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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeWhat least value must be assigned to *, so that the number 54237*8 is divisible by 8?
Correct
Explanation:
For a number to be divisible by 8, the last three digits should be divisible by 8, so
Start with 718 â€“ not divisible by 8, now 728 divisible by 8 so * is 2Incorrect
Explanation:
For a number to be divisible by 8, the last three digits should be divisible by 8, so
Start with 718 â€“ not divisible by 8, now 728 divisible by 8 so * is 2 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeThe area of a circle having diameter 14 m equals the area of rectangle. If the length of rectangle is 8 less than twice the breadth, what is the perimeter of rectangle?
Correct
Explanation:
Ar of rect. = (22/7) *7 * 7 = 154
So (2b8)*b = 154
So 2b^2 â€“ 8b â€“ 154 = 0
Or b^2 â€“ 4b â€“ 77 = 0
Solve, b = 11, so l = 2b8 = 14
So perimeter = 2(11+14)Incorrect
Explanation:
Ar of rect. = (22/7) *7 * 7 = 154
So (2b8)*b = 154
So 2b^2 â€“ 8b â€“ 154 = 0
Or b^2 â€“ 4b â€“ 77 = 0
Solve, b = 11, so l = 2b8 = 14
So perimeter = 2(11+14) 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeA and B have their present ages in the ratio 5 : 3. Also Aâ€™s age 4 years ago and Bâ€™s 4 years hence is same. Find the ratio of Aâ€™s age 2 years hence to Bâ€™s age 2 years ago.
Correct
Explanation:
Let Aâ€™s present age = 5x, and Bâ€™s = 3x
Aâ€™s age 4 years ago to Bâ€™s age 4 years hence is same
So (5x4) = (3x+4)
Solve x = 4
So Aâ€™s present age = 20, and Bâ€™s = 12
Required ratio = (20+2) : (122)Incorrect
Explanation:
Let Aâ€™s present age = 5x, and Bâ€™s = 3x
Aâ€™s age 4 years ago to Bâ€™s age 4 years hence is same
So (5x4) = (3x+4)
Solve x = 4
So Aâ€™s present age = 20, and Bâ€™s = 12
Required ratio = (20+2) : (122) 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeA and B invested in a business as Rs 1600 and Rs 1800 respectively. After 4 months both withdrew this amount and invested Rs 1800 and Rs 1600 respectively. What is the difference in the shares of both after a year if a total profit of Rs 11,730 is obtained?
Correct
Explanation:
A : B
1600* 4 + 1800*8 : 1800*4 + 1600*8
16*4 + 18*8 : 18*4 + 16*8
16 + 18*2 : 18 + 16*2
8 + 18 : 9 + 16
26 : 25
So difference = (2625)/(26+25) * 11730Incorrect
Explanation:
A : B
1600* 4 + 1800*8 : 1800*4 + 1600*8
16*4 + 18*8 : 18*4 + 16*8
16 + 18*2 : 18 + 16*2
8 + 18 : 9 + 16
26 : 25
So difference = (2625)/(26+25) * 11730 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeThe speed of boat is 10 km/hr and speed of stream is 4 km/hr. A person goes 84 km downstream from point A to B and same distance upstream from point B to C. In how much total time will he cover his journey?
Correct
Incorrect

Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeThe value of [{cos^{2}A(sinA + cosA)/cosec^{2}A(sinA – cosA)} + {sin^{2}A(sinA – cosA)/sec^{2}A(sinA + cosA)}] [sec^{2}A – cosec^{2}A]
Correct
Explanation:
[{cos^{2}A(sinA + cosA)/cosec^{2}A(sinA – cosA)} + {sin^{2}A(sinA – cosA)/sec^{2}A(sinA + cosA)}] [sec^{2}A – cosec^{2}A]
= [{cos^{2}A.sin^{2}A(sinA + cosA)/(sinA – cosA)} + {sin^{2}Acos^{2}A(sinA – cosA)/(sinA + cosA)}] [1/cos^{2}A â€“ 1/sin^{2}A]
= [{cos^{2}A.sin^{2}A(sinA + cosA)/(sinA – cosA)} + {sin^{2}Acos^{2}A(sinA – cosA)/(sinA + cosA)}] [{sin^{2}A – cos^{2}A} /cos^{2}A .sin^{2}A]
Take common cos^{2}A.sin^{2}A from first two terms and cancel with denominator of third term.
= [{(sinA + cosA)/(sinA – cosA)} + {(sinA – cosA)/(sinA + cosA)}] [{sin^{2}A – cos^{2}A}]
= [{(sinA + cosA)^{2} + (sinA – cosA)^{2}/(sinA – cosA)(sinA + cosA)}] [(sinA – cosA)(sinA + cosA)]
Solve this gives 2(sin^{2}A + cos^{2}A) = 2*1Incorrect
Explanation:
[{cos^{2}A(sinA + cosA)/cosec^{2}A(sinA – cosA)} + {sin^{2}A(sinA – cosA)/sec^{2}A(sinA + cosA)}] [sec^{2}A – cosec^{2}A]
= [{cos^{2}A.sin^{2}A(sinA + cosA)/(sinA – cosA)} + {sin^{2}Acos^{2}A(sinA – cosA)/(sinA + cosA)}] [1/cos^{2}A â€“ 1/sin^{2}A]
= [{cos^{2}A.sin^{2}A(sinA + cosA)/(sinA – cosA)} + {sin^{2}Acos^{2}A(sinA – cosA)/(sinA + cosA)}] [{sin^{2}A – cos^{2}A} /cos^{2}A .sin^{2}A]
Take common cos^{2}A.sin^{2}A from first two terms and cancel with denominator of third term.
= [{(sinA + cosA)/(sinA – cosA)} + {(sinA – cosA)/(sinA + cosA)}] [{sin^{2}A – cos^{2}A}]
= [{(sinA + cosA)^{2} + (sinA – cosA)^{2}/(sinA – cosA)(sinA + cosA)}] [(sinA – cosA)(sinA + cosA)]
Solve this gives 2(sin^{2}A + cos^{2}A) = 2*1 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThe minimum value of 3 tan^{2}É¸ + 2 cot^{2}É¸ is
Correct
Explanation:
We have formula, Minimum value = 2âˆš(ab)
So Minimum value = 2âˆš(3*2) = 2âˆš6Incorrect
Explanation:
We have formula, Minimum value = 2âˆš(ab)
So Minimum value = 2âˆš(3*2) = 2âˆš6 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeFind the area of the colored portion in the figure below, the radius of the circle being 4 cm.
Correct
Explanation:
Radius = 4
So ar of semi circle = á´¨/2 * 4 * 4 = 8á´¨ sq. cm
Both triangles are isosceles
Area of both triangles = 4*4/2 = 8 sq. cm
So area of both triangles = 2*8 = 16 sq. cm
So area of colored portion = 8á´¨16Incorrect
Explanation:
Radius = 4
So ar of semi circle = á´¨/2 * 4 * 4 = 8á´¨ sq. cm
Both triangles are isosceles
Area of both triangles = 4*4/2 = 8 sq. cm
So area of both triangles = 2*8 = 16 sq. cm
So area of colored portion = 8á´¨16 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeABC is an equilateral triangle with side = 4 cm. With A, B, and C as centres and radius 2 cm three arcs are drawn. Find the area of the region within the âˆ†ABC bounded by the three arcs.
Correct
Explanation:
Each angle of triangle is 60.
Area of 3 sectors = 3 * 60/360 * á´¨ * (2)^2 = 2á´¨
Area of triangle = âˆš3/4 * 4 * 4 = 4âˆš3
So req. area = 4âˆš3 – 2á´¨Incorrect
Explanation:
Each angle of triangle is 60.
Area of 3 sectors = 3 * 60/360 * á´¨ * (2)^2 = 2á´¨
Area of triangle = âˆš3/4 * 4 * 4 = 4âˆš3
So req. area = 4âˆš3 – 2á´¨ 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeFind the least number which when divided by 10, 15, 18 and 30 leaves the same remainder 4 in each case.
Correct
Explanation:
It will be the LCM of these numbers + 4
So 90 + 4Incorrect
Explanation:
It will be the LCM of these numbers + 4
So 90 + 4
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