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) 78 paise | B) 48 paise |

C) 54 paise | D) 62 paise |

Explanation:

Quarter of Kg means 250 gm

Less weight, less price (Direct Proportion)

So,

250 : 200 :: 60 : x => x = 48

So 200 gm will cost 48 paise.

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) 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 ------?

$\Rightarrow \frac{52}{1.14}$= 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)

Therefore, 15:36 ::25:x

=> (15 x X)=(36 x 25) => x = 60

Hence, 15 men can do it in 60 hours.

A) 46 | B) 47 |

C) 48 | D) 49 |

Explanation:

Let the required length be x meters

More men, More length built (Direct proportion)

Less days, Less length built (Direct Proportion)

$\left.\begin{array}{r}\begin{array}{cc}Men& 20:35\end{array}\\ \begin{array}{cc}Days& 6:3\end{array}\end{array}\right\}\vdots \vdots 56:x$

=> (20 x 6 x X)=(35 x 3 x 56)

=> x = 49

Hence, the required length is 49 m.

A) 18 days | B) 21 days |

C) 24 days | D) 30 days |

Explanation:

(2 x 14) men +(7 x 14) boys = (3 x 11) men + (8 x 11) boys

=>5 men= 10 boys => 1man= 2 boys

Therefore, (2 men+ 7 boys) = (2 x 2 +7) boys = 11 boys

( 8 men + 6 boys) = (8 x 2 +6) boys = 22 boys.

Let the required number of days be x.

More boys , Less days (Indirect proportion)

More work , More days (Direct proportion)

$\left.\begin{array}{r}\begin{array}{cc}Boys& 22:11\end{array}\\ \begin{array}{cc}Work& 1:3\end{array}\end{array}\right\}\vdots \vdots 14:x$

Therefore, (22 * 1 * x) = (11 * 3 * 14)

=> x = 21

Hence, the required number of days = 21

A) 3 | B) 5 |

C) 6 | D) 9 |

Explanation:

[(100 x 35) + (200 x 5)]men can finish the work in 1 day

Therefore, 4500 men can finish the work in 1 day. 100 men can finish it in 4500/100= 45 days.

This is 5 days behind Schedule