Khilona earned scores of 97, 73 and 88, respectively in her first three examinations. If she scored $8 0$ in the fourth examination, then her average score will be
A
increased by 1
B
increased by 1.5
C
decreased by 1
✓
decreased by 1.5
Answer
Correct option: D.
decreased by 1.5
(D). decreased by 1.5 $\because$ Average score of Khilona for first three examinations $ =\frac{97+73+88}{3}=86 $ Her average score for all four examinations $ \begin{array}{l} =\frac{97+73+88+80}{4}=\frac{258+80}{4} \\ =84.5 \end{array} $ $ \because 86>84.5 $ $\Rightarrow$ score decreased Difference $=86-84.5=1.5$ Hence, Khilona's average score decreased by 1.5
The mean of three numbers is 40 . All the three numbers are different natural numbers. If lowest is 19, what could be the heighest possible number of remaining two numbers?
✓
81
B
40
C
100
D
71
Answer
Correct option: A.
81
(A). 81 We know, $ \text { Mean }=\frac{\text { Sum of observations }}{\text { Number of observation }} $ Let the remaining two numbers be $x$ and $y$. $\therefore$ Sum of observations $=19+x+y$ Number of observation $=3$ $ \begin{array}{ll} \therefore & 40=\frac{19+x+y}{3} \Rightarrow 19+x+y=120 \\ \Rightarrow & x+y \end{array} $ Let the heighest number be $y$. Then, $y=101-x$ Since, the lowest number is $19, x$ can't be less than 19. i.e. $x>19$ Let $x=20$, then $y=101-20=81$ Therefore, the highest number is 81 .
The mode of the following data is
2,3,5,7,5,4,5,3,3,8,2
A
2 and 3
✓
3 and 5
C
4 and 5
D
only 3
Answer
Correct option: B.
3 and 5
(B). 3 and 5
Given data is
$
2,3,5,7,5,4,5,3,3,8,2
$
From the above data, it is clear that 3 and 5 has occured maximum number of times i.e. 3.
Hence, the mode of data are 3 and 5.
(C). 7 We know that whole numbers are those which starts from zero (0). So, first 15 whole numbers are $ \begin{array}{c} 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14 \\ \therefore \text { Mean }=\frac{\text { Sum of numbers }}{\text { Number of terms }} \Rightarrow \text { Mean }=\frac{105}{15}=7 \end{array} $
If mean and sum of $n$ observations are 29 and 174 respectively, then $n$ is equal to
A
4
✓
6
C
8
D
5
Answer
Correct option: B.
6
(B). 6 We know, $ \begin{aligned} \text { Mean } & =\frac{\text { Sum of all observations }}{\text { Number of observations }} \\ \Rightarrow \quad 29 & =\frac{174}{n} \Rightarrow n=\frac{174}{29}=6 \end{aligned} $ Hence, number of observation $n=6$.
The money saved by a student during first six days of a week are ₹ 46 , ₹ 24, ₹ 29, ₹ 27, ₹ 42 and ₹ 42 . Find the average saving per day.
A
42
B
39
✓
35
D
36
Answer
Correct option: C.
35
(C). 35 Average saving money per day $=\frac{\text { Sum of saved money of all six days }}{\text { Number of days }} $ $=\frac{46+24+29+27+42+42}{6} $ $=\frac{210}{6}=$$$ ₹ $35$ Hence, the average saving of student per day is ₹35.
If mean of 6 observations is 4 , then their sum is
A
20
B
22
✓
24
D
26
Answer
Correct option: C.
24
(C). 24 Let sum of all 6 observations be $x$. $ \begin{array}{l} \text { Then, mean }=\frac{\text { Sum of all observations }}{\text { Number of observations }} \\ \Rightarrow \quad 4=\frac{x}{6} \quad \text { [given, mean }=4 \text { ] } \\ \Rightarrow \quad x=24 \end{array} $