**11.7 * 4.1 - 5.97 = ?**- 48
- 42
- 46
- 39
- None of these

Answer: Option B.

11.7 * 4.1 - 5.97 = 47.97 - 5.97 = 42**( 8649)**^{0.5}- 89
- 97
- 93
- 91
- 99

Answer: Option C.

Find the square root and get the value as 93.

**The value of 1/cosecΘ-cotΘ-1/sinΘ is**- Cosec Θ
- tan Θ
- 1
- cot Θ

Answer: Option D.

Expression = 1/cosec Θ - cot Θ - 1/cosec Θ cosec2Θ-cot2Θ/cosecΘ-cotΘ-cosecΘ =cotΘ (cosec2Θ - cot2Θ-1; 1/sinΘ = cosecΘ)**In a triangle, if three altitudes are equal, then the triangle is**- Right
- Isosceles
- Obtuse
- Equilateral

Answer: Option D.

The only possible case is when the Triangle is equilateral.**A sum was put at simple interest at a certain rate for 2 years. Had it been put at 3% higher rate, it would have fetched Rs 300 more. The sum is**- 5300
- 5500
- 5000
- None of these

Answer: Option C.

Increase of 3% fetched Rs.300 more. It is for 2 years. For 1 year Increase of 3% will fetch Rs.150. So 1 % will fetch Rs.50 100% = 5000.Thomson-Reuters Preparation Links- Sample Aptitude Questions of Thomson-Reuters
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**A sum of Rs 1,550 was lent partly at 5% and partly at 8% per annum simple interest. The total interest received after 3 years was Rs 300. The ratio of money lent at 5% to that lent at 8% is**- 5:8
- 8:5
- 31:6
- 16:15

Answer: Option D.

Partly division of 1550 is x and x - 1550 5 % of x + 8 % of (1550 – x) = 0.05x + 0.08(1550 – x) For 3 years, the total interest is 300. 3[0.05x + 0.08(1550 – x)] = 300 x = 800 and 1550 – x = 750 The ratio of money = 16:15.**The cost price of item B is Rs. 150/- more than the cost price of item A, Item A was sold at a profit of 10% and Item B was sold at a loss of 20%. If the respective ratio of selling price of items A and B is 11:12, what is the cost price of item B?**- Rs. 450/-
- Rs. 420/-
- Rs. 400/-
- Rs. 350/-
- Rs. 480/-

Answer: Option A.

Let us assume cost price of A= X

So that Cost price of B= X+150.

SP of A= X*1.1

SP of B=(X+150)*0.8

Given that

SPA: SPB

11:12

So that 1.1X/(X+150)*0.8= 11/12

X=300

CP of B= 300+150**A vessel contains a mixture of milk and water in the respective ratio of 10 : 3. Twenty-six litres of this mixture was taken out and replaced with 8 litres of water. If the resultant respective ratio of milk and water in the mixture was 5 : 2, what was the initial quantity of mixture in the vessel? (in litres)**- 143
- 182
- 169
- 156
- 130

Answer: Option E.

26 L mixture is taken out.

Quantity of Milk is taken out= 26*(10/13)=20

Quantity of Water is taken out=26*(3/13)=6

As we know that,

(10X-20)/(3X-6+8)=5/2

X=10,

Initial quantity of mixture in the vessel is= 13*X=13*10=130L.**There are 6 consecutive odd numbers. The difference between the square of the average of the first three numbers and the square of the average of the last three numbers is 288. What is the last odd number?**- 31
- 27
- 29
- 25
- 33

Answer: Option C.

Let the 6 consecutive odd no.’s are :

X, X+2, X+4,X+6, X+8, X+10

Avg. of 1st three no’s is X+2.

Avg. of Last three no’s is X+8.

Given that (X+8)2-(X+2)2=288

X=19

Last odd no. is X+10= 29.

Answer is 29.**In a bag there are 6 red balls and 9 green balls. Two balls are drawn at random, what is the probability that at least one of the balls drawn is red?**- 29/35
- 7/15
- 23/35
- 2/5
- 19/35

Answer: Option C.

Probability of atleast one of the balls drawn is red= 1- (9/15)*(8/14)=23/35. Answer is 23/35.