Saturday, June 14, 2014

Previous year Solved math CTET questions - part2

11.A square and a circle have equal perimeters. The ratio of the area of the square to the area circle is

(1) 1:1  (2) 1:4
(3) pi:2 (4) pi:4


Answer:(4) pi:4
Solution:
Step1:Perimeter of square =4s and perimeter of circle = 2xpixr are equal
          => 4s = 2xpixr
          => s/r=pix2
               


Step2:Area of square (sxs) / area of circle (Pi x rxr) = sxs/pixrxr = (s/r)^2 /pi
          but (s/r) = (pi/2)
          => (s/r)^2 /pi = (pi/2)^2/pi = pi/4
           

Therefore, the ratio of the area of square and the circle are in the ratio pi:4


12. ABCD is a square with AB = (x + 16) cm and BC = (3x) cm. The
perimeter (in cm) of the square is :

(1) 16           (2) 24
(3) 32           (4) 96


Answer:(3) 32
Since, ABCD is a square this means AB =BC.
         => (x + 16) = 3x
         =>  2x = 16
         => x = 8 cm.
Therefore, the length of the square is ( 8 + 16) =3x8 = 24cm.
and the perimeter = 8x4 =32 cm

13.The mean of 10 numbers is 0. If 72 and -12 are included in these
numbers, the new mean will be
(1) 0      (2) 5
(3) 6      (4) 60


Answer: (2) 5
Solution:
Number of number = 10 and their mean = 0.
since  two more numbers are added, therefore total numbers = 12.
Using the formula for mean, we have
            mean = ( 0 + 72 -12)/12 = 60/12 = 5


14.The circumference of the base of a right circular cylinder is 44 cm and
its height is 15 cm. The volume (in cm3) of the cylinder is (use pi = 22/7)


(1) 770        (2) 1155
(3) 1540      (4) 2310


Answer:(4) 2310

Solution:
Volume of right circular cylinder = (pixrxr)xl
where pi = 22/7 and r is radius of the circular base and l is the length of the cylinder.
Given that perimeter of circular base = 44
         => 2xpxr = 44
         => 2x(22/7)xr = 44
         => r/7=1
         => r =7
Put these values in the formula for volume to get
        volume = (22/7)x7x7x15 = 2310


15.The value of 0.001 + 1.01 + 0.11 is
(1) 1.111       (2) 1.101
(3) 1.013       (4) 1.121


Answer:  (4) 1.121
Solution:
0.001 + 1.01 + 0.11 = 0.001 + 1.12 = 1.121


16. In 1999, the population of a country was 30.3 million. The number which is the same as  30.3 million is 


(1) 303000000        (2)30300000
(3) 3030000           (4)3030000000


Answer: (2)30300000
Solution:
We know, 1 million = 1000000
Then 30.3 million = 30.3 x 1000000 = 30300000


17. If a^3 = 1 + 7, 3^3= 1 + 7 + b and 4^3 = 1 + 7 + c, where a, b, and c are different positive integers, then the value of a+b+c is
(1)  58       (2) 68

(2)  77       (4) 79

Answer:(4) 79
Solution:
Step1: a^3 = 1 + 7 = 8 = 2^3
        =>   a=2

Step2: 3^3 = 1 + 7 + b 
          => 27 =8 + b
          => b = 21 - 8 =21

Step3: 4^3 = 1 + 7 + c
         => 64 = 8 + c
         => 56 = c

Therefore, a + b+ c = 21 + 2 + 56 =79

 18.We call a number perfect if it is the sum of all its positive divisors, except itself. For example,28 is a perfect number because 28 = 1 + 2 + 4 + 7 +14. Which of the following numbers is a perfect number?

(1) 13        (2) 10
(3)  9         (4) 6


Answer: (4) 6
Solution:
taking each option one by one
(1) 13. There is divisor, other than the number itself, 1. Not a perfect number.
(2) 10. There are three divisors 1,2,5 and their sum = 8.Not a perfect number.
(3) 9. There are two divisors. 1,3 and their sum is 4.Not a perfect number.
(4).6. There are three divisors 1,2,3 and their sum is 6 itself. This is the perfect number we are looking for. 



19. Which of the following number is a perfect square?
(1) 548543213   (2) 548543215
(3) 548543251   (4) 548543241


Answer:(4) 548543241
Solution:
A perfect square cannot end with 3. Rejecting option 1.
Again a perfect that has 5 at the unit digit must have 2 in the tenth place.Therefore, rejecting option 2.
Try the division method on option 4 and will get the perfect square.


20. The product of two whole numbers is 24. The smallest possible sum of these numbers is
(1) 8               (2) 9
(3) 10             (4) 12


Answer:
Solution:
The set of pairs of numbers that gives 24 on multiplication are 
 [(12,2),(8,3),(6,4)]. The sum of these numbers are (14,11,10) respectively.
Therefore the lowest possible sum is 10.





For Solved math Questions on CTET try:

NEW : solved CTET 2015 February - Paper1


1. CTET Math Solved Question Part3

2. CTET Math Solved Questions part4

3. CTET Math Solved Questions Part1

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