Daily GK Update 17th September 2017

1. PV Sindhu beats World Championship winner to win Korea Open:

• Indian shuttler PV Sindhu defeated World Championships gold winner, Japan’s Nozomi Okuhara, to become the first Indian to lift the Korea Open Superseries title .
• She secured a 22-20, 11-21, 21-18 win against Okuhara.
• Sindhu has now won three international titles this year, after claiming the Syed Modi International and India Open titles. This was world number four Sindhu’s fourth win over Okuhara in eight encounters.

2. Chris Gayle becomes first batsman to hit 100 sixes in T20Is:

• Windies’ cricketer Chris Gayle became the first batsman to hit 100 sixes in T20I cricket on Saturday.
• Gayle achieved the feat in his 49th T20I innings during the one-off T20I against England at Chester-le-Street. The 37-year-old West Indian has hit 439 sixes in international cricket and is only behind former Pakistani all-rounder Shahid Afridi, who has hit 476 sixes.

3. PM Modi inaugurates country’s biggest dam, Sardar Sarovar Dam, Called ‘Gujarat’s Lifeline’:

• Prime Minister Narendra Modi today launched the Sardar Sarovar Dam – the world’s second-biggest – on the river Narmada in Gujarat.
• The 1.2-km-long dam, which is 163 metres deep, will irrigate over 18 lakh hectares in the state, according to officials; the water from Narmada will flow into over 9,000 villages through a canal network, Sardar Sarovar Nigam Limited (SSNL).
• After inaugurating the dam, PM Modi will head to Sadhu Bet, an island in the river Narmada where a 182-metre-tall statue of Sardar Vallabhbhai Patel – called the ‘Statue of Unity’.

4. India, Japan sign Open Sky Agreement:

• India and Japan today signed an agreement that will allow their airlines to operateunlimited number of flights between the two countries.
• It opens skies between India and Japan ie. Indian and Japanese carriers can mount now unlimited number of flights to the selected cities of each other’s countries.
• The National Civil Aviation Policy, 2016, allows the government to enter into an ‘open sky’ air services agreement on a reciprocal basis with SAARC nations as well as countries beyond a 5,000 kilometre radius from New Delhi.

5. Advertising Club appoints Vikram Sakhuja as president:

• The Advertising Club has elected Vikram Sakhuja, group CEO – Media and OOH, Madison Communications, as its new president.
• Sakhuja will take over from Raj Nayak, who led the Advertising Club for two consecutive terms.

6. Only Marshal of IAF, the hero of 1965, Arjan Singh Passed Away:

• The sole Marshal of the Indian Air Force, Arjan Singh, passed away at the Army’s Research and Referral Hospital on Saturday. He was 98 years old.
• Arjan Singh had the honour of leading the fly-past of more than a hundred IAF aircraft over the Red Fort on India’s first Independence Day, August 15, 1947.
• In 2002, on the occasion of Republic Day, he was granted the honorary rank of Marshal, the highest military rank attainable, which before him only two Army chiefs, K M Carriappa and Sam Manekshaw, had achieved.

7. Amazon Ties up With Bank of Baroda To Offer Micro-Loans To Sellers:

• Amazon has tied up with Bank of Baroda for the loans, which can go up to Rs. 25 lakh.
• The Indian arm of the US e-commerce giant, Amazon plans to provide unsecured micro-loans, as low as 1 lakh, at interest rates as low as 10.45-11 per cent at a short span of 3-5 days. With the Bank of Baroda tie-up, the company wants to cater to every strata of sellers.

8. Shabana Azmi to be honoured as ‘Icon of Indian Cinema’ at Jagran Film Festival:

• Actres Shaban Azmi will be honoured at the closing ceremony of the Mumbai edition of the eighth Jagran Film Festival next week.
• She will honored as an ‘Icon of Indian Cinema’.The ceremony will take place on September 24.

9. SLC bans Chamara Silva for 2 yrs on fixing charges:

• Former Sri Lanka batsman Chamara Silva has been handed a two-year ban from cricket-related activities for breaching the spirit of the game.
• The announcement came after a seven-month inquiry into events of the third day of the Panadura Cricket Club v Kalutara Physical Culture Club match played from January 23 to 25.

Quantitative Aptitude

Time And Distance Quiz

1. Excluding stoppage, 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

(e) None of these 

Sol.

1. (b); Due to stoppages, it cover 9 km less.

Time taken to cover 9 km=[(9/54)X60)min=10 min

2. Two men starting from the same place walk at the rate of 4 kmph and 4.5 kmph respectively. 
What time will they take to be 8.5 km apart, if they walk in the same direction?

(a) 1 hour

(b) 4 hrs 15 min

(c) 17 hrs

(d) 2 hrs. 55 min.

(e) None of these 

Sol. 

2. (c); To be 0.5 km apart, they take 1 hour

To be 8.5 km apart, they would take=[(1/0.5)X8.5]hrs=17 hrs.

3. Two cyclists start from the same place in opposite directions. One goes towards north at 18 kmph and the other goes towards south at 20 kmph. 
What time will they take to be 47.5 km apart?

(a) 2(1/4) hrs

(b) 1(1/4) hrs

(c) 2 hrs 23 min

(d) 23(1/4) hrs

(e) None of these 

Sol.

3. (b); To be (18+20) km apart, they take 1 hour

To be 47.5 km apart, they would take=[(1/38)X47.5]hrs=1(1/4) hrs.

4. A man covers a certain distance on scooter. Had he moved 3 kmph faster, he would have taken 40 min. 
less. If he had moved 2 kmph slower, he would have taken 40 min. more. The distance (in km) is:

(a) 20

(b) 36

(c) 37.5

(d) 40

(e) None of these 

Sol.

4. (d); Let distance=x km & usual rate=y km/hr.

(x/y)-x/(y+3)=40/60

2y(y+3)=9x…………….(i)

And, x/(y-2)-x/y=40/60

y(y-2)=3x……….(ii)

On dividing (i) by (ii), we get: (2(y+3))/(y-2)=3

y=12

Putting y=12 in (i), we get : x=40

5. If a train runs at 40 kmph, it reaches its destination late by 11 minutes but if it runs at 50 kmph, it is late by 5 minutes only. 
The correct time for the train to complete its journey is :

(a) 13 min.

(b) 15 min.

(c) 19 min. 

(d) 21 min.

(e) None of these 

Sol.

5. (c); Let the correct time to complete the journey be x min.

Then, distance travelled in (x+11) min at 40 kmph

=distance covered in (x+5) min. at 50 kmph.

Therefore, (x+11)X(40/60)=(x+5)X(50/60)

10x=190

x=19 min. 

6. Two buses travel to a place at 45 kmph and 60 kmph respectively. If the second bus takes 5(1/2) hours less than the first for the journey,
 the length of the journey is:

(a) 900 km

(b) 945 km

(c) 990 km

(d) 1350 km

(e) None of these 

Sol.

6. (c); Let the length of journey be x km. Then,

(x/45)-(x/60)=11/2

X=990 km.

7. Walking 5/6 of his usual speed, a man is 10 minutes too late. The usual time taken by him to cover that distance is:

(a) 1 hours

(b) 50 min

(c) 12 min

(d) 8 min. 20 sec.

(e) None of these 

sol 

7. (b); With a speed of 5/6 th of the usual speed, the time taken is 6/5th of the usual time.

Therefore, (6/5 of usual time) – (usual time)=10 min.

Therefore, 1/5 of usual time=10 min. So, usual time=50 min.

8. A car travels a distance of 840 km at a uniform speed. If the speed of the car is 10 kmph more, 
then it takes 2 hours less to cover the same distance. The original speed of the car (in kmph) was:

(a) 45

(b) 50

(c) 60

(d) 75

(e) None of these 

Sol.

8. (c); Let original speed=x kmph.

(840/x)-(840/(x+10))=2

840(x+10)-840x=2x(x+10)

x^2+10x-4200=0

(x+70)(x-60)=0

x=60

Therefore, Original speed=60 kmph.

9. Distance between two towns P and Q is 240 km. A motor cycle rider starts from P towards Q at 8 p.m.
 at a speed of 40 kmph. At the same time another motor cycle rider starts from Q towards P at 50 kmph.
 At what time will they meet?

(a) 9.45 p.m.

(b) 10.30 p.m. 

(c) 10.40 p.m.

(d) 11 p.m.

(e) None of these 

Sol.

9. (c); Suppose they meet x hrs after 8 p.m. Then,

Sum of distance covered by them in x hrs=240  km.

Therefore, 40x+50x=240

x=(240/90) hrs=2 hrs 40 min.

So, they will meet at 10.40 p.m. 

10. A train M leaves Meerut at 5 a.m. and reaches Delhi at 9 a.m. Another train N leaves Delhi at 7 a.m. 
and reaches Meerut at 10.30 a.m. At what time do the two trains cross one another?

(a) 8.26 a.m.

(b) 8 a.m.

(c) 7.36 a.m.

(d) 7.56 a.m.

(e) None of these 

Sol.

10 . (d); Let distance between Meerut & Delhi be x km and let the train meet y hours after 7 a.m.

Clearly, M covers x km in 4 hrs. & N covers x km in 3(1/2) hr.

Therefore, Speed of M=(x/4)kmph & Speed of N=(2x/7) kmph.

[Distance covered by M in (y+2) hrs] [Distance covered by N in y hr]=x 

(x(y+2))/4+(2xy/7)=x

((y+2)/4)+(2y/7)=1

y=(14/15) hrs=((14/15)X60) min=56 min.

So, the trains meet at 7.56 a.m.

Quantitative Aptitude

1.After receiving two successive raises, Gopal’s salary become equal to 15/8 times of his initial salary. By how much percent was the salary raised the first time if the second raise was twice as high (in percent) as the first?

1) 15%

2) 20%

3) 25%

4) 30%

5) 33.33%

Ans.3

2.A Merchant gives 3 consecutive discounts of 10%. 15% and 15% after which he sells his goods at a percentage profit of 30.05% on the C.P. Find the value of the percentage profit that the shopkeeper would have earned if he had given discount of 10% and 15% only.

1) 53%

2) 62.5%

3) 72.5%

4) 68.6%

5) 69.2%

Ans.1

3.Arun sells to Bimal goods at five-thirds the rate of profit at which Bimal has decided to sell it to Chahal. Chahal, on other hand, sells it to Dev at one-third the rate of profit at which Bimal sold it to Chahal. If Dev gives Rs. 2145 to Chahal at 10% profit, how much did Arun buy it for?

1) Rs. 1000

2) Rs. 2000

3) Rs. 1500

4) Rs. 1800

5) None of these

Ans.1

4.A sum of Rs. 1000 after 3 years at compound interest becomes a certain amount that is equal to the amount that is the result of a 3 year depreciation from Rs. 1728. Find the difference between the rates of CI and depreciation. (Given CI is 10% p.a.).

1) 3.33%

2) 0.66%

3) 3.67%

4) 8.33%

5) 1.67%

Ans.5

5.A mixture can be prepared by sugar and salt. Price of sugar is thrice the price of salt. Kunal sells the mixture at Rs. 2160 per 10g, thereby making a profit of 20%. If the ratio of sugar  and salt in the mixture be 2 : 3, find the cost price of sugar.

1) Rs. 210/gm

2) Rs. 300/gm

3) Rs. 120/gm

4) Rs. 354/gm

5) None of these

Ans.2

6.An Irrigation dam has four inlets. Through the fist three inlets, the dam can be filled in 12 minutes; through the second, the third and the fourth inlet, it can be filled in 15 minutes; and through the first and the fourth inlet, in 20 minutes. How much time will it take all the four inlets to fill up the dam?

1) 8 min

2) 10 min

3) 12 min

4) 16 min

5) None of these

Ans.2

7.Two boats go downstream from point X to point Y. The faster boat covers the distance from X to Y 1.5 times as fast as the slower boat. It is known that for every hour the slower boat lags behind the faster boat by 8 km. However, if they go upstream, then the faster boat covers the distance from Y to X in half the time as the slower boat. Find the speed of the faster boat in still water.

1) 12 kmph

2) 20 kmph

3) 24 kmph

4) 25 kmph

5) None of these

Ans.2

Speed Of the faster boat in still water = 20 kmph

8.In an exam, the average was found to be 'x' marks. After deducting computational error, the average marks of 94 candidates got reduced from 84 to 64. The average thus came down by 18.8 marks. The numbers of candidates who took the exam were:

1) 100

2) 90

3) 110

4) 105

5) 120

Ans.1

9.A Statue is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of conical part is 12 cm.

1) 1440 cm2

2) 385 cm2

3) 1580 cm2

4) 770 cm2

5) 880 cm2

Ans.4

10.The probability that a contractor will get a plumbing contract is 2/3 and the probability that he will get an electric contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability that he will get both the contracts?

1) 19/45

2) 13/45

3) 12/35

4) 11/23

5) None of these

Ans.1

Directions (Q. 11-15): Big-bazar Exports produces five types of trousers–A, B, C, D and E, using cloth of three qualities–high, medium and low, and dyes of three qualities–high, medium and low, one trouser require 3 m of cloth. The following table gives specific information about the production in year 2005. It gives information about:

1. The number of trousers (of each category) produced, in thousands.

2. The percentage distribution of cloth quality in each type of trousers, and

3. The percentage distribution of dye quality in each type of trousers.

11.What is the total requirement of cloth?

1) 300,000 m

2) 450,000 m

3) 600,000 m

4) 150, 000 m

5) None of these

Ans.3

1.       (3); Total trousers=200000

So cloth required= 200000*3

12.How many metres of medium quality cloth in consumed?

1) 264,000m

2) 288,000m

3) 312,000m

4) 576, 000m

5) None of these

Ans.2

13.How many metres of low-quality cloth in consumed by C-type trousers?

1) 24,000 m

2) 48,000 m

3) 60,000 m

4) 36,000 m

5) None of these

Ans.4

14.What is the ratio of three qualities of dyes in high-quality cloth?

1) 2 : 3 : 4

2) 1 : 3 : 6

3) 7 : 8 : 12

4) Cannot be determined

5) None of these

Ans.4

No relationship among quality and dye is given

Hence, cannot be determined

15.What is the ratio of low-quality dye used for A-type to that used for B-type trousers?

1) 1 : 3

2) 3 : 8

3) 1 : 2

4) 4 : 9

5) None of these

Ans.4

Quantitative Aptitude

Quant Quiz

Directions : Study the following Grpah carefully and answer the questions given below:

Number of Students Appearing for Aptitude Test from Various Towns (Number in thousands)

1. What is the average number of students appearing for Aptitude test from all the Towns together?

1) 33500 

2) 3350 

3) 17500 

4) 33.5 

5) None of these

2. The number of students appearing for the Aptitude test from Town D is approximately what percent of the number of students appearing for 
the Aptitude test from Town C?

1) 243 

2) 413 

3) 134 

4) 341 

5) 143

3. What is the respective ratio of the number of students appearing for the Aptitude test from Town C and D together to the number of students appearing for 
the Aptitude test from Town A, D and E together?

1) 11 : 13 

2) 20 : 43 

3) 20 : 47 

4) 37 : 20 

5) None of these

4. What is the respective ratio of the number of students appearing for the aptitude test from Town B to Town A?

1) 3 : 4 

2) 13 : 16 

3) 11 : 16 

4) 2 : 3 

5) None of these

5. The number of students appearing for the aptitude test from Town E is approximately what percent of
 total number of students appearing for the Aptitude test from all the Towns together?

1) 15 

2) 17 

3) 19 

4) 21 

5) 23

Directions: In the given pie -charts the distribution of passed students from six different colleges (A, B, C, D, E and F)
 of a city during the year 2012 and 2013 is given.
 In the line graph the percentage of boys passed in year 2012 and the percentage of girls passed in 2013 is shown. 
Answer the following questions based on these graphs.

Total students passed in 2012 = 30000 

Total students passed in 2013 = 40000

6. What is the number of female students passing in 2012 from College C?

1) 2000 

2) 2300 

3) 2400 

4) 2500 

5) 2700

7. What is the number of male students passing in year-2013 from College F?

1) 5760 

2) 5750 

3) 5740 

4) 5730 

5) 5720

8. What is the total number of females passing in 2012 from all colleges?

1) 14645 

2) 15645 

3) 16645 

4) 17645 

5) 18645

9. What is the difference between the number of male students passing in year 2013 and that of female students passing in the same year?

1) 1690 

2) 1680 

3) 1670 

4) 1660 

5) 1650

10. Number of boys passing from College E in 2012 is what percentage of the number of boys passing from College C in year 2013?

1) 70% 

2) 80% 

3) 90% 

4) 100% 

5) 110%

Answers:
1. 1
Reqd avg marks = (40 + 32.5 + 17.5 + 42.5 + 35)/5 thousand
= 16.5 * 1000/5 = 33500

2. 1
Reqd pecentage = (42.5 * 1000)/(17.5 * 1000) * 100%
= 243% approx

3. 1
Required ratio = (17.5 + 42.5)thousand : (40 + 42.5 + 35) thousands
= 60 : 117.5
= 20 : 39.16
= 20 : 39

4. 2
Reqd Ratio = 32500 : 40000

5. 4
Reqd percentage = (35000 * 100)/(40 + 32.5 + 17.5 + 42.5 + 35) * 1000 * 100%
= 35000/1675% = 21% approx

6. 5
50% of (64.8/360 * 30000)
= 50/100 * 5400 = 2700

7. 60% of (86.4/360 * 40000) = 60/100 * 9600
= 5760

8. 2
Total = 30000/(360 * 100) [54 * 40 + 90 * 55 + 64.8 * 50 + 50.4 * 60 + 43.2 * 45 + 57.6 * 60]
30000/(360 * 100) [2160 + 4950 + 3240 + 1944 + 3456]
30000/36000 * 18774 = 15645

9. 2
Total boys = 4000 + 4800 + 1800 + 1600 + 2880 + 5760 = 20840
Total girls = 40000 - 20840 = 19160
Diff = 20840 - 19160 = 1680

10. 5
E-Boys2012 = 43.2/360 * 30000 * 55/100 = 1980
C-Boys2013 = 36/360 * 40000 * 45/100 =1800
Req% = 1980 * 100/1800 = 110%

Quantitative Aptitude

Quant Quiz (Problems On Ages) With Solutions

1. Ashok's age is (1/6) th of his father's age. The father's age will be twice of Kamal's age after 10 years. 
If Kamal's eighth birthday was celebrated 2 years before, what is Ashok's present age?

(a) 24 years

(b) 30 years

(c) 6(2/3) years

(d) None of these

(e) Cannot be determined

Sol. 

1. (d); Kamal's present age=10 years

Kamal's age after 10 years=20 years

Father's age after 10 years =40 years

Father's age now=30 years

Ashok's age now=[(1/6)X30]years=5 years

2. The sum of the ages of a father and son is 45 years. Five years ago the product of their ages was 4 times the father's age at that time. 
The present ages of the father and son respectively are:

(a) 25 years, 10 years

(b) 36 years, 9 years

(c) 39 years, 6 years

(d) None of these

(e) Cannot be determined

Sol.

2. (b); Let son's age=x. Then, father's age=(45-x)

(x-5) (45-x-5)=4(45-x-5)

x-5=4

x=9

Therefore, Father's age now=36 years & son's age=9 years.

3. 10 years ago, Mona's mother was 4 times older than her daughter. After 10 years, the mother will be twice older than the daughter. Mona's present age is:

(a) 5 years

(b) 10 years

(c) 20 years

(d) 30 years

(e) None of these

Sol.

3. (c); Let Mona's age 10 yrs ago=x. Her mother's age then=4x.

2(x+20)=4x+20

2x=20

x=10

Therefore, Mona's present age=(10+10)years=20 years

4. Three years ago, the average age of A and B was 18 years. With C joining them, the average becomes 22 years. How old is C now?

(a) 24 years

(b) 27 years

(c) 28 years

(d) 30 years

(e) None of these

Sol.

4. (a); (A+B)'s age 3 years ago=(18X2)=36 years

(A+B) now=(36+3X2)years=42 years.

(A+B+C) now=(3X22) years=66 years.

C's present age=(66-42) years=24 years.

5. Kamla got married 6 years ago. Today her age is 1(1/4) times her age at the time of marriage. He son's age is 1/10 times her age. Her son's age is:

(a) 2 years

(b) 3 years

(c) 4 years

(d) 5 years

(e) None of these

Sol.

5. (b); Let Kamla's age 6 years ago be x.

therefore, (x+6)=(5/4)x

4(x+6)=5x

x=24

therefore, Kamla's present age=30 years.

therefore, Kamla's son's present age=[(1/10)X30] years=3 years

6. One year ago the ratio of Varun and Veena's ages was 4 : 5. One year hence the ratio of their ages will be 5 : 6. The present age of Veena is:

(a) 9 years

(b) 10 years

(c) 12 years

(d) 6 years

(e) None of these

Sol.

6. (c); Let their ages 1 years ago be 4x and 5x.

[(4x+2)/(5x+2)]=5/6

6(4x+2)=5(5x+2)

x=2 

therefore, Veena's present age=(5x+2)=12 years.

7. A is as much younger to B as he is older to C. If the sum of the ages of B and C is 48 years, what is the present age of A?

(a) 20 years

(b) 24 years

(c) 30 years

(d) 36 years

(e) None of these

Sol. 

7. (b); B-A=A-C

2A=B+C

2A=48

A=24

8. Ten years ago A was half of B in age. If the ratio of their present ages is 3 : 4, what is the sum of their present ages?

(a) 8 years

(b) 20 years

(c) 35 years

(d) 45 years

(e) None of these

Sol.

8. (c); Let their present ages be 3x and 4x. Then,

2(3x-10)=4x-10

2x=10

x=5

therefore, Sum of their present ages=7x=35 years.

9. The ratio of Laxmi's age to the age of her mother is 3 : 11. The difference of their ages is 24 years. The ratio of their ages after 3 years will be:

(a) 1 : 3

(b) 2 : 3

(c) 3 : 5

(d) 2 : 5

(e) None of these

Sol.

9. (a); Let their ages be 3x & 11x.

11x-3x=24

x=3

Their present ages are 9 years & 33 years. 

Ratio of their ages after 3 years = 12 : 36 =1 : 3

10. The age of a man is 4 times that of his son. Five years ago, the man was nine times as old as his sone was at that time. The present age of the man is:

(a) 28 years

(b) 32 years

(c) 40 years

(d) 44 years

(e) None of these

Sol.

10. (b); Let the son's age be x. Then, father's age=4x.

therefore, (4x-5)=9(x-5)

5x=40

x=8

therefore, Present age of the man=32 years.

Quantitative Aptitude

Quant Quiz

Directions (Q. 1-5): Each of the questions below consists of a question and two statements numbered I and II given below it: 
You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and Give answer—

1) if the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
2) if the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
3) if the data in Statement I alone or in Statement II alone are sufficient to answer the question.
4) if the data, in both the Statements I and II together are not sufficient to answer the question.
5) if the data in both the Statements I and II together are necessary to answer the question.

1. What are the values of ‘x’, ‘y’ and ‘z’?
I. x + y + 3z = 5
II. x + y = 2

2. Find the area of a rectangle.
I. Its length is 5m.
II. Its breadth is 2m.


3. What are the speeds of trains ‘A’ and ‘B’?
II. The speed of train ‘A’ is twice the speed of ‘B’.
II. Train A passes train B in 9 sec and its length is 100 m.

4. Find the ages of son and father.
I. The father’s age is twice the son’s age.
II. The sum of their ages is 60 years.

5. What is the number of students present in a class?
I. The average age of students in a class is 12 years and the total age of all the students in the class is 300 years.
II. The age of the oldest student is 16 years and that of the youngest student is 10 years.

Directions (Q. 6-10) : Study the following table carefully to answer these questions.

Population (in lakhs) of three states over the years

6. What is the average population of state B for all the years together?
1) 3.5 lakhs
2) 3.75 lakhs
3) 3.85 lakhs
4) 3.6 lakhs
5) None of these

7. What is the percentage increase in population of state A from 2005 to 2006?
1) 5
2) 5 2/3
3) 6
4) 6 2/3
5) 5 3/4

8. What was the difference between the total population of all three states in 2007 and 2008 respectively?
1) 1,10,000
2) 90,000
3) 1,15,000
4) 95,000
5) None of these

9. What was the average population of all three states in 2009?
1) 5.25 lakh
2) 4.5 lakh
3) 5.5 lakh
4) 4.2 lakh
5) None of these

10. What is the ratio between total population of all three states in 2007 and 2008 respectively?

1) 155 : 144
2) 15 : 13
3) 13 : 15
4) 133 : 144
5) 144 : 155

Answers:

1. 4

2. 5; Area = length × breadth

3. 4

4. 5

Let the age of son be x

father’s age = 2x

x + 2x = 60

x = 20, and 2x = 40

5. 1; 

No. of students = 300/12 = 25

‘II’ is not required.

6. 2

Average population of the state B = (3.2 + 3.6 + 3.4 + 3.8 + 4.1 + 4.4)/6 lakh

= 22.5/6 lakh

= 3.75 lakh

7. 4

Required percentage increase = (4.8 - 4.5)/4.5 * 100 = 6 2/3%

8. 1

Required difference

= [(5.4 – 5.2) + (3.8 – 3.4) + (6.3 – 5.8) ] lakh

= 1.1 lakh = 110000

9. 3

Average population of three states in 2009

= {(5.8 + 4.1 + 6.6)/3} lakh = 16.5/3 lakh = 5.5 lakh

10. 5

Required ratio = (5.2 + 3.4 + 5.8) : (5.4 + 3.8 + 6.3)

= 14.4 : 15.5 = 144 : 155