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Question 1 of 15
1. Question
1 pointsCategory: Quantitative Aptitude2 men or 4 women can complete a work in 11 days working 8 hours each day. In how many days 6 men and 10 women will complete a work twice as large working together 4 hours each day?
Correct
Explanation:
2 m = 4 w
So 1m = 2 w
6m + 10 w = 6 * (2w) + 10w = 22 w
So we have to find the number of days for 22 women
4 women do 1 work in 11 days working 8 hours, let 22 women do twice work in x days working 4 hrs each day, then
W1*D1*H1*W2 = W2*D2*H2*W1
4*11*8*2 = 22*x*4*1
Solve, x = 8 daysIncorrect
Explanation:
2 m = 4 w
So 1m = 2 w
6m + 10 w = 6 * (2w) + 10w = 22 w
So we have to find the number of days for 22 women
4 women do 1 work in 11 days working 8 hours, let 22 women do twice work in x days working 4 hrs each day, then
W1*D1*H1*W2 = W2*D2*H2*W1
4*11*8*2 = 22*x*4*1
Solve, x = 8 days 
Question 2 of 15
2. Question
1 pointsCategory: Quantitative AptitudeA rectangular park has dimensions 180 m × 85 m. If there is a 2.5 m wide path around the sides of the park inside park, then what will be the cost of painting this path at 60 paise per sq. meter?
Correct
Explanation:
Area of garden = 180*85 = 15300 m^{2}
Area of path = (1805)*(855) = 14000 m^{2}
So area of path = 15300 – 14000 = 1300 m^{2}
So cost = 1300*(60/100) = Rs 780Incorrect
Explanation:
Area of garden = 180*85 = 15300 m^{2}
Area of path = (1805)*(855) = 14000 m^{2}
So area of path = 15300 – 14000 = 1300 m^{2}
So cost = 1300*(60/100) = Rs 780 
Question 3 of 15
3. Question
1 pointsCategory: Quantitative AptitudeHow many words of 5 letters with or without meaning can be made from the letters of the word ‘CURRENCY’, when repetition of letters is not allowed?
Correct
Explanation:
Repetition is not allowed, so we take unique letters – C U R E N Y – 6 letters
5 letter word is to be formed.
For first letter there are 6 choices, since repetition is not allowed, for second, third, fourth and fifth letter also we have 5, 4, 3 and 2 choices resp., so total of 6*5*4*3*2 ways = 720 ways.
OR
5 letters to be chosen from 6 letters (C U R E N Y), so ways to choose are 6C5, and then these 5 letters are to be arranged, so total 6C5 * 5! = 6*120 = 720Incorrect
Explanation:
Repetition is not allowed, so we take unique letters – C U R E N Y – 6 letters
5 letter word is to be formed.
For first letter there are 6 choices, since repetition is not allowed, for second, third, fourth and fifth letter also we have 5, 4, 3 and 2 choices resp., so total of 6*5*4*3*2 ways = 720 ways.
OR
5 letters to be chosen from 6 letters (C U R E N Y), so ways to choose are 6C5, and then these 5 letters are to be arranged, so total 6C5 * 5! = 6*120 = 720 
Question 4 of 15
4. Question
1 pointsCategory: Quantitative AptitudeA piece of work which can be completed by 10 men in 16 days is started by 15 men. After working for 3 days, 8 more men joined. In how many days is the work completed?
Correct
Explanation:
Shortcut:
10 men can do in 16 days. 15 men did for 3 days, and then (15+8)=23 men completed in x days. so
10*16 = 15*3 + 23*x
Solve, x = 5 days.
So total 5+3
Or
10 men in 16 days, so 15 men in (10*16)/15 = 32/3 days
So in 3 days 15 completed = (3/32)*3 = 9/32 of work
Work left = 1 – 9/32 = 23/32
Now (15+8) = 23 men have to complete 23/32 work in x days
10 men in 16 days, so 23 men in (10*16)/23 days
So in x days 23 completed = (23/(10*16))*x which equals 23/32
So (23/(10*16))*x = 23/32
Solve, x = 5 days
So total 5+3 daysIncorrect
Explanation:
Shortcut:
10 men can do in 16 days. 15 men did for 3 days, and then (15+8)=23 men completed in x days. so
10*16 = 15*3 + 23*x
Solve, x = 5 days.
So total 5+3
Or
10 men in 16 days, so 15 men in (10*16)/15 = 32/3 days
So in 3 days 15 completed = (3/32)*3 = 9/32 of work
Work left = 1 – 9/32 = 23/32
Now (15+8) = 23 men have to complete 23/32 work in x days
10 men in 16 days, so 23 men in (10*16)/23 days
So in x days 23 completed = (23/(10*16))*x which equals 23/32
So (23/(10*16))*x = 23/32
Solve, x = 5 days
So total 5+3 days 
Question 5 of 15
5. Question
1 pointsCategory: Quantitative AptitudeA box contains 8 blue and 4 yellow balls. Three balls are drawn at random. What is the probability all balls are same in color?
Correct
Explanation:
Case 1: all balls blue
Prob. = ^{8}C_{3} / ^{12}C_{3} = 14/55
Case 2: all balls yellow
Prob. = ^{4}C_{3} / ^{12}C_{3} = 1/55
Add both casesIncorrect
Explanation:
Case 1: all balls blue
Prob. = ^{8}C_{3} / ^{12}C_{3} = 14/55
Case 2: all balls yellow
Prob. = ^{4}C_{3} / ^{12}C_{3} = 1/55
Add both cases 
Question 6 of 15
6. Question
1 pointsCategory: Quantitative AptitudeA bag contains 50 paise, Re 1, and Rs 2 coins in the ratio 2 : 5 : 7 respectively. If the total value of money in bag is Rs 200, what is the total number of coins in the bag?
Correct
Explanation:
Coins – 2x, 5x, 7x, so
Value of 50 paise coins = (50/100)*2x = x, of re 1 coin = 1*5x = 5x, of Rs 2 coins = 2*7x = 14x
So total value is x+5x+14x = 20x which equals 200, so x = 10
Total coins = 2x+5x+7x = 14x = 140Incorrect
Explanation:
Coins – 2x, 5x, 7x, so
Value of 50 paise coins = (50/100)*2x = x, of re 1 coin = 1*5x = 5x, of Rs 2 coins = 2*7x = 14x
So total value is x+5x+14x = 20x which equals 200, so x = 10
Total coins = 2x+5x+7x = 14x = 140 
Question 7 of 15
7. Question
1 pointsCategory: Quantitative AptitudeRs 6000 is lent in two parts in 2 schemes. Scheme A offers 4% rate of interest per annum and scheme B offers 6% rate of interest per annum. The total simple interest received is Rs 1520. If the part in scheme A is lent for 4 years and the part in scheme B is lent for 5 years, what is the part lent in scheme A?
Correct
Explanation:
Let part in scheme A = Rs x, then in scheme B is Rs (6000x)
So [x*4*4/100] + [(6000x)*6*5/100] = 1520
Solve, x = 2000Incorrect
Explanation:
Let part in scheme A = Rs x, then in scheme B is Rs (6000x)
So [x*4*4/100] + [(6000x)*6*5/100] = 1520
Solve, x = 2000 
Question 8 of 15
8. Question
1 pointsCategory: Quantitative AptitudeHow many litres of milk at Rs 42 per litre be mixed with 12 litres of milk at Rs 50 per litre so that on selling the mixture at Rs 54 per litre, there is a gain of 20%?
Correct
Explanation:
Let x litres of milk at Rs 42 per litre be mixed
SP = 54, gain% = 20%, so CP = (100/120) * 54 = Rs 45
So
1st milk(x kg)………………. 2nd milk(5 l)
42…………………………………50
…………………..45
5…………………………………..3
So x/12 = 5/3
Solve, x = 20Incorrect
Explanation:
Let x litres of milk at Rs 42 per litre be mixed
SP = 54, gain% = 20%, so CP = (100/120) * 54 = Rs 45
So
1st milk(x kg)………………. 2nd milk(5 l)
42…………………………………50
…………………..45
5…………………………………..3
So x/12 = 5/3
Solve, x = 20 
Question 9 of 15
9. Question
1 pointsCategory: Quantitative AptitudeA sum of money is lent at 20% p.a. compound interest for 2 years. The difference between the amounts received when compounded semiannually and compounded annually is Rs 60.25. What is the sum of money lent?
Correct
Explanation:
P[1 + (r/2)/100]^{4} – P[1 + r/100]^{2} = 60.25
P[1 + 10/100]^{4} – P[1 + 20/100]^{2} = 60.25
P [ (11/10)^{4} – (6/5)^{2}] = 60.25
Solve, P = 2,500Incorrect
Explanation:
P[1 + (r/2)/100]^{4} – P[1 + r/100]^{2} = 60.25
P[1 + 10/100]^{4} – P[1 + 20/100]^{2} = 60.25
P [ (11/10)^{4} – (6/5)^{2}] = 60.25
Solve, P = 2,500 
Question 10 of 15
10. Question
1 pointsCategory: Quantitative AptitudeA boat takes 20 min less to travel 10 km downstream than to travel the same distance upstream. The speed of the stream is 4 km/hr. What is the upstream speed?
Correct
Explanation:
Let speed of boat in still water = x km/hr
So speed upstream = x4, and speed downstream = x+4
Now given:
Time to travel 10 km downstream = time to travel 10 km upstream – 20/60
So 10/(x+4) = 10/(x4) – 1/3
10/(x4) – 10/(x+4) = 1/3
10x+40 – (10x40)/(x^{2} – 16) = 1/3
80/(x^{2} – 16) = 1/3
x^{2} – 16 = 240
solve, x = 16
so upstream speed = 164Incorrect
Explanation:
Let speed of boat in still water = x km/hr
So speed upstream = x4, and speed downstream = x+4
Now given:
Time to travel 10 km downstream = time to travel 10 km upstream – 20/60
So 10/(x+4) = 10/(x4) – 1/3
10/(x4) – 10/(x+4) = 1/3
10x+40 – (10x40)/(x^{2} – 16) = 1/3
80/(x^{2} – 16) = 1/3
x^{2} – 16 = 240
solve, x = 16
so upstream speed = 164 
Question 11 of 15
11. Question
1 pointsCategory: Quantitative AptitudeDirections (1115): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –
x^{2} + 5x – 14 = 0, 2y^{2} – 13y + 21 = 0
Correct
Explanation:
x^{2} + 5x – 14 = 0
x^{2} + 7x – 2x – 14 = 0
Gives x = 7, 2
2y^{2} – 13y + 21 = 0
2y^{2} – 6y – 7y + 21 = 0
Gives y = 3, 7/2
Put on number line
7…… 2 ….. 3 …… 7/2Incorrect
Explanation:
x^{2} + 5x – 14 = 0
x^{2} + 7x – 2x – 14 = 0
Gives x = 7, 2
2y^{2} – 13y + 21 = 0
2y^{2} – 6y – 7y + 21 = 0
Gives y = 3, 7/2
Put on number line
7…… 2 ….. 3 …… 7/2 
Question 12 of 15
12. Question
1 pointsCategory: Quantitative AptitudeDirections (1115): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –
4x^{2} – 5x – 6 = 0, 4y^{2} – 4y – 3 = 0
Correct
Explanation:
4x^{2} – 5x – 6 = 0
4x^{2} – 8x + 3x – 6 = 0
Gives x = 3/4, 2
4y^{2} – 4y – 3 = 0
4y^{2} + 2y – 6y – 3 = 0
Gives y = 1/2, 3/2
Put on number line
3/4……. 1/2……. 2… 3/2
Alternate valuesIncorrect
Explanation:
4x^{2} – 5x – 6 = 0
4x^{2} – 8x + 3x – 6 = 0
Gives x = 3/4, 2
4y^{2} – 4y – 3 = 0
4y^{2} + 2y – 6y – 3 = 0
Gives y = 1/2, 3/2
Put on number line
3/4……. 1/2……. 2… 3/2
Alternate values 
Question 13 of 15
13. Question
1 pointsCategory: Quantitative AptitudeDirections (1115): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –
2x^{2} – 5x – 12 = 0, 4y^{2} + 28y + 33 = 0
Correct
Explanation:
2x^{2} – 5x – 12 = 0
2x^{2} – 8x + 3x – 12 = 0
Gives x = 3/2, 4
4y^{2} + 28y + 33 = 0
4y^{2} + 22y + 6y + 33 = 0
Gives y = 11/2, 3/2
Put on number line
11/2… 3/2……. 4Incorrect
Explanation:
2x^{2} – 5x – 12 = 0
2x^{2} – 8x + 3x – 12 = 0
Gives x = 3/2, 4
4y^{2} + 28y + 33 = 0
4y^{2} + 22y + 6y + 33 = 0
Gives y = 11/2, 3/2
Put on number line
11/2… 3/2……. 4 
Question 14 of 15
14. Question
1 pointsCategory: Quantitative AptitudeDirections (1115): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –
x^{2} – 8x + 15 = 0, 2y^{2} + 31y + 65 = 0
Correct
Explanation:
x^{2} – 8x + 15 = 0
x^{2} – 3x – 5x + 15 = 0
Gives x = 3, 5
2y^{2} + 31y + 65 = 0
2y^{2} + 26y + 5y + 65 = 0
Gives y = 13, 5/2
Put on number line
13…………. 5/2………. 3……. 5Incorrect
Explanation:
x^{2} – 8x + 15 = 0
x^{2} – 3x – 5x + 15 = 0
Gives x = 3, 5
2y^{2} + 31y + 65 = 0
2y^{2} + 26y + 5y + 65 = 0
Gives y = 13, 5/2
Put on number line
13…………. 5/2………. 3……. 5 
Question 15 of 15
15. Question
1 pointsCategory: Quantitative AptitudeDirections (1115): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer –
x^{2} = 225, y^{2} – 12y – 45 = 0
Correct
Explanation:
x^{2} = 225
Gives x = 15, 15
y^{2} – 12y – 45 = 0
y^{2} – 15y + 3y – 45 = 0
Gives y= 3, 15
Put on number line
15……. 3…… 15
When x = 15, x < both values of y (3 and 15) But when x= 15 x > y=3, and x=y(15)
So relation cent be determined.Incorrect
Explanation:
x^{2} = 225
Gives x = 15, 15
y^{2} – 12y – 45 = 0
y^{2} – 15y + 3y – 45 = 0
Gives y= 3, 15
Put on number line
15……. 3…… 15
When x = 15, x < both values of y (3 and 15) But when x= 15 x > y=3, and x=y(15)
So relation cent be determined.
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