### Practice Aptitude Questions (Arithmetic) With Solutions

Practice Aptitude Questions (Arithmetic) With Solutions Set-58:
Dear Readers, Important Aptitude Questions for SSC/FCI Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1).A book vendor sold a book at a loss of 10% . Had he sold it for Rs. 108 more, he would have earned a profit of 10%. Find the cost of the book.
a)    Rs. 442
b)    Rs. 540
c)    Rs. 648
d)    Rs. 740

2).A sells a cycle to B at a profit of 5% and B sells it to C at a profit of 10%. If C pays Rs. 2310 for it, the cost price of A is
a)    Rs. 2000
b)    Rs.2100
c)    Rs. 1900
d)    Rs. 2010

3).A number is divided into two parts in such a way that 80% of 1st part is 3 more than 60% of 2nd part and 80% of 2nd part is 6 more than 90% of the 1st part. Then the number is
a)    125
b)    130
c)    135
d)    145

4).A reduction of 10% in the price of wheat enables a man to buy 50 gram of wheat more for a rupee. How much wheat could originally be had for a rupee?
a)    400 grams
b)    500 grams
c)    450 grams
d)    350 grams

5).A train starts from a place A at 6 a.m and arrives at another place B at 4.30 p.m. on the same day. If the speed of the train is 40 km per hour, find the distance travelled by the train?
a)    420 km
b)    230 km
c)    320 km
d)    400 km

6).P travels for 6 hours at the rate of 5 km/hr and for 3 hours at the rate of 6 km/hr. the average speed of the journey in km / hr is
a)    3(1 / 5)
b)    5(1 / 3)
c)    1(2 / 9)
d)    2(2 / 5)

7).A certain sum of money at simple interest amounts to Rs. 1012 in 2 ½ years and to Rs. 1067.20 in 4 years. The rate of interest per annum is
a)    2.5%
b)    3%
c)    4%
d)    5%

8).A certain amount of money at r% compounded annually after two and three years becomes Rs. 1440 and RS. 1728 respectively r is
a)    5
b)    10
c)    15
d)    20

9).The edge of a rectangular box are in the ratio 1 : 2 : 3 and its surface area is 88 cm2 The volume of the box is
a)    24 cm3
b)    48 cm3
c)    64 cm3
d)    120 cm3

10).For a triangle, base is 6√3 cm and two base angles are 30° and 60°. Then the height of the triangle is
a)    3√3 cm
b)    4.5 cm
c)    4√3 cm
d)    2√3 cm

1). b) 2).a ) 3). c) 4).c ) 5). a) 6).b ) 7). c) 8).d ) 9). b) 10). b)

Solution:
1). C.P of article = Rs. x
First S.P = Rs. 9x / 10
∴ (9x / 10) + 180 = 110x / 100
(11x / 10) - (9x / 10) = 108
(2x / 10) = 108
(108 × 10) / 2 = Rs. 540
2). If the C.P for A be Rs. x, then
x × (105 / 100) × ( 110 / 100) = 2310
x = (2310 × 100 × 100) / (105 × 110)
= Rs. 2000
3). First part = Rs. x and Second part = Rs. y
∴( x × 80) / 100 = [( y × 60) / 100] + 3
4x / 5 = (3y / 5) + 3
4x – 3y = 15                                       ….. ⅰ
Again,
4y / 5 = (9x / 10) + 6
8y = 9x + 60
8y – 9x = 60                                       …..ⅱ
By equation (ⅰ) × 8 and (ⅱ) × 3,
32x – 24 y = 120
24y – 27x = 180
_________________
5x = 300
x = 60
from equation (ⅰ)
4 × 60 – 3y = 15
3y = 240 – 15 = 225
y = 225 / 3 = 75
∴ x + y = 60 + 75 = 135
4). Original price of wheat = Rs. x / kg
New price = Rs. (9x  / 10) kg
∴ [1 / (9x/10)] – 1/x = 50 / 1000 = 1 / 20
∴(10 / 9x) – (1 / x) = 1 / 20
1 / 9x = 1 / 20
9x = 20
x = 20 / 9
∴ Original price = Rs. 20 / 9 kg
Rs. 20 / 9 = 1000gm
Rs. 1 = (1000 × 9 ) / 20 = 450 gm
5). Time = 10 (1 / 2) hours = 21 / 2 hours
Speed = 40 kmph
Distance = speed × Time
= 40 × (21 / 2) = 420
6). Total distance = 5 × 6 + 3 × 6
= 30 + 18 = 48 km
Total time = 9 hours
Average speed = 48 / 9 = 16 / 3 = 5(1 /3) kmph
7). If the principle be Rs. P and rate of interest by r% p.a then
1012 = P + [ ( P × r × 5) / (100 / 2)              …ⅰ
Interest in 3 / 2 years
= 1067.20 – 1012 = Rs. 55.20
∴ P = (55.20 × 100) / [(3 / 2) × r] = 3680 / r
From equation ⅰ
1012 = (3680 / r) + (3680 × 5) / 200
1012 = (3680 / r) + 92
(3680 / r) = 1012 – 92 = 920
r = 3680 / 920 = 4% per annum
8).  If the principle be RS. P, then
A = P [( 1 + (r / 100)]T
1440 = P [( 1 + (r / 100)]2                                   ….ⅰ
1728 = P [( 1 + (r / 100)]3                                       .....ⅱ
on dividing equation ⅱ  by ⅰ
1728 / 1440 = 1 = (r / 100)
∴ r / 100 = (288 × 100) / 1440 = 20% per annum
9). Sides of the box = x, 2x and 3x cm
∴ 2 (x × 2x + 2x × 3x + 3x × x) = 88
11x2  = 44
x2  = 4
x = 2
∴ volume of the box = x × 2x × 3x
= 6x3 = 6 × 8 = 48 cu.cm
10).

Sin 30° = AC / BC
1 / 2 = AC / 6√3
AC = 3√3
Sin 60° = AD / AC
√3 / 2 = AD / 3√3
AD = (3√3 × √3) / 2 = 4.5 cm