StudyMathOnline: Previous year Solved math CTETQuestions

41.If the cost prize of 10 candles is equal to the selling price of 8 candles, the gain/loss percent is
(1) 25% gain   (2) 20% loss
(3) 20% gain   (4) 25% loss

Answer:(1) 25% gain
Solution:
Let the cost price be X and the selling price be y. Then
     10x=8y
=> y/x = 10/8
=> y/x - 1 = 10/8 - 1
=> (y-x)/x = 2/8 = 1/4

In percentage it is 25% gain.

42.The mean of median and mode of the data 7,6,7,9,8,8,10,8 is
(1) 5.5    (2) 8
(3) 8.5    (4) 9

Answer:
Solution:
Arranging the above data in ascending order we get
6,7,7,8,8,8,9,10
Median for even number of observation is the mean of the n/2 - 1 and he n/2 +1 value.
Here, n=10 => mean of 4th and 5th value 0r (8+8)/2 = 8 is the median of the above observation.
Mode is the observation that is repeated most time. here it is 8.

There mean of mode and median = ( 8 + 8)/2 = 8

43. The area of trapezium-shaped field is 720 m^2, the distance between the two parallel sides is 20 m and the length of one of the parallel sides is 35m. The length of the other parallel side is
(1) 35m    (2) 36m
(3) 37m    (4) 40m

Answer:(3) 37m
Solution:
Area of trapezium is equal to the half of the sum of parallel sides x distance between the parallel sides.
Or area = 1/2( a + b) xh
given that, area = 720 m^2
height = 20 m
               one parallel side = 35 m
to find the other parallel sides.Let it be s
=> 720 = 1/2(35 + s) x 20
=> 1440/20 = 35+s
=> 72 = 35 +s
=> s =37

44. 4 - ( 2 - 9)^0 + 3^2 /1 + 3 is equal to
(1) 17   (2) 16
(3) 15   (4) 12

Answer: (3) 15
Solution:
STep1 : 4 - 1 + 9 + 3
step2 : 3 + 9 + 3
step3 : 15

45. Which of the following fractions is the least?
(1) 24/25      (2) 10/11
(3) 99/100     (4) 68/69

Answer:(3) 99/100
Solution:
we know,
1/2 => 50%
2/3 => 67%
3/4=> 75%
and so on.
Going by this rule as both denominator and numerator are increasing by the same value, then the higher the numbers the higher the value of fraction.
Going by this rule the highest number among these is 99/100.

46. When the number 3^98 is divided by 5, the remainder is
(1) 1     (2) 2
(3) 3     (4) 4

Answer: (4) 4
Solution:
We know,
3^1 =         3
3^2 =              9
3^3 =       27
3^4 =       81
3^5 =     243
3^6 =   729
3^7 =   2187

3^8 =   6561
3^9 = 19683

and so on.
Look at the pattern. You will observe that digits at the unit place are repeating themselves after every fourth power. Taking cue from this pattern, first we will find the digit at the unit's place in the value of 3^98.

Step1: divide 98 by 4. We will get a remainder of 2.
Step2: look at the pattern above. We can infer that the unit digit of 3^98 will be 9.
Step3: When we divide any number that has digit 9 at the unit place, then we will get 4 as remainder.

47.Two positive numbers x and y are inversely proportional. If x increases by 10%, then y decreases by
(1) 10%         (2) 2/11%
(3) 100/11%     (4) 10/11%

Answer: (3) 100/11%
Solution:
For inversely proportional numbers, the product of the number must always be constant, if all other conditions are the same.
Let x and y be both 10, then
   xy=Constant
=> 10x10= Constant

If x increases by 10%, then the new value of x will be 11. And the product of x and y will be 110. But product of x and y should be constant( 100 in this case.) Then y should be decrease to keep the product constant.Now,
xy =100
11y=100
y=100/11
And in percentage terms

[(10 - 100/11)/ 10] x100 = [10/110]x100 = 100/11 %

48. How many times will I be writing 2 if I wrote down numbers from 11 to 199?
(1) 36   (2) 37
(3) 38   (4) 39

Answer:(2) 37
Solution:
From 11 to 100, count numbers such as 12,32,42,52,62,72,82,92 which are 8 in numbers plus numbers from 20,21,22,23,24,25,26,27,28,29, have digits 2, 11 times. Therefore there are 19, 2 digit numbers from 11to 100.
Going by the same rule from 101 to 199, 2 occur 19 times plus 1. ( 2 from 102).
And total 19 + 20 = 39

49.Given n numbers, n >1, of which, one is (1 - 1/n) and all others are 1's. The mean if the n numbers is
(1) 1   (2) n - (1/n^2)
(3) n - (1/n)    (4) 1 - 1/n^2

Answer:1 - 1/n^2
Solution:
Given that there are n numbers, out of which one number is (1-1/n) and others are all one.
Step1: Add all the 'n' numbers
(1 - 1/n) + 1+1+1+..1 = n - 1/n
Step2: use the mean forumula
we have ,
   (n -1/n)/n = 1- (1/n^2)

50.A large basket of fruits contains 3 oranges, 2 apples and 5 bananas. If a piece of fruit is chosen at random, what is the probability of getting an orange or a banana?
(1) 4/5
(2) 1/2
(3) 7/8
(4) 1/5

Answer:(1) 4/5
Solution:
Number of oranges = 3
Number of bananas = 5
Total number of fruits = 10
Therefore,
Probability of getting an orange = 3/10
Probability of getting an banana = 5/10

Therefore the probability of getting either an orange or banana = 3/10 + 5/10
or 8/10 = 4/5