1. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?

A. Rs. 4462.50 B. Rs. 8032.50 C. Rs. 8900 D. Rs. 8925 E. None of these

Answer: Option D
Explanation: Principal = Rs. [latex]\frac{100 \times 4016.25}{9 \times 5}[/latex] = Rs. [latex]\frac{401625}{45}[/latex] = Rs. 8925.

1. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

A. 720 B. 900 C. 1200 D. 1800

Answer: Option C
Explanation: 2(15 + 12) x h = 2(15 x 12) h = [latex]\frac{180}{27}[/latex] m = [latex]\frac{20}{3}[/latex] m. Volume = ([latex]{15 \times 12 \times \frac {20}{3}}[/latex]) m[latex]^{3}[/latex] = 1200 m[latex]^{3}[/latex].

1. If log10 5 + log10 (5x + 1) = log10 (x + 5) + 1, then x is equal to:

A. 1 B. 3 C. 5 D. 10

Answer: Option B
Explanation: [latex]log_{10}[/latex] 5 + [latex]log_{10}[/latex] (5x + 1) = [latex]log_{10}[/latex] (x + 5) + 1 [latex]log_{10}[/latex] 5 + [latex]log_{10}[/latex] (5x + 1) = [latex]log_{10}[/latex] (x + 5) + [latex]log_{10}[/latex] 10 [latex]log_{10}[/latex] [5 (5x + 1)] = [latex]log_{10}[/latex] [10(x + 5)] 5(5x + 1) = 10(x + 5) 5x + 1 = 2x + 10 3x = 9 x = 3.

1. A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:

A. 30 birds B. 60 birds C. 72 birds D. 90 birds

Answer: Option A
Explanation: Let the total number of shots be x. Then, Shots fired by A = [latex]\frac {5}{8}[/latex]x Shots fired by B = [latex]\frac {3}{8}[/latex]x Killing shots by A = [latex]\frac {1}{3}[/latex] of [latex]\frac {5}{8}[/latex] x = [latex]\frac {5}{24}[/latex]x Shots missed by B = [latex]\frac {1}{2}[/latex] of [latex]\frac {3}{8}[/latex] x = [latex]\frac {3}{16}[/latex]x [latex]\frac {3 \times x}{16}[/latex] = 27 or x = [latex]\frac {27 \times 16}{3}[/latex] = 144. Birds killed by A = [latex]\frac {5x}{24}[/latex] = [latex]\frac {5}{24}[/latex] x 144 = 30.

1. A farmer traveled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance traveled on foot is:

A. 14 km B. 15 km C. 16 km D. 17 km

Answer: Option C
Explanation: Let the distance traveled on foot be x km. Then, distance travelled on bicycle = (61 -x) km. So, [latex]\frac {x}{4}[/latex] + [latex]\frac {(61 -x)}{9}[/latex] = 9 9x + 4(61 -x) = 9 x 36 5x = 80 x = 16 km.

1. Three flags each of different colors are available for a military exercise, Using these flags different codes can be generated by waving I. Single flag of different colors II. Any two flags in a different sequence of colors. III. three flags in a different sequence of colors. The maximum number of codes that can be generated is.

A. 6 B. 9 C. 15 D. 18

Answer: Option C
Explanation: This type of question becomes very easy when we assume three colors are red(R) blue(B) and Green(G). We can choose any color. Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, the number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colors, the number of ways is six i.e.., RBG, BGR, GBR, RGB, BRG, GRB. Hence total number of ways by changing flag = 3+ 6 +6 = 15.

1. Rohan and Mohan invest in the ratio 5: 2. 10% of the profit is donated to a hospital before dividing it between the two. Rohan gets Rs. 6840. What is Mohan's share?

A. Rs. 2736 B. Rs. 3800 C. Rs. 4788 D. Rs. 6840

Answer: Option A
Explanation: Ratio of investment = Ratio of Profit Ratio of Profit of Rohan to Mohan = 5:2 Rohan's Share = Rs. 6840 = [latex]\frac {5}{5+2}[/latex] x (90% Total Profit) ----------> 10% given to hospital ∴ 6840 = [latex]\frac {5}{7}[/latex] x [latex]\frac {90}{100}[/latex] x Total Profit ∴ Total Profit = Rs. 10,640/- Vijay's Share = 10,640 - 6840 - 1064 = Rs. 2736/-

1. A drink vendor has 80 liters of Maaza, 144 liters of Pepsi and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required?

A. 35 B. 37 C. 42 D. 30

Answer: Option B
Explanation: The number of liters in each can = HCF of 80, 144 and 368 = 16 liters. Number of cans of Maaza = [latex]\frac{80}{16}[/latex] = 5 Number of cans of Pepsi = [latex]\frac{11}{16}[/latex] = 9 Number of cans of Sprite = [latex]\frac{368}{16}[/latex] = 23 The total number of cans required = 5 + 9 + 23 = 37 cans.