1. Since, six technicians working at the same rate completely work of one server in 10 H. Hence, total work= 10 x 6 = 60 man hours. Now, from 11.oo am to 5 pm total man hours= 6 x 6 = 36 From, 5 pm to 6 pm total man hours = 7 From, 6 pm to 7 pm total man hours =8 From, 7 pm to 8 pm total man hours =9 Total: 60 Hours. Hence, the work will be completed at 8 pm. ( D) |

2. Firstly, make a table taking Persons(p1, p2, p3….p6) as rows and tasks(t1, t2, t3…) as columns. Go through the question again and draw the final table considering all the statements. First task can be done in 3 ways by 3 persons. Second task can be given to two person i.e. ( 3 and 4) Number of ways=2 Third task can be done by 4 person i.e. 4 ways. Similarly, for fourth, fifth and sixth tasks, number of ways are 3, 2, 1 and respectively. Total number of ways= 2 x 3 x 4 x 3 x 2 x 1 = 144 ( A) |

3.(D) |

4. Let the number of students in the front row be x and number of rows be n. Hence, number of students in the next rows would be (x – 3), (x – 6), (x – 9),….. and so on… If n, i.e. number of rows be 3, then number of students =x + (x – 3) + (x – 6) = 630 i.e. 3x = 639 i.e. x= 213 (Thus n=3 possible) Likewise, if n=4, then x + (x – 3) + (x – 6) +(x – 9) = 630 i.e. 4x – 18 = 630 i.e. x= 162 (Thus, n=4 is also possible) If n=5, then x + (x – 3) + (x – 6) + (x – 6) + (x – 9) + ( x – 12) = 630 i.e. 5x- 30= 630 5x = 660 i.e. x= 132 (Thus, n=5 is also possible) If n=6 then 6x= 675 In this case value of x cannot be a integer. Therefore, n=6 is not possible. ( D) |

5. The shaded portion represents the area which shows people who do not watch DD channel. Since 80% watch DD hence 20% do now watch DD. Let those who watch BBC and CNN only x% then . 12+10-x=20 Hence x= 2 |

6. Let x, y and z be the hundredth, tens and unit digits of the original number. According to question, (100z + 10y + x) – ( 100x + 10y + z) = 594 99(z-x)= 594 (z-x)=6 So, the possible values of (x, z) are (1, 7), (2, 8), and (3, 9) Again, the tens digit can have any values from 0, 1, 2, 3, … 9 Hence. Minimum value of their sum= x + y + z = 1 + 0 + 7 = 8 ( C) |

7. 22 ^{3 }+ 23^{3 }+ 24^{3 }+ 25^{3 }+……+ 87^{3 }+ 88^{3
}On rearranging = (22 ^{3 }+88^{3}) + (23^{3 }+ 87^{3}) + (24^{3 }+ 86^{3})+……..(54^{3 }+ 56^{3}) + 55^{3
}Now, we know that a^{n} + b^{n} is divisible by (a + b) when n is odd number.Therefore, all the terms except 55^{3}, is divisible by 110.Now, the remainder when 553 is divided by 110 is 55 Hence the required remainder when whole expression is divided by 110 is 55. |

8. Total number of passenger in the Rajdhani Express=10 x 20=200 So, in the 9 boggies the minimum number of total passengers =12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 144 Hence minimum number of passengers in one boggie can be (200 – 144) = 56 ( C) |

9. Let the number of men be 100. Then, Men x Time= Work 100 x 1 = 100 unit. Amount of work increased by 50% New work= 150 unit as the planned time remains same i.e. 1 Then men required will be 150 i.e. 50 more workers. but since new workers are 25% efficient i.e. 5/4 times efficient as existing workers. Actual number of workers= 50/(5/4)= 40 men Required percent(%)= (40/1oo) x 100 = 40% |

10. For a given volume i.e. 64 cm ^{3}Cube has minimum surface are of length of edge=4 cm. Its diagonal= 4 x ( sq root 3). ( C) |

Sample Question Paper Quantitative 02

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how total 60 hrs in the first question

qus no 2_=- Answer would be 192

T1= 4 ways

T2= 2 ways

T3= 4 ways

T4=3 ways

T5= 2 ways

T6=1 way

toatal ways= 4*2*4*3*2*1

= 192

it should be 144 ways,not 192 because task 1 can be done by only 3 ways (worker 1 ,2 cant do,either 3 or 4 busy in task 2,so one of them can come to do task1 with 5th and 6th worker)

Q3: 100 forward

96 back (because next stone is at 96 , 2m away from first that is at 98)

96 forward

94 back

94 forward

92 back

92 forward

90 back

90 forward

total= 844

dude the stone starts at 100 meter itself ..as stated in question that “starting from X”.

qus no 2_=- Answer would be 192

T1= 3 ways because t2 has already fixed person either 3 or 4. so left with one from 3 or 4 and 5, 6

T2= 2 ways

T3= 4 ways

T4=3 ways

T5= 2 ways

T6=1 way

total ways= 3*2*4*3*2*1

= 144

solution of 5 how it be 2percent?

Anish and Abish, you are absolutely correct. For those who still have doubt, let me make this simple.

It has been mentioned in the question itself that “Every person is to be assigned one task”. So for task 1, you cannot choose P1 and P2. Among P3 and P4, one must be assigned with task 2. When one is taken for task 2 among P3 and P4, the other one is available for task 1. P5 and P6 are already available for task 1. So for task 1, there are 3 ways of assigning and not 4. So to sum it up, we get

T1= 3, T2= 2, T3= 4, t4=3, t5=2, t6=1

so, total ways= 3*2*4*3*2*1= 144 ways.