Solution of Lekhpal 2015 Exam Paper Continues

Solution of Lekhpal 2015 Exam continues............

Q9. The value of [15612 + {154 + (225)^1/2 }^1/2 ]^1/2

(a) 110                    (b) 115                 (c) 125                (d) 99           

Sol. [15612 + {154 + (225)^1/2 }^1/2 ]^1/2

= [15612 + {154 + 15 }^1/2 ]^1/2

= [15612 + {169 }^1/2 ]^1/2

= [15612 + 13 ]^1/2

= [15625 ]^1/2

= 125.

Option (c)

Q10. Two numbers are in the ratio of 3:5 and their LCM is 300. The smaller number is

(a) 45                     (b) 75                    (c) 60                  (d) 90

Sol. Let the numbers be 3x and 5x

Then their LCM = 15x

But A.T. Q:-  15x = 300

→ x = 20

Hence the smallest number = 3x = 3 × 20 = 60.

Option (c)

Q11. If the mean and coefficient of standard deviation of a set of data are 10 and 5 respectively, then the standard deviation of the data set is

(a) 10                    (b) 50                     (c) 5                   (d) 1

Sol. Formula :- Standard Deviation = Mean of data × Coefficient of standard deviation

So, standard deviation = 10 × 5 = 50

Option (b)

Q12. The largest four digit number which can be divided by 24, 30, 36 each is

(a) 9620                (b) 9360                (c) 9960              (d) 9840

Sol.  First calculate the LCM of 24, 30, 36

LCM = 360

Now largest four – digit number = 9999

Divide 9999 by 360 and determine the remainder

Remainder = 279

Hence required largest number = 9999 – 279 = 9720.

But this is not present in given option so we try another method to get the answer

2^nd Method Hit & Trial method

In this method we find that 9360 in the given options is the only number which is fully divisible by 24, 30, 36

Hence required number is 9360.

Option (b) 

Q13. If [x² + (1/x²)] = 23, then the value of [x⁴ + (1/x⁴)] is

(a) 526                  (b) 429                 (c) 629                (d) 527

Sol. Formula :- (a + b)² = a² + b² + 2ab

Now  [x² + (1/x²)] = 23

Squaring both sides, :- [x² + (1/x²)]²  = (23)²

x⁴ + (1/x⁴) + 2 × x² × (1/x²) = 529

x⁴ + (1/x⁴) + 2 = 529

x⁴ + (1/x⁴) = 527

Option (d)

Q14. If the sides of a triangle are in the ratio of 3 :4 :5  and its area is 216 cm², then the perimeter of the triangle is

(a) 72 cm               (b) 76 cm            (c) 84 cm             (d) 80 cm

Sol. Let the sides of the triangle 3x, 4x, 5x

According to the concept it is clear that this triangle is a right – angled triangle with

Base = 3x , Height = 4x and Hypotenuse = 5x

Now Area of Right–angled Triangle = (1/2) × base × height

216 = (1/2) × (3x) × (4x)

6x² = 216

→ x = 6

Now the Perimeter of the triangle = 3x + 4x + 5x = 12x = 12 × 6 = 72cm

Option (a)

Q15. A researcher has collected the following sample data :-

5, 12, 6, 8, 5, 6, 7, 5, 12, 4, 1

The median is

(a) 5                       (b) 6                    (c) 7                       (d) 8.

Sol. Arrange the given data in descending order we get the arranged data as:-

12, 12, 8, 7, 6, 6, 5, 5, 5, 4, 1

Total number of terms = 11

Now median of the given data = middle term of arranged data = (Total term + 1)/2

Median = 6^th term = 6

Option (b)

Q16. The average age of 15 students in a class is 15 years. Among this the average age of 5 students is 14 years and the average age of 9 other students is 16 years. The age of the 15^th student is

(a) 15 years          (b) 14 years        (c) 11 years            (d) 15 whole (2/7) years.

Sol. Total number of students = 15,         Average = 15

Total sum of ages of all 15 students = 15 × 15 = 225 years

Among this group the average age of 5 student = 14

Then the total sum of ages of those 5 students = 15 × 5 = 70 years

Average of another 9 students from the same group = 16

Then the total sum of ages of those 9 students = 16 × 9 = 144 years

Now the total sum of ages of 14 students = 70 + 144 = 214 years

Hence the age of the 15^th student = 225 – 214 = 11 years.

Option (c)

Q17. The LCM and HCF of two numbers are 520 and 4 respectively. If one of these numbers is 52, then the other number is

(a) 52                     (b) 40                 (c) 42                       (d) 50.

Sol.  Let the numbers be a = 52 and b = ?

Their LCM = 520 and their HCF = 4

Formula :- a × b = LCM × HCF

52 × b = 520 × 4

b = 40.

Option (b)

Solution will be continued on this blog soon, just keep visiting.....................