Tag: ways to multiply and divide

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