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Quant Marathon by Gaurav Sharma - Set 4

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

Q1) If 9(x^2 – 2) – 3(x^2 – 2) = 6, then the sum of all possible values of x is

  1. 0
  2. Root(3)
  3. -3
  4. 2root(3)

9^(x^2 – 2) – 3^(x^2 – 2) = 6

Let 3^(x^2 – 2) = y

y^2 – y – 6 = 0

(y-3) (y+2) = 0

y = 3 and y = -2

But y cannot be a negative number, so y = 3

x^2 – 2 = 1 => x^2 = 3 => x = root(3) or – root(3)

Sum = root(3) + (- root(3)) = 0

Q2) log0.6(5x + 6) > log0.6(x^2 + 10) then

  1. -6/5 < x < 1 or x > 4
  2. 1 < x < 4
  3. 0 < x < 0.6
  4. 0 < x < 4

log0.6(5x + 6) > log0.6(x^2 + 10)

Also base 0.6 < 1

5x + 6 < x^2 + 10

x^2 – 5x + 4 > 0

(x – 1)(x – 4) > 0

x < 1 or x > 4

Also, log of negative numbers is not possible

So 5x + 6 > 0 => x > 6/5

And x^2 + 10 > 0 for all values

Hence range will be -6/5 < x < 1 or x > 4

Q3) In an isosceles triangle ABC with AB = AC = 20 cm and BC = 24 cm. Circle with center 0 touches BC at P, CA at Q and AB at R. Find the area of ARPQ

AC = 20, PC = 12

So AP = 16 (Pythagoras theorem)

Now, PC = QC = 12

AQ = 8ARQ ~ ABC

AQ/AC = RQ/BC

8/20 = RQ/24 => RQ = 48/5

ARPQ is a Kite (AR = AQ, RP = QP)

Area = ½ x d1 x d2 = ½ x AP x RQ = ½ x 16 x 48/5

= 76.8 sq cm

Q4) x, y and z are integers that are sides of an obtuse – angled triangle. If xy = 6, find z

  1. 2
  2. 4
  3. 3
  4. Both (a) & (b)

If xy = 6

Then possible cases are: 2 x 3 = 6; 1 x 6 = 6

Case 1:

When the sides of triangle are 1, 6 and z

Here (6 – 1) < z < (6 + 1) => 5 < z < 7

Hence z = 6 but (1, 6, 6) does not form an obtuse triangle.

Case 2: Sides are 2, 3 and z

Here (3 – 2) < z < (3 + 2) => 1 < z < 5

z = 2, 3, 4

Triangle 1: (2, 3, 2)

3^2 > 2^2 + 2^2, hence it is obtuse => z = 2 is possible

Triangle 2: (2, 3, 3)

3^2 # 2^2 + 3^2, hence it is not an obtuse triangle

z = 3 is not possible

Triangle 3 (2, 3, 4)

4^2 > 2^2 + 3^2 hence it is obtuse. Z = 4 is possible.

Option d is correct

Q5) the product of digits of a five digit number is 600. How many such numbers are possible?

600 = 2^3 x 3 x 5

Leaving aside 5^2

Arrangements of 2^3 x 3 = 24 are

( 1 x 8 x 3 ), ( 1 x 4 x 6), (2 x 4 x 3), (2 x 2 x 6)

Note: We will not consider [1 x 2 x 12] as 12 is a 2 digit number

Now possible cases for

(1, 8, 3, 5, 5) are 5!/2! = 60

(1, 4, 6, 5, 5) are 5!/2! = 60

(2, 4, 3, 5, 5) are 5!/2! = 60

(2, 2, 6, 5, 5) are 5!/2!2! = 30

Total = 60 + 60 + 60 + 30 = 210

Q6) If all the four digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6, 7 and 9 without repetition are arranged in ascending order, what will be the rank of number 5293 ?

  1. 897
  2. 898
  3. 908
  4. None of these

For four digit numbers from (1, 2, 3, 4, 5, 6, 7, 9)

Starting with 1 – 7C3 x 3! = 35 x 6 = 210

Similarly for 4-digit numbers starting with 2, 3 and 4 we have 210 cases each.

Total = 210 x 4

4 digit numbers starting with 5 1 - - = 6C2 x 2! = 30

4 digit numbers starting with 5 2 1 - = 5C1 = 5

Similary for 4 digit numbers starting with 5 2 3 - , 5 2 4 - , 5 2 6 - , 5 2 7 - : we will have 5 cases each

Total 25 cases

Now the next number will be 5291 – 1 case

Required number is 5293 – again 1 case

Total cases = 840 + 30 + 25 + 2 = 897

Hence rank of 5293 is 897th.

Q7) Bells A and B ring 8 times and 38 times in a minute respectively. If they start ringing simultaneously after how much time (in seconds) will B ring exactly 10 times more than A ?

  1. 8
  2. 10
  3. 15
  4. 20

B will ring 38 – 8 times more than A in a minute

Hence to ring 10 times more it will take 10/30 minutes

10/30 minutes = 10 x 60/30 = 20 seconds

Q8) The areas ( in cm^2) of two circles are 576 Pi and 729 Pi. If the distance between their centres is 54 cm what is the number of common tangents?

  1. 4
  2. 3
  3. 2
  4. 1

The areas of two circles are 576 Pi and 729 Pi

The radii of the two circles are 24 cm and 27 cm respectively.

Sum of radii = 24 + 27 = 51

And 51 < 54 (distance between the centres)

When the sum of radii ot two circles is less than the distance between their centres, then they do not intersect. So the two circles do not intersect each other. Hence there are 4 common tangents.

Q9) x has 4 factors and xy has 6 factors. How many factors does y have?

If x has 4 factors -> x is of the form a^3 or ab

xy has 6 factors

If x = a^2 then xy = a^5

Means y = a^2, number of factors will be 3

If x = ab then xy = a^2b or ab^2

Here y = a or b. hence number of factors will be 2

So number of factors of y can be 2 or 3.

Q10) A solid right circular cone with base radius r units and height h units is melted and re-molded into a cube of side a units. Which of the following cannot be true?

  1. h > a > r
  2. r > a > h
  3. a > r > h
  4. h > r > a

Volume of the right circular cone = 1/3 Pi r^2 h

Volume of the cube = a^3

It is given that 1/3 Pi r^2 h = a^3

We know, Pi/3 > 1

1/3 Pi r^2 h  > r^2 h

a^3 > r^2 h

Hence the following can be deducted

  1. Exactly 1 of r and h is less and a ‘OR’
  2. Both r and h are less than a

So, h > r > a cannot be true.