Let required fraction be then

............. (i)

and ............ (ii)

Solving (i) and (ii) we get x=3, y=5

Required fraction =

A) 264 | B) 284 |

C) 215 | D) 302 |

Explanation:

Let the number be 'x'. Then, from given data

x/2 + x/3 + x/4 = x+22

13x/12 = x+22

x = 264

A) 55 | B) 105 |

C) 215 | D) 148 |

Explanation:

First line will cut all other 14, similarly second will cut 13, and so on

Total = 14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 105.

A) K=2 & L=3 | B) K=1 & L=4 |

C) K=1 & L=2 | D) K=3 & L=3 |

Explanation:

As per divisibility rule a number is divisible by 6 means it should be divisible by 2 and 3

case 1 : the divisibility rule for 2 is the number should be end with even number

in this case R =2 & R=4

case 2 : The divisibility rule for 3 is the sum of numbers should be divisible by 3

so if we take option (2) Q=1 & R=4 the sum is 39

if we take option (3) Q=1 & R=2 the sum is 37

So Option (2) is correct 39 is divided by 3

A) k=8 & l=2 | B) k=7 & l=2 |

C) k=8 & l=3 | D) k=7 & l=1 |

Explanation:

If a number to be divisile by 88, it should be divisible by both "8" and "11"

Check for '8' :

For a number to be divisible by "8", the last 3-digit should be divisible by "8"

Here 72x23y --> last 3-digit is '23y'

So y=2 [ (i.e) 232 is absolutely divisible by "8"]

Chech for '11' :

For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"

(7 + x + 3) - (2 + 2 + y)

(7 + x + 3) - (2 + 2 + 2)

(10 + x) - 6 should be divisible by "11"

for x = 7

=> 17 - 6 = 11 [ which is absolutely divisible by "11"]

So x = 7 , y= 2.

A) p times | B) 2p times |

C) (p + 4) times | D) (2p + 3) times |

Explanation:

Let the two no's be a and b;

Given product of the no's is p = ab;

If the each nos is increased by 2 then the new product will be

(a+2)(b+2) = ab + 2a + 2b + 4

= ab + 2(a+b) + 4

= p + 2(a+b) + 4

Hence the new product is (p+4) times greater than twice the sum of the two original numbers.