a. 99

b. 132

c. 144

d. 169

**Answer:** c. 144

**Explanation:**

Let 1^{st} number be X.

∴ 2^{nd} number = | X | and 3^{rd} number = | X |

4 | 5 |

Average = | X + X/4 + X/5 | = 87 |

3 |

∴ | 29X | = 87 |

20 * 3 |

The smallest number is 3

a. 60 and 40

b. 50 and 70

c. 70 and 90

d. 80 and 100

**Answer:** b. 50 and 70

**Explanation:**

We see that, all the options satisfy the first given condition of interchanging number of coins. Now, let's check for the second condition.

**Option 1** → 60 - 10 = 50 and 40 + 10 = 50 ∴ 2^{nd} condition not satisfied

**Option 2** → 80 - 10 = 70 and 100 + 10 = 110 ∴ 2^{nd} condition not satisfied

**Option 3** → 70 - 10 = 60 and 90 + 10 = 100 ∴ 2^{nd} condition not satisfied

**Option 4** → No need to calculate. Directly mark this as other 3 are wrong.

But if you wish you can check.

50 - 10 = 40 and 70 + 10 = 80

80 = 2 times 40 **∴ 2 ^{nd} condition satisfied**

a. 99

b. 297

c. 369

d. 396

**Answer:** b. 297

**Explanation:**

Here let digit in hundred's place be H

∴ In Unit's place we have 2H and Ten's place we have 3H

**∴ Original number =** 100H + 10 (3H) + 2H **= 132H**

If we exchange unit's and ten's digits, we get

**New Number =** 100H + 10 (2H) + 3H **= 123H**

Also, 132H - 123H = 27

∴ H = 3

Original Number = 132H = 396

**¾ of 396 = 297**

a. 96

b. 108

c. 125

d. 169

**Answer:** b. 108

**Explanation:**

Let numbers be A and B

∴ (A - B)^{2} = 9

∴ A^{2} - 2AB + B^{2} = 9

Further, A^{2} + B^{2} = 225

∴ 225 - 2AB = 9

**∴ AB = 108 = product of the two numbers**

Going further if the question asks you to find out the values of two numbers

Consider equation (A+B)^{2} = A^{2} + 2AB + B^{2}

∴ (A+B)^{2} = 225 + 2 x 108 = 441

A + B = 21 ---------- (1)

We know, (A - B)^{2} = 9

∴ A - B = 3 ---------- (2)

Adding (1) and (2) we get, 2A = 24

∴ A = 12

Put the value of A in equation 2. We get B = 9

a. 550

b. 600

c. 725

d. 850

**Answer:** b. 600

**Explanation:**

Let there be 100 children.

8% have 0 books and 2 % have 5 books.

So, that makes 10% = 10 children

Remaining are 100 - 10 = 90 children

Of these 90, 27% have one book and 18% have 4 books each.

So that is 27 + 18 = 45% out of 90

So remaining children (100 - 45%) = 55% out of 90 = have 2 or 3 books

∴ Remaining children = | 55 | x 90 = 49.5 children |

100 |

For ? children : 297 children have 2 or 3 books

∴ ? = | 297 x 100 | = 600 children are there in the school. |

49.5 |