Quantitative questions a day 51 to 100

Aarav

I'm starting a new thread as the discussions will happen on this new thread for next 50 questions and the sectional test

Happy solving

Apple

Question:51
Allwin went to the market and bought some chikoos, mangoes, and bananas. Allwin bought 42 fruits in all. The number
of bananas is less than half the number of chikoos; the number of mangoes is more than one-third the number of
chikoos and the number of mangoes is less than three-fourths the number of bananas.
How many more/less bananas did Allwin buy than mangoes?
(a) 3 (b) 6 (c) 11 (d) none of the foregoing

I got (a) . Chikoo = 23 banana=11 Mangoes =8.
Cheers Apple

siddhesh_j

my answer to the qn # 51 :
c=23 , b=11 , m =8

the eqn comes to c/2>3/4b>m>c/3

only soln is above one ; so option (a) 3 is the correct option

crazyfootballer · 09:37 AM

C+M+B=42
B<C/2 M<3/4B M>C/3

C/3 < M < 3C/8 1)
4C/9 < B < C/2 2)

Adding 1) and 2)
7C//9 < B+M < 7C/8

C + 7C/8 = 42 (taking extreme)
=> C = 22.xx But B+M is less than 7C/8 hence C>22.xx

C + 7C/9 = 42 (taking other extreme)
=> C = 23.xx But B+M is greater than 7C/9 hence C<23.xx

=>C=23
therefore, B=11, M=8
Hecne answer is 3.

P.S - I am not sure whether my method is correct or not. Someone pls comment on this.

IdiotR

given c+b+m=42--------1)
and b<1/3*c--------------2)
m>1/3*c-----------3)
m<3/4*b-----------4)
from 1)

m=42-b-c>42-c/2-c (from eqn 2)i.e 42-3c/2
i.e 2m>84-3c
i.e 2m>84-9m(from eqn 3) i.e m>7.6
similarly
c=42-b-m
>42-5c/6
i.e c>22.2

similarly
b>10.5
hence possible vals for m,b,c are 8,11,23 to satisfy c+b+m=42
hance option 3 as 11-3=8
Not sure of itz correctness,somebody comment on ths...

chetna

Quote:

Originally Posted by IdiotR

given c+b+m=42--------1)
and b<1/3*c--------------2)
m>1/3*c-----------3)
m<3/4*b-----------4)
from 1)

m=42-b-c>42-c/2-c (from eqn 2)i.e 42-3c/2
i.e 2m>84-3c
i.e 2m>84-9m(from eqn 3) i.e m>7.6
similarly
c=42-b-m
>42-5c/6
i.e c>22.2

similarly
b>10.5
hence possible vals for m,b,c are 8,11,23 to satisfy c+b+m=42
hance option 3 as 11-3=8
Not sure of itz correctness,somebody comment on ths...

i also solved it this way...

when i got c>22.2...i assumed it to be 23...and it worked fine for the other case

is there any better method....?????

brij23

_____________________
Quant Question # 51
_____________________

Allwin went to the market and bought some chikoos, mangoes, and bananas. Allwin bought 42 fruits in all. The number of bananas is less than half the number of chikoos; the number of mangoes is more than one-third the number of chikoos and the number of mangoes is less than three-fourths the number of bananas.

How many more/less bananas did Allwin buy than mangoes?

(a) 3 (b) 6 (c) 11 (d) none of the foregoing

Soln:

Given:
b < 1/2 c m > 1/3 c
m < 3/4 b So m < 3/8 c

We know, number of mangoes is more than one third of cheekoos and less than three eighth of of chocklates. If we take number of cheekoos as 24 both the conditions cannot be satisfied. So number of mangoes has to be 23 and number of mangoes has to be 8.
The remaining number of fruits will be bananas and it will be 11

Cheekoos Bananas Mangoes
21 11 8

So Allwin bought three more bananas than mangoes.

Hence the correct answer choice would be [a].

rani145541

Quote:

Originally Posted by brij23

_____________________
Quant Question # 51
_____________________

Allwin went to the market and bought some chikoos, mangoes, and bananas. Allwin bought 42 fruits in all. The number of bananas is less than half the number of chikoos; the number of mangoes is more than one-third the number of chikoos and the number of mangoes is less than three-fourths the number of bananas.

How many more/less bananas did Allwin buy than mangoes?

(a) 3 (b) 6 (c) 11 (d) none of the foregoing

Soln:

Given:
b < 1/2 c m > 1/3 c
m < 3/4 b So m < 3/8 c

We know, number of mangoes is more than one third of cheekoos and less than three eighth of of chocklates. If we take number of cheekoos as 24 both the conditions cannot be satisfied. So number of mangoes has to be 23 and number of mangoes has to be 8.
The remaining number of fruits will be bananas and it will be 11

Cheekoos Bananas Mangoes
21 11 8

So Allwin bought three more bananas than mangoes.

Hence the correct answer choice would be [a].

I am also getting 23 11 and 8 as solution so ans is A -3
Hey i dont receive the questions.
Please tell me how to subscribe to question list so that i also get questions along with u guys.

Rani

dhruvpratap

Quote:

Originally Posted by chetna

i also solved it this way...

when i got c>22.2...i assumed it to be 23...and it worked fine for the other case

is there any better method....?????

I also solved it this way, nothin wrong with it

amar_kashyap

I am getting (d)..ie none of the foregoing..
b=c/2-k becoz bananas are less than half the chikoos, similarly
m=c/3+p
m=3b/4-x=3c/8-3k/4-x
Now c , m and b have to be integer. c must be of the form 24n.So c must be=24
Now m+b=18 becoz c+m+b=42
Now b must be 4,8,12,or 16.But as b is less than half of c it has to be less than 12 ie b=4 resulting into m=14 which is not possible becoz as indicated from third eqauation m<b

So the given scenario is not possible
hope i make some sense..