AMCAT Aptitude Question -10
AMCAT Quantitative Ability Previous Papers-10
91)Ritu has 3 shirts in shades of red , 4 in yellow shades and 5 in green shades .Three shirts are picked at random. The probability that all of those are in red shades is:
A)1/12 B)1/660 c)1/66 D)1/4
Solution: Given 3red, 4yellow, 5green shirts
Total shirts = 3+4+5 = 12
Probability of getting Red = 3/12 = 1/4
92)(1.0816)1/2=?
A)0.14 B)1.4 c)1.004 D)1.04
Solution: It written as square root of 1.0816
By trail method, check all options
Like 1.04 * 1.04
1.0816
93)An intern can a job in 15 days . The manager and senior manager are busy with other priorities and thus take 25 and 40 days respectively to complete the task. How long will they task to finish the task if all of them work together?
A)7 b)722/29 c)8 D)821/29
Solution: It can be written as …
Work together = 1/15 + 1/25 +1/40
= 1/5[1/3 + 1/5 + 1/8]
= 1/5[40+24+15/120]
=[79/5*120]
It can be written..
ð 600/79 = 7.59 (Answer is in fractions..in this options solution is not present)
94)if a + b =6, ab = 5, then the value of a-b is:
A)4 B)5 C)6 D)7 E)9
Solutions : By using formulae…..
(a+b)2 - (a-b)2 = 4ab
36 – (a-b)2 = 20
=> (a-b)2 = 16
=> (a-b) = 4
95)At a certain party ratio the ratio of gents and ladies was 1:2 . But when 2 gents and 2 ladies left the party . The ratio became 1:3. How many people were initially present in the party?
A)12 B)15 C)18 D)24
Solutions: Ratio of G & L =1:2 => x : 2x
Two G & Two L are left i.e =>x-2 : 2x-2
=> x-2 : 2x-2 = 1 : 3
=> x-2/2x-2 = 1/3
=>3x-6 = 2x-2
=> X=4 (Gents)
=> 2x=8 (Ladies)
:. Total people = 12
96)The SI on Rs 10 for 4 months at the rate of 3 paise per rupee per month is :
A)Rs 1.20 B)Rs 1.60 C)Rs 2.40 D)rs.3.60/-
Solution : By using Formulae…
SI = P * T* R/100
=>10*4*3/100
=>120/100 => Rs 1.20
97)In a miniature wonderland,three countries Austria,America and Germany are on display.If Austria and America occupied 3/7 and 4/9 of the display area respectively,what is the display area occupied by Germany?
a.3/21 b.5/63 c.8/63 d.1/21
Solution: Let 3/7 + 4/9 + G =1
ð 27+28/63 + G = 1
ð G = 1 – 55/63
ð G = 8/63
98)Find the value of p which satisfies the relation log2(p-1) + 2 =log2(3p+1).
a.1 b.3 c.5 d.7
Solution: log2(p-1) + 2 = log2(3p+1)
ð log2(p-1) + log2(22) = log2(3p+1) (:. Log2 2 2= 2 )
ð log2 22 (p-1) = log2 (3p+1) (:. Log a+ log b = logab )
ð 4p-4 = 3p+1
ð P = 5
99) 28 * 22 =
a.410 b.210 c.216 d.416
Solution : am * an = am+n
Using this formulae = 28 * 22 = 210
100)Mauli purchased a designer saree from mumbai at 8/9th of its MRP.When she came back to Delhi,her neighbour coaxed Mauli to sell the saree to her.She was even ready to pay 9% more than its MRP.What would Mauli’sgain percentage be,if she decides to sell the saree to her neighbour?
a.15.59% b.16.61% c.20.36% d.22.625%
Solution: cost price = 8/9 x
Selling price = 9% more => x + 9/100 x
=>109x/100
Formulae = ( s.p-c.p/c.p) * 100
ð {[(109x/100 )– (8x/9) ]/(8/9)}x100
ð By Solving we get :22.625%
Category: Amcat, Amcat Aptitude
0 comments