Mental Mathematics - Class VI

We know that amongst two consecutive natural numbers, one is even and one is odd.

$ The\quad units\quad place\quad numbers\quad are\quad 4,\quad 8,\quad 7\quad and\quad 3.\ 4\times 8=32\longrightarrow \quad units\quad place\quad number\quad is\quad 2\ 2\times 7=14\longrightarrow \quad units\quad place\quad number\quad is\quad 4\ 4\times 3=12\longrightarrow \quad units\quad place\quad number\quad is\quad 2\ Answer-\quad Option\quad A $

$(-1)\times$$(-1)\times$$(-1)$$.....1001$ times 

Unit digits of both numbers are $6$ and $8$. 

Place value of a digit increases by $10$ times as it moves from right to left.

156 $\times$ 0 = 0

The resultant product of $3$ negative signs would be negative.

$
-{ \frac { 1 }{ 4 }  }^{ -3 }{ \times \frac { 1 }{ 4 }  }^{ 4 }\div { 4 }^{ -2 }{ \quad =\quad -4 }^{ (10x+1) }\ \ -{ 4 }^{ 3 }{ \times 4 }^{ -4 }\times { 4 }^{ 2 }{ \quad =\quad -4 }^{ (10x+1) }\ { -4 }^{ 3-4+2 }{ \quad =\quad -4 }^{ (10x+1) }\ { -4 }^{ 1 }{ \quad =\quad -4 }^{ (10x+1) }\ 1\quad =\quad 10x\quad +1\ 10x\quad =\quad 0\ x\quad =\quad 0
$

Total age of 6 family members four year ago =$25\times 6=150$years
Total age of 6 family member now=150+$4\times 6=150+24=174$years
Total age of 6 family member and child =$25\times 7=175$year
Age of child =175-174=1 year

45 % of 1500 + 35 % of 1700 = ? % of 3175
Or $\frac{45}{100}\times 1500+\frac{35}{100}\times 1700= \frac{x}{100}\times 3175$
Let x is new %

Or $45\times 1500+35\times 1700=3175\times x$
OR 67500+59500=3175x
Or x=$\frac{127000}{3175}=40%$

$(100 + 2) \times (100 + 3)$
           [Using $(x +a) (x+b)=x^2+(a+b)x+ ab$]
$ = (100)^2 + (2 + 3) \times 100 + 2 \times 3$
$ = 10000 + 5 \times 100 + 6 = 10000 + 500 + 6 = 10506$

Let the number of pineapple and watermelon be x and y
Then 7x+5y=38 (given cost of pineapple and watermelon is Rs 7and Rs 5
Or 5y=38-7x
Or $y=\frac{38-7x}{5}$ 
clearly y is whole number only when (38-7x) is divisible by 5
This happens when x=4

Putting x=2, we get $2^2\left(2^2-1\right)= 12. So, x^2\left(x^2-1\right)$ is always divisible by 12

Suresh travelled 1200 km by air which form (2/5) of his trip. 
Then total distance of journey=$\frac{5}{2}\times 1200=3000 km$
Suresh travelled $\frac{1}{3}$of journey by car =$\frac{1}{3}\times 3000=1000 km$
Then suresh travelled distance by train =3000-1200-1000=800 km
If whole journey by train in 8 hours 
So speed of train=$\frac{3000}{8}$=375 km/Hr

Three lightships flash simultaneously at 6;00 a.m. The first lightship flashes every 12 seconds, the seconds lightship every 30 seconds and the third lightship every 66 seconds. 
Then we take common multiple of 12,30 or 66 
That is $2\times 6\times 5\times 11=660$
So  the three lightships next flash together=660 seconds =$\frac{660}{60}=11$mints
The time next flash is after 11 mints=6.11 A.M 

$\left(2^2 + 4^2 + 6^2 + ... + 20^2\right) = \left(1 \times 2\right)^2 + \left(2\times 2\right)^2 + \left(2 \times 3\right)^2 + ... + \left(2 \times 10\right)^2$

$= \left(2^2 \times 1^2\right) + \left(2^2 \times 2^2\right) + \left(2^2 \times 3^2\right) + ... + \left(2^2 \times 10^2\right)$

$= 2^2 \times \left [1^2 + 2^2 + 3^2 + ... + 10^2\right]$

$\left(1^2+2^2+3^2+....+n^2\right) = \frac{1}{6}n\left(n+1\right)\left(2n+1\right)$

$=\left(4\times\frac{1}{6}\times 10\times 11\times 21\right)$

$=\left(4\times 5\times 77\right)$

$= 1540$

In every 15 min there was a break of 5 min
If total 60 min break in all movie then number of commercial break =$\frac{60}{5}=12$
If 12 breaks are in movie then the duration of movie=$12\times 15+15=195    minutes$

Let, $\frac {2x}{3}=96$
$x=\frac {96\times 3}{2}=144$
$\therefore \frac {3}{4}\times 144=3\times 36=108$

$\displaystyle \left ( \left ( 3^2 \right )^{4/3} \div \left ( 3^3 \right )^{2/3} \right ) \times 3^{3/2}$
$= 3^{8/3} \div 3^2 \times 3^{3/2}$
$= \displaystyle 3^{8/3 - 2 + 3/2} $
$= \displaystyle  3^{ \displaystyle \frac{16 - 12 + 9}{6}}$
$= 3^{13/ 6}$

$1+9\times 10 = 91$

$\displaystyle\frac{0.24\times0.35}{0.02\times0.14\times0.15}= \frac{0.084}{0.00042}=200$

Let the smaller number is x 

Let the divisor number is x

Let the number is x

$46)57804(1256$
$\underline {46}$
$118$
$\underline {92}$
$260$
$\underline {230}$
$304$
$\underline {276}$
$\underline {28}$

178*0=0

$7986$
$\underline {\times 548}$
63888
319440
$\underline {3993000}$
$\underline {4376328}$

14 apples = 140
1 apple = $\displaystyle \frac{140}{14}$ = Rs. 10

$796\xrightarrow {\text {rounded to nearest tens}}800$
$19\xrightarrow {\text {rounded to nearest tens}}20$
$\therefore 800\times 20=16,000$

Sum of odd place digits $=2+1+2=5$
Sum of even place-digits $=8+5+3=16$
Difference of above two sums $=16-5=11$
The difference 11 is divisible by 11
$\therefore$ 3,25,182 is also divisible by 11.

Each basket contained $=85$ oranges

Given,

These numbers are divisible by 5 and 7..

1 $\displaystyle \div $ 28 = $\displaystyle \frac{1}{28}$

64$\displaystyle \div $ 1 = 64

238 $\displaystyle \div $ 238 = 1

(98+ 14) $\displaystyle \times $ 0 = 0

$\Rightarrow$  $1$ symbol represents $5$ balloons.

Multiplying unit digit of each number we get 672 hence 2 is the unit digit of the product.

The value of $7.86 \times 4.6 $

$3897\times 999=3897\times (1000-1)$
$=3897\times 1000-3897\times1$
$=3897000-3897$
$=3893103$

$\ \ \ \ \ \ \ \ 34 \ \times\ \ \   56 \ ----$

$16974\div 23\div 2$

$\ \ \ \ 12340\ \times \ \ \ \ \ \ 23 \ ---- \ \ \ \ \ \ \ 37020 \ + 246800 \ ---- \ \ \ \ \ 283820$

Any number multiplied by $0$ gives $0$ itself.

$0$ when divided by anything results as $0$.

$959877\times 10=9598770$

Statement 1-If $7$ friends divide $76895$ beads, each friend will get $(76859\div 7)$ beads i.e., $10985$ beads
So, statement 1 is false
Statement 2: Each family will pay
Rs. $(216352\div 16)=Rs.13552$

$\Rightarrow$  Diesel consumes by generator in $1,hour$ = $2.875,liters.$

$132820\div 20=6641$
and $6641\times 16=106256$

A machine produces $2800$ screws in a day.

Number of government employees in Delhi $= 4200$
Now, $7 \times \square = 4200$
$\Rightarrow \square = \dfrac{4200}{7} = 600$

Given,
$6 \times 3 (3 - 1)$

We have to find the total cost of the trip.

There are $470988$ books and are to be arranged in shelves.

Repeated addition is also known as multiplication. If the same number is repeated then in short we can write that in the form of multiplication.
For example : $3+3+3+3+3$

$B + C = B$

$C = 0$

AA.B $=$ BB $\Rightarrow $ (10 A + A) B $ = $BB= 10 B $+$ B

$AB = B,A = 1$

$ B+A = 3 $

$\Rightarrow B + 1 =3, \Rightarrow B = 2$

1 1 

11

12

22

110

132

$\Rightarrow$  $\spadesuit\spadesuit\spadesuit=18$         [Given]

$ Let\quad x=a(a+1)(a+2)(a+3)\ Now\quad whether\quad a\quad is\quad even\quad or\quad odd,\quad the\quad four\quad consecutive\quad number's\quad product\quad \ contain\quad 2,\quad 3,\quad 4\quad as\quad factors.\ \therefore \quad x\quad is\quad divisible\quad by\quad 2\times 3\times 4=24----(1)\ Option\quad A\longrightarrow Four\quad consecutive\quad numbers\quad contain\quad two\quad even\quad numbers.\ Therefore\quad product\quad is\quad even.\therefore \quad x+1\quad should\quad be\quad odd.\ Option\quad A\quad is\quad correct.\ Option\quad B\longrightarrow x\quad is\quad divisible\quad by\quad 24\quad (from\quad 1)\ Now\quad there\quad exists\quad no\quad prime\quad of\quad the\quad form\quad 24p+1\quad where\quad p\quad is\quad a\quad natural\quad number.\ e.g.\quad \quad 29=24\times 1+5\ \quad \quad \quad \quad \quad 31=24\times 1+7\ \quad \quad \quad \quad \quad 37=24\times 1+13\ With\quad increasing\quad count\quad the\quad second\quad term\quad increases\quad because\quad difference\quad between\ primes\quad increase\quad with\quad higher\quad count.\quad Therefore\quad x+1=n\quad is\quad not\quad a\quad prime\quad under\quad the\quad \ given\quad condition.\quad For\quad a\quad expanded\quad number\quad to\quad be\quad square,\quad the\quad first\quad and\quad last\quad term\ should\quad be\quad square\quad number.   Option\quad C\longrightarrow x=a(a+1)(a+2)(a+3)\ when\quad a\quad is\quad even\quad a=2p\ \therefore \quad x=2p(2a+1)(2a+2)(2a+3)\ \quad \quad \quad =4p(2p+1)(p+1)(2p+3)\ The\quad product\quad is\quad 4\times 1\times 1\times 3=12\ If\quad we\quad add\quad 1\quad then\quad last\quad term\quad of\quad the\quad product\quad is\quad 13\ which\quad is\quad not\quad a\quad square\quad number.\ So\quad x+1=n\quad is\quad not\quad a\quad square\quad number\quad when\quad a\quad is\quad even.\ (2)\quad When\quad a\quad is\quad odd-\ x=(2p+1)(2p+2)(2p+3)(2p+4)\ The\quad last\quad term\quad of\quad the\quad product\quad is\quad 1\times 2\times 3\times 4=24.\ 24+1=25\quad is\quad a\quad square\quad term.\ \therefore \quad n=x+1\quad is\quad a\quad square\quad number\quad when\quad a\quad is\quad odd.\ \therefore \quad Option\quad C\quad is\quad correct\quad when\quad a\quad is\quad odd.\ Option\quad D\longrightarrow \quad Obviously\quad option\quad D\quad is\quad not\quad correct. $

Let x be the number of 50 paise coins. Then the number of 25 paise coins is x + 20. The amount in 50 paise is 0.5x, and in 25 paise is 0.25(x + 20). Given 0.5x = 0.25(x + 20) + 4, we solve 0.5x = 0.25x + 5 + 4, so 0.25x = 9, x = 36. Total amount = 0.5(36) + 0.25(56) = 18 + 14 = 32.

To find the sum of natural numbers between 200 and 500 that are multiples of 7, 3, or both, we use the principle of inclusion-exclusion. Calculate the sum of multiples of 3 plus the sum of multiples of 7, minus the sum of their least common multiple multiples (21), within the specified range, yielding 45149.

We have,

When the number is divided by $14$ it gives a remainder of $8$,

The number $= 14N + 8 (14N$ is divisible by $14)$

When same number is divided by $7$ it will give remainder $1.$

number should be of the form $4k+1$

$\begin{matrix} \dfrac { { 20 } }{ { 100 } } \times 170000=20\times 1700 \ =34000, , , Ans. \  \end{matrix}$

$Speed\>of\>train\>=\>60\>(\frac{Km}{hr})\\=(\frac{60\cdot\>1000\>m}{3600\>sec})\\=(\frac{50}{3})m/sec\\length\>crossed\>in\>9\>seconds\>=\>(\frac{50}{3})\cdot\>9\>=\>150\>m\\\therefore\>length\>of\>train\>=\>150m$

Let $x $ be multiplied by $-23$ to get $575$

The product (n-1)! is divisible by n if n is composite and n > 4 (Wilson's Theorem related logic). For n in [12, 40], (n-1)! is not divisible by n only if n is prime. The primes between 12 and 40 are 13, 17, 19, 23, 29, 31, 37. There are 7 such numbers.

$4 \times 378 \times 25$

Now,

Consider whole number is $a$

odd Natural Number

a(b x c ) = (a x b) x c is an example of associative property.

We know that for any number $n$ multiplied by $0$ is always $0$ that is $n\times 0=0$.

$0$ should be replaced M because
$304\times 4 = 1216$

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

On dividing we get,
$\frac { 9217 }{ 88 } =104\frac { 65 }{ 88 } $

Therefore,

Required number 

$= 9217 + (88 - 65)$ ,Because (88 - 65) < 65.

$= 9217 + 23$

$= 9240$

Now as $A \times B$ gives 2 in the product, the combinations could be

$2\times 1$ or $1 \times 2, 6 \times 2$ or $2\times 6, 3 \times 4$

or $4 \times 3, 6 \times 7$ or $7 \times 6$ or $8 \times 4$ or

$4\times 8, 9 \times 8$ or $8\times 9$. But to get 2 in hundred's

place and 5 in ten's place in the product, B has to be 6 and A has to be

7. Then $6\times 7=42$.
Write 2 and carry over 4. then $7\times 3= 21$ and add 4 to get 2 and 5 in the product.
The value of B is 6 and A is 7.

$35=5\times7$
$45=3\times3\times5$
$55=5\times11$
LCM of (35,45,55)=$3\times3\times5\times\times7\times11=3465$
Since difference between divisor and remainder is 10
Hence least number is $3465-10=3455$

If a number is multiplied by 2 and 5 respectively. Then number should be divided by their L.C.M i.e  by 10.
Clearly, only number divisible by 10 is 30.
Hence, option D is correct.

$a=(2^{-2}-2^{-3}),b=(2^{-3}-2^{-4})and: c=(2^{-4}-2^{-2})$
So,  $3abc=3\times (2^{ -2 }-2^{ -3 })\times (2^{ -3 }-2^{ -4 })\times (2^{ -4 }-2^{ -2 })$
$3abc=3\times (1/4-1/8)\times (1/8-1/16)\times (1/16-1/4)$
$3abc=3\times (1/8)\times (1/16)\times (-3/16)$
$3abc=\frac { -9 }{ 2048 } $
Answer (C) $\displaystyle \frac{-9}{2048}$

$
\frac { { 9 }^{ \frac { 5 }{ 2 }  }-{ 3\times 7 }^{ 0 }{ \quad -\frac { 1 }{ 81 }  }^{ -\frac { 1 }{ 2 }  } }{ { 27 }^{ \frac { 2 }{ 3 }  }-(\frac { 8 }{ 27 } )^{ \frac { 2 }{ 3 }  } } \quad \quad \ \ NR\quad =\quad { 3 }^{ 2\times \frac { 5 }{ 2 }  }-3-(\frac { 1 }{ 81 } )^{ -\frac { 1 }{ 2 }  }\quad =\quad { 3 }^{ 5 }-3-({ 3 }^{ -4\times \frac { -1 }{ 2 }  })\quad =243-3-9=\quad 231\quad \ Dr\quad =\quad { 3 }^{ 3\times \frac { 2 }{ 3 }  }-(\frac { 2 }{ 3 } )^{ 3\times \frac { 2 }{ 3 }  }\quad =\quad 9\quad -\quad \frac { 4 }{ 9 } \quad =\quad 77/9\ \frac { Dr }{ Nr } \quad =\quad \frac { 231 }{ 77/9 } \quad =\quad 3\times 9\quad =\quad 27
$

$20\times 60\div 40-20+10=$
After putting the true sign, we get
$20+ 60- 40\times20\div10=$
$20+60-80=$
$0=0$
Answer (B) 0

$15\frac{2}{3}\times 3\frac{1}{6}+6\frac{1}{3}= 11\frac{7}{18}+x$
Or $\frac{47}{3}\times \frac{19}{6}+\frac{19}{3}= \frac{205}{18}+x$
Or $\frac{893}{18}+\frac{19}{3}= \frac{205}{18}+x$
Or 893+114=205+18x
Or 18x=893+114-205=$\frac{802}{18}=44\frac{5}{9}$

Invited people=125
people attended birthday party=50
As per question, 36% brought gifts
$G=50\times \frac { 36 }{ 100 } $
$G=18$
Answer (C) 18

$\displaystyle\frac{2^{n+4}-2\times2^{n}}{2\times2^{(n+3)}}+2^{-3}$
$\frac { 2^{ n+4 }-2^{ n+1 } }{ 2^{ (n+4) } } +2^{ -3 }$
$\frac { 2^{ n }(2^{ 4 }-2^{ 1 }) }{ 2^{ n }\times { 2 }^{ 4 } } +2^{ -3 }$
$\frac { (2^{ 4 }-2^{ 1 }) }{ { 2 }^{ 4 } } +2^{ -3 }$
$\frac { (16-2) }{ 16 } +\frac { 1 }{ 8 } $
$\frac { 14 }{ 16 } +\frac { 1 }{ 8 } $
$\frac { 14+2 }{ 16 } =1$
Answer (D) 1

As t term is 3t/4,So to find t we need to divide is by 3/4.
Answer (B) Divide both sides by 3/4

average = 12.4 runs per wicket, 
Let's assume total  number of wickets taken by him till the last match=x
 total  number of runs taken by him till the last match=y
$y/x=12.4$
$y=12.4x
$(y+26)/(x+5)=12$
$12.4x+26=12x+60$
$0.4x=34$
$x=85$
The number of wickets taken by him till the last match =85
Answer (D) 85

Given that $4m=5K$ and $6n=7K$
Now divide those two equations, we get $\dfrac { 4m }{ 6n } =\dfrac { 5K }{ 7K } $
Which gives $ \dfrac { m }{ n } =\dfrac { 5 }{ 7 } \times \dfrac { 6 }{ 4 } =\dfrac { 15 }{ 14 } $

$3i\left( 4+3i \right) \left( 52i \right) \ =\quad 156{ i }^{ 2 }\left( 4+3i \right) \ =\quad -156\left( 4+3i \right) \ =\quad -624-468i$

We know that any number multiplied by $0$ will always result in $0$. For example: The equation $0\times 2 = 0$ can be read as 'Zero times Two equals Zero'. That means, you are taking $'2'$ zero times, effectively taking nothing. So, multiplying any number by zero will always be zero.

Option $B$ is correct.
The multiplicative inverse will include $1$ as numerator.

We know that any number multiplied by $1$ will always result in that number itself. For example: The equation $2\times 1 = 2$ can be read as 'two times one equals one'. That means, you are taking $'2'$ one times, effectively taking nothing but multiplying by its unity. So, multiplying any number by one will always be the number itself.

By calculations,

We know that any number divided by $1$ will always result in that number itself. For example: $2\div 1 = 2$ and 

We know that any number multiplied by $1$ will always result in that number itself. Therefore, we only multiply $44$ and $567$ as follows:

21774 / 19 = 1146. Then 1146 / 3 = 382.

Let $x$ be the multiplicative inverse

Sum of all $2$ digit number divided by $4$ yields unity as remainder

$(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab$.

803642 will be divided by 9, if 8+0+3+6+4+2 =23 is divided by 9.
$\Rightarrow$ needs 4 more for that.

Initial quantity of milk and water is Milk  $=\dfrac{4}{5} \times $ $35 = 28 $litres
Water $=\dfrac{1}{5} \times $ $35 =7 $litres.
Let $x$ liters of water is added, then, 
$\dfrac{28}{7 + x} = \dfrac{2}{3}$
$x = 35$
Hence , 35 liters of water is added.

Required average $= \dfrac {12 + 18 + 24 + .... + 78}{12} = 45$.

$6729 = 6720 + 9 = 35\times 192 + 9$
Hence the remainder is $9$.

$15289$ is estimated as $15300$ and $587$ is estimated as $600$.