Quant by Arun Sharma

MBA CAT 2024
2078
Please help me to do these questions from number system thanks:


Amitesh buys a pen,pencil and an eraser for RS 41. if the least price is of pen that is 12 and eraser cost more than the pencil and pen costthe least.



Answer the following questions:

If it is known that the eraser cost is not dividible by 4 then the cost of the pencil is :

a)12 b)13 c)14 d)15 The answer is a)


Is the question correct? You mentioned that the least price of a pen is Rs. 12, and the pen is the cheapest among the three. So how can the answer be a pencil costing Rs. 12?? A pencil should atleast cost Rs. 13.

If we go exactly by your question, my answer to this question would be a pencil would cost Rs. 14.

We have Pen = 12

So, if Pen = 12, Pencil + E = 29, which could either be 13+16 or 14+15. No other pair would satisfy all the conditions

If Pen = 13, Pencil + E = 28, which could be 13+15 or 14+14. Both don't satisfy, as we can't have two objects at the same price.

Thus, Pen has to be at Rs. 12. Since the cost of an eraser is not divisible by 4, the pair of Pencil, Eraser would be 14,15.


A man sold 38 pieces of clothing.Ih he sold at least 11 pieces each item and he sold more shirts than trousers and more trousers than ties then the number of ties that he sold :


a) Exactly 11 b) Atleast 11 c)At least 12 d) Cannot be determined The answer is a) but i am getting b)


We'll have to proceed in exactly the same way.

Shirts > Trousers > Ties, S + Tr + Ties = 38

If Ties = 11, S + Tr = 27, which could be 12+15 or 13+14.
If Ties = 12, S + Tr = 26, which could be 12+14 or 13+13. Again, we can't have two objects at the same price.

Thus the answer is correct, exactly 11.



Find the least sqaure number divisible by 6,8 and 15

The answer is 3600


The least number should atleast be divisible by 6, 8 and 15. So we need to have atleast those prime factors which make up all of these numbers.

6 = 2*3
8 = 2*2*2
15 = 3*5

So the least number (just plain number, not square) which is divisible by 6, 8 and 15 is 2*3*2*2*5

This number contains 2*3, as well as 2*2*2, as well as 3*5

Now to make it a square, we just need to add one factor each of 2, 3 and 5, since there are only odd numbers of them.

Thus, the square would be: 2*3*2*2*5*2*3*5 = 3600


Find the least number by which 30492 must be multiplied or didvided so as to make a perfect square:

a)11 b)7 c)3 d) 2


the answer is b


Just find the factors of 30492 and find which of them is in odd multiples.
The factors are : 2*2*3*3*7*11*11

There's only one 7, so either multiply or divide by 7 to make the number a square.
2079
2080

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The power of 45 that will divide 123! is

a)28 b)30 c)31 d)29


The answer is 31 but i am getting 28 as 123/5+ 123/5^2


Find the smallest natural number n such that n! is divisible by 990
a)3 b) 5 c)11 d) None of these


The answer is d)


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I think above question were missed.

Regarding the first question i ave copied the exact question:-)
2081
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I think above question were missed.

Regarding the first question i ave copied the exact question:-)

If the first question is correct, then I think my answer is correct 😃

I missed the other questions because I don't know how to solve them :lookround:
2082

1/(rt2+rt1) +1/(rt2+rt3) +......+ 1/(rt120+ rt121)=???
got the answer

2083

product of the fourth term and the fifthe term of an arithmetic progression is 456. division of the ninth term by the fourth term of the progression gives quotient as 11 and the remainder as 10. find the first term of the progression.

2084
p2w Says
product of the fourth term and the fifthe term of an arithmetic progression is 456. division of the ninth term by the fourth term of the progression gives quotient as 11 and the remainder as 10. find the first term of the progression.

I don't know how to go about it algebraically, but I did manage to solve it.

Break down 456 into its factors:
2*2*2*3*19

From these, the 4th and 5th terms could be any of the following combinations:
2*228, 4*114, 6*76, 8*57, 12*38 and 19*24.
The order signifies the 4th and 5th number, respectively.

Since we know that it's a positive AP, the other combinations need not be considered.

Now, it's given that the 9th number is 10 more than 11 times than the 4th number. So out of the combinations which form 456 as their product, the first four are immediately ruled out.

2*228 = 9th digit should be 32
4*114 = should be 54
6*76 = should be 76
8*57 = should be 98

Even the last combination is ruled out, as 19*24 = 9th digit should be 219, and there's no way we can reach that number by using d = 5

So the only one remaining:

12, 38, 64, 90, 116, 142 (from 4th to 9th)

4th * 5th = 456
9th/4th = 11x + 10

So the 1st term:
12 - 26*3 = -66
2085
p2w Says
product of the fourth term and the fifthe term of an arithmetic progression is 456. division of the ninth term by the fourth term of the progression gives quotient as 11 and the remainder as 10. find the first term of the progression.

the ans is -66...
2086
p2w Says
product of the fourth term and the fifthe term of an arithmetic progression is 456. division of the ninth term by the fourth term of the progression gives quotient as 11 and the remainder as 10. find the first term of the progression.

==>(a+3d)*(a+4d)=456....(i)
==>(a+8d)=11(a+3d)+10..(ii)
solving above equations..
a=-66
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If you are asking the cost,then you can not afford..:cheers:
2087

the first and third terms of an arithmetic progression are equal,respectively, to the first and third term of a geometric progression, and the second term of the AP exceeds the 2nd term of the GP by 0.25. calculate the sum of the first five terms of the arithmetic progression if its first term is equal to 2.??????

2088

1.how many 2 digit nos less than 50 have product of factorail of their digits less than or equal to the sum of the factorials of their digits?

2.How many integer values of x and y are there such that 4x+7y=3 ,while x

2089
1.how many 2 digit nos less than 50 have product of factorail of their digits less than or equal to the sum of the factorials of their digits?

2.How many integer values of x and y are there such that 4x+7y=3 ,while x
1. Any number which either has a 0 or 1 as one of its digit will suffice. Reason is that 0! = 1! = 1, and 1*x is always less than 1+x. Other than this, only 22 will suffice the condition. So the numbers would be:

10-19
20-22
30, 31
40, 41

Total 17 numbers.
2090
1.how many 2 digit nos less than 50 have product of factorail of their digits less than or equal to the sum of the factorials of their digits?

2.How many integer values of x and y are there such that 4x+7y=3 ,while x
(1) Refer Above Post

(2) 4x + 7y = 3

First value which satisfies :- x = -1 , y = 1 (for x = -ve and y +ve as both can't be -ve or +ve)

Now 'x' will jump with a factor of (-7) and 'y' will jump with a factor of (4)

So next value x = -8 and y = 5 and so on .........

So values of x will fall early under 500 (-1,-8,-15........upto -498 )

It is an AP no. of terms = 72

Now for x = +ve and y = -ve first value x = 6 and y = -3

Hereon 'x' will jump with a factor of (7) and 'y' with a factor of (-4) so next value x = 13 and y = -7 and so on

x = (6,13,20 ....upto 496) No. of terms = 71

So total 143
2091
p2w Says
the first and third terms of an arithmetic progression are equal,respectively, to the first and third term of a geometric progression, and the second term of the AP exceeds the 2nd term of the GP by 0.25. calculate the sum of the first five terms of the arithmetic progression if its first term is equal to 2.??????


AP -> 2 , (2 + d) , (2 + 2d) ........

GP -> 2 , 2r , 2r^2 ..........

Also (2 + 2d) = 2r^2 and

(2 + d) = 2r + 0.25

So we have .....

16(d^2) - 8d - 15 = 0

d = (-3/4) and (5/4) , So two values for sum upto five terms ....

For d = (-3/4) => Sum = 2.5

For d = (5/4) => 22.5
2092

1.Remainder when 33^34^35 is divided by 2.

2. Sm denotes sum of the squares of the first m natural nos .For how many values of m

2093
1.Remainder when 33^34^35 is divided by 2.

2. Sm denotes sum of the squares of the first m natural nos .For how many values of m
1) 33^34^35 is divided by 2.remainder will be 1
2) sm = m(m+1)(2m+1)/6
now for m can be even/odd
m+1 will be odd/even
2m+1 will always be odd
now when m is odd, m+1 will be even and 2m+1 will be odd hence only m+1 will be even so for values where m+1 = 4k (where k = 2 to 25) total = 24 values
similarly
when m is even then m = 4k (where k = 2 to 25) total = 24
hence total = 48 values?
2094
1.Remainder when 33^34^35 is divided by 2.

2. Sm denotes sum of the squares of the first m natural nos .For how many values of m
1==>remainder is 1.
2==>m not sure but i think its 19..
let me calculare it..
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the winner stands alone..
2095
1) 33^34^35 is divided by 2.remainder will be 1
2) sm = m(m+1)(2m+1)/6
now for m can be even/odd
m+1 will be odd/even
2m+1 will always be odd
now when m is odd, m+1 will be even and 2m+1 will be odd hence only m+1 will be even so for values where m+1 = 4k (where k = 2 to 25) total = 24 values
similarly
when m is even then m = 4k (where k = 2 to 25) total = 24
hence total = 48 values?

i m sorry..the question was Remainder when 33^34^35 is divided by 7.
2096
ankur123xyz Says
i m sorry..the question was Remainder when 33^34^35 is divided by 7.

is the answer 1????
2097

1.Company produces z units of output at a total cost of R. where R=(1/10)*z^3-5z^2+10z+5. At what level of output will the avg variable cost attain its minimum?

2.H1,H2..Hn are n harmonic means between a and b then value of ((H1+a)/(H1-a))+((Hn+b)/(Hn-b)) is equal to?\
a) n+1 b) 2n c)2n+3 d) n-1