L&T Infotech Aptitude Questions and Solutions

Question 1. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
A. Rs. 4991
B. Rs. 5991
C. Rs. 6001
D. Rs. 6991

Answer: Option A

Explanation:

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.

Required sale = Rs. [ (6500 x 6) – 34009 ]

= Rs. (39000 – 34009)

= Rs. 4991.

Question 2. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?
A. 23 years
B. 24 years
C. 25 years
D. None of these

Answer: Option A

Explanation:

Let the average age of the whole team by x years.

11x – (26 + 29) = 9(x -1)

11x – 9x = 46

2x = 46

x = 23.

So, average age of the team is 23 years.

Question 3. The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:
A. 3500
B. 4000
C. 4050
D. 5000

Answer: Option B

Explanation:

Let P, Q and R represent their respective monthly incomes. Then, we have:

P + Q = (5050 x 2) = 10100 …. (i)

Q + R = (6250 x 2) = 12500 …. (ii)

P + R = (5200 x 2) = 10400 …. (iii)

Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 …. (iv)

Subtracting (ii) from (iv), we get P = 4000.

P’s monthly income = Rs. 4000.

Question 1. A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
A. 12 days
B. 15 days
C. 16 days
D. 18 days

Answer: Option B

Explanation:
A’s 2 day’s work = (1/20) x 2 = (1/10) .
(A + B + C)’s 1 day’s work = (1/20) + (1/30) + (1/60) = (6/60) = (1/10) .
Work done in 3 days = (1/10) + (1/10) = (1/5) .
Now, (1/5) work is done in 3 days.

Whole work will be done in (3 x 5) = 15 days.

Question 2. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
A. 4 days
B. 5 days
C. 6 days
D. 7 days

Answer: Option A

Explanation:

Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.
Then, 6x + 8y = 1(/10) and 26x + 48y = (1/2) .
Solving these two equations, we get : x = (1/100) and y = (1/200) .
(15 men + 20 boy)’s 1 day’s work = (15/100) + (20/200) = (1/4) .

15 men and 20 boys can do the work in 4 days.

Question 3. A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?
A. 8 hours
B. 10 hours
C. 12 hours
D. 24 hours

Answer: Option C

Explanation:
A’s 1 hour’s work = 1/4 ;
(B + C)’s 1 hour’s work = 1/3 ;
(A + C)’s 1 hour’s work = 1/2 .
(A + B + C)’s 1 hour’s work = ( 1/4 + 1 /3 ) = 7 /12 .
B’s 1 hour’s work = ( 7 – 1 ) = 1 .
12 2 12

Therefore B alone will take 12 hours to do the work.

Question 4. A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in:
A. 15 days
B. 20 days
C. 25 days
D. 30 days

Answer: Option C

Explanation:
(A + B)’s 1 day’s work = (1/10)
C’s 1 day’s work = (1/50)
(A + B + C)’s 1 day’s work = (1/10) + (1/50) = (6/50) = (3/25) . …. (i)

A’s 1 day’s work = (B + C)’s 1 day’s work …. (ii)
From (i) and (ii), we get: 2 x (A’s 1 day’s work) = 3/25
A’s 1 day’s work = (3/50) .
B’s 1 day’s work (1/10) – (3/50) = (2/50) = (1/25) .

So, B alone could do the work in 25 days.

Question 1. In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?
A. 30%
B. 70%
C. 100%
D. 250%

Answer: Option B

Explanation:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 – 125) = Rs. 295.
Required percentage = (295/420) x 100 % = (1475/21) % = 70% (approximately).

Question 2. The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?
A. Rs. 2000
B. Rs. 2200
C. Rs. 2400
D. Data inadequate

Answer: Option A

Explanation:

Let C.P. be Rs. x.
Then, ((1920 – x)/x) * 100 = ((x – 1280 )/x)* 100

1920 – x = x – 1280

2x = 3200

x = 1600
Required S.P. = 125% of Rs. 1600 = Rs. (125/100) * 1600 = Rs 2000.

Question 3. Sam purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?
A. 3.5
B. 4.5
C. 5.6
D. 6.5

Answer: Option C

Explanation:
Cost Price of 1 toy = Rs. (375/12) = Rs. 31.25

Selling Price of 1 toy = Rs. 33

So, Gain = Rs. (33 – 31.25) = Rs. 1.75
Profit % = (1.75/31.25) x 100 % = (28/5) % = 5.6%

Question 4. A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:
A. No profit, no loss
B. 5%
C. 8%
D. 10%
E. None of these

Answer: Option B

Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.
Gain = (80/1600) x 100 % = 5%.

Question1. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:
A. 39, 30
B. 41, 32
C. 42, 33
D. 43, 34

Answer: Option C

Explanation:

Let their marks be (x + 9) and x.
Then, x + 9 =( 56/100) (x + 9 + x)

25(x + 9) = 14(2x + 9)

3x = 99

x = 33

So, their marks are 42 and 33.

Question 2. A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:
A. 588 apples
B. 600 apples
C. 672 apples
D. 700 apples

Answer: Option D

Explanation:

Suppose originally he had x apples.

Then, (100 – 40)% of x = 420.
60 x x = 420
100
x = 420 x 100 = 700.
60

Question 3. In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is of the number of students of 8 years of age which is 48. What is the total number of students in the school?
A. 72
B. 80
C. 120
D. 150
E. 100

Answer: Option E

Explanation:

Let the number of students be x. Then,

Number of students above 8 years of age = (100 – 20)% of x = 80% of x.
80% of x = 48 + 2 of 48
3
80 x = 80
100

x = 100.

Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
A. Rs. 200
B. Rs. 250
C. Rs. 300
D. None of these

Answer: Option B

Explanation:

Let the sum paid to Y per week be Rs. z.

Then, z + 120% of z = 550.
z + 120 z = 550
100
11 z = 550
5
z = 550 x 5 = 250.
11

Question 1. The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
A. 3
B. 4
C. 9
D. Cannot be determined
E. None of these

Answer: Option B

Explanation:

Let the ten’s digit be x and unit’s digit be y.

Then, (10x + y) – (10y + x) = 36

9(x – y) = 36

x – y = 4.

Question 2. The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?
A. 69
B. 78
C. 96
D. Cannot be determined
E. None of these

Answer: Option D

Explanation:

Let the ten’s digit be x and unit’s digit be y.

Then, x + y = 15 and x – y = 3 or y – x = 3.

Solving x + y = 15 and x – y = 3, we get: x = 9, y = 6.

Solving x + y = 15 and y – x = 3, we get: x = 6, y = 9.

So, the number is either 96 or 69.

Hence, the number cannot be determined.

Question 3. In a two-digit, if it is known that its unit’s digit exceeds its ten’s digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:
A. 24
B. 26
C. 42
D. 46

Answer: Option A

Explanation:

Let the ten’s digit be x.

Then, unit’s digit = x + 2.

Number = 10x + (x + 2) = 11x + 2.

Sum of digits = x + (x + 2) = 2x + 2.

(11x + 2)(2x + 2) = 144

22×2 + 26x – 140 = 0

11×2 + 13x – 70 = 0

(x – 2)(11x + 35) = 0

x = 2.

Hence, required number = 11x + 2 = 24.

Question 4. Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.
A. 3
B. 10
C. 17
D. 20

Answer: Option A

Explanation:

Let the number be x.
Then, x + 17 = 60
x

x2 + 17x – 60 = 0

(x + 20)(x – 3) = 0

x = 3.

Question 1. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:
A. 50 km
B. 56 km
C. 70 km
D. 80 km

Answer: Option A

Explanation:

Let the actual distance travelled be x km.

Then, x/10 = (x + 20)/14

14x = 10x + 200

4x = 200

x = 50 km.

Question 2. A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
A. 100 kmph
B. 110 kmph
C. 120 kmph
D. 130 kmph

Answer: Option C

Explanation:

Let speed of the car be x kmph.

Then, speed of the train = (150/100) x = (3/2) x kmph.

(75/x) – (75/(3/2)x) = (125/10*60)

(75/x) – (50/x) = (5/24)

x = (25 *24)/5 = 120 kmph.

Question 3. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:
A. 1 hour
B. 2 hours
C. 3 hours
D. 4 hours

Answer: Option A

Explanation:

Let the duration of the flight be x hours.
Then, (600/x) – (600/(x+(1/2)) = 200
(600/x) – (1200/(2x+1)) = 200

x(2x + 1) = 3

2×2 + x – 3 = 0

(2x + 3)(x – 1) = 0

x = 1 hr. [neglecting the -ve value of x]

Question 4. Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
A. 8 kmph
B. 11 kmph
C. 12 kmph
D. 14 kmph

Answer: Option C

Explanation:

Let the distance travelled by x km.
Then, x/10 – x/15 = 2

3x – 2x = 60

x = 60 km.
Time taken to travel 60 km at 10 km/hr = (60/10) hrs = 6 hrs.

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
Required speed = 60 kmph. = 12 kmph.

Question 1. In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
A. 75 cu. m
B. 750 cu. m
C. 7500 cu. m
D. 75000 cu. m

Answer: Option B

Explanation:

1 hectare = 10,000 m2

So, Area = (1.5 x 10000) m2 = 15000 m2.
Depth = (5/100) m = (1/20) m.
Volume = (Area x Depth) = 15000 x (1/20) m3 = 750 m3.

Question 2. 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:
A. 84
B. 90
C. 168
D. 336

Answer: Option A

Explanation:

Let the length of the wire be h.
Radius = 1/2 mm = 1/20 cm. Then,
22/7 * 1/20 * 1/20 * h = 66.
h = (66 x 20 x 20 x 7)/22 = 8400 cm = 84 m.

Question 3. A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is:
A. 90 cm
B. 1 dm
C. 1 m
D. 1.1 cm

Answer: Option B

Explanation:

Let the thickness of the bottom be x cm.

Then, [(330 – 10) * (260 – 10) * (110 – x)] = 8000 * 1000

320 * 250 * (110 – x) = 8000 * 1000
(110 – x) = (8000 * 1000)/(320 * 250) = 100

x = 10 cm = 1 dm.

Question 4. A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is:
A. 12 kg
B. 60 kg
C. 72 kg
D. 96 kg

Answer: Option B

Explanation:
Volume of water displaced = (3 * 2 * 0.01) m³

= 0.06 m³.
Mass of man = Volume of water displaced * Density of water

= (0.06 * 1000) kg

= 60 kg.

Question 1. Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
A. 2 : 5
B. 3 : 5
C. 4 : 5
D. 6 : 7

Answer: Option C

Explanation:

Let the third number be x.

Then, first number = 120% of x = (120/100)x = (6/5)x

Second number = 150% of x = (150/100)x = (3/2)x

Ratio of first two numbers = (6/5)x : (3/2)x = 12x : 15x = 4 : 5.

Question 2. Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
A. 2 : 3 : 4
B. 6 : 7 : 8
C. 6 : 8 : 9
D. None of these

Answer: Option A

Explanation:

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.

Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).

(140/100) * 5x , (150/100) * 7x and (175/100) * 8x

7x, (21/2)x and 14x.

The required ratio = 7x : (21/2)x : 14x

14x : 21x : 28x

2 : 3 : 4.

Question 3. Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit’s salary?
A. Rs. 17,000
B. Rs. 20,000
C. Rs. 25,500
D. Rs. 38,000

Answer: Option D

Explanation:

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, (2x + 4000)/(3x + 4000) = 40/57

57(2x + 4000) = 40(3x + 4000)

6x = 68,000

3x = 34,000

Sumit’s present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.

Question 4. In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?
A. 50
B. 100
C. 150
D. 200

Answer: Option C

Explanation:

Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.
Then, sum of their values = Rs. (25/100)x + (10 * 2x)/100 + (5 * 3x)/100 = Rs. (60/100)x
(60/100)x = 30

= x = (30 * 100)/60 = 50.

Hence, the number of 5 p coins = (3 x 50) = 150

Question 1. An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is:
A. 21.6 m

B. 23.2 m
C. 24.72 m

D. None of these

Answer: Option A

Explanation:

Let AB be the observer and CD be the tower.

Draw BE perpendicular to CD.

Then, CE = AB = 1.6 m,

BE = AC = 20*3^1/2 m.

(DE/BE) = tan 30° = 1/3^1/2

DE = (20*3^1/2)/(3^1/2 )m = 20 m.

CD = CE + DE = (1.6 + 20) m = 21.6 m.

Question 2. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 173 m
B. 200 m
C. 273 m
D. 300 m

Answer: Option C

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100 m, angle ACB = 30° and angle ADB = 45°.
(AB/AC) = tan 30° = 1/3^1/2 AC = AB x 3^1/2 = 100*3^1/2 m.
(AB/AD) = tan 45° = 1 AD = AB = 100 m.
CD = (AC + AD) = (100*3^1/2 + 100) m

= 100(3^1/2 + 1)

= (100 x 2.73) m

= 273 m.

Question 1. Look at this series: 2, 1, (1/2), (1/4), … What number should come next?
A. (1/3)
B. (1/8)
C. (2/8)
D. (1/16)

Answer: Option B

Explanation:

This is a simple division series; each number is one-half of the previous number.

In other terms to say, the number is divided by 2 successively to get the next result.

4/2 = 2
2/2 = 1
1/2 = 1/2
(1/2)/2 = 1/4
(1/4)/2 = 1/8 and so on.

Question 2. Look at this series: 21, 9, 21, 11, 21, 13, 21, … What number should come next?
A. 14
B. 15
C. 21
D. 23

Answer: Option B

Explanation:
In this alternating repetition series, the random number 21 is interpolated every other number into an otherwise simple addition series that increases by 2, beginning with the number 9.

Question 3. Look at this series: 3, 4, 7, 8, 11, 12, … What number should come next?
A. 7
B. 10
C. 14
D. 15

Answer: Option D

Explanation:
This alternating addition series begins with 3; then 1 is added to give 4; then 3 is added to give 7; then 1 is added, and so on.

Question 4. Look at this series: 8, 6, 9, 23, 87 , … What number should come next?
A. 128
B. 226
C. 324
D. 429

Answer: Option D

Explanation:

8 x 1 – 2 = 6
6 x 2 – 3 = 9
9 x 3 – 4 = 23
23 x 4 – 5 = 87
87 x 5 – 6 = 429 …

Question 1. A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?
A. 10
B. 20
C. 21
D. 25

Answer: Option C

Explanation:

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = 7x – (7/12) * 9 litres = 7x – 21/4 litres.

Quantity of B in mixture left = 5x – (5/12) x 9 litres = 5x – 15/4 litres.

(7x – 21/4)/((5x – 15/4) + 9)
= 7/9

28x – 21 = 7
20x + 21 9

252x – 189 = 140x + 147

112x = 336

x = 3.

So, the can contained 21 litres of A.

Question 2. A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
A. 26.34 litres
B. 27.36 litres
C. 28 litres
D. 29.16 litres

Answer: Option D

Explanation:
Amount of milk left after 3 operations =[ 40 (1 – 4/40) ³] litres
= 40 * (9/10) * (9/10) * (9/10) = 29.16 litres.

Question 3. 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65. How much wine did the cask hold originally?
A. 18 litres
B. 24 litres
C. 32 litres
D. 42 litres

Answer: Option B

Explanation:

Let the quantity of the wine in the cask originally be x litres.

Then, quantity of wine left in cask after 4 operations = [x( 1 – 8/x)⁴ ] litres.

[(x(1 – (8/x))⁴)/x ]= 16/81

(1 – 8/x)⁴ = (2/3) ⁴

(x – 8)/x = 2/3

3x – 24 = 2x

x = 24.

Question 1. A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?
A. 3.6%
B. 4.5%
C. 5%
D. 6%
E. None of these

Answer: Option E

Explanation:

Let the original rate be R%. Then, new rate = (2R)%.

Note:
Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. year(s).

(725 * R * 1)/100 + (362.50 * 2R * 1)/(100*3) = 33.50

(2175 + 725) R = 33.50 * 100 * 3

(2175 + 725) R = 10050

(2900)R = 10050

R = 10050/2900 = 3.46

Original rate = 3.46%

Question 2. A certain amount earns simple interest of Rs. 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?
A. Rs. 35
B. Rs. 245
C. Rs. 350
D. Cannot be determined
E. None of these

Answer: Option D

Explanation:

We need to know the S.I., principal and time to find the rate.

Since the principal is not given, so data is inadequate.

Principal = Rs. (815 – 117) = Rs. 698.

The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is:
A. 2
B.
2 1
2
C. 3
D. 4

Answer: Option A

Explanation:

Amount = Rs. (30000 + 4347) = Rs. 34347.

Let the time be n years.

Then, 30000 (1 + 7/100) ⁿ = 34347

(107/100) ⁿ = (34347/300001) = (11449/10000) = (107/100) ²

n = 2 years.

Question 4. If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same at the same rate and for the same time?
A. Rs. 51.25
B. Rs. 52
C. Rs. 54.25
D. Rs. 60

Answer: Option A

Explanation:

Sum = Rs. (50 * 100)/(2 /* 5) = Rs. 500.

Amount
= Rs. [500 x (1 + 5/100 )²]

= Rs. [500 * (21/20) * (21/20)]
= Rs. 551.25

C.I. = Rs. (551.25 – 500) = Rs. 51.25

Question 1. The reflex angle between the hands of a clock at 10.25 is:
A.180°
B.192 (1/2) °
C.195°
D.197 (1/2) °

Answer: Option D

Explanation:

Angle traced by hour hand in (125/12) hrs = ((360/12) * (125/12)) °= 312 (1/2) ° .

Angle traced by minute hand in 25 min = (360/60) * 25 ° = 150°.

Reflex angle = 360° – 312 (1/2) – 150 ° = 360° – 162 (1/2) ° = 197 (1/2)° .

Question 2. How many times are the hands of a clock at right angle in a day?
A. 22
B. 24
C. 44
D. 48

Answer: Option C

Explanation:

In 12 hours, they are at right angles 22 times.

In 24 hours, they are at right angles 44 times.

What was the day of the week on 17th June, 1998?
A. Monday
B. Tuesday
C. Wednesday
D. Thursday

Answer: Option C

Explanation:

17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3) 1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan. Feb. March April May June
(31 + 28 + 31 + 30 + 31 + 17) = 168 days

168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given day is Wednesday.

Question 4. How many days are there in x weeks x days?
A. 7x²
B. 8x
C. 14x
D. 7

Answer: Option B

Explanation:

x weeks x days = (7x + x) days = 8x days.

Question 1. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
A. 1/2
B. 3/4
C.
3
8
D.
5
16

Answer: Option B

Explanation:

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),
(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),
(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(E) = 27.

P(E) = n(E)/n(S) = 27/36 = 3/4 .

Question 2. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
A.21/46
B.25/117
C.1/50
D.3/25

Answer: Option A

Explanation:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25
= 25_C3 `

= (25 x 24 x 23)/(3 x 2 x 1)

= 2300.

n(E) = (10_C1 x 15_C2)

= [10 x ((15 x 14)/(2 x 1))]

= 1050.

P(E) = n(E)/n(S) = 1050/ 2300 = 21/46

Question 3. Two dice are tossed. The probability that the total score is a prime number is:
A. 1/6
B. 5/12
C. 1/2
D. 7/9

Answer: Option B

Explanation:

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),
(5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

P(E) = n(E)/n(S) = 15 /36 = 5/12 .

Question 1. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
A. 4 years
B. 8 years
C. 10 years
D. None of these

Answer: Option A

Explanation:

Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50

5x = 20

x = 4.

Age of the youngest child = x = 4 years.

Question 2. Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?
A. 24
B. 27
C. 40
D. Cannot be determined
E. None of these

Answer: Option A

Explanation:

Let the present ages of Sameer and Anand be 5x years and 4x years respectively.

Then, (5x + 3)/(4x + 3) = 11/9

9(5x + 3) = 11(4x + 3)

45x + 27 = 44x + 33

45x – 44x = 33 – 27

x = 6.

Anand’s present age = 4x = 24 years.

Question 3. At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun’s age will be 26 years. What is the age of Deepak at present ?
A. 12 years
B. 15 years
C. 19 and half
D. 21 years

Answer: Option B

Explanation:

Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,

4x + 6 = 26 4x = 20

x = 5.

Deepak’s age = 3x = 15 years.

Question 4. The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
A. 8, 20, 28
B. 16, 28, 36
C. 20, 35, 45
D. None of these

Answer: Option B

Explanation:

Let their present ages be 4x, 7x and 9x years respectively.

Then, (4x – 8) + (7x – 8) + (9x – 8) = 56

20x = 80

x = 4.

Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.