# LIC ADO Numerical Ability

5 Steps - 3 Clicks

# LIC ADO Numerical Ability

### Introduction

Scheme of Preliminary Examination for recruitment to the post of Apprentice Development Officer (ADOs) in LIC is as follows:

• Preliminary Examination consisting of the objective test will be conducted online.

• The test will have three sections (with separate timings for each section).

### Pattern

S.NO Name of the Test Number of Questions Maximum Marks Duration
1. Reasoning Ability 35 35 20 Minutes
2. Numerical Ability 35 35 20 Minutes
3. English 30 30** 20 Minutes
Total 100 70 1 hour

### Samples

Percentages:

1. A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?

A. 45%
B. 45$$\frac{5}{11}$$ %
C. 54$$\frac{6}{11}$$ %
D. 55%

Explanation:
Number of runs made by running = 110 – (3 x 4 + 8 x 6)
= 110 – (60)
= 50.
Required percentage =($$\frac{50}{110}$$ * 100) = 45$$\frac{5}{11}$$ %

2. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks were 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

Explanation:
Let their marks be (x + 9) and x.
Then, x + 9 =$$\frac{56}{100}$$(x + 9 + x)
25(x + 9) = 14(2x + 9)
3x = 99
x = 33
So, their marks are 42 and 33.

3. 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

Explanation:
Suppose originally he had x apples.
Then, (100 – 40)% of x = 420.
$$\frac{60}{100}$$*x= 420
x = ($$\frac{420*100}{60}$$) = 700

4. What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?

A. 1
B. 14
C. 20
D. 21

Explanation:
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such numbers =14
Required percentage = ($$\frac{14}{70}$$ * 100)% = 20%

Profit & Loss:

1. Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is:

A. 4$$\frac{4}{7}$$%
B. 5$$\frac{5}{11}$$%
C. 10%
D. 12%

Explanation:

Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.
Selling Price (S.P.) = Rs. 5800.
Gain = (S.P.) – (C.P.) = Rs.(5800 – 5500) = Rs. 300.
Gain % =($$\frac{300}{5500}$$ * 100) % = 5$$\frac{5}{11}$$%

2. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:

A. 15
B. 16
C. 18
D. 25

Explanation:
Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.
S.P. of x articles = Rs. 20.
Profit = Rs. (20 – x).
($$\frac{20-x}{x}$$ * 100 = 25)
2000 – 100x = 25x
125x = 2000
x = 16.

3. If the selling price is doubled, the profit triples. Find the profit percent.

A. 66$$\frac{2}{3}$$
B. 100
C. 105$$\frac{1}{3}$$
D. 120

Explanation:
Let C.P. be Rs. x and S.P. be Rs. y.
Then, 3(y – x) = (2y – x) y = 2x.
Profit = Rs. (y – x) = Rs. (2x – x) = Rs. x.
Profit % =($$\frac{x}{x}$$ * 100)% = 100%

4. 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%

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 =($$\frac{295}{420}$$ * 100)% = 70%

Time & Work:

1. A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is :

A. $$\frac{1}{4}$$
B. $$\frac{1}{10}$$
C. $$\frac{7}{15}$$
D. $$\frac{8}{15}$$

Explanation:

A’s 1 day’s work =$$\frac{1}{15}$$
B’s 1 day’s work =$$\frac{1}{20}$$
(A + B)’s 1 day’s work =($$\frac{1}{15}$$ + $$\frac{1}{20}$$) = $$\frac{7}{60}$$
(A + B)’s 4 day’s work = ($$\frac{7}{60}$$ * 4 ) = $$\frac{7}{15}$$
Remaining work =(1- $$\frac{7}{15}$$) = $$\frac{8}{15}$$

2. A can lay railway track between two given stations in 16 days and B can do the same job in 12 days. With the help of C, they did the job in 4 days only. Then, C alone can do the job in:

A. 9$$\frac{1}{5}$$ Days
B. 9$$\frac{2}{5}$$ Days
C. 9$$\frac{3}{5}$$ Days
D. 10

Explanation:
(A + B + C)’s 1 day’s work = $$\frac{1}{4}$$
A’s 1 day’s work =$$\frac{1}{16}$$
B’s 1 day’s work =$$\frac{1}{12}$$
C’s 1 day’s work =$$\frac{1}{4}$$ – ($$\frac{1}{16}$$ + $$\frac{1}{12}$$) = $$\frac{5}{48}$$
So, C alone can do the work in $$\frac{48}{5}$$ = 9$$\frac{3}{5}$$ Days

3. 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

Explanation:
A’s 2 day’s work =($$\frac{1}{20}$$ * 2) = $$\frac{1}{10}$$
(A + B + C)’s 1 day’s work =($$\frac{1}{20}$$ +$$\frac{1}{30}$$ = $$\frac{1}{60}$$)=$$\frac{1}{10}$$
Work done in 3 days =($$\frac{1}{10}$$ + $$\frac{1}{10}$$) = $$\frac{1}{5}$$Days
Now,$$\frac{1}{5}$$ work is done in 3 days.
Whole work will be done in (3 x 5) = 15 days.

4.A is thrice as good as workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:

A. 20 days
B. 22$$\frac{1}{2}$$ days
C. 25 days
D. 30 days

Explanation:
Ratio of times taken by A and B = 1 : 3.
The time difference is (3 – 1) 2 days while B takes 3 days and A takes 1 day.
If the difference of time is 2 days, B takes 3 days.
If the difference of time is 60 days, B takes ($$\frac{3}{2}$$ * 60 ) = 90 Days
So, A takes 30 days to do the work.
A’s 1 day’s work =$$\frac{1}{30}$$
B’s 1 day’s work =$$\frac{1}{90}$$
(A + B)’s 1 day’s work =($$\frac{1}{30}$$ + $$\frac{1}{90}$$) = $$\frac{2}{45}$$
A and B together can do the work in $$\frac{45}{2}$$ =22$$\frac{1}{2}$$ days

Time & Distance:

1. A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

A. 3.6
B. 7.2
C. 8.4
D. 10

Explanation:
Speed = $$\frac{600}{5*60}$$m/sec.
= 2 m/sec.
Converting m/sec to km/hr
= 2*$$\frac{18}{5}$$km/hr.
=7.2 km/hr.

2. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is:

A. 50 km
B. 56 km
C. 70 km
D. 80 km

Explanation:
Let the actual distance travelled be x km.
Then,$$\frac{x}{10}$$ = $$\frac{x+20}{14}$$
14x = 10x + 200
4x = 200
x = 50 km.

3. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?

A. 9
B. 10
C. 12
D. 20

Explanation:
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km =($$\frac{9}{54}$$ * 60)
= 10 min.

4. The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:

A. 70 km/hr
B. 75 km/hr
C. 84 km/hr
D. 87.5 km/hr

Explanation:
Let the speed of two trains be 7x and 8x km/hr.
Then, 8x =$$\frac{400}{44}$$ = 100
x= $$\frac{100}{8}$$ = 12.5
Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.

Decimal Fractions:

1. What decimal of an hour is a second ?

A. .0025
B. .0256
C. .00027
D. .000126

Explanation:
Required decimal = $$\frac{1}{60*60}$$
= $$\frac{1}{3600}$$
= .00027

2. $$\frac{.009}{?}$$ = .01

A. .0009
B. .09
C. .9
D. 9

Explanation:
Let $$\frac{.009}{x}$$ = .01
Then x =$$\frac{.009}{.01}$$ = .9

3. The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to:

A. 0.02
B. 0.2
C. 0.04
D. 0.4

Explanation:
Given expression = $$(11.98)^{2}$$+ $$(0.02)^{2}$$ + 11.98 x x.
For the given expression to be a perfect square, we must have
11.98 x x = 2 x 11.98 x 0.02 or x = 0.04

4. 3889 + 12.952 – ? = 3854.002

A. 47.095
B. 47.752
C. 47.932
D. 47.95

Explanation:
Let 3889 + 12.952 – x = 3854.002.
Then x = (3889 + 12.952) – 3854.002
= 3901.952 – 3854.002
= 47.95.

Ratio & Propotion:

1. The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?

A. 8 : 9
B. 17 : 18
C. 21 : 22
D. Cannot be determined

Explanation:
Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
= ($$\frac{120}{100}$$*7x) and ($$\frac{110}{100}$$*8x)
= ($$\frac{42x}{5}$$) and ($$\frac{44x}{5}$$)
The required ratio =($$\frac{42x}{5}$$):($$\frac{44x}{5}$$) = 21 : 22.

2. 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

Explanation:
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then,$$\frac{2x + 4000}{3x + 4000}$$ = $$\frac{40}{57}$$
57(2x + 4000) = 40(3x + 4000)
6x = 68,000
3x = 34,000
Sumit’s present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.

3. If 0.75 : x :: 5 : 8, then x is equal to:

A. 1.12
B. 1.2
C. 1.25
D. 1.30

Explanation:
x * 5 = (0.75 x 8)
x = $$\frac{6}{5}$$
x = 1.20

4. The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5 : 8, then the second number is:

A. 20
B. 30
C. 48
D. 58

Explanation:
Let the three parts be A, B, C. Then,
A : B = 2 : 3 and B : C = 5 : 8 = (5 *$$\frac{3}{5}$$) : (8 * $$\frac{3}{5}$$) = 3 : $$\frac{24}{5}$$
A : B : C = 2 : 3 :$$\frac{24}{5}$$
= 10 : 15 : 24
B = 98 * $$\frac{15}{49}$$ = 30

Averages:

1. 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

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.

2. 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

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.

3. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:

A. 35 years
B. 40 years
C. 50 years
D. None of these

Explanation:
Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.
Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years.
Husband’s present age = (90 – 50) years = 40 years.

4. In Arun’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother doest not agree with Arun and he thinks that Arun’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?

A. 67 kg.
B. 68 kg.
C. 69 kg.
E. None of these

Explanation:
Let Arun’s weight by X kg.
According to Arun, 65 < X < 72
According to Arun's brother, 60 < X < 70.
According to Arun's mother, X <= 68
The values satisfying all the above conditions are 66, 67 and 68.
Required average = ($$\frac{66+67+68}{3}$$) = ($$\frac{201}{3}$$) = 68kgs.

LCM & HCF:

1. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

A. 4
B. 7
C. 9
D. 13

Explanation:
Required number = H.C.F. of (91 – 43), (183 – 91) and (183 – 43)
= H.C.F. of 48, 92 and 140 = 4.

2. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A. 276
B. 299
C. 322
D. 345

Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.

3. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A. 4
B. 10
C. 15
D. 16

Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together $$\frac{30}{2}$$ + 1= 16 times.

4. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

A. 4
B. 5
C. 6
D. 8

Explanation:
N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Area:

1. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

A. 15360
B. 153600
C. 30720
D. 307200

Explanation:
Perimeter = Distance covered in 8 min. =($$\frac{12000}{60}$$ * 8) =1600 m.
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320)$$m^{2}$$ = 153600 $$m^{2}$$.

2. An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

A. 2%
B. 2.02%
C. 4%
D. 4.04%

Explanation:
100 cm is read as 102 cm.
A1 = (100 x 100)$$cm^{2}$$ and A2 (102 x 102) $$cm^{2}$$.
(A2 – A1) = $$(102)^{2}$$) – ($$(100)^{2}$$
= (102 + 100) x (102 – 100)
= 404 $$cm^{2}$$
Percentage error =($$\frac{404}{100*100}$$ * 100) = 4.04%

3. The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

A. 16 cm
B. 18 cm
C. 24 cm
E. None of these

Explanation:
$$\frac{2(l + b)}{b}$$ = $$\frac{5}{1}$$
2l + 2b = 5b
3b = 2l
b = $$\frac{2}{3}$$ l
Then, Area = 216 $$cm^{2}$$
l x b = 216
l * $$\frac{2}{3}$$ l = 216
$$l^{2}$$ = 324
l = 18 cm.

4. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A. 2.91 m
B. 3 m
C. 5.82 m
D. None of these

Explanation:
Area of the park = (60 x 40)$$m^{2}$$ = 2400 $$m^{2}$$
Area of the lawn = 2109 $$m^{2}$$ .
Area of the crossroads = (2400 – 2109) $$m^{2}$$ = 291 $$m^{2}$$ .
Let the width of the road be x metres. Then,
60x + 40x – $$x^{2}$$ = 291
$$x^{2}$$ – 100x + 291 = 0
(x – 97)(x – 3) = 0
x = 3.

Simple Interest:

1. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

A. Rs. 650
B. Rs. 690
C. Rs. 698
D. Rs. 700

Explanation:
S.I. for 1 year = Rs. (854 – 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 – 117) = Rs. 698.

2. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?

A. Rs. 4462.50
B. Rs. 8032.50
C. Rs. 8900
D. Rs. 8925
E. None of these

Explanation:
Principal = Rs.($$\frac{100 x 4016.25}{9 x 5}$$)
= Rs.($$\frac{401625}{45}$$)
= Rs. 8925.

3. How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

A. 3.5 years
B. 4 years
C. 4.5 years
D. 5 years

Explanation:
Time = $$\frac{100 x 81}{450 x 4.5}$$years = 4 years.

4. Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

A. 3.6
B. 6
C. 18
D. Cannot be determined
E. None of these

Explanation:
Let rate = R% and time = R years.
Then,$$\frac{1200 x R x R }{100}$$ = 432
12 $$R^{2}$$ = 432
$$R^{2}$$ = 36
R = 6.

Compound Interest:

1. What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.?

A. Rs. 9000.30
B. Rs. 9720
C. Rs. 10123.20
D. Rs. 10483.20
E. None of these

Explanation:
Amount = Rs.[25000 * (1 +$$\frac{12}{100})^{3}]$$
= Rs.25000 *$$\frac{28}{25}$$ * $$\frac{28}{25}$$ * $$\frac{28}{25}$$
= Rs. 35123.20
C.I. = Rs. (35123.20 – 25000) = Rs. 10123.20

2. At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?

A. 6%
B. 6.5%
C. 7%
D. 7.5%

Explanation:
Let the rate be R% p.a.
Then, 1200 x[1 + $$\frac{R}{100}]^{2}$$ = 1348.32
[1 + $$\frac{R}{100}]^{2}$$ = $$\frac{11236}{10000}$$
[1 + $$\frac{R}{100}]^{2}$$ = [$$\frac{R}{100}]^{2}$$
R = 6%

3. The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:

A. 3
B. 4
C. 5
D. 6

Explanation:
P[1 + $$\frac{20}{100}]^{n}$$ > 2P
=$$\frac{6}{5}^{n}$$> 2.
Now,($$\frac{6}{5}^{n}$$ * $$\frac{6}{5}^{n}$$ * $$\frac{6}{5}^{n}$$ * $$\frac{6}{5}^{n}$$)> 2.
So, n = 4 years.

4. Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?

A. Rs. 8600
B. Rs. 8620
C. Rs. 8820
D. None of these

Amount = Rs.8000 x[1 + $$\frac{5}{100}]^{2}$$
= Rs.8000 x $$\frac{21}{200}$$ * $$\frac{21}{200}$$