DRDO Exam Solved Papers

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Sample paper :
DRDO Exam Solved Paper
1)A train covers a distance in 50 min ,if it runs at a speed
of 48kmph on an average.The speed at which the train must run
to reduce the time of journey to 40min will be.
1. Solution::
Time=50/60 hr=5/6hr
Speed=48mph
distance=S*T=48*5/6=40km
time=40/60hr=2/3hr
New speed = 40* 3/2 kmph= 60kmph
2)Vikas can cover a distance in 1hr 24min by covering 2/3 of
the distance at 4 kmph and the rest at 5kmph.the total
distance is?
2. Solution::
Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km
3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr.
the usual time is.
3. Solution::
Usual speed = S
Usual time = T
Distance = D
New Speed is ¾ S
New time is 4/3 T
4/3 T – T = 5/2
T=15/2 = 7 ½
4)A man covers a distance on scooter .had he moved 3kmph
faster he would have taken 40 min less. If he had moved
2kmph slower he would have taken 40min more.the distance is.
4.Solution::
Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x ————–1
x/y-2 – x/y = 40/60 hr y(y-2) = 3x —————–2
divide 1 & 2 equations
by solving we get x = 40
5)Excluding stoppages,the speed of the bus is 54kmph and
including stoppages,it is 45kmph.for how many min does the bus
stop per hr.
5.Solution::
Due to stoppages,it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min
6)Two boys starting from the same place walk at a rate of
5kmph and 5.5kmph respectively.wht time will they take to be
8.5km apart, if they walk in the same direction
6.Solution::
The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmph
Distance between them is 8.5 km
Time= 8.5km / 0.5 kmph = 17 hrs
7)2 trains starting at the same time from 2 stations 200km
apart and going in opposite direction cross each other ata
distance of 110km from one of the stations.what is the ratio of
their speeds.
7. Solution::
In same time ,they cover 110km & 90 km respectively
so ratio of their speed =110:90 = 11:9
8)Two trains start from A & B and travel towards each other at
speed of 50kmph and 60kmph resp. At the time of the meeting the
second train has traveled 120km more than the first.the distance
between them.8. Solution::
Let the distance traveled by the first train be x km
then distance covered by the second train is x + 120km
x/50 = x+120 / 60
x= 600
so the distance between A & B is x + x + 120 = 1320 km
9)A thief steals a ca r at 2.30pm and drives it at 60kmph.the
theft is discovered at 3pm and the owner sets off in another car
at 75kmph when will he overtake the thief
9. Solution::
Let the thief is overtaken x hrs after 2.30pm
distance covered by the thief in x hrs = distance covered by
the owner in x-1/2 hr
60x = 75 ( x- ½)
x= 5/2 hr
thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm
10)In covering distance,the speed of A & B are in the ratio
of 3:4.A takes 30min more than B to reach the destion.The time
taken by A to reach the destinstion is.
10. Solution::
Ratio of speed = 3:4
Ratio of time = 4:3
let A takes 4x hrs,B takes 3x hrs
then 4x-3x = 30/60 hr
x = ½ hr
Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr
Modal Paper For Aptitude
1) The average ages of three persons is 27 years. Their ages are in the proportion of 1:3:5.What is the
age in years of the youngest one among them.
Sol: Let the age of three persons be x, 3x and 5x
-- > 9x/3 = 27 -- > x = 9
2) The average of 11 numbers is 50. If the average of first 6 numbers is 49 and that of last 6 is 52.Find
the 6th number.
Sol: The total sum of 11 results = 11 * 50 = 550
The total sum of first 6 results = 6 * 49 = 294
The total sum of last 6 results = 6 * 52 = 312
Sixth result = 294 + 312 – 550 = 56
3) Find L.C.M of 852 and 1491.
852) 1491 (1
852
639) 852 (1
639
213) 639 (3
639
0
H.C.F of 852 and 1491 is 213
: . L.C.M = 852*1491/213 = 5964
4) The smallest number which when divided by 20, 25, 35, 40 leaves the remainder 6 When divided by
14, 19, 23 and 34 respectively is the difference between divisor and The corresponding remainder is 6.
: . Required number = (L.C.M of 20, 25, 35, 40) – 6
= 1400-6 = 1394
5) The least multiple of 7 which leaves a remainder 4 when divided by 6,9,15 and 18 is
L.C.M of 6,9,15 and 18 is 90.
Let x be the least multiple of 7, which when divided by 90 leaves the remainder 4.
Then x is of the form 90k + 4.
Now, minimum value of k for which 90k + 4 is divisible by 4.
: . x = 4 * 90 + 4 = 364
6) Sum of three even consecutive numbers is 48, and then least number is
1) 16
2) 18
3) 20
4) 14
Sol: 4) Let the numbers be 2n, 2n+2 and 2n+4
2n + (2n+2) + (2n+4) = 48
6n = 48-6 = 42, n = 7
Hence the numbers are -- > 14, 16 and 18
The least number is 14.
7) It being given that √ 15 = 3.88, the best approximation to √5/3 is
1) 0.43
2) 1.89
3) 1.29
4) 1.63
__ ___ __ __ __
Sol: 3) x = √5/3 = √5*3/3*3 = √15 /√ 9 = √15/3 = 3.88/3 = 1.29
8) Of the two-digit numbers (those from 11 to 95, both inclusive) how many have a Second digit
greater than the first digit?
1) 37
2) 38
3) 36
4) 35
Sol: 3) 12 to 19 -- > 8
23 to 29 -- > 7
34 to 39 -- > 6
45 to 49 -- > 5
56 to 59 -- > 4
67 to 69 -- > 3
78 to 79 -- > 2
89 -- > 1
9) The Value of √24 + 3√64 + 4√28 is
Sol: 24*1/2 + 43*1/3 + 28*1/4
-- > 4 + 4 + 4 -- > 12
10) 3 ¼ - 4/5 of 5/6 / 4 1/3 / 1/5 – ( 3/10 + 21 1/5 ) is equal to
Sol: 13/4 – 4/5 * 5/6 / 13/3 / 1/5 – ( 3/10 + 106/5 ) (use BODMASRULE)
-- > 13/4 – 4/6 / 13/3 / 1/5 – 215/10 -- > 31/12 / 13/3 * 5 – 215/10
-- > 31/12 / 65/3 – 43/2 -- > 31/12 / 130 – 129/6 -- > 31/12/1/6 = 31/12 * 6/1
-- > 31/2 = 15 1/2
1) 13 sheeps and 9 pigs were bought for Rs. 1291.85.If the average price of a sheep be Rs. 74. What
is the average price of a pig.
Sol: Average price of a sheep = Rs. 74
: . Total price of 13 sheeps
= (74*13) = Rs. 962
But, total price of 13 sheeps and 9 pigs
= Rs. 1291.85
Total price of 9 pigs
= Rs. (1291.85-962) = Rs. 329.85
Hence, average price of a pig
= (329.85/9) = Rs. 36.65
12) A batsman in his 18th innings makes a score of 150 runs and there by increasing his Average by 6.
Find his average after 18th innings.
Sol: Let the average for 17 innings is x runs
Total runs in 17 innings = 17x
Total runs in 18 innings = 17x + 150
Average of 18 innings = 17x + 150/18
: . 17x + 150/18 = x + 6 -- > x = 42
Thus, average after 18 innings = 42
13) . Find the H.C.F of 777 and 1147.
777) 1147 (1
777
370) 777 (2
740
37) 370 (10
370
0
: . H.C.F of 777 and 1147 is 37.
14) The L.C.M of two numbers is 2310 and their H.C.F is 30. If one number is 210 the Other is
The other number
= L.C.M * H.C.F/given number
= 2310*30/210 = 330
15) The average of 50 numbers is 38. If two numbers namely 45 and 55 are discarded, The average of
remaining numbers is?
Total of 50 numbers = 50 * 38 = 1900
Average of 48 numbers = 1900-(45+55) / 48
= 1800 / 48 = 37.5
16) Divide 50 in two parts so that the sum of reciprocals is (1/12), the numbers are
1) 20,30
2) 24,36
3) 28,22
4) 36,14
Sol: 1) Let the numbers be x and y then
X+y = 50. ………(i)
1/x + 1/y = 12
1/x + 1/50-x = 1/12..From (i) y = 50 – x
-- > 50-x+x/x(50-x) = 1/12
-- > x2-50x+600 = 0
-- > (x-30) (x-20) = 0
&am
Aptitude Material
1) Five years ago the average age of a family of 3 members was 27 years. A child has Been born, due
to which the average age of the family is 25 years today. What is the Present age of the child?
Sol: Average age of the family of 3 members
5 years ago = 27 years
Sum of the ages of the 3 members now
= (27 + 5) * 3 = 96 years
Average age of the family of 4 members now
= 25 years
Sum of the ages of the 4 numbers now
= 25*4 = 100 years
Age of child = 100 – 96 = 4 years
2) In a class of 20 students in an examination in Mathematics 2 students scored 100 Marks each, 3 get
zero each and the average of the rest was 40. What is the average Of the whole class?
Sol: Total marks obtained by a class of 20 students
= 2 * 100 + 3 * 0 + 15 * 40
= 200 + 600 = 800
: Average marks of whole class = 800/20 = 40
3). The greatest number, which can divide 432, 534 and 398 leaving the same remainder 7 in each, is
Required number is the H.C.F of
(432-7), (534-7) and (398-7)
i.e., H.C.F. of 425, 527, 391
Required number = 17
4) The sum of two numbers is 216 and their H.C.F is 27. The numbers are
. Let the numbers be 27a and 27b
Then 27a + 27b = 216
a+b = 8
Value of co-primes a and b are (1,7) (3,5)
: . Numbers are (27*1, 27*7) = (27,189)
5) The greatest number of 4 digits which is divisible by each one of the number 12,18,21 and 28 is
Greatest number of 4 digits is 9999
L.C.M of 12, 18, 21, 28 = 252
On dividing 9999 by 252, the remainder is 171
: . Required number is (9999-171) = 9828
6) Four prime numbers are arranged in ascending order according to their magnitude.Product of first
three is 385 and the product of last three is 1001. The greatest number is.
1) 11
2) 13
3) 17
4) 19
Sol: 2) 385) 1001(2
770
231) 385(1
231 (1
154)231(1
154
77) 154(2
154
0
Hence the product of the middle terms = 77
Greatest prime number = 1001 / 77 = 13.
7) If the square root of 55625 is 75, then
____ ____ ______
√5625 + √56.25 + √0.5625 is equal to
1) 82.25
2) 83.25
3) 80.25
4) 79.25
____ _____ ______
Sol: 2) √5625 = 75; √56.25 = 7.5; √. 5625 = .75
-- > 75+7.5+0.75 = 83.25
8) Which of the following integers has most number of divisors?
1) 176
2) 182
3) 99
4) 101
Sol: 2) 176 = 2,4,8,11,16,22,44,88
182 = 2,7,13,14,26,91
99 = 3, 9, 11, 33
101= 101
9) 10) A boy was asked to find the value of 3/8 of a sum of money. Instead of multiplying The sum by
3/8 he divided it by 3/8 and then his answer exceeded by Rs. 55. Find the Correct be x.
Sol: Let amount be x
8/3* - 3/8 * = 55
-- > 64x – 9x/24 = 55 -- > 55x/24 = 55
-- > x = 24*55/55 = 24
: . 3/8 of x = 3/8 * 24 = Rs.9
10) A boy was asked to find the value of 7/12 of a sum of money. Instead of multiplying The sum by
7/12 he divided it by 7/12 and thus his answer exceeded the correct Answer By Rs.95. Find the correct
answer.
Sol: Let sum = Rs. K
: . 12/7 k – 7k/12 = 95
-- > 144k – 49k/84 = 95 -- > k = 84
:. 7/12 k -- > 7/12 * 84 = Rs. 49
11) In a boat 25 persons were sitting. Their average weight increased one kilogram when One man
goes and a new man comes in. The weight of the new man is 70kgs. Find the Weight of the man who is
going.
Sol: Weight increased per person is 1 kg.
Total increase in weight = 25 kgs
Weight of new man is 70 kgs,
(Which means his weight is 25 kgs heavier)
The weight of the old man was 70 – 25 = 45 kgs
13) What is the greatest possible length that can be used to measure exactly the following Lengths 7m,
3m 85cm, 12m 95cm?
The length to be measured is
700cm, 385cm, 1295cm.
The required length in cm is the H.C.F of 700, 385, and 1295, which is 35 cm.
14) The product of two-digit number is 2160 and their H.C.F is 12. The numbers are
Let the number are 12a and 12b
Then 12a * 12b = 2160
ab = 15
Value of co-primes a and b are (1, 15) (3,5)
: . The two digit numbers are (3*12, 5*12)
= (36, 60)
15) The least number of 6 digits which it exactly divisible by 12, 15 and 18 is
Least number of 6 digits is 100000
L.C.M of 12, 15, 18, is 180.
On dividing 100000 by 180, the remainder is 100
. Required number = 100000 + (180-100) = 100080
16) Two third of three fifth of one fourth of a number is 24. What is 40% of that number?
1) 96
2) 72
3) 120
4) 156
Sol: 1) let the number be x, then
X of 2/3 of 3/5 of ¼ = 24
X * 2/3 * 3/5 * ¼ = 24, x = 240.
Hence 40% of 240 = 40/100 * 240 = 96
17) Which of the following has the fractions in asce
Model paper 3
1.A coffee shop blends 2 kinds of coffee,putting in 2 parts of a 33p. a gm. grade to 1
part of a 24p. a gm.If the mixture is changed to 1 part of the 33p. a gm. to 2 parts of
the less expensive grade,how much will the shop save in blending 100 gms.
a) Rs.90
b) Rs.1.00
c) Rs.3.00
d) Rs.8.00
Ans.C
2.There are 200 questions on a 3 hr examination. Among these questions are 50
Mathematics problems. It is suggested that twice as much time be spent on each
maths problem as for each other question. How many minutes should be spent on
Mathematics problems
a) 36
b) 72
c) 60
d) 100
Ans.B
3.In a group of 15,7 have studied Latin, 8 have studied Greek, and 3 have not
Studied either. How many of these studied both Latin and Greek
a) 0
b) 3
c) 4
d) 5
Ans.B
4.If 13 = 13w/(1-w), then (2w) 2 =
a) 1/4
b) 1/2
c) 1
d) 2
Ans.C
5.If a and b are positive integers and (a-b)/3.5 = 4/7, then
a) b < a
b) b > a
c) b = a
d) b >= a
Ans. A
6.In June a baseball team that played 60 games had won 30% of its game played.
After a phenomenal winning streak this team raised its average to 50% .How many
games must the team have won in a row to attain this average?
a) 12
b) 20
c) 24
d) 30
Ans. C
7.M men agree to purchase a gift for Rs. D. If three men drop out how much more
Will each have to contribute towards the purchase of the gift
a) D/(M-3)
b) MD/3
c) M/(D-3)
d) 3D/(M2-3M)
Ans. D
8.A company contracts to paint 3 houses. Mr.Brown can paint a house in 6 days
while Mr.Black would take 8 days and Mr.Blue 12 days. After 8 days Mr.Brown
goes on vacation and Mr. Black begins to work for a period of 6 days.
How many days will it take Mr.Blue to complete the contract?
Here is the attachment.