Quant by Arun Sharma

MBA CAT 2022
994
jigar_p_civil Says
but menacebhai 50 and 100 both have 2 times 5


multiply them individually

50*100 = 5000

we are not going to consider number of 5's

unless it is

50! ,here number of zeros wld be + = 10 + 2 = 12
995
multiply them individually

50*100 = 5000

we are not going to consider number of 5's

unless it is

50! ,here number of zeros wld be + = 10 + 2 = 12


ok i got it

but if question like 2,4,6,.........1000

then how can we judge?
996
multiply them individually

50*100 = 5000 * some even no = 10000

we are not going to consider number of 5's

unless it is

50! ,here number of zeros wld be + = 10 + 2 = 12


But we have other even nos which on mulltiplication gives one more zero...see it in bold
997
sumitbce Says
But we have other even nos which on mulltiplication gives one more zero...see it in bold


Maafi 😃

we get 12 zeros at the end

coz, extra zero is contributed by 50
998
ok i got it

but if question like 2,4,6,.........1000

then how can we judge?


single zeros, 10,20,...,90,110120,...190,..like wise till 990 we get 99 zeros

Double zeros contributed by 100,200 ...till 900 we get 9 more

so, 99+9 = 108

3 zeros contibuted by 1000 so,108+3= 111

and there are few more zeros contributed by

50,150,250,350...950 so 10 more

total will be 111 + 10 = 121 zeros in all
999
single zeros, 10,20,...,90,110120,...190,..like wise till 990 we get 99 zeros

Double zeros contributed by 100,200 ...till 900 we get 9 more

so, 99+9 = 108

3 zeros contibuted by 1000 so,108+3= 111

and there are few more zeros contributed by

50,150,250,350...950 so 10 more

total will be 111 + 10 = 121 zeros in all



Double zeros contributed by 100,200 ...till 900 we get 9 more

its 18 zero not 9
1000
Double zeros contributed by 100,200 ...till 900 we get 9 more

its 18 zero not 9


No, we have already considered 1 zero, while counting 10,20...90,100,..200..


so counting only single zero from 100, 200, 300 ..an so on
1001
ok i got it

but if question like 2,4,6,.........1000

then how can we judge?


My method was more like a trial and error, errors wld creep in easily

Else we can convert it in to factorials, tht wld be easy

2^500 * 500!

now, + + = 100 + 20 + 4 = 124 zeros

2*4*6*..*100

2^50*50!

+ = 10 + 2 = 12 zeros
1002

Hi ,
I tried to solve this problem in Time and work but I couldn't.Could someone help me out

It takes six days for 3 women and 2 men working together to complete a work.Three men would do the same work 5 days sooner than nine women.How many times does the output of the man exceed that of a women?

a) 3 times b)4 times c)5 times d)6 times

I need explanation too not answer alone:

1003
Hi ,
I tried to solve this problem in Time and work but I couldn't.Could someone help me out

It takes six days for 3 women and 2 men working together to complete a work.Three men would do the same work 5 days sooner than nine women.How many times does the output of the man exceed that of a women?

a) 3 times b)4 times c)5 times d)6 times

I need explanation too not answer alone:


Need to form three equation

3W + 2M = 6 DAYS---(1)

9W = X DAYS
1W= 9X DAYS----(2)
1W per day = 1/9X
3W per day = 3/9X

3M = X-5 Days
1M = 3(X-5) Days --(3)
1M per day= 1/3(X-5)
2M per day = 2/3(X-5)

3/9X + 2/3(X-5)= 1/6 -- From (1)
solving this u will get X=10 days

from 2 -- 1W=90 Days
from 3 --- 1M = 3(10-5) = 15 days

1M=6W
1004
Hi ,
I tried to solve this problem in Time and work but I couldn't.Could someone help me out

It takes six days for 3 women and 2 men working together to complete a work.Three men would do the same work 5 days sooner than nine women.How many times does the output of the man exceed that of a women?

a) 3 times b)4 times c)5 times d)6 times

I need explanation too not answer alone:

My take is 6 times
(2m+3w)*6 = work ....(1)
where m is the amount of work done by one man in one day and w is the amount of work done by one women in one day.

Also Let t be the the time taken by 9 women to complete the task so 3 man will take t-5 days to complete the same task

since 3 man will take t-5 days to complete 1 work
so one man will take 3(t-5) days to complete the task
similarly one women will take 9t days to complete the task

Hence the amount of work by one man in one day = work/3(t-5) = m
amount of work by one women in one day = work/9t = w

Putting it in equation 1 we will get t = 10( 1 discarded)

So m/w = 90/15 = 6
1005

While going through Arun Sharma I got some doubts regarding the explanations given in the book from the following Problems. Can someone plz help me out, by explaining the logic ......

LOD II

Q.4. If a and b are positive integers then 2^1/2 lies ib between:

(a) (a+b)/(a-b) and ab

(b) a/b and (a + 2b)/(a + b)

(c) a and b

(d) ab/(a+b) and (a-b)/ab

Q.11. The Lucknow Indore express without its rake can go 24 km/h , and the spped is diminished by quantity that varies as the square root of the number wagons attached . If it is known that with 4 wagons the speed of the train is 20 km/hr , the greatest number of wagons with which the train engine can just move is:

(a) 144 (b) 140 (c) 143 (d) 124

1006

Q.4. If a and b are positive integers then 2^1/2 lies ib between:

(a) (a+b)/(a-b) and ab

(b) a/b and (a + 2b)/(a + b)

(c) a and b

(d) ab/(a+b) and (a-b)/ab


Option c can be directly ruled out...
For other options, u can check by putting values for A,B like 1,1 and 2,1
Option A: can be ruled out by putting 3,2 as range wud be 5-6
Option D can be ruled out by 1,1
Option B is only 1 left plus it holds true on checking for some trivial values...

Q.11. The Lucknow Indore express without its rake can go 24 km/h , and the speed is diminished by quantity that varies as the square root of the number wagons attached . If it is known that with 4 wagons the speed of the train is 20 km/hr , the greatest number of wagons with which the train engine can just move is:

(a) 144 (b) 140 (c) 143 (d) 124

Let speed be s=24 - k*sqrt(n)...K=const, n=no of wagons
for n=4,s=20...on substitution, this gives us k=2...
so
s = 24 - 2*sqrt(n)
Just move means 's' > 0...
Implies 24 > 2*sqrt(n)...
ie sqrt(n) ie n n so max wagons is 143...Option c

Revert back if not clear...
1007

Hey vivek can u plz explain this part elaborately,it would b of great help :

Let speed be s=24 - k*sqrt(n)...K=const, n=no of wagons
for n=4,s=20...on substitution, this gives us k=2...
so
s = 24 - 2*sqrt(n)

in detail I have some doubt in it.........

1008
Hey vivek can u plz explain this part elaborately,it would b of great help :

Let speed be s=24 - k*sqrt(n)...K=const, n=no of wagons
for n=4,s=20...on substitution, this gives us k=2...
so
s = 24 - 2*sqrt(n)

in detail I have some doubt in it.........

Reduction is directly proportional to sqrt of no of wagons. So reduction can be denoted as say 'r' = k * sqrt(n)

Now actual speed will 24 - r
so s = 24 - k*sqrt(n)

Now just substitute n=4 and s=20 (as we know that for 4 wagons attached, speed is 20)
So this will give us 20 = 24 - k*sqrt(4) = 24-2k
this gives us k=2
So, s = 24 - 2*sqrt(n)

Hope this clear...
1009

The avg. salary of workers in AMS careers is 2000/-.The avg., salary of faculty being 4000/- and the Management trainees being 1250/.The total no. of workers could be.

a)450 b)300 c)110 d)500


2) In an exam the avg, was found to be X marks.After deducting a computational error the avg, of 94 members got reduced from 84 to 64.The average thus come down by 18.8 marks. The no. of candidates who took the exam were

a)100 b)90 c)110 d)105

solve and explain
thanks

1010
The avg. salary of workers in AMS careers is 2000/-.The avg., salary of faculty being 4000/- and the Management trainees being 1250/.The total no. of workers could be.

a)450 b)300 c)110 d)500


2) In an exam the avg, was found to be X marks.After deducting a computational error the avg, of 94 members got reduced from 84 to 64.The average thus come down by 18.8 marks. The no. of candidates who took the exam were

a)100 b)90 c)110 d)105

solve and explain
thanks


1) 2000( x + y) = 4000x + 1250y
Or x/y = 3/8 or x + y = 11k.
Hence 110, option c.

2) The total reduction in marks = 94 (84 - 64) = 1880.
If these marks have been distributed across a total of x people, where x is total number of students the,
1880/ x = 18.8 or x = 100. Hence 100, option a.
1011

HI,
pls solve these following

1) A land owner increased the length & breadth of a rectangular plot by 10% and 20% respectively.find the percentage change in cost of plot assuming land prices are uniform through out his plot
a) 33% b) 35% c) 22.22% d) none
2) The height of triangle increased by 40%.What can be the maximum increase in length of the base, so that the increase in area is restricted to a maximum of 60%
a) 50 b)20 c) 35 d)25
3) The price of sugar is reduced by 25% but inspite of the decrease Aayush ends up increasing his expenditure on sugar by 20%.What is the % change in his monthly consumption of sugar?
a) +60% b)-10% c)+33.33% d)50%
4) Vickys salary is 75% more than Ashus.Vicky got a raise of 40% on his salary while ashu got a raise of 25% in his salary.By what % Vickys salary is more than Ashus?
a) 96% b) 51.1% c) 90% d) none
solve and explain me
thanks

1012
1) A land owner increased the length & breadth of a rectangular plot by 10% and 20% respectively.find the percentage change in cost of plot assuming land prices are uniform through out his plot
a) 33% b) 35% c) 22.22% d) none

Let length be 20 m and bredth be 5 m ( Length and bredth is taken 20 and 5 to facilitate calculation of % which is asked in the question , to find out )

Area is 100m^2

Let cost per m^2 be Re 1 .

So cost of 100 m^2 is Rs 100

Now length is increased by 10 % so Length becomes (110*20/100)22 m

Similarly

Bredth is increased by 20% so bredth becomes (120*5/100) 6 m

Therefore the new area becomes (22*6) 132 m ^2

Cost of the land becomes Rs( 1 * 132 m ^2 )132

So it can easily b seen that change in land price is Rs 132-100 = Rs 32%

As none of the option matches the option is definitely choice(d)

Hope it is clear now.......
1013
3) The price of sugar is reduced by 25% but inspite of the decrease Aayush ends up increasing his expenditure on sugar by 20%.What is the % change in his monthly consumption of sugar?
a) +60% b)-10% c)+33.33% d)50%

Expenditure on sugar =
Price of sugar * consumption of sugar

Price Quantity Expenditure

Original 10 10 100

Now 7.5 x(let) 120 (20% increase in exp)


Now the story is as clear as water 7.5 *x=120

Solving x we get x= 16

So percentage change is 6/10 or 60%

HEnce option a follows.