Topic - Quant Mixed Bag

Solved ? : Yes

Source : CAT Prep Forums ]]>

Topic - Quant Mixed Bag

Solved ? : Yes

Source : CAT Prep Forums ]]>

(2a+1) + (2b+1) ... + (2e+1) = 35

2a + 2b +....+ 2e = 30

a + b + ... + e = 15

Whole number Solution = 19c4 ]]>

No. Of co primes => [(2 * 3 + 1) ( 2 * 2 + 1) - 1]/2

(7 * 5 -1)/2

=> 17

If N is of the form a^x * b^y, Number of co-primes = [(2x +1) (2y+1) - 1]/2

]]>1 winner 115 loser.

In 1 match 1 loser is decided. So 115 matches to decide 115 losers and 1 winner. ]]>

1 digit : 3

2 digit: _ _ =>9

3 digit : _ _ _ =>27

Total 39 so far

1 _ _ _ => 27

3 _ _ _ => 27

Total 93 so far

5 1 1 _ => 3

5 1 3 _ => 3

So 99 done

100th will be 5 1 5 1 Ans ]]>

8^6 * 27/64 >= (xy)^3

2^12 * 3^3 >= (xy)^3

xy = 2^4 * 3 = 48 ]]>

Means if you check for 24! =>

24/5 => 4

25/5 => 5

5/5 => 1

So 5 +1 => 6

Same for 26!

So

98! And 100!

99! And 101!

Similarly

148! --> 150!

149! --> 151!

And

173! --> 175!

174! --> 176!

Remember you don't have to consider 125! (As it results into increment of 3 zeroes )

So 6 possibilities

a) (23!)^2 > 23^23

b) (20!∗19!∗18!) < 57!

c) (33!)^4 < 33^60

d) None of these ]]>

A is right coz

Lhs=> 23!^2 = (23 * 1)(22 * 2)(21 * 3)...(2 * 22)(3 * 23)

However in rhs each terms is =>23 * 23 * 23...

Certainly lhs > rhs

Now in 3rd

33!^2 > 33^33

So 33!^2 * 33!^2 > 33^33 * 33^33

33!^4 > 33^66 hence 3rd is false

OA:C

]]>But 13 straight lines out of the given 37 straight lines pass through the same point A.

Therefore instead of getting (13c2) points, we get only one point A.

Similarly 11 straight lines out of the given 37 straight lines intersect at point B.

Therefore instead of getting (11c2) points, we get only one point B.

Hence the number of intersection points of the lines is = 37c2 - 13c2 -11c2 + 2 = 535 ]]>

And out of which

E --> 1

X --> 1

M -->1

T ---> 1

O ---> 1

(A) --> 2

I --> 2

N --> 2

8 distinct letters.

- 4 letters selected, which are all distinct: 8C4 = 70
- 2 letters alike, and 2 distinct

(eg: AAEM)

= 3c1 x 7c2 = 63 - 2 letters alike, and 2 letters alike

(eg: NNII)

= 3c2 = 3

So answer is, 70 + 63 + 3 = 136

]]>_ : 6 numbers ; _ _ : 6 * 7 = 42 numbers ; _ _ _ : 6 * 7 * 7 = 294 numbers

342 numbers added till 999

1 _ _ _ : 7 * 7 * 7 = 343 added = 685 numbers till now. Remaining 1000 - 685 = 315

3 0 _ _ : 49 added

3 1 _ _ : 49 added

3 3 _ _ : 49 added

3 4 _ _ , 3 7 _ _ , 3 8 _ _ = 49*3 = 147

Total 294 more. Remaining 21

3 9 0 _ : 7

3 9 1 _ : 7

3 9 3 _ : 7

3939

Method:2

Base method approach

0 -> 0

1 -> 1

2 -> 3

3 -> 4

4 -> 7

5 -> 8

6 -> 9

1000th term : 1000 in base 7 = 2626 < - > 3939

AB is a tunnel

And suppose person is at point C

AC : CB = 1:3

Now the time at which bus reaches A

Vikram would reach from point C to point A

And the time at which bus reaches to B

Vikram reaches from C to B

Let assume

Length of tunnel is 4km

So if Vikram moves 1 km forward towards B

Bus would reach at A

So remaining distance covered by Vikram is 2km

And bus covered the whole distance 4km(length of tunnel) in the same time with a speed of 80km/hr

So speed of Vikram = 80/2 => 40km/hr ]]>

Last two digits

(41)^28

=> (81)^14

=> (61)^7

=> 21

Or (41)^28

=> (4 * 8)1

=> 21

So 25-21 = 04

digit at tenth place = 0

]]>