Quant Quiz On Data Interpretation

Dear reader here we are providing some question on data interpretation which can be helpful in your upcoming exam

Direction (1-5): Study the following table carefully and answer the questions given below it.
Number of Executives recruited by six different organizations over the years.

1.    What is the total number of executives recruited by all the organizations together in the year 2006?
(1) 2927                 (2) 3042                       (3) 2864
(4) 3143                 (5) None of these

2.    What is the ratio of total number of executives recruited by organization U in the years 2007 and 2009 together to the total number of executives recruited by organization P in the same years?
(1) 436:517            (2) 499:522                  (3) 517:436
(4) 522:499            (5) None of these

3.    What is the average number of executives recruited by organization S over all the years together? (rounded off to the nearest integer)
(1) 494                   (2) 482                         (3) 514
(4) 506                   (5) 478

4.    What is the per cent increase in the number of executives recruited by organization R in 2005 from the previous year?( rounded off to two digits after decimal)
(1)18.67                 (2) 12.92                      (3) 16.48
(4)13.21                 (5) None of these

5.    The number of executives recruited by organization T in the year 2008 forms approximately what per cent of the total number of executives recruited by all the organizations together in that year?
(1) 11                     (2) 31                           (3) 18
(4) 26                     (5) 23

Direction (6-10): Study the following graph carefully and answer the questions given below it.
Production and sale of printers of various companies in a month.

 

6.    What is the average number of units sold by all the companies together?
(1) 360                   (2) 390             (3) 375
(4) 410                   (5) None of these

7.    Which company had the highest percentage of sale with respect to its production?
(1) D                      (2) B                (3) E
(4) A                      (5) None of these

8.    What is the average number of units produced by all the companies together?
(1) 675                   (2) 650             (3) 625
(4) 600                   (5) None of these

9.    The total units sold by the companies A, B and C together is approximately what per cent of the total units produced by these companies?
(1) 62                     (2) 50               (3) 76
(4) 84                     (5) 58

10.  What is the ratio of the total production of companies D and E to the total sale of the same companies?

(1) 28:15                (2) 9:5              (3) 15:11
(4) 2:3                    (5) None of these

ANSWER WITH SOLUTION

1.    (5)
Total executives recruited were 2953.

2.    (4)
Reqd. ratio = 1044:998
= 522:499

3.    (1)
Reqd. Average = 2965/6 ≈ 494

4.    (2)
Reqd. percentage increase = (54/418) × 100= 12.919≈12.92%

5.    (3)
Reqd. percentage = (510/2854) × 100 ≈18%

6.    (3)
Reqd. average = (650+300+150+450+300+400)/6 = 375 

7.    (4)
It is clear from the graph. 

8.    (2)
Reqd. average = 3900/6=650

9.    (5)
Reqd. percentage = (1100/1900) ×100≈58%

10.  (1)

Reqd. ratio = 1400:750=28:15

00:23 - By Unknown 0

Quant Quiz for SSC Exam

Dear reader here we are providing some question on quant which can helpful in your upcoming exam

1.           A sum of Rs 731 is distributed among A, B and C, such that A receives 25% more than B and B receives 25% less than C. What is C’s share in the amount?
(1) Rs 172                                  (2) Rs 200                  

(3) Rs 262                                  (4) Rs 272      

           

2.           Manoj sold an article for Rs 15000. Had he offered a discount of 10% on the selling price, he would have earned a profit of 8%. What is the selling price?
(1) Rs12,500                              (2) Rs13,500              

(3) Rs12,250                              (4) Rs13,250  

        

3.          What will be the compound interest accrued on an amount of Rs 10,000 @20p.c.p.a. in two years if the interest is compounded half-yearly?
(1) Rs 4,400                               (2) Rs 4,600               

(3) Rs 4,641                               (4) Rs 4,680   

        

4.         The average weight of 4 men is increased by 3kg when one of them who weighs 120 kg is replaced by another man. What is the weight of the new man?

(1) 125 kg                                   (2) 150 kg

(3) 122 kg                                   (4) 132 kg

5.           B and C together can complete a piece of work in 8 days. A and B can complete the same work in 12 days and A and C can complete the same work in 16 days. In how many days can A, B and C together can complete the work?
(1) 3³⁄₉                                        (2) 7⁵⁄₁₃                      

(3) 7⁵⁄₁₂                                      (4) 3⁵⁄₁₂          

        

6.           The average speed of a car is 1⁴⁄₅ times the average speed of the bus. A tractor covers 575km in 23 hours. How much distance the car cover in 4 hours of the speed of the bus is twice the speed of the tractor?

(1) 340 km                                 (2) 480 km     

(3) 360 km                                 (4) 450 km     

        

7.           A pump can fill a tank with water in 2 hours. Because of the leak, it took 2¹⁄₃ hours to fill the tank. The leak can drain all the water of the tank in:
(1) 4¹⁄₃ hours                              (2) 7 hours                  

(3) 8 hours                                  (4)14 hours     

        

8.            A shopkeeper mixed two verities of rice at Rs. 20/kg and Rs. 30/kg in the ratio 2 : 3 and sell the mixture at 10% profit. Find the price per kg at which he sold the mixture?
(1) Rs. 26                                   (2) Rs. 28.8                

(3) Rs. 28                                   (4) Rs. 28.6    

        

9.           The ratio of the present ages of Anju and Sandhya is 13:17 respectively. Four years ago the ratio of their respective ages was 11:15. What will be the respective ratio of their ages six years hence?
(1) 3:4                                         (2) 7:8                         

(3) 5:4                                         (4) 4:5 

           

10.         A shopkeeper bought 30kg of wheat at the rate of Rs45 per kg. He sold 40% of the total  quantity at the rate of Rs50 per kg. Approximately, at what price per kg should he sell the  remaining quantity to make a profit of 25% overall?
(1) Rs 54                                    (2) Rs 52                    

(3) Rs 50                                    (4) Rs 60                                

Answers with Explanation:

1. (4)
Let C's share be Rs x.

Then, B gets =0.75 x.

A gets = 1.25×0.75x

x+0.75x+0.9375x=731

2.6875x=731

x=⁷³¹⁄_2.6875 =272

2. (1)
SP=Rs15000

SP after 10% discount= 15000-10%of15000= 15000-1500=Rs13500

Profit on SP= 8%

CP= ¹³⁵⁰⁰⁄₁₀₈ ×100=Rs 12500

3. (3)
CI=10000[{(¹¹⁄₁₀ )×(¹¹⁄₁₀ )×(¹¹⁄₁₀ )×(¹¹⁄₁₀ )}-1]

4.        (4)

If the average is increased by 3 kg, then sum of weights increases by 3 x 4 = 12 kg .

And this increase in weight is due to the extra weight included due to the inclusion of new person.

Weight of new man = 120 + 12 = 132 kg.

5. (2)
(B+C)'s 1 day work= ¹⁄₈  ...(i)

(A+B)'s 1 day's work= ¹⁄₁₂  ...(ii)

(A+C)'s 1 day's work = ¹⁄₁₆  ...(iii)

On adding all these three equations,

2(A+B+C)'s one day's work = ¹⁄₈ +¹⁄₁₂ ¹⁄₁₆ = ^(6+4+3)⁄₄₈ =¹³⁄₄₈ 

(A+B+C)'s one day's work= ¹³⁄₉₆ 

∴A, B and C can complete the work in ⁹⁶⁄₁₃ =7 ⁵⁄₁₃ 

6. (3)

Speed of the tractor= ^distance/ _time 

=⁵⁷⁵⁄₂₃ =25kmph

∴speed of bus= 50kmph

∴speed of car= ⁹⁄₅ ×50=90kmph

∴distance covered by the car in 4 hours= 4×90=360km

7. (4)

Work done by the leak in 1 hour = (¹⁄₂ -³⁄₇) = ¹⁄₁₄ 

∴leak will empty the tank in 14 hours.

8. (4)

CP of mixture = ^(20*2+30*3)⁄₂₊₃ =26

SP of the mixture = 1.1×26=28.6

9.(4)

Let the present ages of Anju and Sandhya be 13x and 17x years respectively. 

∴^13x-4⁄_17x-4 =¹¹⁄₁₅ 

=>187x-44=195x-60

=>195x-187x=60-40

=>x=2

∴Required ratio= (13×2+6):(17×2+6)

=4:5

10.       (4)

CP of wheat= 30×45=Rs1350

40% of 30kg =12kg

SP of 12kg = 12×50 = Rs 600

For 25% profit, total SP of all the wheat is 1350×¹²⁵⁄₁₀₀ =Rs 1687.5

Remaining wheat (30-12) = 18kg

Rate of remaining wheat = ^1087.5⁄₁₈ ≈Rs60

05:15 - By Unknown 0

SSC QUIZ ON BOAT AND STREAMS

 

Dear reader here we are providing some question on boat and streams which can be help full in your upcoming Exam:

1.   A boat can run 13 km/hr in still water and the speed of current 4 km/hr. in how many hours the boat will covered 68 km down streams?

(1) 2 hours                               (2) 3 hours

(3)  4 hours                              (4) 5 hours

2.   A man rows downstream 32 km and 14 km upstream. If he takes 6 hours to cover each distance, then the Velocity (in km/h) of current is:

(1) 1/2                                      (2) 3/2

(3) 2                                         (4) 5/2

3.   A man can row at 5 km/h in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place?

(1) 1.5 km                               (2) 2.4 km

(3) 3.5   km                             (4) 1 km

4.   A man can row 40 km upstream and 55 km downstream in 13 hours. Also, he can row 30 km upstream and 44 km downstream in 10 hours. Find the speed of the man in still water and the speed of the current.

(1) 5                                          (2) 3

(3) 6                                          (4) 4

5.   In a stream running at 2 km/h, a motorboat goes 6 km upstream and back again to the starting point in 33 minutes. Find the speed of the motorboat in still water.

(1) 22 km/hr                            (2) 20 km/hr

(3) 25 km/hr                            (4) 18 km/hr

6.   A man takes 3 hours 45 minutes to row a boat I5 km downstream of a river and 2 hours 30 minutes to cover a distance of 5 km upstream. Find the speed of the river current in km/hr.

(1) 1 km/hr                              (2) 2 km/hr

(3) 3 km/hr                              (4) 4 km/hr

7.   A person can row a boat d km upstream and the same distance downstream in 5 hours 15 minutes. Also, he can row the boat 2d km upstream in 7 hours. How long will it take to row the same distance 2d km downstream?

(1) 3/2 hours                            (2) 7 hours

(3) 7 (¼)                                  (4) 7/2 hours

8.   A man swimming in a stream which flows 1.5 km/h, finds that in a given time he can swim twice as fast with the stream as he can against it. At what rate does he swim?

(1) 4 km/hr                              (2) 4.5 km/hr

(3) 6 km/hr                              (4) 7 km/hr

9.   A motorboat in still water travels at a sped of 36 km/h. It goes 56 km upstream in 1 hour 45 minutes. The time taken by it to cover the same distance down the stream will be:

(1) 2 hours 25 minutes            (2) 3 hours

(3) 1 hour 24 minutes              (4) 2 hours 21 minutes

10.  A man can row 6 km/h in still water. If the speed of the current is 2 km/h, it takes 3 hours more in upstream  than in the downstream for the same distance.

(1) 30 km                                (2) 24 km

(3) 20 km                                (4) 32 km

 ANSWER WITH EXPLANATION: 

1.   (3)

The speed downstream = 13+4 = 17 km/hr

∴Time taken to covered 68 km = 68/17 = 4 hours

2.   (2)

Rate downstream = 32/6 km/h; Rate upstream = 14/6 km/h

 Velocity of current = 1/2 (32/6 - 14/6) km/h

= 3/2 km/h

3.   (2)

-Speed downstream = (5 + 1) km/h = 6 km/h.

Speed upstream = (5 – 1) km/h = 4 km/h.

Let the required distance be x km.

Then, x/6 + x/4 = 1

2x + 3x = 12

5x = 12

x = 2.4 km.

4.    (2)

Let rate upstream = x km / hr and rate downstream = y km / hr.

Then,  40/x+ 55/y   =13   ...(i)

and   30/x+44/y=10       ...(ii)

Multiplying (ii) by 4 and (i) by 3 and subtracting, we get : 

11/y=1                   or                     y=11.

Substituting y = 11 in (j), we get : x = 5.
\threrefore Rate in still water =1/2(11+5) km/h=8km/h.

Rateofcurrent = 1/2(11−5) km/h=3km/h.

5. (1)

 Let the speed of the motorboat in still water be x km/h. Then,
Speed downstream = (x + 2) km/h; Speed upstream = (x - 2) km/h.
 6/x+2+6/x−2    =    33/60

<=>  11x2−242x+2x−44=0

«=>   (x - 22) (11x + 2) = 0  <=>   x = 22. Hence, speed of motorboat in still water = 22 km/h.

6.    (1)
Rate downstream=15334 km/hr

=15×415km/hr=4km/hr.

Rate upstream=(5212) km/hr

=(5×25)=2km/hr

Speed of current=12(4−2)km/hr=1km/hr.

7.    (4)

 Let the speeds of boat and stream be x and y km/hr respectively.

Then, Rate downstream = (x + y) km/hr and Rate upstream = (x - y) km/hr

Given d / (x+y) + d / ( x-y ) =5 hrs 15 minutes = 21/4 hours

and 2d / (x+y) = 7 => d / (x+y)=7/4 => 2d / (x+y)=7/2

Hence, he takes 7/2 hours to row 2d km distance downstream.

8.     (2)

 Rate of stream =1.5 km/hr

Let speed of man in still water = x km/hr and Distance = d

Upstream speed = (x - 1.5) km/hr Downstream speed = (x + 1.5) km/hr

Given, 2d / (x+1.5) = d/ (x-1.5) => 2x - 3 = x + 1.5 => x = 4.5 km/hr

9.   (3)

 Speed of the motorboat upstream

= 56km/1 ¾ hours

=32 km/h

Let the speed of the current = x km/h

36 – x    = 32

à 36 – 32  = 4 km/h

speed of motorboat downstream = 36 + 4 = 40 km/h

Time taken to cover 56 km at 40 km/h =     56/40 = 7/5 hours

= 1 hour 24 minutes

10.  (2)

Let he required distance be x km.

x/6-2 + x/ 6 + 2 = 3

à x/4 – x/8 = 3

à 2x –x/8 = 3

= x=3 x 8 = 24 km.

05:02 - By Unknown 0

QUANT QUIZ ( SHORTCUTS FOR FRACTIONS PROBLEMS )

You have to Compare the given fractions in a number of problems in Data Interpretation and Quantitative Ability. Let us study some of the common methods of identifying out the largest or smallest of a given set of fractions which are useful for Mental Maths. 

Type – 1

When the numerators are same and the denominators are different, the fraction with the largest denominator is the smallest.

Have a look at the following example.

Example : Which of the following fractions is the smallest?

(3/5) , (3/7) , (3/13), (3/8)

Here, 13 is the largest denominator, so, (3/13) is the smallest fraction. 5 is the smallest denominator, hence (3/5) is the largest fraction.

Here logic is very simple, 

Situation: (i) Assume that you are 5 Children in your family. Your Dad brought an Apple and mom cut it into 5 pieces and distributed among all the children including you. (1/5)

Situation (ii) : Assume that you are 8 Children in your family. Your Dad brought an Apple and mom cut it into 8 pieces and distributed among all the children including you. (1/8)

Type 2 :

When the numerators are different and the denominators are same, the fraction with the largest numerator is the largest. Have a look at the following example. 

Example : Which of the following fractions is the smallest?

(7/5) , (9/5), (4/5), (11/5)

As 4 is the smallest numerator, the fraction 4/5 is the smallest.

As 11 is the largest numerator, the fraction 11/5 is the largest.

Here too logic is very simple,

Situation 1 : Assume that you are 4 Children in your family. Your Dad brought 8 Apples and mom distributed them among all the children including you. (8/4)

Situation 2 :  Assume that you are 4 Children in your family. Your Dad brought 12 Apples and mom distributed them among all the children including you. (12/4)

Type 3

The fraction with the largest numerator and the smallest denominator is the largest.

Example: Which of the following fractions is the largest?

(19/16), (24/11), (17/13), (21/14), (23/15)

Solution : As 24 is the largest numerator and 11 is the smallest denominator, 24/11 is the largest fraction. 

Type 4 : 

When the numerators of two fractions are unequal, we try and equate them by suitably cancelling factors or by suitably multiplying the numerators. Thereafter we compare the denominators as in TYPE 1. Have a look at the following examples.

Example: Which of  the following fractions is the largest?

(64/328), (28/152), (36/176), (49/196)

Solution : 64/328 = 32/164 = 16/82 = 8/41 this is approximately equal to 1/5

Note : In these type of problems, approximate values will be enough. No need to get EXACT values.

25/152 = 14/76 = 7/38 this is approximately equal to 1/5.5

36/176 = 18/88 = 9/44 this is approximately equal to 1/5

49/196 = 7/28 = ¼

As all the numerators are 1 and the least denominator is 4, the fraction 49/196 is the largest

Example: Which of the following fractions is the largest?

(71/181), (214/519), (429/1141)

Solution : (71/181) = (71 X 6) / (181 X 6) = 426/1086

(214/519) = (214 X 2) / (519 X 2) = 428/1038

The numerators are now all ALMOST equal (426, 428 and 429). The smallest denominator is 1038. 

So, the largest fraction must be 428/1038  that is 214/519 :)

Type 5 : 

For a fraction Less than 1 :

If the difference between the numerator and the denominator is same then the fraction with the larger values of numerator and denominator will be the largest. Have a look at the following example.

Example: Which of the following fractions is the largest?

(31/37), (23/29), (17/23), (35/41), (13/19) 

Solutions:  difference between the numerator and the denominator of each fraction is 6.... So the fraction with the larger numerals i.e., 35/41 is the greatest and the fraction with smaller numerals i.e., 13/19 is the smallest.

Type 6 :

For a fraction Greater than 1

If the difference between the numerator and denominator is same, then the fraction with the smaller values will be the largest.

Example : Which of the following fraction is largest ?

(31/27), (43/39), (57/53), (27/23), (29/25)

Solution : As the difference between the numerator and the denominator is same, the fraction with the smaller values i.e., 27/23 is the largest.

01:07 - By Unknown 0

Quant Quiz on Percentage With

EXPLANATION)

  1.    Concept of percentage or Percentage formula

To express x% as a fraction: We have, x%
x/100 Thus, 30%=30/100= 3/10

In simple terms we can conclude that percentage symbol "%" means 1/100

In case we are having a fraction and we have to calculate its percentage then we will multiply it by 100.
that is, a/b=(a/b)×100)%

2.    Increase or decrease in percentage with price

(1) If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure will be = [R/ (100+R)×100]%

( 2) If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure will be = [R/(100−R)×100]

3.    Results on population
Formula's on result of population are very important when we have to calculate the population n years after or n years before,
Let the population of a town be P now and suppose it increases at the rate of R% per annum, then
(1) Population after n years =P(1+R/100)ⁿ

(2)  Population before n years =P/(1+R/100)ⁿ

4.    Results on Depreciation
We know that value of a machine depreciate with time, so it will decrease with the time. To calculate the value of machine after n years or before n years, we use these formula's

Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum.
(1) Population after n years   = P(1−R/100)ⁿ

(2) Population before n years =P/(1−R/100)ⁿ

5.    Important results for Percentage
The formulas we are going to mention below is same as of increase or decrease in consumption with increase or decrease in the commodity price, just here we are in a bit different context.
(1) If A is R% more than B, then B is less than A by = [R/(100+R)×100]%

(2) If A is R% less than B, then B is more than A by = [R/(100−R)×100]%

QUESTIONS :

1.    Two numbers are less than third number by 30% and 37% respectively. How much percent is the second number less than by the first

(1) 20%                 (2) 30 %

(3) 10%                 (4) 40%

2.    In an examination, 34% of the students failed in mathematics and 42% failed in English. If 20% of the students failed in both the subjects, then find the percentage of students who passed in both the subjects.

(1) 40%                 (2) 41%

(3) 43%                 (4) 44%

3.    In the new budget , the price of kerosene oil rose by 25%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase ?

(1) 20%                 (2) 30%

(3) 22%                 (4) 24%

4.    The value of a machine depreciates at the rate of 10% per annum. If its present is Rs.1,62,000.? What was the value of the machine 2 years ago ?

(1) 200000             (2) 20000

(3) 2000                 (4) None of these

5.    A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation?

(1) 60%                 (2) 62%

(3) 64%                 (4) 65%

 ANSWERS WITH EXPLANATIONS:

1.    (3) : Let the third number is x.
then first number = (100-30)% of x
= 70% of x = 7x/10

Second number is (63x/100)
Difference = 7x/10 - 63x/100 = 7x/10
So required percentage is, difference is what percent of first number
=> (7x/100 * 10/7x * 100 )% = 10%

2.    (4) : Failed in mathematics, n(A) = 34
Failed in English, n(B) = 42

n(AUB) = n(A) + n (B) – n (A∩ B )

= 34 + 42- 20 = 56
Failed in either or both subjects are 56

Percentage passed = (100−56)%=44%

3.    (1) : Reduction in consumption = [((R/(100+R))*100]%
? [(25/125)*100]%=20%.

4.    (1): Value of the machine 2 years ago
= Rs.[162000/(1-(10/100)²)]=Rs.[162000*(10/9)*(10/9)]= Rs.200000

5.    (3) : Let the number be x

Then, error =  (5/3)x – (3/5)x =(16/15)x

Error% = [(16x/15) /(5x/3)] * 100% = 64%

04:52 - By Unknown 0