# DMRC/CIL ONLINE PRACTICE SET-APTITUDE/NUMERICAL ABILITY WITH EXPLANATION1

1).Find the difference between the 30th
and the 34th term of the series 16, 32, 48, 64….

a)
72
b)
66
c)
68
d)
64

2).The LCM of two numbers is 55 times their
HCF. The sum of the LCM and the HCF is 6272. If one number is 280, then the
other number is
a)
2464
b)
3226
c)
6272
d)
2326

3).The total earning of A, B and C is Rs 95000
and they save 30%, 25% and 20% of their incomes respectively. If savings are in
the ratio of 4: 5: 6, then what is the difference between the amount of
earnings of B and C?
a)
Rs. 21000
b)
Rs. 9000
c)
Rs. 18000
d)
Rs. 15000

4).The numbers of boys and girls in a college
are in the ration of 4: 1. If 50% of the boys and 75% of the girls are adults,
the percentage of students who are not adults is?
a)
40%
b)
45%
c)
30%
d)
50%

he bought it at 10% more and sold it for Rs. 70 more, he would have gained 25%.
Find the cost price of the goods.
a)
Rs. 800
b)
Rs. 200
c)
Rs. 400
d)
Rs. 600

6).The difference between a discount of 40% on
Rs. 1000 and two successive discount of 36% and 5% on the same amount is:
a)
Rs. 8
b)
Rs. 4
c)
Rs. 16
d)
Rs. 24

7).At the certain rate of simple interest a
sum becomes double in 14 years. In how many years will it become 4 times itself?
a)
58 years
b)
42 years
c)
28 years
d)
56 years

8).A sum of money becomes Rs. 2500 in 2 years
and Rs. 5184 in 4 years on compound interest. Find the rate of compound
interest.
a)
44%
b)
50%
c)
40%
d)
36%

9).In a military camp, ration was stored for
660 militants at the rate of 800 gramps per militant for 45 days. After 10 days
some more militants came from other camp and the remaining ration got finished
only in 15 days at the rate of 1400 gramps per militant. Find the number of
militants coming in.
a)
240
b)
248
c)
240
d)
220

10).16 men or 20 boys can do a piece of work
in 30 days. The number of days that 15 men and 15 boys would take to finish the
same work is
a)
13×7/9
b)
17×1/7
c)
15×5/9
d)
17×7/9

Check here below for the detailed
solution of the above aptitude questions
:
1.)
Given series 16, 32, 48, 64 ….
Common
difference (d) = 32 – 16 = 16
First
term = (a) = 16
30th
term = t30 = a + (n – 1) d
=
16 + (30 – 1) 16
=
16 + 29 × 16 = 480
Now,
34th term = t34= a + (n-1) d
=
16 + (34 – 1) 16
=
16 + 33 × 16 = 544
Required
difference = 544 – 480 = 64

2).
Let the HCF be x.
So,
LCM =55x
According
to the question,
55x
+ x = 6272
Or,
56x = 6272
X
= 112
Now,
LCM × HCF = First number × Second number
Or,
55x × x = 280 × second number
Or,
55 × 112 × 112 = 280 × second number
Second
number = (55×112×112)/280
=
2464

3).
Saving percentages of A, B and C are 30%, 25% and 20% respectively.
Now,
(A×30)/100: (B×25)/100: (C×20)/100 = 4: 5: 6
Or,
3A/10: B/4: C/5 = 4: 5: 6
Or,
6A: 5B: 4C = 4/6: 5/5: 6/4
A:
B: C = 2/3:1:3/2
Multiplying
each ration by 6,
A:
B: C = 4:6:9
Now
the sum of the ratios = 4 + 6 + 9 = 19
Earning
of B = (6/19) × 95000 = Rs. 30000
Earning
of C = (9/19) × 95000 = Rs. 45000
Required
difference = (45000 – 30000)
=
Rs. 15000

4).
Let the total number of boys and girls are 100.
Now,
number of adult boy and girls students
=
[(4/5) × 100 × (1/2) + (1/5) × (3/4) × 100)] = 40 + 15 = 55
Number
=
(100 – 55) = 45
Required
percentage = 45%

5).
Let the cost price be Rs. 100
SP
at 20% profit = Rs. 120
CP
on 10% more = Rs. 110
SP
on 25% profit = (110 × 125)/100 = Rs. 137.5
Difference
of selling price = (137.5 – 120) = Rs. 17.5
On
Rs. 17.5 more the CP is Rs. 100
On
Rs. 70 more the CP is (100/17.5×70)
=
(100×10×70)/175 = Rs. 400

6).
SP at 40% discount = (1000×60)/100 = Rs. 600
SP
at 36% and 5% successive discounts
=
1000 × (64/100) × (95/100) = Rs. 608
Required
difference = (608 – 600) = Rs. 8

7).
Let the principal be Rs. 100
Time
= 14 years
Amount
= Rs. 200
Interest
= (200 – 100) = Rs. 100
Rate
= [(Interest × 100)/(Principal × time)] = [(100×100)/(100×14)] %
=
50/7%
Now,
amount = Rs. 400
Interest
= (400 – 100) = Rs. 300
Time
= (Interest × 100/Principal × Rate) = (300 × 100/100 × 50) × 7
=
42 years

8). If a sum of money on compound interest becomes
Rs. x in t1 years and Rs. y in t2 years,  (1 + r/100)t2-t1
= y/x
Hence,
y = Rs. 5184
X
= Rs. 2500
t2
= 4 years
t1
= 2 years
So, (1 + r/100)4-2= 5184/2500
Or,
(1+ r/100)2 = 1296/625
Or,
1 + r/100 = 36/25
Or,
r/100 = (36/25) – 1
Or,
r = (11/25) × 100
r
= 44%

9).
Remaining time = (45 – 10)
=
35 days
Present
militants = 660
Let
the number of militants coming in be x
Total
militants = (660 + x)
If
x militants had not come, 660 militants would have eaten the remaining ration
at the rate of 800 grams per militant in 35 days    ….. (i)
Now,
(660 + x) militants are the remaining ration at the rate of 1400 gram/militant
in 15 days.
Then,
660 × 800 × 35 = (660 + x) × 1400 × 15
Or,
(660 + x) = (660×800×35)/(1400×15)
Or,
660 + x = 880
Or,
x = 880 – 660
X
= 220

10).
16 men = 20 boys
1
man = 20/16 = 5/4 boys
15
men = 75/4 boys
Total
number of boys = (75/4) + 15 = (75 + 60)/4
=
135/4 boys
20
boys can do a piece of work in 30 days.
1
boy can do a piece of work in 30 × 20 days.
135/4
boys can do a piece of work in (30 × 20 × 4)/135 days = 17 × (7/9) days