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## Number System: Level 2 Test - 6

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Question 1 |

The total tractor production in a state is 294000, out of which 150000 are made by Mahindra & Mahindra. While Out of every 1000 Mahindra tractors, 98 are red in color, but only 53% of the total tractor production is red. Find the total percentage of non-Mahindra tractors that are red out of total non-Mahindra tractors.

50.25% | |

81.25% | |

98.30% | |

61.40% |

Question 1 Explanation:

Total production of red tractor = 294000 x 53% = 155820

Number of red tractors of Mahindra make = 150000/1000x98 = 14700

Number of non-Mahindra tractors = 294000 -150000 = 144000

Therefore, Number of non-Mahindra red tractors = 155820-14700 = 141120

Hence, required percentage = (141120/144000) x 100 = 98%

Number of red tractors of Mahindra make = 150000/1000x98 = 14700

Number of non-Mahindra tractors = 294000 -150000 = 144000

Therefore, Number of non-Mahindra red tractors = 155820-14700 = 141120

Hence, required percentage = (141120/144000) x 100 = 98%

Question 2 |

The highest score in an innings is 2/9

^{th}of the total score in the inning and the next highest score is 2/9 of the remainder score. The difference of two scores is 8 runs. Find the total score in the innings?152 | |

162 | |

142 | |

132 |

Question 2 Explanation:

Let the total score in the innings be x.

Then, highest score = 2/9 x

then Next highest score= 2/9 (x -2/9 x)

=2/9x-2/9 (x-2/9x) =8

= x- x+2/9x = 8x9/2

x=8x9x9/2x2 = 162

Then, highest score = 2/9 x

then Next highest score= 2/9 (x -2/9 x)

=2/9x-2/9 (x-2/9x) =8

= x- x+2/9x = 8x9/2

x=8x9x9/2x2 = 162

Question 3 |

In a class, 50 students play Hockey, 20 students Football , while 10 students play both hockey Football. How many student play at least one of these two games?

10 | |

80 | |

50 | |

60 |

Question 3 Explanation:

Number of students who play Hockey = 50

Number of students who play Football = 20

Number of students who play both games = 10

Number of students who play only hockey = 50 -10= 40

Number of students who play only Football = 20 -10 = 10

:. Number of students who play at least one game =40+ 10+10=60

Number of students who play Football = 20

Number of students who play both games = 10

Number of students who play only hockey = 50 -10= 40

Number of students who play only Football = 20 -10 = 10

:. Number of students who play at least one game =40+ 10+10=60

Question 4 |

The natural numbers are divided into groups of (1), (2, 3), (4, 5, 6), (7, 8, 9, 10) and so on. Find the sum of numbers in the 50th group of this series.

1225 | |

24505 | |

62525 | |

52650 |

Question 4 Explanation:

Let S = 1 + 2 + 4 + 7 + ... +

First number of 50th term

= 50 x 49/2+ 1 = 1226

Sum of numbers of 50th term = 1226 + 1227 +…………….. Upto 50th term

= 50/2 [2 x 1226 + (50 - 1) x 1]

= 25 x 2501= 62525

*Tn or**=1+2+4+ ... +Tn_1+Tn*

Subtracting, we get*0*=1 + [1 + 2+ 3 + ... (n -1)] -*Tn**T*= n (n -1)/2 + 1_{N}=1+2+3+………….. +(n-l)+1First number of 50th term

= 50 x 49/2+ 1 = 1226

Sum of numbers of 50th term = 1226 + 1227 +…………….. Upto 50th term

= 50/2 [2 x 1226 + (50 - 1) x 1]

= 25 x 2501= 62525

Question 5 |

A man decided to pay off a debt of Rs 3600 by 40 annual instalments that are in AP. When 30 of the installments are paid he dies leaving one-third of the debt unpaid. The value of the 8th installment is

Rs 35 | |

Rs 50 | |

Rs 65 | |

None |

Question 5 Explanation:

Let the first instalment be

Again, 2400 = 30

=> 160 = 2a +

Solving Eqs. (i) And (ii), 20 =

Therefore, 180 = 2a + 39 x 2 =>

Value of 8th instalment

=51+ (8-1) x2=51+14 =Rs 65

*'a'*and the common difference between any two consecutive instalments be 'd’*.*Using the formula for the sum of an AP, S_{n}= n/2*[2a*+*(n -1) d]*we have,*3600=*40/2*[2a+ (40-1) d] =20(2a+ 39d)*

=> 180 =*2a+ 39d*... (i )Again, 2400 = 30

*[2a*+ (30*-l) d]*=*15(2a*+*29d)*=> 160 = 2a +

*29d*... (ii)Solving Eqs. (i) And (ii), 20 =

*10d*=> d = 2Therefore, 180 = 2a + 39 x 2 =>

*2a=102=>a=51*Value of 8th instalment

=51+ (8-1) x2=51+14 =Rs 65

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