# Numbers Questions

A) 30,400 | B) 30,200 |

C) 34,600 | D) 38,400 |

Explanation:

we can treat every two consecutive terms as one.

So, we will have a total of 100 terms of the nature:

(2 + 5) + (7 + 6) + (12 + 7).... => 7, 13, 19,....

We know the sum of n terms $\frac{n\left(n+1\right)}{2}$

Now, a= 7, d=6 and n=100

Hence the sum of the given series is

S= 100/2 x[2 x 7 + 99 x 6]

=> 50[608]

=> 30,400.

A) 7 | B) 5 |

C) 4 | D) 6 |

Explanation:

Sum of decimal places = 7.

Since the last digit to the extreme right will be zero (since 4 x 5 = 20), so there will be 6 significant digits to the right of the decimal point.

A) Rs. 13 and Rs. 17 | B) Rs. 17 and Rs. 19 |

C) Rs. 23 and Rs. 17 | D) Rs. 13 and Rs. 19 |

Explanation:

If fares for cities L and M from K are x and y respectively, then

2x + 3y = 77 or 6x + 9y = 231...(1)

3x + 2y = 73 or 6x + 4y = 146...(2)

subtracting 2 from 1, we get

5y = 85

y = Rs. 17 => x = Rs. 13

A) L / (k-3) | B) k / (L-3) |

C) 2K / 3L-K | D) 3L / K(K-3) |

Explanation:

Intial contribution = L/K

After 3 men drop, then contribution =L/K-3

The amount more to pay in contribution = $\frac{L}{K-3}-\frac{L}{K}$

= $\frac{3L}{K(K-3)}$

A) 1 | B) 2 |

C) 3 | D) 4 |

Explanation:

On dividing 627349 by 15, we get remainder = 4.

Therefore, the obtained remainder is the least number to be subtracted from the given number so that the the mnumber is divisible by 15.

Here 4 is the least number to be subtracted from 627349 so that it is divisible by 15.

A) 147 | B) 168 |

C) 162 | D) 182 |

Explanation:

The pattern is -4, +9,-16, +25, -36

The next number=119+49=168

A) 546, 3901 | B) 415, 3770 |

C) 668, 4023 | D) 404, 3759 |

Explanation:

Let the smaller number be k.

Then, larger number = (3355 + k)

Therefore 3355 + k = (6k + 15)

‹=› 5k = 3340

‹=› k = 668.

Therefore, the smaller number k = 668 and now the larger number = 3355+668 = 4023.

A) 4 | B) 1 |

C) 5 | D) 3 |

Explanation:

(3x + 2) (2x - 5) = $a{x}^{2}+kx+n$ ...............(1)

But (3x + 2)(2x - 5) = $6{x}^{2}-11x-10$........(2)

so by comparing (1) & (2),

we get a= 6, k= -11 , n= -10

(a - n + k) = 6 + 10 - 11 = 5.