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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

If the price of wheat is reduced by 5%. How many kilograms of wheat a person can buy with the same money which was earlier sufficient to buy 38 kg of wheat?

Correct

Explanation:
Let the original price = 100 Rs per kg
Then money required to buy 38 kg = 38*100 = Rs 3800
New price per kg is 95% of Rs 100 = 95
So quantity of wheat bought in 3800 Rs is 3800/95 = 40 kg

Incorrect

Explanation:
Let the original price = 100 Rs per kg
Then money required to buy 38 kg = 38*100 = Rs 3800
New price per kg is 95% of Rs 100 = 95
So quantity of wheat bought in 3800 Rs is 3800/95 = 40 kg

Question 2 of 10

2. Question

1 points

Category: Quantitative Aptitude

20 litres of the water is drawn out of a jar and filled with milk. This operation is performed 1 more times. If the ratio of the quantity of water left in jar to that of milk in jar is 25 : 11, what was the initial quantity of water present in the jar?

Correct

Explanation:
Let initial quantity of milk = x litres
After two times, quantity of milk left in jar = x [1 – 20/x]^{2}
So x [1 – 20/x]^{2} / x = 25/25+11
[1 – 20/x]^{2} = 25/36
Square root both sides, so [1 – 20/x] = 5/6
Solve, x = 120

Incorrect

Explanation:
Let initial quantity of milk = x litres
After two times, quantity of milk left in jar = x [1 – 20/x]^{2}
So x [1 – 20/x]^{2} / x = 25/25+11
[1 – 20/x]^{2} = 25/36
Square root both sides, so [1 – 20/x] = 5/6
Solve, x = 120

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

A solid cone of height 10 cm and diameter of base 20 cm is cut out from a solid sphere of radius 10 cm. Find the percentage of wood wasted.

Correct

Explanation:
Volume of sphere = (4/3)ᴨr^{3}
Volume of cone = (1/3)ᴨr^{2}h = (1/3)ᴨr^{3} [since r = h]
Both radius are same and height is also equal to radius, so
Wood wasted = (4/3)ᴨr^{3} – (1/3)ᴨr^{3} = ᴨr^{3}
So % of wood wasted = [ᴨr^{3}]/ [(4/3)ᴨr^{3}] * 100
= (3/4)*100 = 75

Incorrect

Explanation:
Volume of sphere = (4/3)ᴨr^{3}
Volume of cone = (1/3)ᴨr^{2}h = (1/3)ᴨr^{3} [since r = h]
Both radius are same and height is also equal to radius, so
Wood wasted = (4/3)ᴨr^{3} – (1/3)ᴨr^{3} = ᴨr^{3}
So % of wood wasted = [ᴨr^{3}]/ [(4/3)ᴨr^{3}] * 100
= (3/4)*100 = 75

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

If a^{2} + b^{2} + c^{2} = ab + bc + ac, then find the value of (a+b)/c

Correct

Explanation:
a^{2} + b^{2} + c^{2} = ab + bc + ac
2a^{2} + 2b^{2} + 2c^{2} = 2ab + 2bc + 2ac
a^{2} – 2ab + b^{2} + b^{2} – 2bc + c^{2} + a^{2} – 2ac + c^{2} = 0
(a-b)^{2} + (b-c)^{2} + (a-c)^{2} = 0
So (a-b)^{2} = 0, (b-c)^{2} = 0 and (a-c)^{2} = 0
So, a = b, b = c, a = c or a= b = c
So (a+b)/c = (a+a)/a = 2

Incorrect

Explanation:
a^{2} + b^{2} + c^{2} = ab + bc + ac
2a^{2} + 2b^{2} + 2c^{2} = 2ab + 2bc + 2ac
a^{2} – 2ab + b^{2} + b^{2} – 2bc + c^{2} + a^{2} – 2ac + c^{2} = 0
(a-b)^{2} + (b-c)^{2} + (a-c)^{2} = 0
So (a-b)^{2} = 0, (b-c)^{2} = 0 and (a-c)^{2} = 0
So, a = b, b = c, a = c or a= b = c
So (a+b)/c = (a+a)/a = 2

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

In equilateral ∆ABC, 2 point P and Q are marked on AB and AC respectively such that PQ is parallel to BC. If length of PQ segment is 8 cm, find the area of ∆APQ.

Correct

Explanation:

PQ is parallel to BC
So angle APQ = ABC = 60 and angle AQP = ACB = 60
So ∆APQ is also equi. Triangle.
So its area is √3/4 * 8 * 8

Incorrect

Explanation:

PQ is parallel to BC
So angle APQ = ABC = 60 and angle AQP = ACB = 60
So ∆APQ is also equi. Triangle.
So its area is √3/4 * 8 * 8

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

A train having length 300 m crossed a platform in 20 seconds. With same speed, the train crossed a man standing on the platform in 12 seconds. What is the length of the platform?

Correct

Explanation:
Speed of train = 300/12 = 25 m/s
So (300+x)/25 = 20
Solve, x = 200

Incorrect

Explanation:
Speed of train = 300/12 = 25 m/s
So (300+x)/25 = 20
Solve, x = 200

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

A and B can complete a work in 18 days and 24 days respectively. They start the work and A left after some days. Now C who can alone complete the same work in 10 days joined with B and they completed the remaining work in 5 days. For how many days did B work?

Correct

Explanation:
B worked for 5 days + the days after which A left.
Let A left after x days. so
(1/18)*x + (1/24)*(5+x) + (1/10)*5 = 1
Solve, x = 3
So B worked for 5+3 days

Incorrect

Explanation:
B worked for 5 days + the days after which A left.
Let A left after x days. so
(1/18)*x + (1/24)*(5+x) + (1/10)*5 = 1
Solve, x = 3
So B worked for 5+3 days

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

What annual installment will discharge a debt of Rs 6000 due in 5 years at 10% per annum simple interest?

Correct

Explanation:
I = 100A/ [100t + rt(t-1)/2]
So I = 100*6000/ [100*5 + 10*5*(5-1)/2] = 100*6000/ [500 + 100]

Incorrect

Explanation:
I = 100A/ [100t + rt(t-1)/2]
So I = 100*6000/ [100*5 + 10*5*(5-1)/2] = 100*6000/ [500 + 100]

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

The sum of 6% of a number and 3% of other number is 3/7 of the sum of 2% of the first number and 9% of second. The ratio of first number to second number is