A) Rs. (xy/d) | B) Rs. (xd) |

C) Rs. (yd) | D) Rs. (yd/x) |

Explanation:

Cost of *x* metres = Rs. d.

Cost of 1 metre=Rs.(d/x)

Cost of y metres=Rs.[(d/x).y]=Rs.(yd/x)

A) 14th element | B) 9th element |

C) 12th element | D) 7th element |

Explanation:

If we consider the third term to be ‘x”

The 15th term will be (x + 12d)

6th term will be (x + 3d)

11th term will be (x + 8d) and

13th term will be (x + 10d).

Thus, as per the given condition, 2x + 12d = 3x + 21d.Or x + 9d = 0.

x + 9d will be the 12th term.

Thus, 12th term of the A.P will be zero.

A) 18 | B) 14 |

C) 22 | D) 24 |

Explanation:

1 2 3 4 5 6(Shankar) 7(Nitu) 8(Althaf) 9 10 11 12 13 14

Here, Althaf is 8th from front, Shankar is 9th from rear end and Nitu is between them

So minimum no. of boys standing in the queue = 14

A) Rs. 175 | B) Rs. 325 |

C) Rs. 340 | D) Rs. 260 |

Explanation:

Assume first child (the youngest) get = Rs. x

According to the question ;

each son having Rs. 30 more than the younger one

Second child will get = Rs. x + 30

Third child will get = Rs. x + 30 + 30 = x + 60

Forth child will get = Rs. x + 30 + 30 + 30 = x + 90

Fifth child will get = Rs. x + 30 + 30 + 30 + 30 = x + 120

Total amount they got = Rs. 2000

x + (x+30) + (x+60) + (x+90) + (x+120) = 2000

5x + 300 = 2000

5x = 1700

x = Rs. 340

So the youngest child will get Rs. 340.

A) 29.32 sec | B) 42.51 sec |

C) 39.25 sec | D) 45.61 sec |

Explanation:

Given loom weaves 1.14 mts of cloth in one second then 52 mts of cloth can be weaved by loom in,

1.14 ----- 1

52.0 ------?

= 45.61 sec

A) 40 | B) 50 |

C) 60 | D) 70 |

Explanation:

Let the required no of hours be x. Then

Less men , More hours (Indirct Proportion)

15:36 ::25:x (15 x X)=(36 x 25)

Hence, 15 men can do it in 60 hours.