Aptitude 1

Q1. When n is divided by 6, the remainder is 4. When 2n is divided by 6, the remainder is

1
2
0
4

( b) n = 6 x K + 4 2n = 2 ( 6k + 4)
= 12 k + 8
= 12 k + 6 + 2
= 6 ( 2k + 1 ) + 2

Q2. The smallest among the numbers 2²⁵⁰, 3¹⁵⁰, 5¹⁰⁰ and 4²⁰⁰ is

2²⁵⁰
4²⁰⁰
5¹⁰⁰
3¹⁵⁰

(c) : 2²⁵⁰ =( 2⁵)⁵⁰ = 3250
3¹⁵⁰ = 27⁵⁰
5¹⁰⁰ = 25⁵⁰
4²⁰⁰ = 256⁵⁰
SMALLEST NUMBER = 25⁵⁰

Q3.If A : B = 3 :4 and B :C = 6 : 5, then C : A is

9 : 8
10 : 9
9 : 10
8 : 9

(b)

Q4. The difference of two numbers is 5 and the difference of their squares is 135. The product of the numbers is:

176
178
180
182

(a) Let the two numbers are x and y => x - y = 5 and x2 – y2 = 135
(x – y) (x + y ) = 135 => 5 ( x + y) = 135  x + y = 27
Solving x – y = 5 and x + y = 27 We get x = 16 and y = 11
So product of the number = 16x11= 176

Q5. 91 + 4 92 + 4 93 + 4 94 is divisible by

17
13
11
3

(a) 4 ⁹¹ + 4 ⁹² + 4 ⁹³ + 4 ⁹⁴ = 4 ⁹¹ ( 1 + 4 + 4² + 4 ³ ) = 4⁹¹ x 85
= is divisible by 17

Q6. A sum of Rs. 395 was divided among A, B and C in such a way that B gets 25% more than A and 20% more than C. What is the share of A?

Rs. 195
Rs. 180
Rs. 98
Rs. 120

(d) Let each one’s share is A,B and C respectively, then
B =125A/100 = 120C/100
A =100B/125 =4/5B, C = 100B/120 = 5/6B 4/5B +B +5/6B =395, 79B/30 =395, B =395*30/79 =150
A = 4*150/5 = 120

Q7. The amount of work in a leather factory is increased by 50% By what per cent is it necessary to increase the number of workers to complete the new amount of work in previously planned time, if the productivity of the new labor is 25% more.

60%
66.66%
40%
33.33%

(c) Men x Time = Work 100 x 1 = 100 unit
150 x 1 = 150 unit
Extra man power required = 50
But since new workers are 5/4 times as efficient as existing workers.
Actual no. of workers = (50) / (5/4) = 40 men Hence, required percentage = (40 /100) x 100
= 40 %

Q8. In an army there are 2100 Soldiers. They have food reserves for 50 days. Some people among them left on leave. Now they can utilize their food reserves for 75 days. How many soldiers left on leave?

800
700
600
28

(b) : Assume that the soldiers who left on leave are K
So,
SoldiersDays
2100 50
2100 - k 75
if the number of soldiers reduces, the number of days increases. Then,
75 : 50 :: 2100 : (2100 - k) (75/50) = ((2100/ (2100-k)) 4200 = 6300 - 3k
=>3k = 2100 => k = 2100/3 = 700

Q9. Two buses A and B leave the same bus depot, A towards the North and B towards the East. The bus A travels at a speed of 5 km/hr more than that of the bus B. If after four hours the distance between the two buses is 100 km, find the speed of the bus A. a) 60 kmph b) 40 kmph c) 20 kmph d) 15 kmph

60 kmph
40 kmph
20 kmph
15 kmph

(c) : Let the speed of the bus B be x km/hr. Then the speed of the bus A will be ( x +5) kmph
Distance traveled in 4 hours is 4x and 4(x + 5) for the two buses respectively
(4x)² + {4 (x + 5) }² = (100)²
16x² + 16x² + 160x + 400 = 10000 32x² +160x –9600 = 0
x² + 5x –300 = 0 (x-15) (x +20 ) = 0 x = 15
:. Speed of the bus A = 15 +5 = 20 kmph

Q10. If the radius of a circle is diminished by 10%, then its area is diminished by :

10%
19%
20%
36%

(b) Let the original radius be R cm.
New radius = (90% of R) cm =[(90/100)*R] cm
= 9R/10 cm.
Original area = πR².
Diminished area = [πR²-π(9R/10)²] cm² = [(1-(81/100) πR²] cm^{2 = ((19/100 πR2) cm2. Decrease% = [(19πR2/100)*(1/πR2)*100]% = 19%.}