Quant by Arun Sharma

MBA CAT 2024
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dd017 Says
Hey people I want free ebook of Quant by Arun Sharma...Please post me some links

its not avalaible...try searching fr Quant by R.S.Agarwal...its avalaible fr download
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Q: Find the unit digit of the expression 55^725+73^5810+22^853
options: a.4 b.0 c.6 d.5

can anyone please explain what is the shortest way to solve this kind of problems?


To find the unit digit divide the expression by 10
=> (55^725 + 73^5810 + 22^853) / 10
55^725 / 10 will give unit digit as 5
73^5810 % 10 --> 3^5810 % 10 --> 9^2905
Now 9^(odd no) ends with 9 itself
=> 73^5810 = 9
22^853 % 10 --> 2^853 --> (2^4)^213 * 2 --> (unit digit is 6) * 2 --> 2
=> 5 + 9 + 2 / 10
=> 16/10
=> 6 unit digit
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To find the unit digit divide the expression by 10
=> (55^725 + 73^5810 + 22^853) / 10
55^725 / 10 will give unit digit as 5
73^5810 % 10 --> 3^5810 % 10 --> 9^2905
Now 9^(odd no) ends with 9 itself
=> 73^5810 = 9
22^853 % 10 --> 2^853 --> (2^4)^213 * 2 --> (unit digit is 6) * 2 --> 2
=> 5 + 9 + 2 / 10
=> 16/10
=> 6 unit digit


I am sorry I made a mistake while calculating... it should be zero...the approach posted above is perfect 😃
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waiting fr the solution as i feel by my approach it should be 6

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my take on 1st ques-option (B)
2nd ques-option (A)
let me know are they correct then i will post my approach.

answer to the first question is A and second is correct i.e A
kindly post the approach for the 2nd one
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3. A man buys milk at rs. 8.5 per litr and dilutes it with water. He sells the mixture at the same rate and thus gains 11.11% .Find the quantity of water mixed by himin every litre of milk
a. 0.111 litres
b.0.909 litr
c. 0.1 litre
d. 0.125 litre

pls explain with approach


puys can somebody solve this question??
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3. A man buys milk at rs. 8.5 per litr and dilutes it with water. He sells the mixture at the same rate and thus gains 11.11% .Find the quantity of water mixed by himin every litre of milk
a. 0.111 litres
b.0.909 litr
c. 0.1 litre
d. 0.125 litre



my take of this question is option (C)
look let us assume X is the cost price of milk he has mixed...and now by profit formula we know
(111.11/100)X=8.50..because question has said he is selling at the same price...
now we will get X=7.65rs..this means the milk which is present in the solution cost this much amount...
and by unitary method we can calculate in this much amount there will be .9 liter milk...hence remaning is water...that is option (C)
i hope this will help....
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pallavi70 Says
puys can somebody solve this question??

Ok remember this formula whn working on mixtures problems:
If two ingredients are mixed
then (Quantity of cheaper)/(Quantity of dearer)= (C.P. of dearer)-(Mean Price) divided by (Mean Price)-(C.P. of dearer)
thus coming tour problem:
Let the C.P. be Re 1
thus the S.P. = Re 1
Gain = 11.11%
thus C.P. of 1 litre of mixture = (100/111.11)*1 ~ 9/10
by formula: Quantity of milk / Quantity of water = (1-9/10) divided by (9/10 - 0) (since C.P. of water =0)
thus the ration would come out to be 1/9 ~0.111111111111111
thus quantity of water = 0.111111 * 1 = 0.1111 Litres
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my take of this question is option (C)
look let us assume X is the cost price of milk he has mixed...and now by profit formula we know
(111.11/100)X=8.50..because question has said he is selling at the same price...
now we will get X=7.65rs..this means the milk which is present in the solution cost this much amount...
and by unitary method we can calculate in this much amount there will be .9 liter milk...hence remaning is water...that is option (C)
i hope this will help....

the answer to this question is 0.111 so A
i guess there must be some cal mistake on your part...i got the answer by using calculator...the problem in this questin is arising because of the %value which is 11.11%....if the SP is 120 and profit % is 20% we generallly use the concept of multipying factor and we cal CP as 120/1.2 and we get 100rs.. i want to apply same thing here...i knw SP is 8.5 and profit % is 11.11% which is nothing but 1/9 but how to apply the concept of MF in this case??
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the answer to this question is 0.111 so A
i guess there must be some cal mistake on your part...i got the answer by using calculator...the problem in this questin is arising because of the %value which is 11.11%....if the SP is 120 and profit % is 20% we generallly use the concept of multipying factor and we cal CP as 120/1.2 and we get 100rs.. i want to apply same thing here...i knw SP is 8.5 and profit % is 11.11% which is nothing but 1/9 but how to apply the concept of MF in this case??


pallavi ji aap ek choti si cheez miss kar gaye....
that is...yahan par X/9=8.5 nahi hoga...
reason-profit hai yeh...profit hamesha 1+(something) hota hai
hence here it will be (10/9)X=8.5 hoga
i hope now your doubt is clear....
please pm me if there is anything remaining...
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here the c.P. is immaterial...whwtever be the C.P.,if the C.P=S.P. then it doesn't matter...u can use the formula above as posted by me

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ayush2011 Says
here the c.P. is immaterial...whwtever be the C.P.,if the C.P=S.P. then it doesn't matter...u can use the formula above as posted by me

thanks for your expalnation...actually i usually solve these kind of questions through alligation method...so when u are using this approach..CP does matter as that will decide the ratio in which two quantities should be mixed
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Thanks much for the help...your explanation realy helps!!!...
Yes the answer is c.6

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ayush2011 Says
Still waiting fr the solution

Puys come up with the approach fr this problem, its frustating me
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Puys i got stuck on these problems...any suggestions

1.P(x) is a degree 2 polynomial in x. It leaves remainder 5 on division by (x - 1) and remainder 2 on
division by (x + 2). When P(x) is divided by (x - 1) (x + 2) then find the remainder.

2.If zeroes of the polynomial f(x) = x3 - 3px2 + qx - r are in A.P., then find the value of 2p3.

my problem is still unsolved
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ayush2011 Says
Puys come up with the approach fr this problem, its frustating me


Your approach is correct. The answer is 6....sorry for replyn late....
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Q. the sum of two number is 20 and their geometric mean is 20% lower than their arithmetic mean. Find the ratio of the numbers.
options- a.4:1 b.9:1 c.1:1 d.17:3
Guys please mention the approach also.

Thanks in advance....

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Q. the sum of two number is 20 and their geometric mean is 20% lower than their arithmetic mean. Find the ratio of the numbers.
options- a.4:1 b.9:1 c.1:1 d.17:3
Guys please mention the approach also.

Thanks in advance....

option a 4:1
as sum is 20 AM is 10
so GM is 10-2 = 8
16,4 satisfies both the conditions...so answer is 4:1
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Q. the sum of two number is 20 and their geometric mean is 20% lower than their arithmetic mean. Find the ratio of the numbers.
options- a.4:1 b.9:1 c.1:1 d.17:3
Guys please mention the approach also.

Thanks in advance....

Option a it is;
x+y= 20
(xy)^(1/2) = (x+y)/2 * (.8 )
Solve to get x,y...............
Also you can go by options eg:x & 4x etc........

Regards,
Never Back Down
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option a 4:1
as sum is 20 AM is 10
so GM is 10-2 = 8
16,4 satisfies both the conditions...so answer is 4:1


Yes option (a)4:1 is the answer..
thanks much for the approach....