Q1. If a, b, c, d and e are prime numbers where a < b < c < d < e and a2 + b2 + c2 + d2 + e2 = even number then find the value of a5 .
a. 243 b. 32 c. 3125 d. 16807
Q2. If | 6 – r | = 7 and | 4q – 8 | = 12, what is the maximum value of q/r ?
a. 1 b. 13/5 c. 2 d. None of these Q3. If f(x + 1) = x3 + 6x2 + 12 x + 7, then what is f(x) ?
a. x3 + 5x2 + x b. x3 + 2x2 + 5x c. x3 + 6x2 - 7 d. None of these Q4. If a, b and c are distinct positive integers, then find the product of
(a + b)(b + c)(c + a).
a. >8abc b. <8abc c. =8abc d. None of these
Q5. The harmonic mean of two positive real numbers is 4. Their arithmetic mean A and their geometric mean G satisfy the relation 2A + G2 = 27 . Find the two numbers ?
a. 4 and 4 b. 2 and 6 c. 3 and 6 d. 5 and 10/3 |
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a. 243 b. 32 c. 3125 d. 16807
Sol.
the square of even and odd numbers are even and odd respectively.
if the sum of five numbers is even; then 1, 3 or all 5 are even.
but in a set of prime numbers only one number is even which is 2.
Ans. 32
Q2. If | 6 – r | = 7 and | 4q – 8 | = 12, what is the maximum value of q/r ?
a. 1 b. 13/5 c. 2 d. None of these
Sol.
Either r is -1 or 13.
Either q is -1 or 5.
maximum value of q/r is (-1)/(-1) = 1
Q3. If f(x + 1) = x3 + 6x2 + 12 x + 7, then what is f(x) ?
a. x3 + 5x2 + x b. x3 + 2x2 + 5x c. x3 + 6x2 - 7 d. None of these
Sol. go through option,
If f(x) = x3 + 5x2 + x
then f(x +1) = (x +1)3 + 5(x +1)2 +(x +1)
f(x + 1) = x3 + 6x2 + 12 x + 7
Sol.
If a, b and c are distinct positive integers,
(a + b +c)>3(abc)1/3
[(a+b)+(b+c)+(c+a)]>3(8abc)1/3........ (i)
[(a+b)+(b+c)+(c+a)]>3[(a+b)(b+c)(c+a)]1/3........ (ii)
from (i) and (ii)
Cant say which is greater.
a. 4 and 4 b. 2 and 6 c. 3 and 6 d. 5 and 10/3
Sol.
Go through option,
(i)Let numbers 4 and 4, then H is 4.
and 2A+G2 is not equal 27.
(ii)Let numbers 2 and 6, then H is not equal 4.
(iii)Let numbers 3 and 6, then H is 4.
and 2A+G2 is equal 27.
@ Humraj,
or simply put x = x-1 :-)
regards,
shaheen
the best way to solve such pbms is to assume some values of a,b,and c
let a=1,b=2,c=3
(a + b)(b + c)(c + a)= 60
8abc = 48
to be sure chk for one more set,
a=3,b=4,c=5
(a + b)(b + c)(c + a)= 504
8abc = 480
hence (a + b)(b + c)(c + a)> 8abc
regards,
shaheen
let H be harmonic mean and x,y be two numbers
by defn,
2/H = 1/x + 1/y
2/H = (x + y)/xy
now x + y = 2A
and xy = G^2
hence,2/H = 2A/G^2
==> G^2 = A*H
In short,remember G.M. is the geometric mean of A.M. and H.M.
Given,2A + G^2 = 27 and H = 4
G^2/2 + G^2 = 27
hence G^2 = 18
we see that onlu option c) i.e numbers 3 and 6 satisfy this... :-)
regards,
shaheen
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