Topic : Quant Mixed Bag

Solved ? : Yes

Source : HandaKaFunda ]]>

Topic : Quant Mixed Bag

Solved ? : Yes

Source : HandaKaFunda ]]>

C C C C … C

Now, you put in partitions in between them. Let me denote the partitions with P

C C C P C P C C C C C P C C C C C C C C C C C

A gets the chocolates before the first partition, B gets the chocolates between the first two partitions, C gets the chocolates between the second and the third partition and D gets the chocolate after the third partition. In the arrangement shown above A gets 3 chocolates, B gets 1, C gets 5, and D gets 11. Any rearrangement of the above, will lead to a new distribution of chocolates.

The above can be rearranged in 23! / 20! 3! We got this because there are a total of 23 entities out of which 20 Cs are identical and 3 Ps are identical.

Now, to extend this concept, what if we have to distribute ‘n’ chocolates in ‘r’ kids. After putting the n chocolates in a line, we would need r - 1 partitions. This would mean that there will be a total of n+r-1 entities out of which n would be identical (of one type) and the other r - 1 would be identical as well (of another type).

**So, the number of ways in which that would be possible = (n + r - 1)! / n! (r-1)!**

This, in other words, is (n + r - 1) C (r - 1)

And now, we can use this formula to solve any similar questions

a + b + c + d = 20

Case 1: a, b, c, d are non-negative integers.

Number of solutions = (20 + 4 - 1) C (4 - 1) = 23 C 3 = 1771

Case 2: a, b, c, d are positive integers

We allocate at least a value of 1 to a, b, c, d.

So, we can say a = a’ + 1, b = b’ + 1, c = c’ + 1, d = d’ + 1 where a’, b’ c’, d’ are non-negative integers

=> a’ + 1 + b’ + 1 + c’ + 1 + d’ + 1 = 20

=> a’ + b’ + c’ + d’ = 16

=> Number of solutions = (16 + 4 - 1) C (4 - 1) = 19 C 3

Case 3: a, b, c, d are non-negative integers such that a > 5 and b > 2

We allocate at least a value of 5 to a and 2 to b

So, a = a’ + 5 and b = b’ + 2

=> a’ + 5 + b’ + 2 + c + d = 20

=> a’ + b’ + c + d = 13

=> Number of solutions = (13 + 4 - 1) C ( 4 - 1) = 16 C 3

Case 4: a, b, c, d are non-negative integer such that a > b

Let us first consider the situation where a = b

If a = b = 0, c + d = 20. This has 21 solutions

If a = b = 1, c + d = 18. This has 19 solutions

If a = b = 2, c + d = 16. This has 17 solutions

.

.

If a = b = 10, c + d = 0. This has 1 solution

So, the total number of a solutions when a = b is 21 + 19 + 17 … + 1 = 11/2 * (21 + 1) = 121

We know that the number of solutions when a, b, c, and d are non-negative integers is 1771. Out of these 1771 cases, in 121 cases a = b.

So, in 1771 - 121 = 1650 cases a is not equal to b.

In half of the above cases a will be greater than b whereas in the other half of the cases a will be less than b.

So, number of solutions where a > b is 1650/2 = 825

Case 1: When A is coming down

A will take 50 seconds to complete 50 steps.

In 50 seconds, escalator would have moved 50n steps.

Total number of steps on the stationary escalator = 50 + 50n

Case 2: When B is coming down

B will take 25 seconds to complete 75 steps

In 25 seconds, escalator would have moved 25n steps.

Total number of steps on the stationary escalator = 75 + 25n

Total number of steps on the stationary escalator is a constant

=> 50 + 50n = 75 + 25n

=> 25n = 25

=> n = 1

Total number of steps = 50 + 50 = 75 + 25 = 100

]]>Number of people who ate Egg = 86

=> Those who didn't = 100 - 86 = 14

=> 14 people did not eat eggs.

Let us number those P1, P2, P3 ... P14

Similarly, 25 people did not eat bacon

Let us number those P15, P16, P17 ... P39

Similarly, 38 did not have toast

Let us number those people P40, P41, P42 ... P77

Similarly, 18 people did not have coffee

Let us number those people P78, P79, P80 ... P95

Now, we are left with 5 people (P96, P97, P98, P99, P100) who ate / drank all of the four dishes / drink involved. This is the minimum possible configuration.

]]>After that, we can use the property

Last two digits of (...a1)^(...b) will be [Last digit of a * b]1

Let us try and apply this concept in the given question

Last two digits of 123^123!

= Last two digits of 23^123!

= Last two digits of (23^4)^(123!/4)

= Last two digits of (529^2)^(123!/4)

= Last two digits of (...41)^(a large number ending in a lot of zeroes)

= Last two digits of (...01) {Here I have used the concept mentioned above}

= 01 ]]>

The condition given is d > e > f

Let us first select three digits for a, b, and c.

From the given 6 digits (1,2,3,4,5,6), 3 digits for a,b, and c can be selected in 6C3 = 6!/3!3! = 20 ways.

These three selected digits can be arranged on three positions of a, b, and c in 3! = 6 ways

The remaining 3 digits are automatically selected for d, e, and f.

So, we have 1 way of selecting them.

The biggest digit will be allocated to 'd'.

The second biggest digit will be allocated to 'e'.

The third biggest digit will be allocated to 'f'

So, we have 1 way of arranging them on the three positions of d, e, and f.

Total ways = 20 * 6 * 1 * 1 = 120

]]>2! + 4! + 6! + .... 98!

Let us look at the units digit for all of them

2! = 1 * 2 = 2; unit's digit is 2

4! = 1 * 2 * 3 * 4 = 24; unit's digit is 4

6! = 1 * 2 * 3 * 4 * 5 * 6 = 720; unit's digit is 0

8! = multiple of 6!, divisible by 10; unit's digit is 0

10! = multiple of 6!, divisible by 10; unit's digit is 0

12! = multiple of 6!, divisible by 10; unit's digit is 0

.

.

.

98! = multiple of 6!, divisible by 10; unit's digit is 0

Unit's digit of 2! + 4! + 6! + .... 98!

= Sum of unit's digits of individual terms

= 2 + 4 + 0 + 0 + 0 ..... 0

= 6

Total amount of food = N * 4x = 4Nx

Food consumed in 20 days = 20 * 4x = 80x

Food left after 20 days = 4Nx - 80x

The remaining food was consumed by 3x men in N days

=> Food left was = N * 3x = 3Nx

=> 4Nx - 80x = 3Nx

=> Nx = 80x

=> N = 80 days ]]>

S(2) = 1.(2)^1 + 2.(2)^2 = 2 + 8 = 10

S(3) = 1.(2)^1 + 2.(2)^2 + 3.(2)^3 = 2 + 8 + 24 = 34 = 2 + 2.(2)^4

S(4) = 1.(2)^1 + 2.(2)^2 + 3.(2)^3 + 4.(2)^4 = 2 + 8 + 24 + 64 = 98 = 2 + 3.(2)^5

and so on.

So, S(n) = 2 + (n-1).2^(n+1)

So, S(100) = 2 + 99.2^101 ]]>

Via Flipkart: Website offers a 30% discount bringing down the price to 70 Rs. The flipkart app offers another 20% discount ( on 70 Rs.) i.e. a discount of 14 Rs. This reduces the price to 56 Rs.

Via Amazon: Website offers a 20% discount bringing down the price to 80 Rs. The Amazon app offers another 30% discount (on 80 Rs.) i.e. a discount of 24 Rs. This reduces the price to 56 Rs.

As you can see, both of them are offering it to you at the same price. So go ahead, buy it from Amazon - because it is better.

]]>In this question, we know that the first term is 7

=> a = 7

The last term in the series is 448

=> ar^(n-1) = 448

=> r^(n-1) = 64

Now, this could be 2^6 or 4^3 or 8^2

To figure this part out, we also have one more information that the sum of the series is 889

=> a(r^n - 1)/(r - 1) = 889

Now, we can solve the above equations to get the answer. Or we can just assume some values and solve this out

Let us assume that the common ratio, r = 2

=> Sum of the GP = 889

=> 7 + 14 + 28 + 56 + 112 + 224 + 448 = 889

This fits. So, the common ratio is 2 ]]>

The trick to solving these questions is to find out higher powers one by one.

Basically, we should try and find out the values of x^n + 1/x^n

When n = 2

To get the value of n = 2, we need to square the original equation

(x + 1/x) = 3

=> (x + 1/x)^2 = 3^2

=> x^2 + 1/x^2 + 2 = 9

=> x^2 + 1/x^2 = 7

When n = 3

To find the value, we need to multiply the equations where n = 1 and n = 2

(x + 1/x) * (x^2 + 1/x^2) = 3*7

=> x^3 + 1/x^3 + x + 1/x = 21

=> x^3 + 1/x^3 + 3 = 21

=> x^3 + 1/x^3 = 18

When n = 4

To find the value, we can square the equation we got at n = 2

(x^2 + 1/x^2)^2 = 7^2

=> x^4 + 1/x^4 + 2 = 49

=> x^4 + 1/x^4 = 47

When n = 5

To find the value, we can multiply the equations we got at n = 1 and n = 4

(x + 1/x) * (x^4 + 1/x^4) = 3*47

=> x^5 + 1/x^5 + x^3 + 1/x^3 = 141

=> x^5 + 1/x^5 + 18 = 141

=> x^5 + 1/x^5 = 123

So, the value we were trying to find out was 123

Alternatively, you can just multiply the values that you have got at n = 2 and 3 to get the answer.

(x^2 + 1/x^2) * (x^3 + 1/x^3) = 7*18

=> x^5 + 1/x^5 + x + 1/x = 126

=> x^5 + 1/x^5 +3 = 126

=> x^5 + 1/x^5 = 123

Dearr(n) = n!(1/0! - 1/1! + 1/2! - 1/3!.... 1/n!)

In this question, we need to place 5 objects (letters) in 5 positions (addresses) such that none of them is correct. This can be done in

Dearr(5) = 5! (1/0! - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!)

=> Dearr(5) = 120 (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)

=> Dearr(5) = 60 - 20 + 5 - 1 = 44

So, we can send the 5 letters, such that all are delivered at wrong adresses, in 44 different ways.

]]>