Quantitative
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CAT Analysis Ready Reckoner Model Papers B-Schools accepting CAT
Model Papers B-Schools accepting XAT Sin A = OD/OA Sin 45 = r/OA 1/√2 = r/OA OA = r√2 AE = OA+OE = r√2 + r = r(√2 + 1) Therefore r = R/(√2+1) Area of circle =πr2 = π(R/(√))1+22
= πR2 / (2+1+2√2) = √R2/(3+2√2)
top 15) Area of rectangle = 12 x 18 = 216 cm2 Area of unshaded region = Area of (rt. Triangles AED+BEF) = (1/2)(9x12) + (1/2)(9x6) = 81 cm2 Therefore area of shaded region = 216- 81 = 135 cm2. Therefore ratio is 135:81 = 15:9 = 5:3
16) By trial and error method, the answer is (c).
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top 17) The radius of the circle is given as 10√2. The diameter of the circle is 202√ The diameter of the circle will become the diagonal of the square The relation between the diagonal and side of square is D = 2√ 202√ = √2Σ S=20 cm
18) 10 kms is represented as 2cms 15.16 sq cms will be equal to 15.16.100 = 6000 sq kms 5
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19) At the rate of 50 kmph, 150 kms can be covered in 3 hrs Then, 20 mins for repair. To cover the remaining 250 kms, it will take 5 hrs In total, man takes 3 hrs + 20 mins + 5 hrs = 8 hr 20 mins
20)
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Distance between the top of two poles = Distance between E and D ED = √[(EC2 + CD2)] = √[(202+42)] = √[(400+16)] = √416 = 4√26m
top
|
Program Delivery
|
Contact Us
|
Sitema p
|
CAT Analysis
Admission
| Notifications
Classroom Correspondence
CAT Analysis Ready Reckoner Model Papers B-Schools accepting CAT
Model Papers B-Schools accepting XAT Sin A = OD/OA Sin 45 = r/OA 1/√2 = r/OA OA = r√2 AE = OA+OE = r√2 + r = r(√2 + 1) Therefore r = R/(√2+1) Area of circle =πr2 = π(R/(√))1+22
= πR2 / (2+1+2√2) = √R2/(3+2√2)
top 15) Area of rectangle = 12 x 18 = 216 cm2 Area of unshaded region = Area of (rt. Triangles AED+BEF) = (1/2)(9x12) + (1/2)(9x6) = 81 cm2 Therefore area of shaded region = 216- 81 = 135 cm2. Therefore ratio is 135:81 = 15:9 = 5:3
16) By trial and error method, the answer is (c).
top
top 17) The radius of the circle is given as 10√2. The diameter of the circle is 202√ The diameter of the circle will become the diagonal of the square The relation between the diagonal and side of square is D = 2√ 202√ = √2Σ S=20 cm
18) 10 kms is represented as 2cms 15.16 sq cms will be equal to 15.16.100 = 6000 sq kms 5
top
19) At the rate of 50 kmph, 150 kms can be covered in 3 hrs Then, 20 mins for repair. To cover the remaining 250 kms, it will take 5 hrs In total, man takes 3 hrs + 20 mins + 5 hrs = 8 hr 20 mins
20)
top
top
Distance between the top of two poles = Distance between E and D ED = √[(EC2 + CD2)] = √[(202+42)] = √[(400+16)] = √416 = 4√26m
top
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