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Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating question sample in Quadratic Equations and Miscellaneous which is common for SBI, IBPS, RBI, LIC, NICL and other competitive exams.
Q (1-5). In the following questions, two equations numbered I & II are given. You have to solve both equations and mark one of the options given below.
- I.49X²-196X + 195 = 0
II.27Y² – 84Y + 65 =0
1. X >Y
2. X <Y
3. X ≥ Y
4. X ≤ Y
5. X = Y or relationship cannot be establishedAnswer – 1. X >Y
Explanation :
X = 15/7, 13/7
Y =5/3, 13/9 - I.X²-√15 (√5+√3) X +√3375 = 0
II.Y² +3(2√3-√5) Y -9√60 =0
1. X >Y
2. X <Y
3. X ≥ Y
4. X ≤ Y
5. X = Y or relationship cannot be establishedAnswer – 3. X ≥ Y
Explanation :
X = 5√3, 3√5
Y = 3√5, -6√3 - I.110X²+63X+9 = 0
II.195Y² +56 Y+4=0
1. X >Y
2. X <Y
3. X ≥ Y
4. X ≤ Y
5. X = Y or relationship cannot be establishedAnswer – 2. X <Y
Explanation :
X = -3/11, -3/10
Y = – 2/13, -2/15 - I.X²+35 X +306 = 0
II.Y² +Y -272 =0
1. X >Y
2. X <Y
3. X ≥ Y
4. X ≤ Y
5. X = Y or relationship cannot be establishedAnswer – 4. X ≤ Y
Explanation :
X= -17,-18
Y= -17, 16 - I.6X²-(9√2+4√3) X +√216 = 0
II.√6Y² – (2√2+3√3) Y+6 =0
1. X >Y
2. X <Y
3. X ≥ Y
4. X ≤ Y
5. X = Y or relationship cannot be establishedAnswer – 5. X = Y or relationship cannot be established
Explanation :
Both the equations are same take √6 common in the first equation. - A thief saw a police at a distance of 250m, coming towards him at 36 Km/hr. Thief takes 5 seconds to realize that he is police coming to catch him and started running away from police at 54 Km/hr. But police realize after 10 seconds when the thief is running away, that he is actually a thief and starts chasing him at 66 km/hr. After police realizing in what time did he catch the thief?
1. 50 sec
2. 75 sec
3. 100 sec
4. 120 sec
5. Cannot be determinedAnswer – 2. 75 sec
In 5 seconds police traveled 50 m then distance is 200 m
After thief realizes he travels 150 m in 10 sec and police travels 100 m
Then distance = 200+150-100 = 250
x/54 = x+250/72
x= 1125 m
t = 1125/15 = 75 sec - The average age of 100 employees in an office in the year 2002 is 25 years. In 2004, 20 employees left the job, whose average age was 30 years. In the year 2007, 30 new employees joined, whose average age was 21 years. The average age of all the employees in the year 2010 is?
1. 26
2. 27
3. 28
4. 29
5. 30Answer – 5. 30
Explanation:
In 2002: 100*25 =2500
In 2004: 2500+100*2 – 20*30 = 2100
In 2007: 2100+80*3+30*21 = 2970
In 2010: 2970+3*110 = 3300
Average = 3300/110 = 30 - Satish lends an equal sum of money at the same rate of interest to Ravi and Ramu. The money lend to Ravi becomes thrice in 8 years at SI. Satish lends to Ramu for the first two years at CI and for the rest 6 years at SI. If the, difference between amounts of Ravi and Ramu after 8 years is Rs.13920. What is the sum lent by Satish to Ramu initially?
1. 14440
2. 14920
3. 15360
4. 16890
5. NoneAnswer – 3. 15360
Explanation:
Ravi:
3x = x+ x*8*R/100
R =25%
Amount =3x
Ramu:
1st two years CI = x(1+25/100)2
25x/16
Next 6 years SI = 25x/16*6*25/100 = 75x/32
Amount = 25x/16 + 75x/32 = 125x/32
Difference = 125x/32 – 3x = 13920
x =15360 - The interest received on Rs.(4000+x) at 10% SI per annum for 2 years is same as interest received on Rs.(4200-x) at 11% CI per annum for the same period. What is the value of x?
1. 50
2. 100
3. 200
4. 250
5. Cannot be determinedAnswer – 3. 200
(4000+x)*10*2/100 = (4200-x)*(1+10/100)2 – (4200-x)
x =200 - Two vessels contain a mixture of Milk and water. In the first vessel the ratio of Milk to water is 5:2 and in the second vessel, the ratio is 7:1.Another vessel of 45 liters is filled from these vessels to contain a mixture of Milk and water in the ratio 4:1. How many liters are taken from the second vessel?
1.21
2. 22
3. 23
4. 24
5. NoneAnswer – 4. 24
Water in 45-liter vessel is 9 liters
9 = 2/7*n+ (45-n)*1/8
n =21
Second vessel = 45-21 =24
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