Algebra
1.If n is the sum of two consecutive odd integers and less than 100, what is the greatest possibility of n?
Solution:
Taking greatest possibility
we take 51 + 49 but 51 + 49 = 100,
we need x + y < 100
so we take
47 + 49 = 96
2. If the roots of the equation (x + 1) (x + 9) + 8 = 0 are a and b, then the roots of the equation (x + a) (x + b) - 8 = 0 are
Solution:
(x + 1) (x + 9) + 8 = 0
=> (x^2) + 10x + 17 = 0
Hence a+b = -10 and ab=17
(x + a) (x + b) - 8 = 0
=> (x^2) + (a+b)x + ab - 8 = 0
(x^2) + 10x + 17 - 8 = 0
=> x = 1,9
3. If F(X) = AF(X)4 + B(X)2 + X + 5 , F(– 3) = 2, then F(3) = ?
Solution:
f(-3)=af(-3)4+bf(-3)+(-3)+5
2=a*2*4+b*2+2
8a+4b=0
f(3)=af(3)4+bf(3)2+3+5
f(3)[1-4a-2b]=_8
multiply by 2 both sides
f(3)[2-(8a+4b)]=16
f(3)=8
(or)
81a+9b-3+5=2
=>81a+9b+5=5
now f(3)
81a+9b+3+5
=5+3=8
4. 4^2 + 2* 5^2 + 3* 6^2 +4* 7^2+...........+27*30^2 ?
Solution:
∑ r^2 = n(n+1)(2n+1)/6
∑ r^3 = [n(n+1)/2]
∴ 4^2+(5^3+6^3+......30^3)-3(5^2+6^2+.....30^2)
=16+(216225-100)-3*(9455-30)
=187866
5. a, b, c are non negative integers such that 28a+30b+31c=365. Then a+b+c is?
Solution:
ans is always 12.
here a=1 i.e february.
b=4 (april,june,sept,nov)
c=7 (jan,march,may,july,aug,oct,dec)
6. 77!*(77!-2*54!)^3/(77!+54!)^3 + 54!*(2*77!-54!)^3/(77!+54!)^3
Solution:
after evaluating the given problem we get
(77!-54!)*(77!+54!)^3/(77!+54!)^3=(77!-54!)
let 77!=a;
and 54!=b;
it becomes a simple alzebra.
7. Let f(x)= 1+x+x^2+....x^6. what is the remainder when f(x^7) is divided by f(x).
Solution:
geometric progression formula= sum = a( r^m -1)/ (r-1)
in given series f(x)= 1+x+x^2+x^3+x^4+x^5+x^6
m=7
a=1
r=x
so, f(x)= x^7-1/(x-1)
f(x)(x-1)=x^7-1
f(x)(x-1)+1= x^7
f[f(x)(x-1)+1]= f(x^7)
f(x^7)mod f(x) = f[f(x)(x-1)+1] mod f(x)
= f(1)
= 1+1+1+1+1+1+1= 7 (ans)
8. If f(x) = 2x+2 what is f(f(3))?
Solution:
f(x) = 2x+2
f(3) = 2*3+2=8
f(f(3))= 2*f(3)+2
= 2*8+2 = 18
9. If f(x)=x^4+ax^3+bx^2+cx+d such that
f(1)=f(2)=f(3)=f(4), find the value of b.
Solution:
ans b=35
f(1)=1+a+b+c+d
f(2)=16+8a+4b+2c+d
f(3)=81+27a+9b+3c+d
f(4)=256+64a+16b+4c+d
given
f(1)=f(2)=f(3)=f(4)
firstly
f(1)=f(2)
1+a+b+c+d=16+8a+4b+2c+d
7a+3b+c=-15...........equ(1)
then
f(1)=f(3)
we get
13a+4b+c=-40..........equ(2)
then
f(1)=f(4)
we get
21a+5b+c=-85.........equ(3)
solve equ(2)-equ(1)
we get
6a+b=-25..............equ(4)
solve equ(3)-equ(2)
we get
8a+b=-45...............equ(5)
solve equ(4) and equ(5) and put the value in other equactions for all the values
we get
a=-10 ,b=35,c=-50
proof
=put the value of a ,b,c
we get
f(1)=f(2)=f(3)=f(4)
10. What is the value of x?
x - 2y = 7
4x – 28 = 8y
Solution:
both are same equations
4X(x-2y=7) is 4x-8y=28
so no solution exists
11. 2^x/(1+2^x) = 1/4 so 8^x/(1+8^x) = ?
Solution:
(1+2^x)/2^x=4
1/2^x+1=4
2^x=1/3
(2^3)^x/(1+(2^3)^x)
(2^x)^3/((2^x)^3+1)
by substituting 2^x=1/3
we get 1/28
12. if the equation A+B+C+D+E=FG is the two digit number whose value is 10F+G and the letters A,B,C,D,E,F and G each represent different digits. If FG is as large as possible, what is the value of G?
Solution:
FG is as large as possible and all the 7 numbers should be different.
Let’s try out a few possibilities..
9 + 8 + 7 + 6 + 5 = 35…5 is getting repeated twice.
9 + 8 + 7 + 6 + 4 = 34…4 is getting repeated
9 + 8 + 7 + 5 + 4 = 33…3 repeats
9 + 8 + 6 + 5 + 4 = 32
None of the numbers repeat in the above case and 32 is the maximum number FG can have. The value of G is 2.
13. How many different sums can be formed with the coins,5 rupee,1 rupee,50 paisa,25 paisa,10 paisa and 1 paisa?
Solution:
total no. of combination = 6C1+6C2+6C3+6C4+6C5+6C6
=> 2^6-1 (As nC0+nC1+......+nCn = 2^n i.e including zero)
=63
14. Find the no of zeros in the product of 1^1*2^2*3^3*.....*49^49?
Solution:
consider the case of multiples of 5
5^5 x 10^10 x 15^15 x 20^20 x 25^25 x 30^30 x 35^35 x 40^40 x 45^45
or
5^5 x 10^10 x 15^15 x 20^20 x (5^2)^25 x 30^30 x 35^35 x 40^40 x 45^45
or
5^5 x 10^10 x 15^15 x 20^20 x 5^50 x 30^30 x 35^35 x 40^40 x 45^45
total zeros at the end of product 5^5 x 10^10 x 15^15 x 20^20 x 25^25 x 30^30 x 35^35 x 40^40 x 45^45 will be
5+10+15+20+50+30+35+40+45
= 250 zeros
15. In the question A^B means A raised to the power B,f(x)=ax^4-bx^2+x+5,f(-3)=2,then f(3)=?
Solution:
Ans is:8
f(-3)=2
2=a(-3)^4-b(-3)^2-3+5
2=81a-9b+2
81a-9b--->1
Then f(3)
81a-9b=81a-9b+3+5-->2
Solve the eq..we can get 8
16. The number of natural numbers n such that (n+1)^2/(n+7) is an integer, is
Solution:
(n+1)^2/n+ 7 = k
k= (n+7)^2 /n+7 – (12n + 48)/n+7
if k is integer then (12 + 48)/n+7 has to be an integer.
12n +48/n+7=12(n+4)/n+7.
= 12 (n+7 -3)/n+7
36/n+7 has to be an integer.
n+7 = 9 ,12 , 18 , 36
n can take 4 values
17. find the last digit of the following expression:
4+9^2+4^3+9^4+4^5+9^6+....+9^100
Solution:
There are two series here: series 1 and series 2
In the series 1:-
4+4^3+4^5+.....+4^99
last digit is 0
In the series 2:-
9^2+9^4+9^6+.....+9^100
last digit is also 0 here.
18. Function f satisfies the equation f(x)+2*f(6-x)=x for all real numbers x. Find the value of f(1)
Solution:
let x=1 then, f(1)+2f(5)=1-------(1)
let x=5 then, f(5)+2f(1)=5-------(2)
from (2) f(5)=5-2f(1) sub in (1)
f(1)+2(5-2f(1))=1
f(1)+10-4f(1)=1
-3f(1)=-9
f(1)=3
ans=3
19. N is an integers and N>2 at most how many integers among N+2,N+4,N+5,N+6 and N+7 are prime integers?
Solution:
For N=3 we have - 5,7,8,9,10 so 2 prime
For N=4 we have - 6,8,9,10,11 so 1 prime
For N=5 we have - 7,9,10,11,12 so 2 prime
For N=6 we have - 8,10,11,12,13 so 2 prime
& so on....
So max no of Primes are 2
20. (98*98*98 – 73*73*73)/( 98*98*98 – 73*73*73)=?
Solution:
its a Dummy question Dont attend this type of questions
21. x^y+y^x=46
find x,y
Solution:
take x = 45
y = 1
45^1 + 1^45 = 46
22. Given, a(i) = a(i-1)+a(i-2) where i>=3
Now if a10=322 then what is the value of a5?
hints:- a10=a9+a8, a9=a8+a7 and so on
Solution:
We can solve this by using the relation as
a3=a2+a1,
a4=a3+a2
and the put the value of a3.....
similarly upto a10 which is equal to
a10=34a2+21a1 and a10=322,
so by putting
a1=4 and
a2=7 we get the result as
a5=29
23. If (3a+6b)/(5a+12b) =(12/23) determine the value of (3a2 +5b2 )/(abs)
Solution:
3a+6b/5a+12b=12/23
=>after calculating we gt
a/b=2/3
now 3a^2+5b^2/ab
=> 3a/b + 5b/a
=> 3 *2/3 + 5*3/2
=> 2+15/2
=> 19/2
24. Assume that f(1)=0 and f(m+n) = f(m) +f(n) +4(9mn-1) for all natural numbers (integers<0) m & n. what is the value of f(17)?
Solution:
f(1+1)= 0+0+4*(9-1)=32
f(2+2)=32+32+4*35=204
f(4+4)=204+204+4*143=980
f(8+8)=980+980+4*(9*64-1)=4260
f(1+16)=4260+0+4(9*16-1)=4832(ans)
25. sum of the digits in the product (16^100)*(125^135)
Solution:
(16^100)*(125^135)= (2^400)*(5^405)
which means (2*5)^400*(5^5)
3+1+2+5+0+0+0..........=11
26. F(x)=ax+b where a and b are real numbers if f(f(f(x)))=8x+21 what is the value of a+b?
Solution:
f(x)=ax+b;
f[f(x)]=a^x+ab+b
f{f[f(f)]}=xa^3+a^2b+ab+b=8x+21
xa^3=8x ==>a=2
ba^2+ab+a=21 ,, put value of a=2
then b=3
ANS:=
a+b=5
27. In the polynomial f(x) =2*x^4 - 49*x^2 +54, what is the product of the roots, and what is the sum of the roots (Note that x^n denotes the x raised to the power n, or x multiplied by itself n times)?
Solution:
let x1 and x2 be two roots of the equation then
sum of roots is x1+x2= (-b)/a= 49/2
and product is given by x1 * x2
= c/a=54/2
=27
28. how many integer x satisfies the equation (x^2-x-1)^x+2 = 1
Solution:
ans is 4
if x=2,-2,-1,0
29. Find the value of "n" where 348 + 31996 +33943+33n.
Solution:
(3^16 ) ^3+3^1996+3^3943+(3^n)^3
So it is equivalent to a^3+3a^2b+3ab^2+b^3
So,3*(3^16)^2*3^n=3^1996
3^(33+n)=3^1996
n=1963
30. x,y,z are positive integers if x(y+z)=63, y(z+x)=40, z(y+x)=33. what is the value of x+y+z?
Solution:
xy+xz=63
yz+yx=40
zy+zx=33
2(xy+xz+yz)=63+33+40
xy+yz+xz=136/2
xy+z(y+x)=68
xy=68-33
xy=35=7*5
xz=28=7*4
x=7
y=5
z=4
x+y+z=16
31. In this question A^B means A is raise to the power B.let f(x)=1+x+x^2+x^3+.........+x^6.The reminder when f(x^7) is divided by f(x) is?
Solution:
(1+x^7 + x^14 + x^21 .........+ x^42)/ (1+x+x^2+x^3.......x^6)
=(sum of gp with first term 1 and common ratio x^7) /
(sum of go with first term 1 and common ratio x)
=[1*(x^49 - 1)/(x-1)] / [1*(x^7 - 1)/(x-1)]
=(x^49 - 1)/(x^7 - 1)
According to remainder theorem,
Deno=0
x^7 - 1=0 , x=1
Put in numerator,
Remainder= 1^49 -1=0
32. f a(b+c)=33
b(c+a)=40
c(a+b)=49
then find abc
Solution:
a(b+c)=33
so a=11 or 3
b+c=11 or 3
c(a+b)=49
c=7
a+b=7
so a=3
and b=4
and c=7
abc=84
33. If f(x) is 7x+12.then,f-1(x) (f inverse of x)is
Solution:
let f(x)=7x+12=y
=> 7x=y-12
=> x=(y-12)/7
=> f-1(x)=(x-12)/7
34. If (3a+6b)/(5a+12b)=12/23 determine the value of 3a2+5b2/ab
Solution:
(3a+6b)/(5a+12b)=12/23---->eq 1
sol eq--->1 the v vil get a/b=2/3
div 'ab 'to the num and den for 3a2+5b2/ab
therfore, 3*(a/b)+5*(ba)
sub a/b=2/3 in ab eqtn
v get 19/2
35. Suppose that the function f is defined on natural numbers (integers>0) such that f(1)=1 &f(x+y)=f(x)+f(y)+3xy+3 for all x,y.whart is the value of f(5)+f(7)?
Solution:
f(1)=1
f(2)=f(1)+f(1)+3*1*1+3=8
f(3)=f(1)+f(2)+3*1*2+3=1+8+6+3=18
f(4)=f(1)+f(3)+3*1*3+3=1+8+6+3=1+18+9+3=31
f(5)=f(1)+f(4)+3*1*4+3=47
f(7)=f(4)+f(3)+3*4*3+3=88
Hence f(5)+f(7)=135
36.(1-1/3^2) (1-1/4^2) (1-1/5^4)----------(1-1/2015^2) = x/2015 then x value
1+1/3)(1-1/3)(1+1/4)(1-1/4)......(1-1/2015)(1+1/2015)=x/2015
2/3*4/3*3/4*5/4.......................2014/2015*2016/2015=x/2015
2/3*2016/2015=x/2015
after solving
x=1344.
37. the number 2^32 -1 has several divisors greater than 1 and less than 100.the sum of these divisors is clue:2^2n+1 is a prime for n=1,2,3,4
Solution:
To solve this qn we have use this fact--
(2^2n)+1 = prime for n=1,2,3,4 & the formula
a2-b2 = (a+b)(a-b)
Now
(2^32 )-1 = (2^16 -1) (2^16+1)
= (2^8 -1) (2^8+1)(2^16+1)
= (2^4 -1) (2^4+1) (2^8+1)(2^16+1)
Repeting the formula again & again we get:
(2^32 )-1 = (2 -1) (2 +1)(2^2 +1) (2^4+1) (2^8+1)(2^16+1)
= 1 x 3 x 5 x 17 x (factor larger than 100) x (factor larger than 100)
So factors smaller than 100 are 3,5,17,15,51,85
So sum
= 3+5+17+15+51+85
= 176
38. a+b=4,ab=3,find the sum of receiprocals of a and b
Solution:
1/a+1/b=a+b/ab,
so 1/a+1/b=a+b/ab
=4/3
39. G(0)=-1, G(1)=1, G(N)=G(N-1) - G(N-2), G(5)= ?
Solution:
G(2)=G(1)-G(0)=1-(-1)=2
G(3)=G(2)-G(1)=2-1=1
G(4)=G(3)-G(2)=1-2=-1
G(5)=G(4)-G(3)=-1-1=-2
40. Let F(m,n) =45*m+36*n,where m and n are integers (+ve or -ve) What is the minimum positive value for f(m,n) (This may be achived for various values of m and n)?
Solution:
ans.9 bcz f(m,n)=45*m+36*n
putting m=1,n=-1 in above eqn we get
f(m,n)=45*1+36*(-1)
=9
41. abc+abc+abc=abm,,,m=c than a,b???
Solution:
put a= b/1.5 & b= a/2 & c=m
now, abc+abc+abc
= 3abc
= 3*(b/1.5)*(a/2)*m
= 3*(abm/3)
= abm
42. 3X/5Y = 5Y/3X. Find the value of X/Y
Solution:
3x*3x=5y*5y or (3x)^2=(5y)^2 or 3x=5y
x/y=5/3
43. Possible value of x for equation
(x^2-x-1)^(x+2)=1 ?
Solution:
ans are 0,1,2,-2
44.
Solution:
Taking greatest possibility
we take 51 + 49 but 51 + 49 = 100,
we need x + y < 100
so we take
47 + 49 = 96
2. If the roots of the equation (x + 1) (x + 9) + 8 = 0 are a and b, then the roots of the equation (x + a) (x + b) - 8 = 0 are
Solution:
(x + 1) (x + 9) + 8 = 0
=> (x^2) + 10x + 17 = 0
Hence a+b = -10 and ab=17
(x + a) (x + b) - 8 = 0
=> (x^2) + (a+b)x + ab - 8 = 0
(x^2) + 10x + 17 - 8 = 0
=> x = 1,9
3. If F(X) = AF(X)4 + B(X)2 + X + 5 , F(– 3) = 2, then F(3) = ?
Solution:
f(-3)=af(-3)4+bf(-3)+(-3)+5
2=a*2*4+b*2+2
8a+4b=0
f(3)=af(3)4+bf(3)2+3+5
f(3)[1-4a-2b]=_8
multiply by 2 both sides
f(3)[2-(8a+4b)]=16
f(3)=8
(or)
81a+9b-3+5=2
=>81a+9b+5=5
now f(3)
81a+9b+3+5
=5+3=8
4. 4^2 + 2* 5^2 + 3* 6^2 +4* 7^2+...........+27*30^2 ?
Solution:
∑ r^2 = n(n+1)(2n+1)/6
∑ r^3 = [n(n+1)/2]
∴ 4^2+(5^3+6^3+......30^3)-3(5^2+6^2+.....30^2)
=16+(216225-100)-3*(9455-30)
=187866
5. a, b, c are non negative integers such that 28a+30b+31c=365. Then a+b+c is?
Solution:
ans is always 12.
here a=1 i.e february.
b=4 (april,june,sept,nov)
c=7 (jan,march,may,july,aug,oct,dec)
6. 77!*(77!-2*54!)^3/(77!+54!)^3 + 54!*(2*77!-54!)^3/(77!+54!)^3
Solution:
after evaluating the given problem we get
(77!-54!)*(77!+54!)^3/(77!+54!)^3=(77!-54!)
let 77!=a;
and 54!=b;
it becomes a simple alzebra.
7. Let f(x)= 1+x+x^2+....x^6. what is the remainder when f(x^7) is divided by f(x).
Solution:
geometric progression formula= sum = a( r^m -1)/ (r-1)
in given series f(x)= 1+x+x^2+x^3+x^4+x^5+x^6
m=7
a=1
r=x
so, f(x)= x^7-1/(x-1)
f(x)(x-1)=x^7-1
f(x)(x-1)+1= x^7
f[f(x)(x-1)+1]= f(x^7)
f(x^7)mod f(x) = f[f(x)(x-1)+1] mod f(x)
= f(1)
= 1+1+1+1+1+1+1= 7 (ans)
8. If f(x) = 2x+2 what is f(f(3))?
Solution:
f(x) = 2x+2
f(3) = 2*3+2=8
f(f(3))= 2*f(3)+2
= 2*8+2 = 18
9. If f(x)=x^4+ax^3+bx^2+cx+d such that
f(1)=f(2)=f(3)=f(4), find the value of b.
Solution:
ans b=35
f(1)=1+a+b+c+d
f(2)=16+8a+4b+2c+d
f(3)=81+27a+9b+3c+d
f(4)=256+64a+16b+4c+d
given
f(1)=f(2)=f(3)=f(4)
firstly
f(1)=f(2)
1+a+b+c+d=16+8a+4b+2c+d
7a+3b+c=-15...........equ(1)
then
f(1)=f(3)
we get
13a+4b+c=-40..........equ(2)
then
f(1)=f(4)
we get
21a+5b+c=-85.........equ(3)
solve equ(2)-equ(1)
we get
6a+b=-25..............equ(4)
solve equ(3)-equ(2)
we get
8a+b=-45...............equ(5)
solve equ(4) and equ(5) and put the value in other equactions for all the values
we get
a=-10 ,b=35,c=-50
proof
=put the value of a ,b,c
we get
f(1)=f(2)=f(3)=f(4)
10. What is the value of x?
x - 2y = 7
4x – 28 = 8y
Solution:
both are same equations
4X(x-2y=7) is 4x-8y=28
so no solution exists
11. 2^x/(1+2^x) = 1/4 so 8^x/(1+8^x) = ?
Solution:
(1+2^x)/2^x=4
1/2^x+1=4
2^x=1/3
(2^3)^x/(1+(2^3)^x)
(2^x)^3/((2^x)^3+1)
by substituting 2^x=1/3
we get 1/28
12. if the equation A+B+C+D+E=FG is the two digit number whose value is 10F+G and the letters A,B,C,D,E,F and G each represent different digits. If FG is as large as possible, what is the value of G?
Solution:
FG is as large as possible and all the 7 numbers should be different.
Let’s try out a few possibilities..
9 + 8 + 7 + 6 + 5 = 35…5 is getting repeated twice.
9 + 8 + 7 + 6 + 4 = 34…4 is getting repeated
9 + 8 + 7 + 5 + 4 = 33…3 repeats
9 + 8 + 6 + 5 + 4 = 32
None of the numbers repeat in the above case and 32 is the maximum number FG can have. The value of G is 2.
13. How many different sums can be formed with the coins,5 rupee,1 rupee,50 paisa,25 paisa,10 paisa and 1 paisa?
Solution:
total no. of combination = 6C1+6C2+6C3+6C4+6C5+6C6
=> 2^6-1 (As nC0+nC1+......+nCn = 2^n i.e including zero)
=63
14. Find the no of zeros in the product of 1^1*2^2*3^3*.....*49^49?
Solution:
consider the case of multiples of 5
5^5 x 10^10 x 15^15 x 20^20 x 25^25 x 30^30 x 35^35 x 40^40 x 45^45
or
5^5 x 10^10 x 15^15 x 20^20 x (5^2)^25 x 30^30 x 35^35 x 40^40 x 45^45
or
5^5 x 10^10 x 15^15 x 20^20 x 5^50 x 30^30 x 35^35 x 40^40 x 45^45
total zeros at the end of product 5^5 x 10^10 x 15^15 x 20^20 x 25^25 x 30^30 x 35^35 x 40^40 x 45^45 will be
5+10+15+20+50+30+35+40+45
= 250 zeros
15. In the question A^B means A raised to the power B,f(x)=ax^4-bx^2+x+5,f(-3)=2,then f(3)=?
Solution:
Ans is:8
f(-3)=2
2=a(-3)^4-b(-3)^2-3+5
2=81a-9b+2
81a-9b--->1
Then f(3)
81a-9b=81a-9b+3+5-->2
Solve the eq..we can get 8
16. The number of natural numbers n such that (n+1)^2/(n+7) is an integer, is
Solution:
(n+1)^2/n+ 7 = k
k= (n+7)^2 /n+7 – (12n + 48)/n+7
if k is integer then (12 + 48)/n+7 has to be an integer.
12n +48/n+7=12(n+4)/n+7.
= 12 (n+7 -3)/n+7
36/n+7 has to be an integer.
n+7 = 9 ,12 , 18 , 36
n can take 4 values
17. find the last digit of the following expression:
4+9^2+4^3+9^4+4^5+9^6+....+9^100
Solution:
There are two series here: series 1 and series 2
In the series 1:-
4+4^3+4^5+.....+4^99
last digit is 0
In the series 2:-
9^2+9^4+9^6+.....+9^100
last digit is also 0 here.
18. Function f satisfies the equation f(x)+2*f(6-x)=x for all real numbers x. Find the value of f(1)
Solution:
let x=1 then, f(1)+2f(5)=1-------(1)
let x=5 then, f(5)+2f(1)=5-------(2)
from (2) f(5)=5-2f(1) sub in (1)
f(1)+2(5-2f(1))=1
f(1)+10-4f(1)=1
-3f(1)=-9
f(1)=3
ans=3
19. N is an integers and N>2 at most how many integers among N+2,N+4,N+5,N+6 and N+7 are prime integers?
Solution:
For N=3 we have - 5,7,8,9,10 so 2 prime
For N=4 we have - 6,8,9,10,11 so 1 prime
For N=5 we have - 7,9,10,11,12 so 2 prime
For N=6 we have - 8,10,11,12,13 so 2 prime
& so on....
So max no of Primes are 2
20. (98*98*98 – 73*73*73)/( 98*98*98 – 73*73*73)=?
Solution:
its a Dummy question Dont attend this type of questions
21. x^y+y^x=46
find x,y
Solution:
take x = 45
y = 1
45^1 + 1^45 = 46
22. Given, a(i) = a(i-1)+a(i-2) where i>=3
Now if a10=322 then what is the value of a5?
hints:- a10=a9+a8, a9=a8+a7 and so on
Solution:
We can solve this by using the relation as
a3=a2+a1,
a4=a3+a2
and the put the value of a3.....
similarly upto a10 which is equal to
a10=34a2+21a1 and a10=322,
so by putting
a1=4 and
a2=7 we get the result as
a5=29
23. If (3a+6b)/(5a+12b) =(12/23) determine the value of (3a2 +5b2 )/(abs)
Solution:
3a+6b/5a+12b=12/23
=>after calculating we gt
a/b=2/3
now 3a^2+5b^2/ab
=> 3a/b + 5b/a
=> 3 *2/3 + 5*3/2
=> 2+15/2
=> 19/2
24. Assume that f(1)=0 and f(m+n) = f(m) +f(n) +4(9mn-1) for all natural numbers (integers<0) m & n. what is the value of f(17)?
Solution:
f(1+1)= 0+0+4*(9-1)=32
f(2+2)=32+32+4*35=204
f(4+4)=204+204+4*143=980
f(8+8)=980+980+4*(9*64-1)=4260
f(1+16)=4260+0+4(9*16-1)=4832(ans)
25. sum of the digits in the product (16^100)*(125^135)
Solution:
(16^100)*(125^135)= (2^400)*(5^405)
which means (2*5)^400*(5^5)
3+1+2+5+0+0+0..........=11
26. F(x)=ax+b where a and b are real numbers if f(f(f(x)))=8x+21 what is the value of a+b?
Solution:
f(x)=ax+b;
f[f(x)]=a^x+ab+b
f{f[f(f)]}=xa^3+a^2b+ab+b=8x+21
xa^3=8x ==>a=2
ba^2+ab+a=21 ,, put value of a=2
then b=3
ANS:=
a+b=5
27. In the polynomial f(x) =2*x^4 - 49*x^2 +54, what is the product of the roots, and what is the sum of the roots (Note that x^n denotes the x raised to the power n, or x multiplied by itself n times)?
Solution:
let x1 and x2 be two roots of the equation then
sum of roots is x1+x2= (-b)/a= 49/2
and product is given by x1 * x2
= c/a=54/2
=27
28. how many integer x satisfies the equation (x^2-x-1)^x+2 = 1
Solution:
ans is 4
if x=2,-2,-1,0
29. Find the value of "n" where 348 + 31996 +33943+33n.
Solution:
(3^16 ) ^3+3^1996+3^3943+(3^n)^3
So it is equivalent to a^3+3a^2b+3ab^2+b^3
So,3*(3^16)^2*3^n=3^1996
3^(33+n)=3^1996
n=1963
30. x,y,z are positive integers if x(y+z)=63, y(z+x)=40, z(y+x)=33. what is the value of x+y+z?
Solution:
xy+xz=63
yz+yx=40
zy+zx=33
2(xy+xz+yz)=63+33+40
xy+yz+xz=136/2
xy+z(y+x)=68
xy=68-33
xy=35=7*5
xz=28=7*4
x=7
y=5
z=4
x+y+z=16
31. In this question A^B means A is raise to the power B.let f(x)=1+x+x^2+x^3+.........+x^6.The reminder when f(x^7) is divided by f(x) is?
Solution:
(1+x^7 + x^14 + x^21 .........+ x^42)/ (1+x+x^2+x^3.......x^6)
=(sum of gp with first term 1 and common ratio x^7) /
(sum of go with first term 1 and common ratio x)
=[1*(x^49 - 1)/(x-1)] / [1*(x^7 - 1)/(x-1)]
=(x^49 - 1)/(x^7 - 1)
According to remainder theorem,
Deno=0
x^7 - 1=0 , x=1
Put in numerator,
Remainder= 1^49 -1=0
32. f a(b+c)=33
b(c+a)=40
c(a+b)=49
then find abc
Solution:
a(b+c)=33
so a=11 or 3
b+c=11 or 3
c(a+b)=49
c=7
a+b=7
so a=3
and b=4
and c=7
abc=84
33. If f(x) is 7x+12.then,f-1(x) (f inverse of x)is
Solution:
let f(x)=7x+12=y
=> 7x=y-12
=> x=(y-12)/7
=> f-1(x)=(x-12)/7
34. If (3a+6b)/(5a+12b)=12/23 determine the value of 3a2+5b2/ab
Solution:
(3a+6b)/(5a+12b)=12/23---->eq 1
sol eq--->1 the v vil get a/b=2/3
div 'ab 'to the num and den for 3a2+5b2/ab
therfore, 3*(a/b)+5*(ba)
sub a/b=2/3 in ab eqtn
v get 19/2
35. Suppose that the function f is defined on natural numbers (integers>0) such that f(1)=1 &f(x+y)=f(x)+f(y)+3xy+3 for all x,y.whart is the value of f(5)+f(7)?
Solution:
f(1)=1
f(2)=f(1)+f(1)+3*1*1+3=8
f(3)=f(1)+f(2)+3*1*2+3=1+8+6+3=18
f(4)=f(1)+f(3)+3*1*3+3=1+8+6+3=1+18+9+3=31
f(5)=f(1)+f(4)+3*1*4+3=47
f(7)=f(4)+f(3)+3*4*3+3=88
Hence f(5)+f(7)=135
36.(1-1/3^2) (1-1/4^2) (1-1/5^4)----------(1-1/2015^2) = x/2015 then x value
1+1/3)(1-1/3)(1+1/4)(1-1/4)......(1-1/2015)(1+1/2015)=x/2015
2/3*4/3*3/4*5/4.......................2014/2015*2016/2015=x/2015
2/3*2016/2015=x/2015
after solving
x=1344.
37. the number 2^32 -1 has several divisors greater than 1 and less than 100.the sum of these divisors is clue:2^2n+1 is a prime for n=1,2,3,4
Solution:
To solve this qn we have use this fact--
(2^2n)+1 = prime for n=1,2,3,4 & the formula
a2-b2 = (a+b)(a-b)
Now
(2^32 )-1 = (2^16 -1) (2^16+1)
= (2^8 -1) (2^8+1)(2^16+1)
= (2^4 -1) (2^4+1) (2^8+1)(2^16+1)
Repeting the formula again & again we get:
(2^32 )-1 = (2 -1) (2 +1)(2^2 +1) (2^4+1) (2^8+1)(2^16+1)
= 1 x 3 x 5 x 17 x (factor larger than 100) x (factor larger than 100)
So factors smaller than 100 are 3,5,17,15,51,85
So sum
= 3+5+17+15+51+85
= 176
38. a+b=4,ab=3,find the sum of receiprocals of a and b
Solution:
1/a+1/b=a+b/ab,
so 1/a+1/b=a+b/ab
=4/3
39. G(0)=-1, G(1)=1, G(N)=G(N-1) - G(N-2), G(5)= ?
Solution:
G(2)=G(1)-G(0)=1-(-1)=2
G(3)=G(2)-G(1)=2-1=1
G(4)=G(3)-G(2)=1-2=-1
G(5)=G(4)-G(3)=-1-1=-2
40. Let F(m,n) =45*m+36*n,where m and n are integers (+ve or -ve) What is the minimum positive value for f(m,n) (This may be achived for various values of m and n)?
Solution:
ans.9 bcz f(m,n)=45*m+36*n
putting m=1,n=-1 in above eqn we get
f(m,n)=45*1+36*(-1)
=9
41. abc+abc+abc=abm,,,m=c than a,b???
Solution:
put a= b/1.5 & b= a/2 & c=m
now, abc+abc+abc
= 3abc
= 3*(b/1.5)*(a/2)*m
= 3*(abm/3)
= abm
42. 3X/5Y = 5Y/3X. Find the value of X/Y
Solution:
3x*3x=5y*5y or (3x)^2=(5y)^2 or 3x=5y
x/y=5/3
43. Possible value of x for equation
(x^2-x-1)^(x+2)=1 ?
Solution:
ans are 0,1,2,-2
44.
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