Operations - class-VI

$Total >roses =4+5=9$

$6$ containers can hold $15600$ amount of oil

Let the $6$ numbers are $x, x+1, x+2, x+3, x+4, x+5$.

True,  we can find the successor of $599$ as follows:

False, as predecessor of  $399$ is $398 (399-1=398)$

False, since we can clearly analyse that predecessor of $10$ is 

False, as successor of $99$ is $100$.

$50342135 - 46380651 = 3961484$.

$1 + 2 - 3 + 4 - 5 + 6 - 7 + 8$

    1 0

  5 0 0

 $99999$
$-10000$
______
$89999$

Let Y closed pipe after a min .
Then part filled by (X+Y) in a min +Part filled by X in (18-x) 
So $a(\frac{1}{24}+\frac{1}{32})+(18-a)\frac{1}{24}= 1$
Multy by 96 
Or 4a+3a+72-4a=96
Or 3x=24
Or x=8 min

$x=\frac{4\sqrt{2}}{\sqrt{2+1}}$
$\frac{1}{\sqrt{2}}\left ( \frac{x+2}{x-2}+\frac{x+2\sqrt{2}}{x-2\sqrt{2}} \right )$
Put the value of x
$\frac{1}{\sqrt{2}}\left ( \frac{\frac{4\sqrt{2}}{\sqrt{2+1}}+2}{\frac{4\sqrt{2}}{\sqrt{2+1}}-2}+\frac{\frac{4\sqrt{2}}{\sqrt{2+1}}+2\sqrt{2}}{\frac{4\sqrt{2}}{\sqrt{2+1}}-2\sqrt{2}} \right )$
=$\frac{1}{\sqrt{2}}\left ( \frac{4\sqrt{2}+2\sqrt{2}+2}{4\sqrt{2}-2\sqrt{2}-2} \right )+\left ( \frac{4\sqrt{2}+4+2\sqrt{2}}{4\sqrt{2}-4-2\sqrt{2}} \right )$
=$\frac{6\sqrt{2}+2}{2\sqrt{2-2}}+\frac{6\sqrt{2}+4}{2\sqrt{2}-4}$
=$\frac{1}{\sqrt{2}}\left ( \frac{32-24\sqrt{2}}{16-12\sqrt{2}} \right )$
=$\sqrt{2}$

$
:Q:R\quad =\quad 6:5:4\ Let\quad P\quad \quad =\quad 6x\ Q=\quad 5x\ R\quad =\quad 4x\ { P }^{ 2 }+{ Q }^{ 2 }{ +\quad R }^{ 2 }\quad =\quad 192500\ { (6x) }^{ 2 }+{ (5x) }^{ 2 }+(4x)^{ 2 }\quad =\quad 192500\ 77{ x }^{ 2 }\quad =\quad 192500\ { x }^{ 2 }\quad =\quad 2500\ x\quad =\quad 50\ \ \frac { P+Q-R }{ 2 } \quad =\quad \frac { 6x+5x-4x }{ 2 } \quad =\quad \frac { 7x }{ 2 } \quad =\quad \frac { 7\times 50 }{ 2 } \quad =\quad 175
$

  1  2
  5  A  9
  3  B  7
  2  C  8
1 1  1  4
$2+A+B+C=11$
$A+B+C=9$
Clearly max value of $C=9$

According to the question
$4 a 3 + 9 8 4 = 1 3 b 7$
$a + 8 = b$
Then$b - a = 8$
According to the question  13b7 is divisible by 11
$(7 + 3) - (b + 1) = (9 - b)$
$(9 - b) = 0$
$b = 9$
$b = 9   and   a = 1$
Then $a+b=9+1=10$

Let student get x correct answer then 50-x student get wrong answer 
Then 4(x)+(50-x)(-1)=120
Or 4x-50+x=120
Or 5x=120+50=170
Or x=34

$(\sqrt 7+\sqrt 5)(\sqrt 7-\sqrt 5)$
$\Rightarrow \left(\sqrt {7}\right)^2-\left(\sqrt{5}\right)^2$
$\Rightarrow 7-5 =2$

Let the two consecutive odd numbers are x and x+2
$x+x+2 = 2004$
$x+x+2-2 = 2004 -2 $ (Subtract 2 from both the sides)
$2x = 2002$
$\frac{2x}{2} = \frac{2002}{2}$ (Divide 2 from both the sides)
$x = 1001$
Hence two numbers are 1001 and 1003

The sum of  375 and 125 $=375+125 = 500$
The difference of 375 and 125 $ = 375-125 = 250$
Hence Sum contains $500 \div 250 = 2 $ times their differences.

The man down stream is 16 km in one hour and he take twice time in upstream 

If a=2 ,b=3 and c=4

Given 125 x cm \second=(x+7)x km\hr 

Given 5 mates in 5 hours makes 5 laborers

The averages weight of seven man diminished by 3 kg .Then total weight diminished by $3\times 7=21 kg $

Let the number to be subtracted is $x$

Greatest - 8 digit no. 99999999
Smallest - 9 digit no. $\underline {100000000}$
                                  $=\underline {199999999}$

Let the certain number be x
As per problem,
$\Rightarrow \frac { x }{ 3 } -2=1$
 $\Rightarrow \frac { x-6 }{ 3 } =1$
 $\Rightarrow x-6=3$
 $\Rightarrow x=9$
The correct answer will be $9\times 3+2=27+2=29$

Stock $=23,898 kg$.
Stock sold in morning $=1,795$
Stock sold in evening $=\underline {2,348}$
                                         $\underline {4,143}$
  Stock left $=23,898-4143$
$=19,755 kg$.

   $18,47,525$
$+\underline {23,05,218}$
  $\underline { 41,52,743}$
$\therefore$ Number of children
$=5,12,10,913-41,52,743$
$=4,70,58,170$

888
088
008
008
$\underline {+8}$
$\underline {1000}$

$35^oC- 25^o C = 10^oC$
$10^o C = \displaystyle \left( 10 \times \frac{9}{5} \right ) + 32$
$= 18 + 32 = 50^o F$.

To add Various numbers with different decimal points we pick the number with the largest decimal point and extend the others to the same decimal point.

The amount of sugar sold on first day = 12.75 Kg

In a movie hall, the vacant seats are $=363$

Each army truck could carry a maximum soldiers $=55$

A balloon seller sold $=236$ red balloons

There are 4 symbols used in the question to depict houses.

Required sum $=(2+4+6+......+30)$
This is an A.P in which $a=2,d=(4-2)=2$ and $l=30$.
Let the number of terms be $n$. Then,
${t} _{n}=30$ $\Rightarrow$ $a+(n-1)d=30$
$\Rightarrow$ $2+(n-1)\times 2=30$
$\Rightarrow$ $n-1=14$
$\Rightarrow$ $n=15$
$\therefore$ ${S} _{n}=\cfrac{n}{2}(a+l)=\cfrac{15}{2}\times (2+30)=240$

Number of chocolates with Seema $=10$

Total number of packets  pencils $=25$

Number of magic pens with Ravi $=15$

Negative numbers lie left side of Zero.

Positive Numbers lie right side of Zero on numberline.

The smallest 6 digit number $=1,00,000$

Age of Raghu $= 21 $ years
Age of Kavita $= 72$ years
Therefore, sum of their ages $= 21+72 =93 $ years

Sum of digit of $2$ digit number $10x+y$ is $15$

$\Rightarrow$  Uncle shyam gives $230$ each apples to Neha, Joy and Karan.

$ a+b=30,ab=176 \ a+b=30 \ \Rightarrow { \left( { a+b } \right) ^{ 2 } }={ \left( { 30 } \right) ^{ 2 } } $

$\begin{matrix} \left( { 308+264 } \right) c{ m^{ 2 } }=572c{ m^{ 2 } }   \ It, , is, , true., , , , Ans. \  \end{matrix}$

$-59, -41, 73, -92, 81, -(-41), -3$.

$5p-11p=-6p$

We know that $0$ is the only number which, when added to some other number then it gives the same number.

Assume the whole numbers

For option A
$40+5\div45\times15-25$
if we change sign according to the question then
$40\times5=45\div15+25$
$200\neq28$
For option B
$40\div5+45\times15-25$
If we change the sign according to the question then
$40=5\times45\div15+25$
$40=5\times3+25$
$40=40$
Hence, option 'B' is correct.

the greatest $7$ digit number is $9999999$

Let us first subtract one from $5$ as shown below:

Let us have a look at the properties of whole numbers under subtraction:

$a^2 - b^2 = 13$
$(a-b)(a +b) = 1 \times 13$
By comparing, $a - b = 1, a + b = 13$
Add both the equations,
$2a = 14$
$a = 7$

Since 3 + B given us 1 in the answer, so B has to be 8 as $3+8=11$.
Now 1 is carried over so, $A+1+2=B$
i.e. $A+1+2=8$. So, A is 5.
Also $6+2=B$ in hundred's place to confirm that B is 8.
The value of A is 5 and B is 8.

Let the number of 50 paise coins =x
So the number of 25 paise coins=2x
Amount with 50 paise coines=0.5x
Amount with 25 paise coines=0.5x
According to question 
0.5x+0.5x=27
Or x=27
So the number of 50 paise coins=x=27
Or the number of 25 paise coines=2x=54 

In the given Arithmetic Progression,
First term $ = a  = 1 $
Common difference $ = 5 - 1 = 4 $

Let $ 4n - 3 $ be the $  k $ th term.

Then $ {x} _{n} = a + (n-1)d $
$ => 4n- 3 = 1 + (k-1)4 $
$ => 4n -1 = 1 +4k-4 $
$ => k = n $

So,  $ 4n - 3 $ is the $ n $ th term.

Now, Sum to 'n' terms of an AP $ = \frac {n}{2} (2a+(n-1)d) = \frac {n}{2} (2+(n-1)4) = \frac {n}{2} (4n-2) =n(2n-1) $

Distance between two points $ ({x} _{1}, {y} _{1}) $ and $ ({x} _{2}, {y} _{2}) $ is $ \sqrt {{({x} _{2}-{x} _{1})}^{2} + {{(y} _{2}-{y} _{1})}^{2} } $

So, distance between $ (-4,-5) ; (-4,-10) $ is $ \sqrt {{(-4+4)}^{2} + {(-10+5)}^{2} } = 5 $

Subtracting 4 from 9 yields 5. This satisfies the condition given in the question.

Given, $ 14y - 4 = 24y + 26 $
$ \Rightarrow  14y-24y = 26 + 4 $
$ \Rightarrow  -10y = 30 $
$ \Rightarrow  y = -3 $
Thus $ 12y = 12 \times (-3) = -36 $

• Grace has $8$ red , $4$ green , $4$ blue jellybeans
• In order to have one of each color , she should take out $7$ red, $3$ green , $3$ blue jellybeans
• So she has to take out $7+3+3=13$ jellybeans

Smallest three digit number $=100$

Greatest three digit number $=999$

Smallest three digit number $=999$

Subtracting 280575 from 370500 results in 89925.

To find the missing number, add 452567 and 253510, which equals 706077.

Highest Number $\rightarrow$ $932$

Population of City $\rightarrow$ Number of Men $+$ Number of Women $+$ Number of Children

(A) Total length of cloth Tina had $=1566$cm $=15.56$m
Length of cloth she cut from it $=13$m $50$cm
$=13.50$m
$\therefore$ Length of cloth left with her $=15.56-13.50$
$=2.06$m
(B) Total weight of vegetables Neetu bought $=500g+250g+700g+500g+100g+300g=2350$g i.e., $2.350$kg
(C) Total amount spent by Rita$=Rs. (117.35+50.75)=Rs. 168.10$
Amount left will Rita $=Rs. (500-168.10)$
$=Rs. 331.90$
(D) Total distance covered by Yashika $=20$km $50$m $=20.050$km
Distance covered by auto $=20.050-(1+15)=20.050-16=4.05$km
So, option (C) is incorrect.

Rita's watch was 10 min fast 

$2+3=5$

• $1$ dollar is equal to $100$ cents
  
• Given $300$ jellybeans is equal to $x$ dollar , which implies $100x$ cents
• For $50$ cents , we get $300/2x = 150/x$ jellybeans