United India Insurance AO Exam 2013 Question Paper

Tanvi invested Rs. 45,000 for 12 months

Anisha invested Rs. 52,000 for 4 months

Ratio of profit received by Tanvi and Anisha

= $$(45000 \times 12) : (52000 \times 4)$$

= $$(45 \times 3) : (52) = 135 : 52$$

Total profit = Rs. 56,165

$$\therefore$$ Anita's share of profit = $$\frac{52}{135 + 52} \times 56165$$

$$\approx 52 \times 300 = Rs. 15,600$$

To find : $$(\frac{5}{6} \times 216) - (\frac{1}{4} \times 428)$$

= $$(5 \times 36) - (107)$$

= $$180 - 107 = 73$$

Let number = $$3x$$

Acc. to ques, => $$\frac{1}{3} \times 3x = 96$$

=> $$x = 96$$

Number = $$3 \times 96 = 288$$

$$\therefore$$ 67% of the number = $$\frac{67}{100} \times 288$$

= $$192.96$$

Number of students = 35 and number of teachers = 6

Sweets received by each student = $$\frac{20}{100} \times 35 = 7$$

Sweets received by each teacher = $$\frac{40}{100} \times 35 = 14$$

$$\therefore$$ Total sweets = $$(35 \times 7) + (6 \times 14)$$

= $$245 + 84 = 329$$

Let the total amount with Prithvi = $$Rs. 100x$$

% amount spent = $$73 \%$$ = $$\frac{73}{100} \times 100x = 73x$$

Total amount spent = 89,745 + 51,291 = 1,41,036

=> $$73x = 141036$$

=> $$x = \frac{141036}{73} = 1932$$

$$\therefore$$ Total amount = $$100 \times 1932 = Rs. 1,93,200$$

C.I. = $$Rs. 5,596.8$$

=> $$P [(1 + \frac{R}{100})^T - 1] = 5596.8$$

=> $$22,000 [(1 + \frac{R}{100})^2 - 1] = 5596.8$$

=> $$[(1 + \frac{R}{100})^2 - 1] = \frac{5596.8}{22000} = 0.2544$$

=> $$(1 + \frac{R}{100})^2 = 0.2544 + 1 = 1.2544$$

=> $$1 + \frac{R}{100} = \sqrt{1.2544} = 1.12$$

=> $$\frac{R}{100} = 1.12 - 1 = 0.12$$

=> $$R = 0.12 \times 100 = 12 \%$$

$$\therefore S.I. = \frac{P \times R \times T}{100}$$

= $$\frac{22000 \times 12 \times 2}{100}$$

= $$220 \times 24 = Rs. 5,280$$

Let the two successive numbers = $$x , (x + 1)$$

Product = $$x \times (x + 1) = 4032$$

=> $$x^2 + x - 4032 = 0$$

=> $$x^2 + 64x - 63x - 4032 = 0$$

=> $$(x - 63) (x + 64) = 0$$

=> $$x = 63$$

$$\therefore$$ Larger number = $$x + 1 = 63 + 1 = 64$$

Let Sonal's age = $$9x$$ years

=> Nitya's age = $$5x$$ years

Acc. to ques, => $$\frac{9x + 8}{5x + 8} = \frac{13}{9}$$

=> $$81x + 72 = 65x + 104$$

=> $$16x = 104 - 72 = 32$$

=> $$x = \frac{32}{16} = 2$$

$$\therefore$$ Difference in ages = $$9x - 5x = 4x$$

= $$4 \times 2 = 8$$ years

Using : $$M_1 D_1 = M_2 D_2$$

30 men can do a piece of work in 16 days

=> $$M_1 = 30 , D_1 = 16$$ and $$D_2 = 20$$

Let $$M_2$$ men finish the same work in 20 days

=> $$30 \times 16 = M_2 \times 20$$

=> $$M_2 = \frac{480}{20} = 24$$

Let C.P. = $$Rs. x$$

Case 1 : If S.P. = $$Rs. 1,516$$

=> Profit = $$S.P. - C.P. = (1516 - x)$$

Case 2 : If S.P. = $$Rs. 1,112$$

=> Profit = $$C.P. - S.P. = (x - 1112)$$

Acc. to ques, => $$1516 - x = x - 1112$$

=> $$2x = 1516 + 1112 = 2628$$

=> $$x = \frac{2628}{2} = Rs. 1,314$$