Quant by Arun Sharma

MBA CAT 2024
1776
please explain how to solve it
what is the remainder when 2(8!) - 21(6!) divides 14(7!) - 14(13!)?
a)1
b)7!
c)8!
d)9!

The answer is 7!.


7! / 7!
7!*14 mod 7!*13 = 7!
7! mod 7!*13 = 0

So remainder = 7!
1777
avinav2712 Says
It's because you can take 38! common in the expression. Is the answer correct?



Ya the answer is right.But when we remove 38! as common from the expression how can we be sure thatthe remaining expression doesnot have the highest power of 3?
1778

Hi puys,
Can you give the solns for the follwing:::: These are from Arun sharma Progressions

1. Find the value of 1-2-3+2-3-4+.......+ upto 100 terms??

a. -694 b.-626 c.-624 d.-549 e. -676

2. If a man saves Rs.1000 each year and invests at the end of the year at 5% compound interest how much will the amount be at the end of 15 years?

a. 21,478 b. 21,578 c. 22,578 3. 22,478 e. 22,178

3. Find the infintie sum of the series 1+1/3+1/6+1/10+/15+......
a. 2 b.2.25 c. 3 d. 4

4. An arithmetic progression P consists of n terms. From the progression three different progressions P1P2 and P3 are created such that P1 is obtained by the 1st,4th,7th....... terms of P, P2 has 2nd ,5th,8th ....... terms of P, P3 has the 3rd,6th,9th.terms of P. It is found that of P1,P2,P3 two progressions have the property that their average itself is a term of the original progression P. Which of the following can be a possible value of n??

a. 20 b.26 c.36 d. Both 1 and 2

5 . Product of the fourth term and fifth term of an AP is 456. Divivsion of the ninth term by the fourth term of the progression gives quotient as 11 and remainder as 10. Find the first term of the AP
a. -52 b. -42. c. -56 d.-66 e.-50

6. Find the sum of series 1.2+2.2^2+3.2^3+........+100.2^100

7. If x,y,z are in GP and a^x,b^y and c^z are in ?
a. AP b. GP c. HP

8. If three +ve real numbers x,y,z are in AP such that xyz =4 ,then what will be the minimum value of y?
a. 2^1/3 b.2^2/3 c.2^1/4 d. 2^3/4 e. 2^4/3

Please give detiled solutions

1779
Hi puys,
Can you give the solns for the follwing:::: These are from Arun sharma Progressions

1. Find the value of 1-2-3+2-3-4+.......+ upto 100 terms??

a. -694 b.-626 c.-624 d.-549 e. -676

2. If a man saves Rs.1000 each year and invests at the end of the year at 5% compound interest how much will the amount be at the end of 15 years?

a. 21,478 b. 21,578 c. 22,578 3. 22,478 e. 22,178


for the second one

is the answer 36

let the series be a,a+d,a+2d,a+3d

then the three series formed will have first term as a,a+d and a+2d

the common differrence to all three will be 3d

when you find sum of all three and then find the average

it will of the form n/18(6a+6d +(n/6-1)9d)

only 36 will satisfy ..i hope i am correct


4. An arithmetic progression P consists of n terms. From the progression three different progressions P1P2 and P3 are created such that P1 is obtained by the 1st,4th,7th....... terms of P, P2 has 2nd ,5th,8th ....... terms of P, P3 has the 3rd,6th,9th.terms of P. It is found that of P1,P2,P3 two progressions have the property that their average itself is a term of the original progression P. Which of the following can be a possible value of n??

a. 20 b.26 c.36 d. Both 1 and 2
is the answer 36

let the series be a ,a+d,a+2d....
all the three series will have common diffrence of 3d

and the number of terms will be n/3

if you find some of these 3 series and and sum all three sums.

find average... only 36 will satisfy(basically it should be divisible by 6)

i hope i am right
and first term a,a+d,a+2d

3. Find the infintie sum of the series 1+1/3+1/6+1/10+/15+......
a. 2 b.2.25 c. 3 d. 4

4. An arithmetic progression P consists of n terms. From the progression three different progressions P1P2 and P3 are created such that P1 is obtained by the 1st,4th,7th....... terms of P, P2 has 2nd ,5th,8th ....... terms of P, P3 has the 3rd,6th,9th.terms of P. It is found that of P1,P2,P3 two progressions have the property that their average itself is a term of the original progression P. Which of the following can be a possible value of n??

a. 20 b.26 c.36 d. Both 1 and 2

5 . Product of the fourth term and fifth term of an AP is 456. Divivsion of the ninth term by the fourth term of the progression gives quotient as 11 and remainder as 10. Find the first term of the AP
a. -52 b. -42. c. -56 d.-66 e.-50

6. Find the sum of series 1.2+2.2^2+3.2^3+........+100.2^100

7. If x,y,z are in GP and a^x,b^y and c^z are in ?
a. AP b. GP c. HP

8. If three +ve real numbers x,y,z are in AP such that xyz =4 ,then what will be the minimum value of y?
a. 2^1/3 b.2^2/3 c.2^1/4 d. 2^3/4 e. 2^4/3

Please give detiled solutions



For

1 > it will consist of three series with first term as 1,-2,-3 and common difference of 1,-1 and -1 .

all three series will have 33 terms

finding the sum of these 3 series will give you sum of 99 terms

100th term will be 34

so the answer will be -626
1780
Hi puys,
Can you give the solns for the follwing:::: These are from Arun sharma Progressions

1. Find the value of 1-2-3+2-3-4+.......+ upto 100 terms??

a. -694 b.-626 c.-624 d.-549 e. -676

2. If a man saves Rs.1000 each year and invests at the end of the year at 5% compound interest how much will the amount be at the end of 15 years?

a. 21,478 b. 21,578 c. 22,578 3. 22,478 e. 22,178

3. Find the infintie sum of the series 1+1/3+1/6+1/10+/15+......
a. 2 b.2.25 c. 3 d. 4

4. An arithmetic progression P consists of n terms. From the progression three different progressions P1P2 and P3 are created such that P1 is obtained by the 1st,4th,7th....... terms of P, P2 has 2nd ,5th,8th ....... terms of P, P3 has the 3rd,6th,9th.terms of P. It is found that of P1,P2,P3 two progressions have the property that their average itself is a term of the original progression P. Which of the following can be a possible value of n??

a. 20 b.26 c.36 d. Both 1 and 2

5 . Product of the fourth term and fifth term of an AP is 456. Divivsion of the ninth term by the fourth term of the progression gives quotient as 11 and remainder as 10. Find the first term of the AP
a. -52 b. -42. c. -56 d.-66 e.-50

6. Find the sum of series 1.2+2.2^2+3.2^3+........+100.2^100

7. If x,y,z are in GP and a^x,b^y and c^z are in ?
a. AP b. GP c. HP

8. If three +ve real numbers x,y,z are in AP such that xyz =4 ,then what will be the minimum value of y?
a. 2^1/3 b.2^2/3 c.2^1/4 d. 2^3/4 e. 2^4/3

Please give detiled solutions



for the second one

is the answer 36

let the series p be a,a+d,a+2d,a+3d

p1 will a,a+3d,a+6d
p2 will be a+d,a+4d,a+7d
p3 will be a+2d,a+5d,a+8d

all will have n/3 terms if you add sums of all three

the average will be of the form n/18(6a+6d +(n/3-1)9d)

so basically n should be divisible by 18

i hope its correct
1781
Hi puys,
Can you give the solns for the follwing:::: These are from Arun sharma Progressions

1. Find the value of 1-2-3+2-3-4+.......+ upto 100 terms??

a. -694 b.-626 c.-624 d.-549 e. -676


Please give detiled solutions


consider 3 nos a time 1-2-3, 2-3-4, 3-4-5..
thus 97th 98th 99th term will be 33-34-35 and 100th term is 34

answer= AP of 1 ...33 - AP of 2...34 - AP of 3....35 +34
= 33(34-36-3/2 +34
33*-20+34
-660+34
=-626

which is option B
1782
Hi puys,
Can you give the solns for the follwing:::: These are from Arun sharma Progressions

1. Find the value of 1-2-3+2-3-4+.......+ upto 100 terms??

a. -694 b.-626 c.-624 d.-549 e. -676

QUOTE]

There are three series following..
1st starting from 1 till 34..
2nd from 2 till 34
and 3rd from 3 till 35..
sum of these three series will result in
595-594-627
= -626
1783
Hi puys,
Can you give the solns for the follwing:::: These are from Arun sharma Progressions



3. Find the infintie sum of the series 1+1/3+1/6+1/10+/15+......
a. 2 b.2.25 c. 3 d. 4

Please give detiled solutions


my take to this question
option a) 2
..
1784
Hi puys,
Can you give the solns for the follwing:::: These are from Arun sharma Progressions


4. An arithmetic progression P consists of n terms. From the progression three different progressions P1P2 and P3 are created such that P1 is obtained by the 1st,4th,7th....... terms of P, P2 has 2nd ,5th,8th ....... terms of P, P3 has the 3rd,6th,9th.terms of P. It is found that of P1,P2,P3 two progressions have the property that their average itself is a term of the original progression P. Which of the following can be a possible value of n??

a. 20 b.26 c.36 d. Both 1 and 2


Please give detiled solutions


my take :
option d) both 1 and 2..
if correct then will post the approach...
:o
1785
Hi puys,
Can you give the solns for the follwing:::: These are from Arun sharma Progressions



5 . Product of the fourth term and fifth term of an AP is 456. Divivsion of the ninth term by the fourth term of the progression gives quotient as 11 and remainder as 10. Find the first term of the AP
a. -52 b. -42. c. -56 d.-66 e.-50


Please give detiled solutions


my take option
c) -66..
if correct will post the approach..
is it the answer??
1786
my take to this question
option b) 2.25

just applied a bit logic the sum of 1st 7, 8 terms is coming out to be 1.9........something..and then the term is getting much more smaller and smaller like .002, .001 and so on..
so it shud be 2.25 ..and it cannot be 2 bcoz there are infinite terms so will result in sumthing greater than 2...
my take would be 2. The nth term for the series will be 2/n(n+1) which can be further written as 2/n - 2/n+1 . For odd number of terms, it will always be 2. For even number of infinite terms it will be very close to 2. What is the official answer?
1787
209598 Says
Ya the answer is right.But when we remove 38! as common from the expression how can we be sure thatthe remaining expression doesnot have the highest power of 3?


Because:
58! - 38! = 38! (58*57*...37 - 1)

Now, 58*57*... will definitely be a factor of 3, and any factor of 3 less one, will never have a power of 3 in it.
1788

3. Find the infintie sum of the series 1+1/3+1/6+1/10+/15+......
a. 2 b.2.25 c. 3 d. 4


rth term = 2/(r+2)(r+1)
=>2/(r+1)(r+2)
=>2
Put r = 1 and so on....
=>1

Another 1 was missing so SUM = 2...
1789
avinav2712 Says
It's because you can take 38! common in the expression. Is the answer correct?



The answer is right
1790

Please help me in solve this:
A,B,C have a total of 80 coins.A triples the no. of coins with the others by giving them some coins from his collection.Next, B repeats the same process.After this B now has 20 coins.Find the no. of coins he had at the beginning?
a)11
b)10
c)9
d)12

1791

Please help me in solve this:
The series of numbers (1,1/2,1/3,....1/1972) is taken.Now 2 numbers are taken from this series say x,y.Then the operation x+y+x.y is performed to get a consolidated no.the process is repeated.what will be the value of the set after all the numbers are consolidated into one number.
a)1970
b)1971
c)1972
d)none

1792
Please help me in solve this:
The series of numbers (1,1/2,1/3,....1/1972) is taken.Now 2 numbers are taken from this series say x,y.Then the operation x+y+x.y is performed to get a consolidated no.the process is repeated.what will be the value of the set after all the numbers are consolidated into one number.
a)1970
b)1971
c)1972
d)none

1972 is the Asnwer
1793
for the second one

is the answer 36

let the series p be a,a+d,a+2d,a+3d

p1 will a,a+3d,a+6d
p2 will be a+d,a+4d,a+7d
p3 will be a+2d,a+5d,a+8d

all will have n/3 terms if you add sums of all three

the average will be of the form n/18(6a+6d +(n/3-1)9d)

so basically n should be divisible by 18

i hope its correct



The answer is both a and b
1794
my take :
option d) both 1 and 2..
if correct then will post the approach...
:o


Hi ...

its correct..tell me the procedure
1795

4. An arithmetic progression P consists of n terms. From the progression three different progressions P1P2 and P3 are created such that P1 is obtained by the 1st,4th,7th....... terms of P, P2 has 2nd ,5th,8th ....... terms of P, P3 has the 3rd,6th,9th.terms of P. It is found that of P1,P2,P3 two progressions have the property that their average itself is a term of the original progression P. Which of the following can be a possible value of n??
a. 20 b.26 c.36 d. Both 1 and 2


Let P = a,a+d,a+2d,a+3d..,a+(n-1)d
P1 = a,a+3d,a+6d,a+9d....,a+(n-3)d
P2 = a+d,a+4d,a+7d,.......,a+(n-2)d
P3 = a+2d,a+5d,a+8d,.....,a+(n-1)d

P1 = n/6 * = n/6 *
P2 = n/6 * = n/6 *
P3 = n/6 * = n/6 *

Now (P1 + P3)/2 = n/12 * = n/6 *
(P1 + P2)/2 = n/6 *
(P2 + P3)/2 = n/6 *

Tk =

I'll go by options now...looking for more elegant approach...When I would know...I'll post....