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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

Two pipes A and B can fill a tank in 20 hrs and 30 hrs respectively. Both pipes A and B and a third pipe C all are opened and the tank gets filled in 60 hours. What is the work done alone by pipe C?

Correct

Explanation:
Since the tank is getting filled in more than 20 and 30 hrs, this means that C is emptying pipe
Let C alone empties in x hrs, then
(1/20 + 1/30 – 1/x) = 1/60
1/x = 1/15
So C empties in 15 hrs

Incorrect

Explanation:
Since the tank is getting filled in more than 20 and 30 hrs, this means that C is emptying pipe
Let C alone empties in x hrs, then
(1/20 + 1/30 – 1/x) = 1/60
1/x = 1/15
So C empties in 15 hrs

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

The length and the breadth of a rectangular card are increased by 1 m and 2 m respectively and due to this the area of the card increased by 66 sq. m. But if the length and breadth are decreased by 2 m and 1 m respectively, area is decreased by 51 sq. m. Find the perimeter of the original card.

Correct

Explanation:
Let original length = l, breadth = b, so area = lb
When l and b increased:
(l+1)(b+2) = lb + 66
Solve, 2l + b = 64
When both decreased:
(l-2)(b-1) = lb – 51
Solve, l + 2b = 53
Now solve both equations, l = 25, b = 14
Perimeter = 2(25+14)

Incorrect

Explanation:
Let original length = l, breadth = b, so area = lb
When l and b increased:
(l+1)(b+2) = lb + 66
Solve, 2l + b = 64
When both decreased:
(l-2)(b-1) = lb – 51
Solve, l + 2b = 53
Now solve both equations, l = 25, b = 14
Perimeter = 2(25+14)

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

An article is marked at Rs 25,000. It was bought by the trader at successive discounts of 20% and 5%. After this he spent Rs 1,000 on its transportation and finally sold the article for Rs 25,000. What is his profit% in the whole transaction?

Correct

Explanation:
Successive discounts of 20% and 5% makes overall discount of (-20) + (-5) + (-20)(-5)/100 = -25 +1 = -24%
So he bought the article for [(100-24)/100] * 25000 = 19,000
Spent 1000 on repairs, so total CP = 1000 + 19000 = 20,000
SP = 25,000
So profit% = (5000/20000) * 100

Incorrect

Explanation:
Successive discounts of 20% and 5% makes overall discount of (-20) + (-5) + (-20)(-5)/100 = -25 +1 = -24%
So he bought the article for [(100-24)/100] * 25000 = 19,000
Spent 1000 on repairs, so total CP = 1000 + 19000 = 20,000
SP = 25,000
So profit% = (5000/20000) * 100

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

A town having population of 1,50,000 has 60% males and rest females. If 80% of males and 70% of females are literate, then what is the total number of illiterates in the town?

Correct

Explanation:
Males = (60/100)*1,50,000 = 90,000 so females = 60,000
Male illiterates = (20/100)*90,000 = 18,000
Female illiterates = (30/100)* 60,000 = 18,000
Total 18+18

Incorrect

Explanation:
Males = (60/100)*1,50,000 = 90,000 so females = 60,000
Male illiterates = (20/100)*90,000 = 18,000
Female illiterates = (30/100)* 60,000 = 18,000
Total 18+18

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

The length of each side of an equilateral triangle is 20√3 cm. What is the area of the incircle?

Correct

Explanation:

BD = DC = 10√3 cm
In rt. angled ∆ADB
AD = √[(AB)^2 – (BD)^2] = √[(20√3)^2 – (10√3)^2]
AD = = √[(20√3 – 10√3) (20√3 + 10√3)] = √[(10√3) (30√3)] = 30
Radius of circle = OD = (1/3)*30 = 10 cm
So area = (22/7)*10*10

Incorrect

Explanation:

BD = DC = 10√3 cm
In rt. angled ∆ADB
AD = √[(AB)^2 – (BD)^2] = √[(20√3)^2 – (10√3)^2]
AD = = √[(20√3 – 10√3) (20√3 + 10√3)] = √[(10√3) (30√3)] = 30
Radius of circle = OD = (1/3)*30 = 10 cm
So area = (22/7)*10*10

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

If 2x + 5y = 6 and 8x^{3} + 125y^{3} = 210, then value of xy is

Correct

Explanation:
Use formula x^{3} + y^{3} = (x+y)(x^{2} + y^{2} – xy)
8x^{3} + 125y^{3} = 210
(2x+5y)( (2x)^{2} + (5x)^{2} – 2x*5y) = 210
Use a^{2} + b^{2} – ab = (a+b)^{2} – 3ab
So
6 * ( (2x+5y)^{2} – 3*2x*5y) = 210
So (6)^{2} – 3*2x*5y = 35
So 1 = 3*2x*5y
So xy = 1/30

Incorrect

Explanation:
Use formula x^{3} + y^{3} = (x+y)(x^{2} + y^{2} – xy)
8x^{3} + 125y^{3} = 210
(2x+5y)( (2x)^{2} + (5x)^{2} – 2x*5y) = 210
Use a^{2} + b^{2} – ab = (a+b)^{2} – 3ab
So
6 * ( (2x+5y)^{2} – 3*2x*5y) = 210
So (6)^{2} – 3*2x*5y = 35
So 1 = 3*2x*5y
So xy = 1/30

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

A’s 4 days work is equal to B’s 6 days work. If A can complete the work in 8 days then to complete the work B will take how many days?

Correct

Explanation:
A completes 1 work in 8 days, so in 4 days, 4/8 work = 1/2 work
So B in 6 days complete 1/2 work, so 1 work in 2*6 = 12 days

Incorrect

Explanation:
A completes 1 work in 8 days, so in 4 days, 4/8 work = 1/2 work
So B in 6 days complete 1/2 work, so 1 work in 2*6 = 12 days

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

Mean of 10 numbers is 40. But it was observed that numbers 12 and 16 are wrongly taken as 21 and 61. The correct mean is

Correct

Explanation:
Difference = 12+16 – 21 – 61 = -54
So correct average = 40 – 54/10 = 34.6

Incorrect

Explanation:
Difference = 12+16 – 21 – 61 = -54
So correct average = 40 – 54/10 = 34.6

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

In ∆ABC, the sides AB, BC and CA are produced to E, F and G respectively. If angle CBE = angle ACF = 125 degree, then the value of angle GAB in degrees is

Correct

Explanation:

CBE = 125, so ABC = 180-125 = 55
Similarly ACB = 55
So GAB = ABC+ACB = 55+55 = 110

Incorrect

Explanation:

CBE = 125, so ABC = 180-125 = 55
Similarly ACB = 55
So GAB = ABC+ACB = 55+55 = 110

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

Two stations A and B are 150 km apart from each other. One train starts from A at 6 AM at a speed of 30 km/hr and travels towards B. Another train starts from station B at 7 AM at a speed of 20 km/hr. At what time will they meet?

Correct

Solution:
30x + 20(x-1) = 150
X = 17/5 or 3 2/5 so 3 hrs and (2/5)*60 minutes = 3 hrs 24 mins
So 6 AM + 3 hrs 24 mins = 9:24 AM

Incorrect

Solution:
30x + 20(x-1) = 150
X = 17/5 or 3 2/5 so 3 hrs and (2/5)*60 minutes = 3 hrs 24 mins
So 6 AM + 3 hrs 24 mins = 9:24 AM