+6x26x6 = ax3+bx2+cx+dax3+bx2+cx+d
Equating the coefficients of like learns on both the sides, a = 6, b = 6, c = -6 and d = -6.
The answer: (C)
Question: (x +1 )(2x - 2)(3x + 3) = ax3+bx2+cx+dax3+bx2+cx+d What are the values of a, b, c and d?
Options:
(A) 2
(B) 7
(C) 6
(D) 1
(E) 9
Rationale: I = (450*8*5)/100 = 180
The answer: (C)
Question: Find the simple interest on Rs.450 for 8 months at 5 paisa per month?
Options:
(A) 277
(B) 270
(C) 180
(D) 266
(E) 121