Quant Marathon by Gaurav Sharma  Set 2

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 1/3^{rd} of what she then had to C and C gave 1/4^{th} 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.
 10
 30
 15
 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?
 44
 67
 38
 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.
 20%
 40%
 75%
 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?
 32 kg
 64 kg
 16 kg
 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 ?
 11.11%
 12.5%
 15%
 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
 174
 260
 338
 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)
 19.5
 20.5
 21
 22.5
Area of polygon with verticles (x1,y1), (x2,y2) … (x6,y6) =
½ [x1y2 – y1x2) + (x2y3 – y2x3) + … + (x6y1 – y1x6)]
Here, area = ½ [(40) + (18 – 8 ) + (28 – 6) + (47) + (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
 8 m/min
 4 m/min
 5 m/min
 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?
 0.50
 0.75
 0.60
 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)
 49/100
 49/200
 51/200
 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)
 8/85
 35/429
 7/85
 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/(n1)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)! – (K1)K!
K(K+1)! – (K1)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!