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Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

Find the least number in which multiplied by $1800$ given a perfect cube, then find the sum of the digits of that number.

  1. $2$
  2. $3$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
factor of $ 1800 - 10 \times 10\times 18 = 3\times 3\times 2\times 2\times 5\times 2\times 5 $

$ = 2^{3}\times 3^{2}\times 5^{2}$

for perfect cube we need $  3\times 5 = 15 $

So, sum of digits $(15) = 1+5=6$

so option (c) is right
Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

The sum of all three-digit natural numbers which leave a remainder $2$ when divided by $3$

  1. $168450$
  2. $168850$
  3. $165840$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
First three digit number leaves remainder $2$ is $101$
The next being $104, 107, 110$,...…
Last $3$ digit number $=998$
$\therefore 101, 104, …… 998$
$a=101, d=3$
$n^{th}$ term $\Rightarrow 998=a+(n-1)d$
$998=101+(n-1)3$
$(n-1)\not{3}={\not{897}} _{299}$
$n=300$.
$S _n=\dfrac{300}{2}(2(101)+(300-1)3)=\dfrac{300}{2}(202+897)$
$=164850$.
Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

Given $n =1 + x$ and x is the product of four consecutive integers. Then which of the following is true?

  1. n is an odd numbers

  2. n is prime

  3. sometimes a perfect square

  4. All the given
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$ 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. $

Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

Kunal has only $25$ paise and $50$ paise coins with him. The total amount in $50$ paise denomination is $Rs. 4$ more that the total amount in $25$ paise denomination. The number of $25$ paise coins is $20$ more than the number of $50$ paise coins. What is the total amount with Kunal?

  1. $Rs. 32$
  2. $Rs. 36$
  3. $Rs. 40$
  4. $Rs. 24$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

The sum of all natural members which multiples of $7$ or $3$ or both and lie between $200$ and $500$ is

  1. $45049$
  2. $40149$
  3. $45149$
  4. $45249$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

A number when divided by $14$ leaves a remainder of $8$, but when the same number is divided by $7$, it will leave the remainder ?

  1. 3

  2. 2

  3. 1

  4. can't be determined
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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 $1.$

hence, this is the answer.

Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

The sum of all two digit numbers which when divided by 4 , yield unity as remainder is 

  1. $1012$
  2. $1201$
  3. $1212$
  4. $1210$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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


Smallest 2 digit number that gives remainder $1$ when divided by $4$ $\Rightarrow$ $13 (when \ k=3)$ first term of A.P

Largest 2 digit number that gives remainder $1$ when divided by $4$ $\Rightarrow$ $97 (when \ k=24)$ last term of AP

Series: $13,17,21,....97$

$97=a+(n-1)d$

$97=13+(n-1)4$

$89=(n-1)4$

$(n-1)=21$

$n=22$

Sum of series $=\cfrac{n}{2}$[first term  + last term]

$=\cfrac{22}{2}[13+97]$

$=11\times (110)$

$=1210$

Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

A train running at the speed of 60$\mathrm { km } / \mathrm { hr }$ crosses a pole in 9 seconds. What is the length of the train

  1. $150$ metres
  2. $180$ metres
  3. $324$ metres

  4. Cannot be determined

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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