Free Practice Questions for Maths Subject, Attempt Now!

Three years ago, average of age of $A, B$ and $C=29$ years
In present, average of age of $A, B$ and $C=(29+3)=32$ years
In present, sum of age of $A, B$ and $C=32 \times 3=96$ years
Five years ago, average of age of $B$ and $C=23$ years
In present, average of age of $B$ and $C=23+5=28$ years
In present, sum of age of $B$ and $C=28 \times 2=56$ years
Therefore, the present age of $A=A+B+C-(B+C)=96-56=40$ years
Hence, option B is correct.

Let the numbers are 2x, 3x and 4x

According to question,

⇒ (2x)² + (3x)² + (4x)² = 2349

⇒ 4x² + 9x² + 16x² = 2349

⇒ 29x² = 2349

⇒ x² = 81

⇒ x = 9

Therefore, the average of three numbers = $=\frac{2 x+3 x+4 x}{3}=\frac{9 \times 9}{3}=27$

Hence, option A is correct.

Let the eight consecutive numbers are $(x-4),(x-3),(x-2),(x-1),(x),(x+1),(x+2),(x+5$
According to question,
$
\Rightarrow \frac{(x-4)+(x-3)+(x-2)+(x-1)+(x)+(x+1)+(x+2)(x+3)}{8}=10.5
$
$
\Rightarrow 8 x-4=84
$
$
\Rightarrow 8 \mathrm{x}=88
$
$
\Rightarrow x=11
$
Therefore, the largest number $=x+3=11+3=14$
Hence, option A is correct.

The average age of 125 students in a group is $16.2$ years.
Sum of average age of 125 students $=125 \times 16.2=2025$ years
$40 \%$ of the students are boys and the rest are girls.
Number of boys $=125 \times \frac{40}{100}=50$
Number of girls $=125-50=75$
The average age of the boys is $20 \%$ more than the average age of the girls.
Let average age of girls $=x$ years
Average age of boys $=\left(x+\frac{20 x}{100}\right)=\left(x+\frac{x}{5}\right)$ years
=$50\left(x+\frac{x}{5}\right)+75(x)=2025$
=$50 x+10 x+75 x=2025$
=$135 x=2025$
=$x=\frac{2025}{135}=15$
Average age of boys $=\left(x+\frac{20 x}{100}\right)=\left(x+\frac{x}{5}\right)$ years $=\left(15+\frac{15}{5}\right)=15+3=18$ years

ATQ,
Sum of the marks scored by three students $=3 \times 38=114$
Sum of the marks scored by five student $=5 \times 40$ $=200$
Maximum marks can be scored by one of the new comer $=45$
$\therefore$ Lowest marks that can be scored by one of the new comer $=200-114-45)=41$

Number of boys $=\frac{84}{(10+11)} \times 10=\frac{84}{21} \times 10=40$
Number of girls $=\frac{84}{(11+10)} \times 11=\frac{84}{21} \times 11=44$
Let the average score of girls $=x$
Average score of boys $=(x) \times \frac{(100-20)}{100}=\frac{80 x}{100}=\frac{4 x}{5}$
Now, according to question,
$
\begin{aligned}
&\Rightarrow 84 \times 95=40 \times \frac{4 x}{5}+44 \times(x) \\
&\Rightarrow 84 \times 95=32 x+44 x \\
&\Rightarrow 84 \times 95=76 x \\
&\Rightarrow x=105
\end{aligned}
$
Therefore, required average of boys $=\frac{4 x}{5}=\frac{4}{5} \times 105=84$

BASIC APPROACH -

Let number of students in class initiall $y=x$
Average weight of $x$ students in a class $=69.5 \mathrm{~kg}$.
It is given that when 10 students of average weight $68 \mathrm{~kg}$ joined the class, and 6 students of average weight $60 \mathrm{~kg}$ left the class, it was noted that the average weight of the new group of students increased by $2 \mathrm{~kg}$.
Now, present number of students in class $=x+10-6=x+4$
New average of the class $=69.5+2=71.5 \mathrm{~kg}$
Average $=\frac{\text { Sum of observations }}{\text { Total number of observation }}$
$
\Rightarrow 69.5+2=\frac{69.5 x+68 \times 10-6 \times 60}{x+4}
$
$
\Rightarrow 71.5=\frac{69.5 x+68 \times 10-6 \times 60}{x+4}
$
$
\Rightarrow 71.5 x+286=69.5 x+680-360
$
$
\Rightarrow 2 x=34
$
$
\Rightarrow x=17
$
Number of students in the class now $=17+4=21$

MATHS MIRROR APPROACH -

When 16 students left the class, they took average weight of 60 kg with them.
So, left average weight = 69.5 – 60 = 9.5 

Total weight left by them = 9.5 × 6 = 57
10 new students brought average weight less by = 71.5 – 68 = 3.5
Total weight given to new students = 3.5 × 10 = 35
Weight remained after distributing to new students = 57 – 35 = 22
Remaining weight that is distributed among old ones is 22 while average weight is increased by only 2
So old students = $\frac{22}{7}=11$
Total number of students for now = 11 + 10 = 21

If a three digits number is abc, then we can write it as $100 a+10 b+c$
So, we can write $2x5$ as $(200+10x+5)$
We can write $x35$ as $(100x+35)$
We can write $63x$ as $(630+x)$
Now, according to question,
$\Rightarrow \frac{335+(200+10 x+5)+(100 x+35)+(630+x)+406}{5}=411$
$
\Rightarrow \frac{1611+111 x}{5}=411
$
$
\begin{aligned}
&\Rightarrow 1611+111 x=2055 \\
&\Rightarrow 111 x=444 \\
&\Rightarrow x=\frac{444}{111} \\
&\Rightarrow x=4
\end{aligned}
$
Therefore, required average of number $x-1, x-3, x+3$ and $x+5=\frac{(4-1)+(4-3)+(4+3)+(4+5)}{4}$
$
\begin{aligned}
&=\frac{3+1+7+9}{4} \\
&=\frac{20}{4} \\
&=5
\end{aligned}
$

Let the seven consecutive odd numbers are (x - 5), (x - 3), (x - 1), (x + 1), (x + 3), (x + 5), (x + 7)

According to question,

Average = 

 27 = 

 189 = 7x + 7

 7x = 189 – 7

 x = 

 x = 26

Last term = x + 7 = 26 + 7 = 33

If the last number is removed from the list, then the sum of remaining numbers = x – 5 + x – 3 + x – 1 + x + 1 + x + 3 + x + 5 = 6x = 6 × 26 = 156

Therefore, the average of the remaining numbers =  = 26

Hence, option A is correct.

ATQ,
W X Y Z
Sum of three smallest number
$
(\mathrm{w}+x+\mathrm{y})=25.5 \times 3
$
Sum of largest three number
$
(x+y+z)=29.5 \times 3
$
Range = largest number - smallest number
$
\begin{aligned}
&=(29.5-25.5) \times 3 \\
&\Rightarrow 4 \times 3 \\
&\Rightarrow 12
\end{aligned}
$

ATQ,
$\Rightarrow a+b+c+d=26 \times 4$
$a+b+c+d=104 \ldots \ldots \ldots$ (i)
$a+b=19.5 \times 2$
$a+b=39 \ldots \ldots .$ (ii)
Putting the value of $(a+b)$ in equation (i)
$c+d=104-39$
$c+d=65$
$
c+d=\frac{65}{2}=32.5
$
Average of $(c+d)$ will be $=32.5$

Given:

Score after 11 innings = 51

Total score of 11 innings = 561

Score after next innings = 51+2 = 53

Total score of 12 innings = 636

Next innings score = (636 - 561)

Next innings score = 75

ATQ,
Let the average marks of 5 persons be $x$ and marks of $6^{\text {th }}$ student $y$,
Total marks obtained by 5 student is $5 x$, And if we add marks of $6^{\text {th }}$ students then the total mark will be come $5 x+y$,
Now the average marks be came
$\Rightarrow \frac{5 x+y}{6}=\frac{x+2}{1}$
$\Rightarrow x=y-12$
$\Rightarrow y=x+12$
Hence the value of 6 th student is 12 marks greater than average of 5 students.

Average age of 15 students $=15$ years

Total age of 15 students $=15 \times 15=225$

Total age of 5 students $=5 \times 14=70$

Total age of 9 students $=9 \times 16=144$

So the age of the 15th student $=225-(70+144)$ $=225-214=11$ year

ATQ,
Let 4 tests raghubir average marks $=x$
$
\begin{aligned}
&25 \times 12=8 \times 23+4 \times x \\
&75-46=x \\
&x=29
\end{aligned}
$

ATQ,
Sum of three number $=16.5 \times 3=49.5$
Sum of first two number $=15.5 \times 2=31$
Sum of last two number $=18.5 \times 2=37$
Sum of first two number and last two number
$
=31+37 \mathrm{}=68
$
II nd number $=68-49.5$ $=18.5$
1 st number $=18.5$
IIIrd number $=49.5-18.5 \times 2$ $=12.5$
Hence, the number is $12.5,18.5,18.5$

Given:

Average of three numbers = 30

Average = Sum of all elements/number of elements

Total = 30 × 3 = 90

3rd number = 90 – 18 – 27

= 90 – 45 = 45

∴ The third number is 45

Hence, option (4) is the correct answer.

ATQ,
Let the sum of age of remaining girls be $x$ year and that of new girls be y years.
$
\Rightarrow \text { Average }=\frac{\text { Total age }}{\text { Total no. of girls }}=\frac{x+18}{20}
$
$\Rightarrow$ New average $=\frac{x+y}{20}$
$
\Rightarrow \frac{x+18}{20}-\frac{x+y}{20}=\frac{2}{12}
$
$
\Rightarrow 18-y=\frac{10}{3}
$
$
\Rightarrow y=\frac{44}{3} \text { year }
$
$y=14$ years 8 month
$\therefore$ The age of new girl is 14 year 8 month

ATQ,
Four number arranged in ascending order
$\mathrm{W}: \mathrm{X}: \mathrm{Y}: \mathrm{Z}$
Sum of smallest three number $=25.5 \times 3$ $=76.5$
Sum of biggest three number $=29.5 \times 3$ $=88.5$
Range $=$ biggest number - smallest number $=88.5-76.5$
$
=12
$

ATQ,
Weight of $x$
$
\begin{aligned}
x &=50+\frac{26}{1}\left(\frac{1}{2}\right) \\
&=50+13 \\
&=63 \mathrm{~kg}
\end{aligned}
$