A) 22 | B) 33 |

C) 35 | D) 28 |

Explanation:

Given

[(180)^{2} ÷ 60×14] ÷ 9 = ? × 24

** **[(180 x 180) ÷ 60×14] ÷ 9 = ? × 24

?= 180×3×14 / 9×24

? = 35.

A) Rs. 10 | B) Rs. 15 |

C) Rs. 20 | D) Rs. 25 |

Explanation:

Let the Oranges be 'O', Bananas be 'B' and Apples be 'A'

From the given data,

2O + 3B + 4A = 15 ...... (I)

3O + 2B + A = 10 ..... (II)

On adding (I) and (II),

5O + 5B + 5A = 25

O + B + A = 5

4O + 4B + 4A = 20.

A) 100 | B) 1 |

C) 4 | D) 0 |

Explanation:

Given

It is in the form of

= a + b + c

Here a = 37, b = 35 & c = 28

=> a + b + c = 37 + 35 + 28 = 100

Therefore, = 100

A) 63 | B) 62 |

C) 64 | D) 65 |

Explanation:

Total strength of the class is given by

15 + 44 + 4 + 3 - 1 = 65

A) 2000 | B) 3000 |

C) 2500 | D) 3500 |

Explanation:

Let 's' be the number of small packets and 'b' the number of large packets sold on that day.

Therefore, s + b = 5000 ... eqn (1)

Each small packet was sold for Rs.150.

Therefore, 's' small packets would have fetched Rs.150s.

Each large packets was sold for Rs.250.

Therefore, 'b' large packets would have fetched Rs.250b.

Total value of sale = 150s + 250b = Rs. 10.5 Lakhs (Given)

Or 150s + 250b = 10,50,000 ... eqn (2)

Multiplying equation (1) by 150, we get 150s + 150b = 7,50,000 ... eqn (3)

Subtracting eqn (3) from eqn (2), we get 100b = 3,00,000

Or b = 3000

We know that s + b = 5000

So, s = 5000 - b = 5000 - 3000 = 2000.

2000 small packets were sold.