is the answer 1.
Please upload few quant exercises in pdf files if possible..
mbajamesbond Says(why d= -5 and not 5 , what if the ap was mix,increasing and dcreasing)
'd' is defined as second term - first term.
So 15-20 = -5.
In a progression, the series either only decreases or only increases, which is why it is called a progression.
Progression
1-Find the common ration of the GP whose 1st and last term are 25 and 1/625 and the sum of the GP is 19531/625.
2-Find the no. of terms in GP whose 1st term is 16 , sum is 1365/64 and common ratio is 1/4.
3-If x
4-If (1^3-t1)+ (2^3-t2)+(3^3-t3)+(4^3-t4)+............(n^3-tn)=n^2(n-3)/4, find tn where t1,t2,t3.....tn are the term of series
1-Find the common ration of the GP whose 1st and last term are 25 and 1/625 and the sum of the GP is 19531/625.
2-Find the no. of terms in GP whose 1st term is 16 , sum is 1365/64 and common ratio is 1/4.
3-If x
4-If (1^3-t1)+ (2^3-t2)+(3^3-t3)+(4^3-t4)+............(n^3-tn)=n^2(n-3)/4, find tn where t1,t2,t3.....tn are the term of series
I came across this quest:
The digits of a three-digit number A are written in the reverse order to form another three-digit
number B. If B > A and BA is perfectly divisible by 7, then which of the following is necessarily
true? (This was asked in CAT 2005)
A.112 B.100 C.106 D.118
Answer Provided: C
Please help me out with a justification.
I came across this quest:
The digits of a three-digit number A are written in the reverse order to form another three-digit
number B. If B > A and BA is perfectly divisible by 7, then which of the following is necessarily
true? (This was asked in CAT 2005)
A.112
B.100
C.106
D.118
Answer Provided: C
Please help me out with a justification.
Let the number A be pqr ie = 100p+10q+r
So, number B is rqp= 100r+10q+p
Therefore, B-A= 99(r-p)
Since, B>A, B-A is divisible by 7, therefore
r-p=7 (since 99 and 7 are coprime)
Hence, two cases arise
r=8, p=1. Minimum value of Number A= 108
r=9, p=2. Maximum value of Number A= 299
So, out of the options provided, only option C matches
Hope it helps.
Progression
1-Find the common ration of the GP whose 1st and last term are 25 and 1/625 and the sum of the GP is 19531/625.
2-Find the no. of terms in GP whose 1st term is 16 , sum is 1365/64 and common ratio is 1/4.
3-If x
4-If (1^3-t1)+ (2^3-t2)+(3^3-t3)+(4^3-t4)+............(n^3-tn)=n^2(n-3)/4, find tn where t1,t2,t3.....tn are the term of series
Posting solution of ques 1
1st term a = 25
last term a*r^(n-1) = 1/625
Therefore, r^(n-1) = 1/(625*25)
Sum = a*(r^n-1)/(r-1) = 19531/625
ie. 25*(r/625*25 -1)/(r-1) = 19531/625
ie. (r - 625*25)/(r-1) = 19531
Now solve for r. It comes out to be 1/5.
Hopefully, you can solve ques 2 using same technique.
Progression
1-Find the common ration of the GP whose 1st and last term are 25 and 1/625 and the sum of the GP is 19531/625.
2-Find the no. of terms in GP whose 1st term is 16 , sum is 1365/64 and common ratio is 1/4.
3-If x|
4-If (1^3-t1)+ (2^3-t2)+(3^3-t3)+(4^3-t4)+............(n^3-tn)=n^2(n-3)/4, find tn where t1,t2,t3.....tn are the term of series
Solving ques 3
Let S = 2+4x+6x^2+8x^3+....
So, xS = 2x+4x^2+6x^3+....
Subtracting 2nd from 1st eqn
S(1-x)=2+2x+2x^2+2x^3+.....
S(1-x)= 2/(1-x) ..... Since |xTherefore S=2/(1-x)^2
Progression
1-Find the common ration of the GP whose 1st and last term are 25 and 1/625 and the sum of the GP is 19531/625.
2-Find the no. of terms in GP whose 1st term is 16 , sum is 1365/64 and common ratio is 1/4.
3-If x
4-If (1^3-t1)+ (2^3-t2)+(3^3-t3)+(4^3-t4)+............(n^3-tn)=n^2(n-3)/4, find tn where t1,t2,t3.....tn are the term of series
Solving ques. 4
(1^3-t1)+ (2^3-t2)+(3^3-t3)+(4^3-t4)+............(n^3-tn)=n^2(n-3)/4
Now, replacing n by (n-1)
(1^3-t1)+ (2^3-t2)+(3^3-t3)+(4^3-t4)+............(n-1^3-tn-1)=(n-1)^2(n-4)/4
Now, subtracting 2nd eqn from 1st
(n^3-tn)= n^2(n-3)/4 - (n-1)^2(n-4)/4
Solve for tn.
Hi Friends,
Please help me with the following questions :
Q1. What is the HCF of (2^100 - 1) and (2^120 - 1) ?
Q2. What is the remainder when (22^33 + 10^35) is divided by 45 ?
Q3. Digital sum of a number = sum of the digits of a number.
Now, What is the digital sum of 19^100 ?
Q4. What is the remainder when 32^32^32 is divided by 7 ?
Q5. When a certain 2 digit number is added to another 2 digit number formed by reversing the digits of the previous number, the sum is a perfect square. Hoe many such 3 digit numbers are there ?
Q6. How many no of zeroes will be there at the end of 12! expressed in base 6 ?
Q7. The smallest natural number which is a perfect square and is of the form abbb, lies in between :-
a)1000 to 2000 b)2000 to 3000
c)3000 to 4000 d)4000 to 5000
Q8. If 126 natural numbers are put side by side in ascending order to create a large number 12345................125126. What will be the remainder when N is divided by 5625 ?
Q9.We are writing all multiples of 111 to 324. How many times will we write the digit 3?
Q10. How many times 6 appears b/w 11 to 400?
Q11. Pallindrome P is of S digits, where S is even, written in base N. Which of the following no definitely divide P ?
Hi Friends,
Please help me with the following questions :
Q1. What is the HCF of (2^100 - 1) and (2^120 - 1) ?
Q2. What is the remainder when (22^33 + 10^35) is divided by 45 ?
Q3. Digital sum of a number = sum of the digits of a number.
Now, What is the digital sum of 19^100 ?
Q4. What is the remainder when 32^32^32 is divided by 7 ?
Q5. When a certain 2 digit number is added to another 2 digit number formed by reversing the digits of the previous number, the sum is a perfect square. Hoe many such 3 digit numbers are there ?
Q6. How many no of zeroes will be there at the end of 12! expressed in base 6 ?
Q7. The smallest natural number which is a perfect square and is of the form abbb, lies in between :-
a)1000 to 2000 b)2000 to 3000
c)3000 to 4000 d)4000 to 5000
Q8. If 126 natural numbers are put side by side in ascending order to create a large number 12345................125126. What will be the remainder when N is divided by 5625 ?
Q9.We are writing all multiples of 111 to 324. How many times will we write the digit 3?
Q10. How many times 6 appears b/w 11 to 400?
Q11. Pallindrome P is of S digits, where S is even, written in base N. Which of the following no definitely divide P ?
Ans 2 - 2 (not sure)
Ans 3 - 10
Ans 4 - 4
Ans 6 - 5
Ans 7 - 1444
Ans 8 - 126
9- Didn't understand the question itself.
Ans 10 - 79
11- Following? Where are the "Following" numbers?
Q1. What is the HCF of (2^100 - 1) and (2^120 - 1) ?
This is how I solved it:
For any number in the form (a^n-b^n) where n is even...is always divisible by (a-b)
We have to convert both the above nos. in a^n-b^n form
so hcf of 120 and 100 is 20.
so nos become: 2^100 -1 = (2^20)^5 -1
and 2^120-1 = (2^20)^6-1
a is 2^20
so no. must be divisible by 2^20-1 i.e 1048575
so this is the hcf
Q1. What is the HCF of (2^100 - 1) and (2^120 - 1) ?
This is how I solved it:
For any number in the form (a^n-b^n) where n is even...is always divisible by (a-b)
We have to convert both the above nos. in a^n-b^n form
so hcf of 120 and 100 is 20.
so nos become: 2^100 -1 = (2^20)^5 -1
and 2^120-1 = (2^20)^6-1
a is 2^20
so no. must be divisible by 2^20-1 i.e 1048575
so this is the hcf
I am still working on questn 2.
For 3rd..this is a pattern based question.
for eg: digit sum of 12*12 will be digit sum of 12(=3) * digit sum12(=3) i.e 9
similarly this is the pattern for every multiplication problem.
digit sum of 19 is 10 (or eventually 1+0=1) , so any power of 19 will have digit sum 1.
hence its 1.
hey ppl,can anyone help me with this sequence series question "12,93,730,5097,30570,?" the options are a) 152835 b)152837 c)152839 d)152841 e)152843
Hi,
Q) The total number of integer pairs (x, y) satisfying the equation x + y = xy is (CAT 2004)
A.0
B.1
C.2
D. none of the above
Hi,
Q) The total number of integer pairs (x, y) satisfying the equation x + y = xy is (CAT 2004)
A.0
B.1
C.2
D. none of the above
xy-x-y=0
=>x(y-1)-(y-1) =1
=>(x-1)(y-1)=1
now, -1*-1 =1; 1*1=1
when x-1=-1; and y-1 =-1 => x,y=0,0
when x-1=1 ;and y-1 =1 => x,y=2,2
so 2 solutions are possible
Hi,
Q) The total number of integer pairs (x, y) satisfying the equation x + y = xy is (CAT 2004)
A.0
B.1
C.2
D. none of the above
x + y = x y
x + y - x y=0
x(1-y)+y=0
add -1 in both side
x(1-y)+y-1=-1
x(1-y)-1(1-y) = -1
(1-y)(x-1) = -1
(x-1)(y-1) = 1
1*1 =1
or
-1 * -1 =1
so x=y=0
or
x=y=2
so possible pair
(0,0 ) ,( 2,2 )
Hi,
Q) The total number of integer pairs (x, y) satisfying the equation x + y = xy is (CAT 2004)
A.0
B.1
C.2
D. none of the above
my take
2Rearranging the terms in the equation. we get
xy-x-y=0
x(y-1) -1(y-1) -1 = 0
(x-1)(y-1) = 1
x = 2, y = 2
x = 0, y = 0
(x,y) pairs are (2,2) and (0,0)
Thanks guys...
This is another problem:
Q) A number N when divided by a divisor D gives a remainder of 52. The number 5N when divided by D
gives a remainder of 4. How many values of D are possible?
A.1
B.3
C.6
D.7