OICL Specialist Officer Finance Exam 2015 Question Paper

Expression : $$\frac{2430}{16}-16.97+\sqrt{(?)}=164$$

=> $$\frac{2432}{16} - 17 + \sqrt{x} = 164$$

=> $$152 - 17 + \sqrt{x} = 164$$

=> $$\sqrt{x} = 164 - 135 = 29$$

=> $$x = 29^2 = 841$$

Expression : $$\frac{9601}{11.98}\times\sqrt{(531)}+95.88=?$$

= $$\frac{9600}{12} \times \sqrt{529} + 96$$

= $$800 \times 23 + 96$$

= $$18400 + 96 = 18496$$

Expression : $$15.99\times9.89-\sqrt{624.89}-17.001\times1.99=?^{2}$$

=> $$(x)^2 = (16 \times 10) - (\sqrt{625}) - (17\times 2)$$

=> $$(x)^2 = 160 - 25 - 34 = 101$$

=> $$x \approx \sqrt{100} = 10$$

Expression : $$?\%of(\frac{5224}{5.001}\times\frac{3}{11})=375.05$$

=> $$\frac{x}{100} \times \frac{5225}{5} \times \frac{3}{11} = 375$$

=> $$\frac{x}{100} \times 95 \times 3 = 375$$

=> $$x = \frac{375 \times 100}{285} \approx 132$$

The pattern is :

1500 - 100 = 1400

1400 - 116 (= 100 + 16) = 1284

1284 - 164 (= 116 + 16*3) = 1120

1120 - 308 (= 164 + 48*3) = 812

812 - 740 (=308 + 144*3) = 72

The pattern followed is :

4 $$\times 2 + 1$$ = 9

9 $$\times 3 + 2$$ = 29

29 $$\times 4 + 3$$ = 119

119 $$\times 5 + 4$$ = 599

599 $$\times 6 + 5$$ = 3599

The pattern followed is :

10 $$\times \frac{3}{2}$$ = 15

15 $$\times \frac{4}{2}$$ = 30

30 $$\times \frac{5}{2}$$ = 75

75 $$\times \frac{6}{2}$$ = 225

225 $$\times \frac{7}{2}$$ = 787.5

Numbers are multiplied by successive natural numbers and then 7 is added

2 $$\times 1 + 7$$ = 9

9 $$\times 2 + 7$$ = 25

25 $$\times 3 + 7$$ = 82

82 $$\times 4 + 7$$ = 335

335 $$\times 5 + 7$$ = 1682

Numbers of the form $$(n^2 - 1)$$ are subtracted, where $$n \geq 2$$

121 $$- (2^2 - 1)$$ = 118

118 $$- (3^2 - 1)$$ = 110

110 $$- (4^2 - 1)$$ = 95

95 $$- (5^2 - 1)$$ = 71

71 $$- (6^2 - 1)$$ = 36

Let length of train B = $$x$$ m

=> Length of train A = $$2x$$ m

Speed of train A = 126 km/h = $$126 \times \frac{5}{18} = 35$$ m/s

Speed of train B = 90 km/h = $$90 \times \frac{5}{18} = 25$$ m/s

Since, the trains are moving in opposite direction, relative speed = 35 + 25 = 60 m/s

Time taken = 9 sec

=> $$\frac{2x + x}{60} = 9$$

=> $$3x = 9 \times 60 = 540$$

=> $$x = \frac{540}{3} = 180$$ m

Thus, length of train A = 360 m

$$\therefore$$ Time taken by train A take to cross a platform of length 690 metres

= $$\frac{360 + 690}{35} = \frac{1050}{35}$$

= $$30$$ seconds