Directions (1-5): The following table shows the number of persons who got injured in earthquake from five different states of India during five different years. Study the table carefully to answer the following questions.

Q1. If total no. of person in Maharashtra who injured due to earthquake in 2004 was 150% more than the total no. of person from same state who injured in earthquake in 2008, then the total no. of person who injured in earthquake in 2005 in Maharashtra is what percent of total no. of persons who injured in 2008 in same state (It is given that total persons who injured in earthquake in Maharashtra throughout all the years is 88,000)?

A. 190% B. 210% C. 200% D. 150% E. 250%

Answer : Option C
Explanation: Let no. of person who injured in Maharashtra in 2004 was x
∴ No. of persons who injured in same state in 2008 = [latex] \frac{100} {250}[/latex] × x
= [latex] \frac{2x} {5}[/latex]
x + [latex] \frac{2x} {5}[/latex] = 88,000 - (20,000 + 18,000 + 15,000) = 35,000
⇒ x = 25,000
[latex] \frac{2x} {5}[/latex] = [latex] \frac{2} {5}[/latex] x 25000 = 10,000
∴ Required percentage = [latex] \frac{20000} {10000}[/latex] x 100 = 200%
Q2. Total no. of person who injured in earthquake in Bihar and Maharashtra together in the year 2005 is what percent more or less than that from Gujarat and Bihar together in 2006?

A. 25% more B. 25% less C. 20% less D. 20% more E. 30% less

Answer : Option B
Explanation: Total persons injured in earthquake from Bihar and Maharashtra together in 2005
= 25000 + 20000 = 45,000
Total person injured in earthquake from Gujarat and Bihar together in 2006
= 40,000 + 20,000 = 60,000
Required percentage = [latex] \frac{60000-45000} {45000}[/latex] x 100
25% less
Q3. If difference between the no. of person who injured in earthquake in Gujarat and Assam together in 2005 is 32,000 and total no. of persons who injured in earthquake in Assam was 25% more than that in Kerala in 2005 then find the total no. of persons who were injured in 2005 due to earthquake in all the states together?

A. 1,05,000 B. 1,50,000 C. 1,10,000 D. 95,000 E. 1,15,000

Answer : Option A
Explanation: No. of persons in Assam who injured in earthquake in 2005 = 8000 x [latex] \frac{125} {100}[/latex] = 10,000
∴ persons injured in Gujarat in 2005 = 32,000 + 10,000 = 42,000
∴ Required answer = 42,000 + 25,000 + 20,000 + 8,000 + 10,000 = 1,05,000
Q4. If 32%, 24% and 18% persons out of total injured persons in state Gujarat, Bihar and Maharashtra respectively died in the year 2006, then find the total no. of person from these three states together who are still alive.

A. 54,160 B. 55,160 C. 58,160 D. 57,160 E. 49,260

Answer : Option B
Explanation: Total number of professors = [latex] \frac{1} {9}[/latex] x [latex] \frac{9} {25}[/latex] x 375 = 15
Q5. If ratio between total no. of persons who were injured in earthquake in states Bihar and Maharashtra in the year 2008 is 5 : 4 and total person who injured in 2008 from all states is 63,000 then total person who injured in 2008 in Bihar and Maharashtra together is what percent of total person who injured in 2008 from all states together?

A. [latex] \frac{400} {7}[/latex]% B. [latex] \frac{300} {7}[/latex]% C. [latex] \frac{200} {7}[/latex]% D. [latex] \frac{100} {7}[/latex]% E. [latex] \frac{500} {7}[/latex]%

Answer : Option C
Explanation: Let total person injured in earthquake in Bihar and Maharashtra is 5x and 4x respectively.
∴ 9x = 63,000 – (30,000 + 2,000 + 4,000) ⇒ 9x = 27,000
Required percentage = [latex] \frac{27,000} {60,000}[/latex] x 100 = [latex] \frac{300} {7}[/latex]%
= 42[latex] \frac{6} {7}[/latex]%

Directions (1-5): The table given below provides the percentage of number of students from 5 different colleges who got placed in various companies during campus placement in year 2016. It was recorded that all students from all colleges got placed.

Q1. In KITM, 320 students were placed in at most 2 companies, which is 136 less than the number of students in HCTM who were placed in at least 5 companies. Find the ratio of total number of students in KITM and HCTM.

A. 3 : 5 B. 4 : 11 C. 2 : 3 D. 2 : 5 E. 3: 4

Answer : Option C
Explanation: Students placed in at most 2 companies = 40% = 320
∴ Total number of students in KITM = [latex] \frac{320} {40} [/latex] x 100 = 800
Students placed in at least 5 companies in HCTM = 320 + 136 = 456 which is equal to 38%
∴ Total students in HCTM = [latex] \frac{456} {38} [/latex] x 100
∴ Required ratio = [latex] \frac{800} {1200} [/latex] = 2 : 3
Q2. Find the difference in number of students who were placed in at least 4 companies and that of in at least 3 companies in college LPU if its total strength is 850.

A. 105 B. 201 C. 160 D. 102 E. 120

Answer : Option D
Explanation: Students placed in at least 4 companies = [latex] \frac{60} {100} [/latex] x 850
Students placed in at least 3 companies = [latex] \frac{72} {100} [/latex] x 850 = 612
∴ Required difference = 102
Q3. Which college records the maximum number of students who were placed in at least 4 companies provided that the strength of students in each college is 1500?

A. LPU B. GITM C. MMU D. HCTM E. KITM

Answer : Option A
Explanation: KITM = 38%
GITM = 41%
MMU = 58%
LPU = 60%
HCTM = 53%
∴ Required answer is LPU
Q4. Total number of students placed in 5 companies in KITM is same as that of in HCTM. If 135 students of HCTM were placed in 1 company, then find total strength of KITM.

A. 2750 B. 2680 C. 2700 D. 2500 E. None of these

Answer : Option C
Explanation: Students placed in 5 companies in HCTM = [latex] \frac{135} {15} [/latex] × 24 = 216
∴ Total students in KITM = [latex] \frac{216} {8} [/latex] x 100 = 2700
Q5. LPU and MMU both has total strength of 1600 students each

A. 448 B. 488 C. 484 D. 450 E. 438

Answer : Option A
Explanation: Required average = [latex] \frac{1} {2} [/latex](28 + 28) x [latex] \frac{1600} {100} [/latex]
= 448

1. The ratio of the volumes of two cubes is 729 : 1331. What is the ratio of their total surface areas?

A. 81 : 121 B. 9 : 11 C. 729 : 1331 D. 27 : 121

Answer : Option A
Explanation: Ratio of the sides = [latex]{\sqrt[3]{729}}[/latex] : [latex]{\sqrt[3]{1331}}[/latex] = 9 : 11 Ratio of surface areas = [latex]{9}^{2}[/latex] : [latex]{11}^{2}[/latex] = 81 : 121
2. The length of a rectangle is two - fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 1225 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units?

A. 140 B. 156 C. 175 D. 214

Answer : Option A
Explanation: Given that the area of the square = 1225 sq.units =>Side of square = [latex]\sqrt{125}[/latex] = 35 units The radius of the circle = side of the square = 35 units Length of the rectangle = [latex]\frac{2}{5}[/latex] * 35 = 14 units Given that breadth = 10 units Area of the rectangle = lb = 14 * 10 = 140 sq.units
3. A cube of side one meter length is cut into small cubes of side 10 cm each. How many such small cubes can be obtained?

A. 10 B. 100 C. 1000 D. 10000

Answer: Option C
Explanation:Along one edge, the number of small cubes that can be cut = [latex]\frac{100}{10}[/latex] = 10 Along each edge 10 cubes can be cut. (Along length, breadth and height). Total number of small cubes that can be cut = 10 * 10 * 10 = 1000
4. An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq m?

A. Rs. 3642.40 B. Rs. 3868.80 C. Rs. 4216.20 D. Rs. 4082.40

Answer : Option D
Explanation: Length of the first carpet = (1.44)(6) = 8.64 cm Area of the second carpet = 8.64(1 + [latex]\frac{40}{100}[/latex]) 6 (1 + [latex]\frac{25}{100}[/latex]) = 51.84(1.4)[latex]\frac{5}{4}[/latex] sq m = (12.96)(7) sq m Cost of the second carpet = (45)(12.96 * 7) = 315 (13 - 0.04) = 4095 - 12.6 = Rs. 4082.40
5. The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm. Find the ratio of the breadth and the area of the rectangle?

A. 1 : 96 B. 1 : 48 C. 1 : 84 D. 1 : 68

Answer : Option A
Explanation: Let the length and the breadth of the rectangle be 4x cm and 3x respectively. (4x)(3x) = 6912 12[latex]{x}^{2}[/latex] = 6912 [latex]{x}^{2}[/latex] = 576 = 4 * 144 = [latex]{2}^{2}[/latex] * [latex]{12}^{2}[/latex] (x > 0) => x = 2 * 12 = 24 Ratio of the breadth and the areas = 3x : 12[latex]{x}^{2}[/latex] = 1 : 4x = 1 : 96