Divisiblity Concepts

Divisiblity Concepts with Example Questions

Concept – Divisibility of a number

Divisibility by 2 — A number is divisible by 2 if the unit digit is zero or divisible by 2.

Ex. – 2, 4, 8, 16, 18 etc. are all divisible by 2.

Divisibility by 3 — A number is divisible by 3 if the sum of digits of the number is divisible by 3. 

Ex. – if 43n25 is divisible by 3 then find the minimum value n

Sol. 4+3+n+2+5 is divisible by 3 

So n=1,4,7

Minimum value of n=1 

Divisibility by 4 — A number is divisible by 4 if the number formed by the last two digits is divisible by 4 .

Ex. – 56248 is divisible by 4 since 48 is divisible by 4. 

Divisibility by 5 — A number is divisible by 5 if unit’s digit is 0 or 5. 

Ex. – 1225, 13750, 245 etc. are all divisible by 5. 

Divisibility by 7 — The unit digit of given number is doubled and then it is subtracted from the number obtained after omitting the unit digit. 

If the remainder is divisible by 7 then the given number is also divisible by 7.

 Ex. -is  343 divisible by 7 or not

Sol. Double unit digit 3 of 343 is 6 

Then, 34 – 6 = 28 which is divisible by 7.

Divisibility by 8 — A number is divisible by 8, if the number formed by the last 3 digits is divisible by 8.

Ex. 21056 is divisible by 8, as the number formed by last three digit i.e. 056 is divisible by 8. 

Divisibility by 9 — A number is divisible by 9 if the sum of its digits is divisible by 9.

 Ex. If 256x5 is divisible by 9, find the maximum  value of x 

Sol.   value of x divisible by 9 ⇒(18+x)/9

x=0,9       maximum value of x =9

Divisibility by 10 — An number is divisible by 10 if its last digit is zero 

 Divisibility by 11 — An number is divisible by 11 if the differences of the sum of the digit at odd places and sum of digit at even places is either zero or divisible by 11.

Ex. If 3x4567 is divisible by 11 then what is the value of x

Sol. 3+4+6-x+5+7=0 or divisible by 11

⇒1-x=0  

x=1  

Divisibility by 25 —  A number is divisible by 25 if the number formed by the last two digits is divisible by 25 or the last two digit are zero. 

Divisibility by 125 — A number is divisible by 125 if the number formed by last three digit is divisible by 125 or the last three digit are zero 

Divisibility by Composite number 

Divisibility of Composite number is the divisibility of its Coprime factor 

For Ex. – 

1. divisibility of 33 is divisibility of 3 and divisibility of 11 

2. divisibility of 165 is the divisibility of 3, 5 and 11. 

Time and Work

Quick Study Notes and Quiz on Time and Work

TIPS FOR SOLVING QUESTIONS RELATED TO TIME AND WORK:

1. The total work is taken as 1.

2. If A can do a piece of work in 'n' days, ,then A's 1 day's work is equal to = 1/n

3. If A's 1 day's work = 1/n, then A can complete the work in 'n' days.

4. If A is thrice as good a workman as B, then

    Ratio of work done by A and B = 3 : 1.

    Ratio of time taken by A and B to finish the work = 1 : 3.

5.If ‘M1’ persons can do ‘W1’ work in ‘D1’ days and ‘M2’ persons can do W2 work in D2, days then we have a very general formula in the relationship of M1 D1 W2 = M2 D2 W1. The above relationship can be taken as a very basic and all-in-one formula. We also derive 

A) More men less days and conversely more days less men. 

B) More men more work and conversely more work more men. 

C) More days more work and conversely more work more days. 

6.If we include the working hours (say T1 and T2) for the two groups then the relationship is 

M1 D1 T1 W2 = M2 D2 T2 Q1
Again, if the efficiency (Say E1 and E2) of the persons in two groups is different then the relationship is

M1 D1 T1 E1 W2 = M2 D2 T2 E2 W1

Now, we should go ahead starting with simpler to difficult and more difficult questions. 

Questions on Above Formulaes

1.A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?

A. Rs. 375

B. Rs. 400

C. Rs. 600

D.Rs. 800

2. Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed?

A. 3 pm

B. 2 pm

C. 1:00 pm

D. 11 am

3. 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 1/2days

C. 25 days

D. 30 days

4. P is 30% more efficient than Q. P can complete a work in 23 days. If P and Q work together, how much time will it take to complete the same work?

A. 9

B. 11

C. 13

D. 15

5. 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

6.40 men can cut 60 trees in 8 hours. If 8 men leave the job how many trees wiil be cut in 12 hours?

A. 74

B. 82

C. 72

D. 73

7.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

8. P and Q can complete a work in 20 days and 12 days respectively. P alone started the work and Q joined him after 4 days till the completion of the work. How long did the work last?

A. 5 days

B. 10 days

C. 14 days

D. 22 days

9. If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600.

A. 12

B. 14

C. 16

D. 18

10. Assume that 20 cows and 40 goats can be kept for 10 days for Rs.460. If the cost of keeping 5 goats is the same as the cost of keeping 1 cow, what will be the cost for keeping 50 cows and 30 goats for 12 days?

A. Rs.1104

B. Rs.1000

C. Rs.934

D. Rs.1210

ANSWERS AND SOLUTION :
1(B)Explanation:
C's 1 day's work = 1/3 - (1/6 + 1/8) = 1/3 - 7/24 = 1/24
A's wages : B's wages : C's wages = 1/6 : 1/8 : 1/24 = 4 : 3 : 1.
C's share (for 3 days) = Rs. (3 x 1/24 x 3200) = Rs. 400.

2(C)Explanation :
Work done by P in 1 hour = 1/8
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the machines were operating.
Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11
which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm.

3(B)Explanation:
Ratio of times taken by A and B = 1 : 3.
The time difference is (3 - 1) 2 days while B take 3 days and A takes 1 day.
If difference of time is 2 days, B takes 3 days.
If difference of time is 60 days, B takes (3/2 x 60) = 90 days.
So, A takes 30 days to do the work.
A's 1 day's work = 1/30
B's 1 day's work = 1/90
(A + B)'s 1 day's work = (1/30 1/90) = 4/90 = 2/45
So  A and B together can do the work in 45/2 = 22 1/2 days.

4(C)Explanation :
Work done by P in 1 day = 1/23
Let work done by Q in 1 day = q
q × (130/100) = 1/23
=> q = 100/(23×130) = 10/(23×13)
Work done by P and Q in 1 day = 1/23 + 10/(23×13) = 23/(23×13)= 1/13
=> P and Q together can do the work in 13 days

5(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
So,  15 men and 20 boys can do the work in 4 days.

6(C) Explanation
By basic formula:
M1 = 40, D1 = 8, W1 = 60(cutting of trees is taken as work)
M2 = 40 - 8 = 32, D = 12, W2 = ?
Putting the values in the formula
M1 D1 W2 = M2 D2 W1
We have , 40 x 8 x W2 = 32 x 12 x  60
or, W2 = (32 x 12 x 60)/(40 x 8) = 72

7(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 = (7/12 - 1/2) = 1/12 .
So, B alone will take 12 hours to do the work.

8(B)Explanation :
Work done by P in 1 day = 1/20
Work done by Q in 1 day = 1/12
Work done by P in 4 days = 4 × (1/20) = 1/5
Remaining work = 1 – 1/5 = 4/5
Work done by P and Q in 1 day = 1/20 + 1/12 = 8/60 = 2/15
Number of days P and Q take to complete the remaining work = (4/5) / (2/15) = 6
Total days = 4 + 6 = 10

9(C)Explanation :
Wages of 1 woman for 1 day = (21600)/(40 x 30)
Wages of 1 man for 1 day = (21600×2)/(40×30)
Wages of 1 man for 25 days = (21600×2×25)/(40×30) = 1080000/1200=900
Number of men = 14400/900 =16

10(A)Explanation :
Assume that cost of keeping a cow for 1 day = c ,
cost of keeping a goat for 1 day = g
Cost of keeping 20 cows and 40 goats for 10 days = 460
Cost of keeping 20 cows and 40 goats for 1 day = 460/10 = 46
=> 20c + 40g = 46
=> 10c + 20g = 23 ---(1)
Given that 5g = c
Hence equation (1) can be written as 10c + 4c = 23 => 14c =23
=> c=23/14
cost of keeping 50 cows and 30 goats for 1 day
= 50c + 30g
= 50c + 6c (substituted 5g = c)
= 56 c = 56×23/14
= 92
Cost of keeping 50 cows and 30 goats for 12 days = 12×92 = 1104

Time and Distance

Short Notes on Time and Distance

TIPS AND FORMULAS ON TIME AND DISTANCE:

1. Speed =  Distance/Time

2. Time =  Distance /Speed

3. Distance = Speed x Time

4. km/hr to m/s conversion:
1 kmph =   5/18 m/s
5. m/s to km/hr conversion:

 1 m/s =   18/5 m/s

6. If the ratio of speeds of train A and B is a : b, then the ratio of time taken by them to cover the same distance = b : a.

7. If a man covers a certain distance at x km/hr and an equal distance at y km/hr. Then,the average speed during the whole journey is
2xy/(x + y) km/h

8. The time taken by a train in passing a pole or standing man is the same as the time taken by the train to cover a distance equal to its own length.

9. The time taken by a train of length 'L' metres in passing a stationary object of length 'B' metres is equal to the time taken by the train to cover a distance equal to (L + B) m.

10. If two trains are moving in the same directions at u m/s and v m/s, where u > v, then their relative speed will be equal to the difference of their speeds i.e. (u - v) m/s.

11. If two trains are moving in the opposite directions at u m/s and v m/s, then their relative speed will be equal to the sum of their speeds i.e. (u + v) m/s.

12.If two trains of length 'a' metres and 'b' metres are moving in the same directions at u m/s and v m/s respectively, then: 

The time taken by the faster train to cross the slower train is

 (a + b)/(u - v) sec.

13.If two trains of length 'a' metres and 'b' metres are moving in the opposite directions at u m/s and v m/s respectively, then:

The time taken by the faster train to cross the slower train is

 (a + b)/(u + v) sec.

14.If two trains start at the same time from points A and B towards each other and after crossing they take 'a' and 'b' hour in reaching B and A respectively, then:

(A's speed) : (B's speed) = Öb : Öa

Questions based on above formula:
1.A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is

A. 11 hrs

B. 8 hrs 45 min

C. 7 hrs 45 min

D. 9 hrs 20 min

2. Two men P and Q start a journey from same place at a speed of  3 1/2 km/hr and 3 km/hr  respectively. If they move in the same direction then what is the distance between them after 4 hours?

A.  3 km

B.  2 1/2 km

C.  2 km

D. 3 1/2 km

3.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

4. 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. 12

B. 11

C. 10

D. 9

5. A student walks from his house at 4 km/hr and reaches his school 5 minutes too late. If his speed had been 5 km/hr, he would have reached 10 minutes too early. The distance of the school from his house is:

A.  5/3 km

B. 5/27 km

C. 5 km

D.  5/9 km

6. In a journey of 160 km, a train covers the first 120 km at a speed of 80 km/h and the remaining distance at 40 km/h. The average speed of the train for the whole journey is:

A.  60 km/h

B.  64 km/h

C.  68 km/h

D.  72 km/h

7. Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart?

A. 17 hr

B. 14 hr

C. 12 hr

D. 19 hr

8. A tractor is moving with a speed of 20 km/h, x km ahead of a truck moving with a speed of 35 km/h. If it takes 20 minutes for the truck to overtake the tractor, then x is equal to:

A.  5 km

B.  10 km

C.  15 km

D.  20 km

9. By walking at 4/5  of his normal speed, a man reaches his office 10 minutes late. How much time he normally takes to reach his office?

A.  40 minutes

B.  45 minutes

C.  50 minutes

D.  60 minutes

10. Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?

A. 1 hr 42 min

B. 1 hr

C. 2 hr

D. 1 hr 12 min

ANSWERS AND SOLUTION:
1(C)Explanation :
Given that time taken for riding both ways will be 2 hours lesser than
the time needed for waking one way and riding back
From this, we can understand that
time needed for riding one way = time needed for waking one way - 2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min

2(C)Explanation:
If two trains are moving in the same directions at u m/s and v m/s, where u>v, then their relative speed will be equal to the difference of their speeds i.e.(u - v) m/s
Let u = 3 1/2 km/hr and v = 3 km/hr
Thus, Relative  speed = 3 1/2 - 3
= 1/2 km/hr
Therefore, Required distance = Speed x Time = 1/2 x 4
= 2 km

3(C)Explanation:
Let speed of the car be x kmph.
Then, speed of the train = 150/100 x = 3/2 x kmph.
So 75/x - 75/(3/2)x = 125/(10 x 60)
=> 75/x - 50/x = 5/24
=> x = (25 x 24)/5 = 120 kmph.

4(C)Explanation :
speed of the bus excluding stoppages = 54 kmph
speed of the bus including stoppages = 45 kmph
Loss in speed when including stoppages = 54 - 45 = 9kmph
=> In 1 hour, bus covers 9 km less due to stoppages
Hence, time that the bus stop per hour = time taken to cover 9 km
=distance/speed = 9/54 hour =1/6 hour = 60/6 min=10 min

5(C)Explanation:
Let the distance of the school from his house be x km.
Time = Distance/Speed
So x/4 - x/5 = 15/60
or, (5x - 4x)/20 - 1/4
or, x = 5 km

6(B)Explanation:
Time = Distance/Speed
Total time taken for the journey = 120/80 + 40/40 = 5/2 hours
So, Average speed = 160 x 2/5
= 64 km/hr

7(A)Explanation :
Relative speed = 5.5 - 5 = .5 kmph (because they walk in the same direction)
distance = 8.5 km
time = distance/speed=8.5/.5=17 hr

8(A)Explanation:
Distance = Speed x Time
Distance covered by the truck in 20 minutes = 35 x 20/60 =  35/3 km
Distance covered by the tractor in 20 minutes = 20 x 20/60 =  20/3 km
So 20/3 + x = 35/5
or, x = 5 km

9(A)Explanation:
Walking at 4/5 of the normal speed means that the time taken would be 5/4 of the normal time.
Let the normal time taken to reach the office be "t" minutes,
So 5/4t - t 10 minutes
or , t/4 = 10
or, t = 40 minutes

10(D)Explanation :
New speed = 6/7 of usual speed
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time - usual time = 12 minutes
=> 1/6 of usual time = 12 minutes
=> usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes

Daily GK Update 15th August 2017

1. Nation celebrates its 71st Independence Day on 15th August 2017

i. The nation celebrates its 71st Independence Day today (15th August 2017) and the main function was held at the Red Fort of the national capital where Prime Minister Narendra Modi hoisted the National Flag. 

ii. The unfurling of the Tricolour was synchronized with the 21 Gun Salute fired by the valiant gunners of the elite 2281 Field Regiment (Ceremonial). 

2. 112 Gallantry Awards to Armed Forces Personnel, members of Paramilitary Forces

i. On the occasion of 71st Independence Day, 112 Gallantry Awards have been approved by the President Ram Nath Kovind for the Armed Forces and Paramilitary Forces. It includes five Kirti Chakras, 17 Shaurya Chakras, 85 Sena Medals (Gallantry), three Nao Sena Medals (Gallantry) and two Vayu Sena Medals (Gallantry). Havaldar Giris Gurung from first Gorkha Rifles who was martyred during a major counter-infiltration operation along the Line of Control in Naugam sector of Kashmir has been awarded Kirti Chakra posthumously. 

ii. Major David Manlun, who was martyred during an encounter with militants in Mon district of Nagaland, CRPF commandant Pramod Kumar who was martyred in a terror attack in Srinagar's Nowhatta hours before last Independence Day has also been awarded Kirti Chakra Posthumously. 

iii. CRPF commandant Chetan Kumar Cheetah who miraculously recovered from the deep coma after being hit by at least nine bullets during a gun battle with terrorists in Jammu and Kashmir has been awarded Kirti Chakra, the 2nd highest peacetime gallantry award.

3. Home Minister launches live twitter wall on ‘Bharat Ke Veer’

i. Home Minister Rajnath Singh has launched the live twitter wall for promoting Bharat Ke Veer portal on the eve of 71st Independence Day. On the occasion, Mr Singh and Ministers of State for Home Affairs Hansraj Gangaram Ahir and Kiren Rijiju contributed their one month salary for Bharat Ke Veer corpus. 

ii. Mr Singh has handed over a cheque of 1.59 lakh rupees to the Union Home Secretary Rajiv Mehrishi in New Delhi. 

Important Takeaways from Above News-

• Bharat Ke Veer portal was launched by Rajnath Singh and actor Akshay Kumar in April 2017.
• ‘Bharat Ke Veer’ portal aims to facilitate online donation directly to the families of paramilitary troopers who have sacrificed their lives for the country in the line of duty from January 1, 2016.
• Rajnath Singh is a Lok Sabha MP from Lucknow Constituency, Uttar Pradesh.
• Bharat Ke Veer portal is technically supported by National Informatics Centre (NIC) and powered by State Bank of India.

4. PM launches website to honour gallantry award winners

i. Prime Minister Narendra Modi has launched a website to honour all the gallantry award winners since Independence. The portal will preserve and tell the stories of bravest men and women, civilians as well as armed forces personnel.

ii. In remembrance of heroes who have been awarded gallantry awards since Independence, launched the site http://gallantryawards.gov.in. PM also welcomed citizens’ participation in the process.

5. Indian export grows nearly 4 percent in July

i. The country's exports grew three point nine percent in July this year (2017), on a yearly basis to 22 (point) 5 billion dollars. Exports rose due to an increase in shipments of petroleum, chemicals and marine products. Official data released today showed that imports rose 15.4 per cent in July, to 34 billion dollars, due to a rise in inward shipments of crude oil and gold. 

ii. The country's trade deficit widened to 11.4 billion dollars during the month under review, from 7.8 billion dollars in July last year (2016). Gold import spurted 95 per cent, to 2.1 billion dollars in July. Oil imports increased 15 per cent in July.

6. Lalit Modi resigns from Rajasthan Cricket Association

i. Lalit Modi has resigned as president of the Nagaur District Association, marking an end to his presence in the Rajasthan Cricket Association (RCA), which was banned by the Board of Control for Cricket in India (BCCI) for allowing him in its fold.

ii. The 50-year-old, who left India after being charged with money laundering, sent his resignation to both the RCA as well as BCCI CEO Rahul Johri.

7. Sankara Rao to be IFCI Chief Executive

i. IFCI Ltd, a government company, is set to have a new Managing Director & CEO in E Sankara Rao, who is currently Chief General Manager at India Infrastructure Finance Company Ltd (IIFCL).

ii. The Appointments Committee of Cabinet has given its nod for the Department of Financial Services (DFS) proposal to appoint Emandi Sankara Rao as Managing Director & Chief Executive Officer of IFCI Ltd. Rao’s appointment will be valid for three years. Rao is also currently a director and chief executive officer at IIFCL Asset Management Company Ltd.

Important Takeaways from Above News-

• Industrial Finance Corporation of India (IFCI) Limited established on 01st July 1948.
• IFCI Limited is the first Development Financial Institution in the country to cater to the long-term finance needs of the industrial sector.

8. Santosh Sharma appointed CMD of Hindustan Copper Limited 

i. Santosh Sharma has appointed as Chairman and Managing Director of Hindustan Coper Limited (HCL). He is at present Director (Operations) in the HCL. 

ii. The Appointments Committee of the Cabinet (ACC) has approved his appointment to the post for a period of five years, an order issued by the Department of Personnel and Training (DoPT). 

9. Indian Government asks Internet Giants and others to Remove Blue Whale links

i. The government has directed internet majors- Google, Facebook, Whatsapp, Instagram, Microsoft and Yahoo - to immediately remove the links of dangerous online game Blue Whale Challenge, which has led to the suicide of children in India and other countries. Mumbai and West Midnapore district have reported deaths linked to this game.

ii. Ministry of Electronics and IT (Meity) instructed the internet majors in a letter to ensure that any such link of this deadly game in its own name or similar game is immediately removed from their platform.The letter was issued following instructions from Law and IT Minister Ravi Shankar Prasad.

10. Nation Tribute to Sri Aurobindo on 145th Birth Anniversary

i. India paid tributes to Sri Aurobindo on his 145th Birth Anniversary. Sri Aurobindo's rich thoughts and grand vision for India continue to be a great source of inspiration.

ii. A spiritual reformer and philosopher, Sri Aurobindo had a major contribution in the Indian freedom struggle.

Important Takeaways from Above News-

• Sri Aurobindo was born in Calcutta on 15th August 1872.
• Sri Aurobindo left his body on 05th December 1950.

11. 10-yr-old Indian youngest to climb 6,153m mountain

i. A 10-year-old mountaineer from Visakhapatnam, Kaamya Karthikeyan, successfully summited one of the highest peaks in the Himalayas, making her one of the youngest people in the world to achieve this feat.

ii. Kaamya completed the expedition three days before she turned 10, on August 7, thereby conquering Mount Stok Kangri. At a height of 6,153 metres (20,187 ft), Stok Kangri is not recommended for beginners and is considered ideal for advanced trekkers.

12.  Cristiano Ronaldo hit with five-match ban

i. World Player of the Year, Cristiano Ronaldo was banned for five-match by the Spanish football federation. He has been banned for pushing the referee after being sent-off in Real Madrid's 3-1 win over Barcelona in the First Leg of Spanish Super Cup. 

ii. Ronaldo received an automatic one-match ban for his dismissal and an extra four games for his reaction towards referee Ricardo de Burgos Bengoetxea. He has also been fined 4,500 US dollars and has 10 days to appeal the ban.

Important Takeaways from Above News-

• Cristiano Ronaldo is a Portuguese professional footballer. 
• Ronaldo plays as a forward for Spanish club Real Madrid and the Portugal national team.
• Cristiano Ronaldo has won Fifa's world player of the year award for the fourth time after winning both major European titles in 2016.

13. Indian boxers clinch 10 medals at Serbian tourney

i. India bagged two gold in the 6th Golden Glove Women's Boxing Tournament at Vojvodina, Serbia. While Jyoti won Gold in 51 kg category, and Vanlalharitpuii bagged it 60 kg category. In all, India won two gold, four silver and four bronze medals in the tournament. 

ii. The silver medals were clinched by Anjali in 48kg, Sakshi in 54kg, Ashtha in 69kg and Anupama in 81kg categories. Mandeep in 57kg, Nisha in 64kg, Shruti in 75kg and Neha in 81+kg category settled for bronze medals.

Quant Quiz

Quant : Quiz

1.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

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

3.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

4. If A = x% of y and B = y% of x, then which of the following is true?

A.  A is smaller than B.

B.  A is greater than B

C.  Relationship between A and B cannot be determined.

D. None of these

5. In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 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.  100

6.A student multiplied a number by 3/5 instead of 5/3.

What is the percentage error in the calculation?

A.  34%

B.  44%

C.  54%

D.  64%

7. In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:

A.  2700

B.  2900

C.  3000

D.  3100

8. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?

A.  57%

B.  60%

C.  65%

D.  90%

9. Helen went to the stationers and bought things worth Euro 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?

A.  Euro 15

B.  Euro 15.7

C.  Euro 19.7

D.  Euro 20

10.The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:

A.  4.37%

B.  5%

C.  6%

D.  8.75%

Answers 
1.C
 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.

2.  D
 Suppose originally he had x apples.
Then, (100 - 40)% of x = 420.
60/100 x x = 420
x = (420 x 100)/60 = 700.

3. C
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 number =14
Required percentage = (14/70 x 100)% = 20%.

4.D
 x% of y = (x/100 x y) = (y/100 x x) = y% of x
A = B.

5.D
 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 + 2/3 of 48
80/100x = 80
x = 100.

6.D
7. A
Number of valid votes = 80% of 7500 = 6000.
Valid votes polled by other candidate = 45% of 6000
= (45/100 x 6000) = 2700.

8. A Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Required percentage = (11628/20400 x 100)% = 57%.

9. C
Explanation: Let the amount taxable purchases be Euro x.
Then, 6% of x = 30/100
x = (30/100 x 100/6) = 5.
Cost of tax free items = Euro [25 - (5 + 0.30)] = Euro 19.70

10.B
Increase in 10 years = (262500 - 175000) = 87500.
Increase% = (87500/175000 x 100)% = 50%.
Required average = (50/10)% = 5%.

Quant Quiz

Quant : Quiz

1. The distance between two stations A and B is 220 km. A train leaves A towards B at an average speed of 80 km/hr. After half a hour, another train leaves B towards A at an average speed of 100 km/hr. The distance of the point where the two trains meet, from A is :
a. 120 km
b. 130 km
c. 140 km
d. 150 km


2.  A cart has to cover a distance of 80 km in 10 hours. If it covers half of the journey in (3/5)th time, what should be its speed to cover the remaining distance in the time left ?
a. 8 km/hr
b. 20 km/hr
c. 6.4 km/hr
d. 10 km/hr.

3. Ravi started cycling along the boundaries of a square field from corner point A.  After half an hour, he reached the corner point C, diagonally opposite to A. If his speed was 8 km/hr. what is the area of the field in square km ?
a. 64
b. 4
c. 8
d. Cannot be determined

4. A man goes uphill with an average speed of 35 km/hr and comes down with an average speed of 45 km/hr. The distance travelled in both the cases being the same, the average speed for the entire journey is :
a. 38.38 km/hr
b. 39.38 km/hr
c. 40 km/hr
d. None of these

5. A man walking at 3 km/hr crosses a square field diagonally in 2 min. The area of the field is :
a. 2500 m2
b. 3000 m2
c. 5000 m2
d. 6000 m2

6. A certain distance is covered at a certain speed. If half of this distance is covered in double the time, the ratio of the two speeds is :
a. 4 : 1
b. 1 : 4
c. 2 : 1
d. 1 : 2

7. If a boy takes as much time in running 10 m as a car takes in covering 25 m, the distance covered by the boy during the time the car covers 1 km. is :
a. 400 m
b. 40  m
c. 250 m
d. 650 m

8. The ratio between the rates of walking of P and Q is 2 : 3. If the time taken by Q to cover a certain distance is 36 minutes, the time taken by P to cover that much distance is :
a. 24 min
b. 54 min
c. 48 min
d. 21.6 min

9. A man, on tour, travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. the average speed for the first 320 km of the tour, is :
a. 35.55 km/hr
b. 71.11 km/hr
c. 36 km/hr
d. 72 km/hr


10. A man travels 35 km partly at 4 km/hr and at 5 km/hr. If he covers former distance at 5 km/hr and later distance at 4 km/hr, he could cover 2 km more in the same time. The time taken to cover the whole distance at original rate is 
a. 9 hours
b. 7 hours
c. 412 hours
d. 8 hours

Answers 


Answers
1.  A

2. D
Distance left = (12×80) km=40 km
Time left = [(1-35)×10]=25×10=4hrs
Speed required = (40÷4) km/hr= 10 km/hr.

3. B
As his speed is 8kmph, in half an hour he covers 4 kmph. But 4 km is equal to two sides of the square.  So side of the square = 2 km
Area of the field = 2 km x 2 km = 4 km


4. B
Average speed = (2×35×45/35+45) km/hr = 39/38 km/hr.

5. C
Speed = (3×58) m/sec = (56) m/sec.
Distance covered in (2×60)sec. = (56×2×60)m=100m
Length of diagonal = 100 m
So, area = 12×(diagonal)2 = (12×100×100)m2 = 5000m2

6. A
Let x km be covered in y hrs. Then,
Ist speed = (xy) km/hr.
2nd speed = (x2÷2y)km/hr =  (x4y) km/hr
Ratio of speed = xy:x4y=1:14=4:1

7.  A
25 : 10 :: 1000 : x or x = 10×100025 = 400 m

8. B
Ratio of times taken = 12:13
12:13 = x : 36  or 12×36:13×x?18=13×x or x = 54 min.

9.B
Average speed  = (2×64×8064+80) km/hr = (2×64×80144)km/hr.
=71.11 km/hr.

10. D
Suppose the man covers first distance in x hrs.and second distance in y hrs. Then,
4x+5y=35 and 5x+4y = 37
Solving the equations,
we get x = 5 and y = 3
Total time taken = (5+3)hrs = 8 hrs

Quant Quiz

Quant Quiz - Part II

Directions (1- 5): The following pie-chart shows the sources of funds to be collected by the National Highways Authority of India (NHAI) for its Phase II projects. Study the pie-chart and answers the question that follow.

Sources of funds to be arranged by NHAI for Phase II projects (in Rs. crores)

1. Nearly about 20% of the funds are to be arranged through:

A. SPVS 

B. External Assistance

C. Annuity 

D. Market Borrowing

2. If NHAI could receive a total of Rs. 9695 crores as External Assistance, by what percent (approximately) should it increase the Market Borrowing to arrange for the shortage of funds?

A. 4.5 % 

B. 7.5 %

C. 6 % 

D. 8 %

3. If the toll is to be collected through an outsourced agency by allowing a maximum 10% commission, how much amount should be permitted to be collected by the outsourced agency, so that the project is supported with Rs. 4910 crores?

A. Rs. 6213 crores 

B. Rs. 5827 crores

C. Rs. 5401 crores 

D. Rs. 5316 crores

4. The central angle corresponding to Market Borrowing is

A. 52° 

B. 137.8°

C. 187.2° 

D. 192.4°

5. The approximate ratio of the funds to be arranged through Toll and that through Market Borrowing is

A. 2:1 

B. 1:6

C. 3:11 

D. 2:5

Directions (6 – 10): The following line graph gives the ratio of the amounts of imports by a company to the amount of exports from that company over the period from 1995 to 2001.

Ratio of Value of Imports to Exports by a Company over the Years.

6. If the imports in 1998 were Rs. 250 crores and the total exports in the years 1998 and 1999 together was Rs. 500 crores, then the imports in 1999 was?

A. Rs. 250 crores 

B. Rs. 300 crores

C. Rs. 357 crores 

D. Rs. 420 crores

7. The imports were minimum proportionate to the exports of the company in the year ?

A. 1995 

B. 1996

C. 1997 

D. 2000

8. What was the percentage increase in imports from 1997 to 1998?

A. 72 

B. 56

C. 28 

D. Data inadequate

9. If the imports of the company in 1996 was Rs. 272 crores, the exports from the company in 1996 was?

A. Rs. 370 crores 

B. Rs. 320 crores

C. Rs. 280 crores 

D. Rs. 275 crores

10. In how many of the given years were the exports more than the imports?

A. 1 

B. 2

C. 3 

D. 4

Answers:
1. Answer: (B)
20% of the total funds to be arranged = Rs. (20% of 57600) crores
= Rs. 11520 crores = Rs. 11486 crores. (Approx)
Rs. 11486 crores is the amount of funds to be arranged through External Assistance

2. Answer: (C)
Shortage of funds arranged through External Assistance = Rs. (11486 - 9695) crores
= Rs. 1791 crores.
Increase required in Market Borrowing = Rs. 1791 crores.
Percentage increase required = {[(1791 / 29952) ] x 100}% = 5.98%  = 6%. (Approx)

3. Answer: (C)
Amount permitted = (Funds required from Toll for projects of Phase II) + (10% of these funds)
= Rs. 4910 crores + Rs. (10% of 4910) crores
= Rs. (4910 + 491) crores
= Rs. 5401 crores.

4. Answer: (C)
Central angle corresponding to Market Borrowing =[(29952/57600)] x 360º = 187.2º

5. Answer: (B)
Required ratio = 4910/29952= 1/6.1 = 1/6 (Approx)

6. Answer: (D)
The ratio of imports to exports for the years 1998 and 1999 are 1.25 and 1.40 respectively.
Let the exports in the year 1998 = Rs. x crores.
Then, the exports in the year 1999 = Rs. (500 - x) crores.
1.25 = 250/x
x = 250 /1.25 = 200 ( Using ratio for 1998
Thus, the exports in the year 1999 = Rs. (500 - 200) crores = Rs. 300 crores.
Let the imports in the year 1999 = Rs. y crores.
Then, 1.40 = y / 300
y = ( 300 * 1.40 ) = 420
Imports in the year 1999 = Rs. 420 crores.

7. Answer: (C)
The imports are minimum proportionate to the exports implies that the ratio of the value of imports to exports has the minimum value.
Now, this ratio has a minimum value 0.35 in 1997, i.e., the imports are minimum proportionate to the exports in 1997.

8. Answer: (D)
The graph gives only the ratio of imports to exports for different years. To find the percentage increase in imports from 1997 to 1998, we require more details such as the value of imports or exports during these years.
Hence, the data is inadequate to answer this question.

9. Answer: (B)
Ratio of imports to exports in the year 1996 = 0.85.
Let the exports in 1996 = Rs. x crores.
Then, 272/x = 0.85
x = 272 / 0.85= 320
Exports in 1996 = Rs. 320 crores.

10. Answer: (D)
The exports are more than the imports imply that the ratio of value of imports to exports is less than 1.
Now, this ratio is less than 1 in years 1995, 1996, 1997 and 2000.