A) Rs.111 crore | B) Rs. 1111 crore |

C) Rs. 11111 crore | D) can't be determined |

Explanation:

also ( M - 11111 = 11111 - S)

A = 11111

A) 14 | B) 16 |

C) 12 | D) 15 |

Explanation:

Let 'K' be the total number of sweets.

Given total number of students = 112

If sweets are distributed among 112 children,

Let number of sweets each student gets = 'L'

=> K/112 = L ....(1)

But on that day students absent = 32 => remaining = 112 - 32 = 80

Then, each student gets '6' sweets extra.

=> K/80 = L + 6 ....(2)

from (1) K = 112L substitute in (2), we get

112L = 80L + 480

32L = 480

L = 15

Therefore, 15 sweets were each student originally supposed to get.

A) 74 | B) 79 |

C) 72 | D) 60 |

Explanation:

Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12

To raise the average by one (from 14 to 15), he scored 12 more than the existing average.

Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.

A) Rs. 404 | B) Rs. 514 |

C) Rs. 340 | D) Rs. 616 |

Explanation:

Let the average expenditure per head be Rs. p

Now, the expenditure of the mess for old students is Rs. 44p

After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3

Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33

Therefore, 59(p-3) = 44p + 33

59p - 177 = 44p + 33

15p = 210

=> p = 14

Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.

A) 79 | B) 77 |

C) 76 | D) 73 |

Explanation:

Assuming Sripad has scored the least marks in subject other than science,

Then the marks he could secure in other two are 58 each.

Since the average mark of all the 3 subject is 65.

i.e (58+58+x)/3 = 65

116 + x = 195

x = 79 marks.

Therefore, the maximum marks he can score in maths is 79.

A) 81 | B) 78 |

C) 80.5 | D) 81.5 |

Explanation:

Let the number of students in classes A, B and C be P, Q and R respectively.

Then, total score of A = 83P, total score of B = 76Q, total score of C = 85R.

Also given that,

(83P + 76Q) / (P + Q) = 79

=>4P = 3Q.

(76Q + 85R)/(Q + R) = 81

=>4R = 5Q,

=>Q = 4P/3 and R = 5P/3

Therefore, average score of A, B, C = ( 83P + 76Q + 85R ) / (P + Q + R) = 978/12 = 81.5