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Quant Marathon by Gaurav Sharma - Set 2

Gaurav Sharma, Director of Genius Classes Of Mathematics (GCM), Karnal ( Haryana ), completed his graduation from Kurukshetra University in Electronics. Being inclined towards Mathematics, he pursued his Masters in Mathematics and also completed his L.L.B. He believes that, when a man quits learning he will be more senile onwards. He is a mentor right from his college days and loves teaching and inspiring students. Team Genius Classes, along with the regular classes, organizes workshops, events and seminars time to time. You can find lot of good stuffs in his Facebook group - CAT Mausi

Each of the four girls A, B, C and D had a few chocolates with her. A first gave ½ of the chocolates with her to B, B gave 1/3rd of what she then had to C and C gave 1/4th of what she then had to D. Finally, all the four girls had an equal number of chocolates. If A had 90 chocolates more than B initially, find the difference between number of chocolates that C and D initially had.

  1. 10
  2. 30
  3. 15
  4. Cannot be determined

Let number of chocolates with A, B, C and D be a, b, c and d respectively. The transaction tables are as follows

We have, a/2 = 2/3 (b + a/2) => a = 4b

And we are given a - b = 90 => a = 120 and b = 30

So everyone had 120 – 120/2 = 60 chocolates at the end.

Putting values of a and b in ¾ (c + 1/3(b + a/2)) = 60

We have c = 50

Also d + ¼(c + 1/3(b+a/2)) = 60

We get d = 40

Hence difference between values of c and d is 10

How many natural numbers when expressed in base 5, base 4 and base 3 form three – digit, four digit and five digit numbers respectively?

  1. 44
  2. 67
  3. 38
  4. 72

Three digit numbers in base 5 are from 5^2 to 5^3 – 1

From 25 – 124

Four digit numbers in base 4 are from 4^3 to 4^4 – 1

From 64 to 225

Five digit numbers in base 3 are from 3^4 to 3^5 – 1

From 81 – 242

So, numbers common to all the intervals are from 81 – 124

Required answer is 124 – 81 + 1 = 44

If fresh grapes contain 90% water and 10% pulp by weight and 10 kg of fresh grapes yield 2.5 kg of dry grapes, then find the percentage of pulp by weight in dry grapes.

  1. 20%
  2. 40%
  3. 75%
  4. 80%

If the weight of fresh grapes is 10 kg then, it contains 9 kgs of water and 1 kg of pulp (by weight).

If 10 kg of fresh grapes yield 2.5 kg of dry grapes, the quantity of pulp would still remain the same.

So in 2.5 kg of dried grapes there is 1 kg of pulp

Percentage of pulp is 100/2.5 = 40%

A trader has one weighing stone each of weights 1, 2, 4, 8, 16 … and 1024 kg. If the trader gave another trader some of the stones, weighing a total of 2007 kg, what is the weight of the heaviest weighing stone remaining with him?

  1. 32 kg
  2. 64 kg
  3. 16 kg
  4. Cannot be determined

The stones weigh 2^0, 2^1… 2^10 kgs

The total weight that can be given as

2^0 + 2^1 + … + 2^10 = 2^11 – 1 = 2047

But he gave 2007 which is 40 less than the total weight

So the maximum weight the trader has with himself will be 2^n

With the largest value of n for which 2^n < 40

So n = 5 and 2^5 = 32

A shopkeeper sells two television sets at rs. 18000 each. On the first television he makes a profit of x% but on the other, he incurs a loss of 10%. If the shopkeeper later figured out that he made neither profit nor loss is the two transactions put together, then find x ?

  1. 11.11%
  2. 12.5%
  3. 15%
  4. 10%

Let the CP of the two TV sets be A and B respectively

Then 90 x B/100 = 18000 => B = 20,000

Also SP of both the TVs = 36000

CP of both the TV is also 36000 (no profit, no loss)

A = 36000 – B

A = 16000

Now profit % on sales of A = [ (18000 – 16000 )/16000 ] x 100 = 12.5%

If x + y + z = 0 and x^2 + y^2 + z^2 = 26, find x^4 + y^4 + z^4

  1. 174
  2. 260
  3. 338
  4. 676

By a simple trial of x = 1, y = 3 and z = -4 or

x = 1, y = 3 and z = -4

Both the equations x + y + z = 0 and x^2 + y^2 + z^2 = 26

So x^4 + y^4 + z^4 = 1^4 + 3^4 + 4^4 = 338

Else we can go with the traditional method

(x + y + z)^2 = x^2 + y^2 + z^2 + 2 (xy + yz + zx)

0 = 26 + 2 (xy + yz + zx)

xy + yz + zx = -13

(xy + yz + zx ) (x^2 + y^2 + z^2) = -13 x 26

x^3 ( x + y + z) – x^4 + y^3 ( x + y + z) – y^4 + z^3 ( x + y + z) – y^4  = -13 x 26

As ( x + y + z) = 0

x^4 + y^4 + z^4 = 338

Find the area of the polygon formed by joining the points (2,0) , (3,2) , (4,6), (1,7), (-1,4), (0,2)

  1. 19.5
  2. 20.5
  3. 21
  4. 22.5

Area of polygon with verticles (x1,y1), (x2,y2) … (x6,y6) =

½ |[x1y2 – y1x2) + (x2y3 – y2x3) + … + (x6y1 – y1x6)]|

Here, area = ½ |[(4-0) + (18 – 8) + (28 – 6) + (4-7) + (-2) + (-4)]| = 20.5 sq units

A man of height 1.2m walks at uniform speed of 16 m/min away from the lamp post of height 6m. The rate at which the length of his shadow increases is

  1. 8 m/min
  2. 4 m/min
  3. 5 m/min
  4. 6 m/min

ED/AB = 1.2/6 = x/(x+y)

5x = ( x + y)

y = 4x => x = y/4

Rate at which x changes = ¼ rate at which y changes

Rate at which x changes = ¼ x 16 = 4

An iron cube of side 4 cm is melted and exhaustively recast into N1 cubes of side 1 cm each and N2 cubes of side 2 cam each and N3 cubes of side 3 cm each ( N1, N2, N3 > 0). What is the probability that N2 is an odd number?

  1. 0.50
  2. 0.75
  3. 0.60
  4. 0.40

N1 x 1^3 + N2 x 2^3 + N3 x 3^3 = 4^3

N1 + 8N2 + 27N3 = 64

Case 1 :

N3 = 1 => N1 + 8N2 = 37

Possible values of N2 are 1,2,3,4

Case 2:

N3 = 2 => N1 + 8N2 = 10

The only possible value of N2 is 1

Hence the required probability is 3/5 = 0.60

Evaluate : 1/(3^2 – 1) + 1/(5^2 – 1) + 1/(7^2 – 1) … + 1/(99^2 – 1)

  1. 49/100
  2. 49/200
  3. 51/200
  4. None of these

S = 1/(3^2 – 1) + 1/(5^2 – 1) + 1/(7^2 – 1) … + 1/(99^2 – 1)

= 1/(2 x 4) + 1/(4 x 6) + 1/(6 x 8) + … + 1/(98 x 100)

= ½ [ ½ - ¼ + ¼ - 1/6 + 1/6 – 1/8 + 1/8 + … + 1/98 – 1/100]

= ½ (1/2 – 1/100)

= ½ ( 49 / 100)

= 49/200

Find the sum of

1/( 1 x 3 x 5) + 1/(3 x 5 x 7) + 1/(5 x 7 x 9) + … + 1/(13 x 15 x 17)

  1. 8/85
  2. 35/429
  3. 7/85
  4. 11/133

Here the series is sum of continued products taken three at a time but in denominator

So here n = 3

Sum is given by 1/(n-1)d [ 1/(1x3) – 1/(15x17)]

Find the sum of 5 x 6 x 7 + 6 x 7 x 8 + … + 16 x 17 x 18

Here the series is sum of continued products taken three at a time. So here n = 3

Sum is given by

1/(n + 1)d [ 16 x 17 x 18 x 19 – 4 x 5 x 6 x 7]

Find the sum of first 100 terms of the series whose general term is given by T(k) = (K^2 + 1)k!

T(K) = (K^2 + 1)K!

= (K(K+1) – (K -1))K!

= K(K+1)! – (K-1)K!

K(K+1)! – (K-1)K!

T(1) = 1x2! – 0

T(2) = 2 x 3! – 1 x 2!

T(3) = 3 x 4! – 2 x 3!

T(100) = 100 x 101! – 99 x 100!

T(1) + T(2) + … T(100) = 100 x 101!