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ANSWERS - HURDLE 26 : MENSURATION

PART A-->

Ans 1. 22,800

Ans 2. Rs 33 per Sq Ft.

Ans 3. 10 Ft

Ans 4. Rs. 27,160

Ans 5. Rs 44 per Sq ft.

Ans 6. 12 Ft.

Ans 7. Rs 30,966

Ans 8. Rs 48 per Sq ft.

Ans 9. 9 ft

Ans 10. Rs 33,020

Ans 11. Rs 38 per Sq ft.

Ans 12. 12 ft. (Note there are 2 doors)

Ans 13. Rs 28,958

Ans 14. Rs 28 per Sq ft.

Ans 15. 11 Ft.

Ans 16. Rs 32,614

Ans 17. Rs 38 per Sq ft.

Ans 18. 11 ft.

Ans19. C , Ans20. D, Ans21. G , Ans22. J, Ans23. K , Ans24. E , Ans25. A

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SOLUTIONS OF SOME QUESTION OF 'PART A':

Area of walls = 2*(Length*Height +Breadth*Height)- Area of windows- Area of doors.

Solution 1.
Area of walls = 2(15*12 +10*12)-20-30 = 550 Sq Ft.
a. Cost of carpeting = 42*15*10= 6300
b. Cost 0f papering =550*22= 12100
c. cost of painting door = 20*25= 500
d. Cost of painting window= 30*35= 1050
e. Cost of painting roof = 19*15*10=2850
So total cost of renovating the room = a+b+c+d+e=22800

Solution2.
Area of walls = 2(15*11 +12*11)-25-35 = 534 Sq Ft.
a. Cost of carpeting = 15*12*X
b. Cost of papering = 23*534= 12282
c. cost of painting door = 28*25= 700
d. Cost of painting window= 24*35= 840
e. Cost of painting roof = 15*12*15= 2700
So total cost of renovating the room = a+b+c+d+e=22462
adding a to e we have 16522+ 180 X= 22462
or 180 X= 5940
so X = 5940/180 = 33

Solution 3.
Area of walls = 2(16*H +13*H)-28-40 = 58H-68
a. Cost of carpeting = 36*16*13= 7488
b. Cost 0f papering =(58H-68)*25= 1450H-1700
c. cost of painting door = 33* 28= 924
d. Cost of painting window= 22*40= 880
e. Cost of painting roof = 23*16*13= 4784
So total cost of renovating the room = a+b+c+d+e=26,876
7488 +1450H-1700+924+880+4784=26876
1450H=14500
so H=10

++++++++++++++++++++++++++++++++

PART B

Ans B1. One sphere will be full and second sphere will have 10737 cubic cms.

Ans B2. 32 cms

Ans B3. 27 cms.

Ans B4. 20 cms

Ans B5. 20 cms

Ans B6. 24 cms

Ans B7. 1 sphere full and 1572 cubic cms in second one.

Ans B8. 22 cms

Ans B9. 14 cms (Kindly note 17,449 is the balance capacity of second sphere)

Ans B10. 32 cms.

Ans B11. 7 cms

Ans B12. 3 sphere full and 36,814 cubic cms in the 4th one.

Ans B13. 40 cms each side.

Ans B14. 22 cms.

Ans B15. 15 cms.

Ans B16. 42 cms

Ans B17. 21 cms

Ans B18. 3 cms

Ans B19. E , Ans B20. G , Ans B21. J , Ans B22. A , Ans B23. D, Ans B24. L

========================================

SOLUTION FOR PART B:

SOLUTION OF SOME OF THE PROBLEMS IS AS FOLLOWS :

Volume of sphere = (4/3).(22/7)R.R.R where R = radius of the
sphere.
Volume of Cube = Side*Side*Side
Volume of box : length* breadth* height
Volume of Cone = (1/3) (22/7)*Radius*Radius *height
Volume of Cylinder = (22/7)*Radius*Radius *height

Solution B1:
a. Volume of cube = 35*35*35= 42875
b. Volume of Box = 12*24*13= 3744
c. Volume of cylinder = (22/7)*7*7*17=2618
d. Volume of Cone = (1/3)*((22/7)*7*2*6= 308
Summing all the above(a+b+c+d) we get 49545.
Volume of sphere = (4/3).(22/7)*21*21*21= 38808
So one (Truncate 49545/38808) such sphere will be full and the
balance will be 49545-38808= 10737 which will be poured in
second sphere.

Solution B4:
a. Volume of cube = 74088
b. Volume of Box = 546.L
c. Volume of cylinder = 54208
d. Volume of Cone = 770
Total Volume (a to d) = 129066 + 546.L
Volume of 3 sphere = 116424
Volume of the Box = 14*17*99= 23562
Total Volume = 139986
so 129066 + 546.L= 139986
so L = 20.

Solution B9:
a. Volume of cube = 24389
b. Volume of Box = 4186
c. Volume of cylinder = (22/7)*19*19*H= 7942.H/7
d. Volume of Cone = 15884
Total Volume (a to d) = 44283 + 7942.H/7
Volume of 1 sphere = 38808
Balance volume of 2nd sphere = 17,449 so it means that it is
(38808-17,449) 21359 cubic cms full.
So total volume of water = 38808 +21359= 60167
so we have :
44283 + 7942.H/7 = 60167
or 7942.H/7= 15884
or H= 15884*7/7942
or H= 14.

Hurdle 26 : Mensuration
There are 2 parts in this Hurdle, Part A and Part B. PART A : Attempt any 8 questions from Q1 to Q18.
Hurdle26.htm

  Hurdle 27 : Ratio
Monthly income of X and Y are in ratio of P:Q and their expenses are also in ratio of P:Q. .
Hurdle27.htm
  Links of Hurdle 1 to 30
Our Hurdles are prepared on the basis of past entrance tests  papers.

Hurdle.htm
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