Free online Quantitative aptitude, Data interpretation questions and answers for all those Freshers who are preparing for different exams or recruitment tests.

## Quantitative Aptitude Topics

##### 1. Problems on Average:

1) What is the average of first five multiples of 12? (Problems on average)

- 36
- 38
- 40
- 42

The correct answer is **A**

**Explanation:**

Average = 12∗(1+2+3+4+5) ∗ 1/5

= 12 ∗ 15∗ 1/5

= 12 ∗ 3= 36

Try to solve more problems on average.

2) Average of five numbers is 20. If each number is multiplied by 2, what will be the new average?

- 30
- 40
- 50
- 60

The correct answer is **B**

**Explanation:**

The new average = Initial average ∗ 2

= 20 ∗ 2 = 40

##### 2. Problems on Number:

1) What is the difference in the place value of 5 in the numeral 754853?

- 49500
- 49950
- 45000
- 49940

The Correct answer is (B)

**Answer with explanation:**

The digit 5 has two place values in the numeral, 5 * 10^{5} = 50,000 and 5 * 10^{1} = 50.

∴Required difference = 50000 – 50 = 49950

2) What should be added to 1459 so that it is exactly divisible by 12?

- 4
- 3
- 5
- 6

The Correct answer is (C)

**Answer with explanation:**

On dividing 1459 by 12, the remainder is 7.

∴The number to be added would be = 12 – 7 = 5

##### 3. Problems on Compound Interest:

1) What is the compound interest on Rs. 2500 for 2 years at rate of interest 4% per annum?

- Rs. 180
- Rs. 204
- Rs. 210
- Rs. 220

The Correct answer is (B)

**Explanation:**

Principal (P) = Rs. 2500

Rate of interest(r) = 4%

Time (t) = 2 years

Compound Interest = Amount – Principal

2) What is the amount for a sum of money Rs.7500 at 6% rate of interest C.I. for 2 years?

- Rs. 8427
- Rs. 8417
- Rs. 8400
- Rs. 8390

The Correct answer is (A)

**Explanation:**

Principal = Rs. 7500

r = 6%

t = 2 years

##### 4. Problems on Partnership:

1) Sohan started a business with a capital of Rs. 80000. After 6 months Mohan joined as a partner by investing Rs. 65000. After one year they earned total profit Rs. 20000. What is share of Sohan in the profit?

- Rs. 5222.2
- Rs. 14222.2
- Rs. 6222.2
- Rs. 6777.7

The Correct answer is (B)

Explanation:

Sohan’s capital be C1 = 80000

Mohan’s capital be C2 = 65000

Sohan’s time be T1 = 12 months

Mohan’s time be T2 = 6 months

Profit earned = 20000

Apply formula:

Therefore, Sohan’s share =(80000*12*20000)/(80000*12+65000*6)

=19200000000/(960000+90000)

=19200000000/1350000

=780000/135

=14222.2

2) A, B and C are partners. They have invested Rs.35000, Rs. 25000 and 10,000 respectively for the same period. If the total profit is Rs. 18000, find the share of A.

- Rs. 9000
- Rs. 9500
- Rs. 8000
- Rs. 8500

The Correct answer is (A)

Explanation:

A’s capital be C1= 35000

B’s capital be C2= 25000

C’s capital be C3= 10000

Profit = 18000

T1=T2=T3

Apply formula: C1*P/(C1+C2+C3)

Share of A= (35000*18000)/(35000+25000+10000)

=630000000/70000

=9000

##### 5. Problems on Ages:

1) A mother is twice as old as her son. If 20 years ago, the age of the mother was 10 times the age of the son, what is the present age of the mother?

- 38 years
- 40 years
- 43 years
- 45 years

The Correct answer is (D)

**Explanation:**

Let the age of son = X years

∴Age of mother would be =2X

As per question 20 years ago;

10 (X -20) = 2X – 20

10X – 200 = 2X – 20

10X – 2X= – 20 + 200

8X = 180

X=180/8= 22.5 years

∴Age of mother = 22.5 * 2 = 45 years

2) Four years ago a man was 6 times as old as his son. After 16 years he will be twice as old as his son. What is the present age of man and his son?

- 34, 9
- 33, 7
- 35, 5
- 36, 6

The Correct answer is (A)

**Explanation:**

Let age of son 4 years ago be = X

So, age of man 4 years ago would be = 6X

As per question after 16 years;

2* age of son = age of man

2(X + 4 +16) = (6X + 4 + 16)

2X +40 = 6X + 20

2X – 6X = 20 – 40

– 4X = – 20

X = 5 years

∴Present age of son = 5 + 4= 9 years

Present age of man = 6X + 4= 6*5 + 4 = 30 +4 = 34 years

##### 6. Problems on Calender:

1) If January 1, 1996, was Monday, what day of the week was January 1, 1997?

- Thursday
- Wednesday
- Friday
- Sunday

The correct option is (B)

**Explanation:**

The year 1996 is divisible by 4, so it is a leap year with 2 odd days.

As per the question, the first day of the year 1996 was Monday, so the first day of the year 1997 must be two days after Monday. So, it was Wednesday.

2) The first republic day of India was celebrated on January 26, 1950. What day of the week was it?

- Wednesday
- Friday
- Thursday
- Tuesday

The correct option is (C)

**Explanation:**

⇒ 1600 years are divisible by 400, so the year 1600 has 0 odd days.

⇒ 300 years have 1 odd day.

⇒ 49 years = (12 leap years + 37 years)

= (12*2, odd days + 37*1, odd days)

= 24 + 37 = 61 odd days

On dividing 61 by 7, we get remainder 5, so 49 years have 5 odd days.

⇒From January 26, 1950, to January 26, 1950, we have 26 days.

26 days = 3 weeks + 5 odd days

So, total number of odd days = 1 + 5 + 5 = 11 days

⇒11 days = 1 week + 4 odd days

4 odd days represent Thursday, so it was Thursday on January 26, 1950.

##### 7. Problems on Boats and Streams:

1) The speed of a boat in still water is 5km/hr. If the speed of the boat against the stream is 3 km/hr, what is the speed of the stream?

- 1.5 km/hr
- 2 km/hr
- 2.5 km/hr
- 1 km/hr

The correct answer is **B**

**Answer with explanation:**

Let the speed of stream = X km/hr

Speed of boat = 5 km/hr

Speed upstream = 3km/hr

**Apply formula:** Speed upstream = speed of boat – speed of stream

∴ 3 = 5 – X

X = 5 – 3 = 2 km/hr

2) A man rows downstream at 20 km/hr and rows upstream at 15 km/hr. At what speed he can row in still water?

- 17.5 km/hr
- 18 km/hr
- 20.5 km/hr
- 22 km/hr

The correct answer is **A**

**Apply formula: Speed in still water = 1/2(speed downstream + speed upstream)**

Speed downstream = 20 km/hr

Speed upstream = 15 km/hr

∴ Required speed = 1/2 (20 + 15) km/hr

= 1/2 ∗ 35 = 17.5 km/hr

Try to solve more problems on boats and streams.

##### 8. Problems on Clocks:

1) How many times the hands of a clock coincide in a day?

- 24
- 22
- 23
- 21

The Correct answer is (B)

**Explanation:**

The hands of a clock coincide only once between 11 O' clock and 1 O' clock, so in every 12 hours, the hands of a clock will coincide for 11 times.

∴ In a day or 24 hours, the hands of a clock will coincide for 22 (11+11) times

2) At what time between 3 and 4, the hands of a clock are straight and point in opposite directions?

- 42min. past 3
- 45min. past 3
- 49min. past 3
- 51min. past 3

The Correct answer is (C)

**Explanation:**

At 3 O' clock, the hands of the clock are 15 minute spaces apart. To be in straight line and in opposite directions, the hands must be 30 minutes spaces apart.

Now, the minute hand is 15 minute spaces behind the hour hand, so it has to gain 45 (15+30) minute spaces over the hour hand.

55 minute spaces are gained in 60 minutes by the minute hand

45 minutes will be gained in =60/55* 45 minutes

= 540/11= 49 1/11 minutes past 3.

##### 9. Problems on Height and Distance:

1) Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse observed from the ships are 30° and 45° respectively. If the lighthouse is 100m high, find the distance between the two ships.

- 155.80 m
- 157.80 m
- 159. 80 m
- 161.80 m

The Correctoption is(B)

**Answer with explanation:**

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

AB = 100 meters, ∠ACB = 45°, ∠ADB = 60°

AC = AB (100m)

AC = 100 m

Distance between ships, CD = AC + AD

2) A ladder is leaning against a wall. It makes a 60° angle with the wall. If the distance between foot of ladder and wall is 5.5 meters, find the length of the ladder.

- 9.5m
- 10m
- 10.5m
- 11m

The Correctoption is(D)

**Answer with explanation:**

Let BC be the wall and AC be the ladder.

∠BAC = 60° and AB = 5.5 meter

AC = 2 * AB

= 2 * 5.5 = 11 meters (Option D)

##### 10. Problems on Percentage:

1) 40 % of 280 =?

- 112
- 116
- 115
- 120

The Correct answer; option (A)

**Explanation:**

x % of a given number 'n' =

x = 40 and n = 280

∴ 40 % of 280 =40/100* 280 = 112

2) Whose 35% is 280?

- 700
- 750
- 800
- 850

The Correct answer; option (C)

**Explanation:**

Let the required value is x.

As per question; 35% of x = 280

35/100 *x = 280

x= 800

Try to solve more problems on percentage.

##### 11. Problems on Pipes and Cisterns:

1) A pipe can fill a tank in 6 hours and another pipe can empty the tank in 12 hours. If both the pipes are opened at the same time, the tank can be filled in

- 10 hours
- 12 hours
- 14 hours
- 16 hours

Correct answer; option (B)

**Answer with explanation:**

Part of the tank filled in one hour =1/6

Part of the tank emptied in one hour =1/12

Net part of the tank filled in one hour;

=1/6 – 1/12

=(2-1)/12 = 1/12

1/12 Part of the tank can be filled in one hour.

∴ The tank will be filled completely in 12 hours.

**Solution 2:**

Apply formula; = XY/(Y-X)

X = 6 hours and Y = 12 hours

∴ (6*12)/(12-6)= 12 hours

2) Three pipes A, B and C can fill a cistern in 8 minutes,12 minutes and 16 minutes respectively. What is the time taken by three pipes to fill the cistern when they are opened together?

- 3.7 minutes
- 4 minutes
- 4.5 minutes
- 5 minutes

Correct answer; option (A)

**Answer with explanation:**

Part of the tank filled by A in one minute = 1/8

Part of the tank filled by B in one minute = 1/12

Part of the tank filled by C in one minute = 1/16

Net part of the tank filled by A+B+C in one minute;

= 1/8 + 1/12 + 1/16

=(6+4+3)/48 = 13/48

13/48 Part of the cistern is filled in one minute.

∴ The whole tank will be filled in 48/13 = 3.7 minutes

##### 12. Problems on Profit and Loss:

1) A shopkeeper sold an article for Rs. 2500. If the cost price of the article is 2000, find the profit percent.

- 23%
- 25%
- 27%
- 29%

The Correct answer is (B)

**Answer with explanation:**

C.P. = Rs. 2000

S.P. = Rs. 2500

Profit or Gain = S.P. -C.P.

= 2500 – 2000 = 500

**Apply formula:** Profit % =__Profit__ ∗100

C.P.

=__500__ ∗100 =25 %

2000

2) A man purchases a TV for Rs. 8000 and sells it at 10% loss. What is the selling price of T.V?

- Rs.7200
- Rs.7000
- Rs.6900
- Rs.6500

The Correct answer is (A)

**Answer with explanation:**

C.P. of the TV = Rs. 8000

S.P. of the TV =?

Loss incurred = 10%

**Apply formula:** Selling Price =(100-loss %)/100 * C.P.

Therefore, S.P. =(100-10)/100 * 8000

= 90/100 * 8000 = 7200

Try to solve more problems on profit and loss.

##### 13. Problems on Speed and Distance:

1) A running man crosses a bridge of length 500 meters in 4 minutes. At what speed he is running?

- 8.5 km/hr
- 7.5 km/hr
- 9.5 km/hr
- 6.5 km/hr

The correct answer is (B)

**Answer with explanation:**

Speed= Distance/Time

- Distance = 500 meters
- Time = 4 minutes ⟶ 4 x 60 = 240 seconds

Speed =^{500}⁄_{240}=^{25}⁄_{12}m/s

We need answer in km/hr:

Speed in km/hr=^{25}⁄_{12}*^{18}⁄_{5}=^{90}⁄_{12}⟶^{30}⁄_{4}= 7.5 km/hr

2) A car running at a speed of 140 km/hr reached its destination in 2 hours. If the car wants to reach at its destination in 1 hour, at what speed it needs to travel?

- 300 km/hr
- 280 km/hr
- 250 km/hr
- 240 km/hr

The correct answer is (B)

**Answer with explanation:**

Speed= Distance/Time

Distance to be covered = Speed x Time

= 140 * 2 = 280 km

Time = 1 hour

Required Speed =^{280}⁄_{1}= 280 km/hr (Option B)

##### 14. Problems on Simple Interest:

1) If Suresh borrows Rs. 36000 from Mahesh at rate of interest 6% S.I, at the end of four years how much interest Suresh has to pay along with principal amount?

- Rs. 12560
- Rs. 12960
- Rs. 13500
- Rs. 14500

The Correct answer is (B)

**Answer with explanation:**

Principal amount = Rs. 36000

Rate of interest = 6

Number of years or time = 6 years

2) If A lends Rs. 4500 to B at 8% per annum and B lends the same sum to C at 10% per annum, find the gain of B in a period of 3 years.

- Rs. 220
- Rs.240
- Rs. 250
- Rs.270

The Correct answer is (D)

**Answer with explanation:**

The gain of B will be equal to the difference between the interest which C pays to D and the interest which B pays to A for the amount borrowed.

= 1350 – 1080 = Rs. 270

##### 15. Problems on Trains:

1) A train moving at speed of 90 km/hr crosses a pole in 7 seconds. Find the length of the train.

- 150 m
- 165 m
- 175 m
- 170 m

The correct option is (C).

**Answer with explanation:**

Length of the train is equal to the distance covered by train to cross the pole. So, we will find the distance travelled by the train in 7 seconds by applying the following formula:

Distance= Speed x Time

Speed is given in Km/hr so we will convert it into m/s as answers are given in meters.

Speed=90* 5/18 = 25 m/s

Time = 7 seconds

Distance = 25 * 7= 175 meters

2) A train of length 200 meters crosses a man running at 10 km/hr in the same direction in 10 seconds. What is the speed of the train?

- 72 km/hr
- 95 km/hr
- 85 km/hr
- 82 km/hr

The correct option is (D).

**Answer with explanation:**

When the train and man are moving in same direction then relative speed will be the difference between their individual speeds. In this problem the other way to find the relative speed is to divide the distance covered (length of train) by the time taken by the train to cross the man.

Relative Speed=200/10

We will convert it into Km/hr

200/10 * 18/5= 72 km/hr

Now, let the speed of the train is X km/hr. So, the relative speed, 72 km/hr = X km/hr – 10 km/hr

X-10=72

X= 72+10

X= 82 km/hr

Try to solve more problems on train.

##### 16. Problems on Time and Work:

1) Worker A completes a task in 8 days, and worker B completes the same task in 10 days. If both A and B work together, in how many days they will complete the task?

- days.
- days.
- days.
- days.

The Correct answer is (B)

**Explanation:**

Worker A completes the task in 8 days. So, in one day, he will complete1/8 part of the task.

So, A's one day work =1/8

Similarly, B's one day work =1/10

∴ (A+B)'s one day work = 1/8 + 1/10 = 9/40

9/40 of the task is completed in one day so both will complete the whole task in 40/9 ^{days}

2) Vikas and Mohan working together can complete a work in 6 days. If Vikas alone completes the same work in 10 days, in how many days Mohan alone can complete the same work?

- 13 days
- 14 days
- 15 days
- 16 days

The Correct answer is (C)

**Explanation:**

Vikas and Mohan together can complete the task in 6 days. So, in one day, they will complete 1/6 part of the task.

Therefore, (Vikas + Mohan)'s one day work will be =1/6

Similarly, Vikas's one day work =1/10

Therefore, Mohan's one day work = 1/6 -1/10

= (5-3)/30 =2/30 =1/15

In one day Mohan completes the 1/15 part of the work so he will complete the entire work in 15 days.

Try to solve more Problems on time and work.

##### 17. Problems on Alligation and Mixture:

1) A 60 liter mixture of milk and water contains 10% water. How much water must be added to make water 20% in the mixture?

- 8 liters
- 7.5 liters
- 7 liters
- 6.5 liters

he Correct answer is (B)

**Answer with explanation:**

The mixture has 10% water, so the milk would be 90% of 60 liters.

Milk = 90/100 * 60 = 54 liters

100

∴Water = 60 – 54 = 6 liters

Let water to be added = x liters

Now, __6+x__ * 100 = 20

60+x

__6+x__ = __1__

60+x 5

30 + 5x = 60+x

30 – 60 = x – 5x

– 30 = – 4x

x =__30__ =7.5 liters

4

2) An alloy contains14 parts of tin and 100 parts of copper. What is the percentage of tin in the alloy ?

- 12.3%
- 13%
- 11.5%
- 11%

The Correct answer is (A)

**Answer with explanation:**

In 114 parts of alloy, tin = 14 parts

∴Percentage of tin = __14__ ∗100=12.3%

114

##### 18. Problems on Surds and Indices:

1) The value of 5^{1/4} x (125)^{0.25} is

- 5
- 5
- 25

**Answer:** C

**Explanation:**

5^{1/4} x (125)^{0.25}

5^{1/4} × (5^{3} )^{1/4}= 5 1/4 + 3/4=5

2) Value of (256)^{0.16} x (256)^{0.09} is

- 64
- 4
- 8
- 16

**Answer:** B

**Explanation:**

(256)^{0.16} x (256)^{0.09}

(256)^{0.16+0.09} = (256)^{0.25} = (256)^{1/4}

= (4^{4})^{1/4} = 4^{4 x 1/4} = 4

##### 19. Problems on HCM and LCM:

1) What is the HCF of 1095 and 1168?

- 37
- 73
- 43
- 83

**Answer:** B

**Explanation:**

By Division Method

Therefore the HCF is 73

2) Find the HCF of 210, 385, and 735.

- 7
- 14
- 21
- 35

**Answer:** D

**Explanation:**

210 = 2 x 5 x 3 x 7 x 1

385 = 5 x 7 x 11 x 1

735 = 5 x 7 x 3 x 7 x 1**HCF** = 5 x 7 = 35

##### 20. Problems on Area:

1) What is the area of a triangle with base 5 meters and height 10 meters?

- 20 square meters
- 35 square meters
- 25 square meters
- 40 square meters

**Answer:** C

**Explanation:**

Area of a triangle = ½ * base * height

So, the area = ½* 5 * 10

=25 square meters

2) The base of a right-angled triangle is 10 and hypotenuse is 20. What is its area?

- 52 meters
- 58 meters
- 68 meters
- 60 meters

**Answer:** D

**Explanation:**

The area of a right angled triangle = ½ * base * height

Base = 10, Hypotenuse = 20

Height^{2} = Hypotenuse^{2} – Base^{2}

= 20^{2} – 16^{2}

= 400 – 256

Height^{2} = 144

Height = 12

Area = ½ * base * height

= ½ * 10 * 12

= 60 meters

##### 21. Problems on Ratio and Proportion:

1) A: B: C is in the ratio of 3: 2: 5. How much money will C get out of Rs 1260?

- 252
- 125
- 503
- None of these

**Answer:** D

**Explanation:**

C's share = [C's ratio/ sum of ratios] * total amount

C's share = (5/10) * 1260

C's share = 630

2) If a: b is 3: 4 and b: c is 2: 5. Find a: b: c.

- 3: 2: 5
- 3: 6: 5
- 3: 4:10
- 2: 3: 4

**Answer:** C

**Explanation:**

The ratio of a: b is 3: 4

The ratio of b: c is 2: 5

**Note: To find the ratio in such questions, multiply a to b, then b to b, and then b to c.**

a: b: c = 3*2: 4*2: 4*5

a: b: c = 6: 8: 20

So, a: b: c = 3: 4: 10

##### 22. Problems on Races and Games:

1) In a kilometer race, A beats B by 40 meters or by 5 seconds. What is the time taken by A over the course?

- 1 minute 57 seconds.
- 2 minutes.
- 1.5 minutes.
- None of these.

**Answer:** B

**Explanation:**

A runs 1km = 1000 m, it means B runs 1000 – 40 = 960 meters

Time is taken by B for 40 meters = 5 seconds

Time is taken by B for 960 meters = 5/40 × 960=120 seconds

The time in which B covered 960 meters, A has covered 1000 meters.

∴ A's time over the course = 120 seconds

= 2 minutes

2) In a race of 600 m, A can beat B by 50 m, and in another race of 500 m, B can beat C by 60 m. Then by how many meters will A beat C in a race of 400 meters?

- 77 m
- 77 m
- 70 m
- None of these.

**Answer:** A

**Explanation:**

When A runs 600 m,

Then B runs (600 – 50) = 550 m

Now, When B runs 500 m,

Then C runs (500 – 60) = 440 m

When B runs 550 m then C runs =(440*550)/500 = 484 m

When A runs 600 m then C runs 484 m,

Now in a 400 meters race, when A runs 400 runs, C will run: 484/600 * 400 = 968/3 m

∴ A beats C in a 400 m race by

(400-(968/3))=(1200-968)/3 =232/3 =77 1/3 m

##### 23. Problems on Probablity:

1) What is the probability of getting an even number when a dice is rolled?

- 1/5
- 1/2
- 1/3
- 1/4

**Answer:** B

**Explanation:**

The sample space when a dice is rolled, S = (1, 2, 3, 4, 5 and 6)

So, n (S) = 6

E is the event of getting an even number.

So, n (E) = 3

So, the probability of getting an even number P (E) =

=

2) What is the probability of getting two tails when two coins are tossed?

- 1/3
- 1/6
- 1/2
- 1/4

**Answer:** D

**Explanation:**

The sample space when two coins are tossed = (H, H), (H, T), (T, H), (T, T)

So, n(S) = 4

The event “E” of getting two tails (T, T) = 1

So, n(E) = 1

So, the probability of getting two tails, P (E) =

=

##### 24. Problems on Volume and Surface Area:

1) The surface area of a cube is 600 cm2. The length of its diagonal is

- cm
- cm
- 10 cm
- 10 cm

**Answer:** C

**Explanation:**

Surface area of cube = 6 a^{2}

600 = 6 a^{2}

a^{2} = 100

a=

Diagonal of cube =

2) The length of the longest rod that can be placed in a room 30 m long, 24 m broad and 18 m high, is

- 30 m
- 15 m
- 30 m
- 60 m

**Answer:** C

**Explanation:**

Length of room = 30 m

Breadth of room = 24 m

Height of room = 18 m

##### 25. Problems on Decimal Fraction:

1) Find the missing term of the given expression:

18.834 + 818.34 -? = 618.43

- 217.644
- 218.744
- 217.744
- 217.844

**Answer:** B

**Explanation:**

18.834 + 818.34 – ? = 618.43

Or, 18.834 + 818.340 – 618.43 =?

837.174 – 618.430 = 218.744

2) What is the value of (25.732)^{2} – (15.732)^{2}?

- 414.64
- 414.256
- 414.128
- 414.52

**Answer:** A

**Explanation:**

We know that, a^{2} – b^{2} = (a+b) (a – b)

Or, (25.732)^{2} – (15.732)^{2} = (25.732 + 15.732) (25.732 – 15.732)

Or, the value = 41.464 * 10 = 414.64

##### 26. Problems on Simplificatin:

##### 27. Problems on Stocks and Shares:

1) What is the market price of a 9% share when a person gets 180 by investing Rs. 4000?

- 150
- 200
- 250
- 300

**Answer:** B

**Explanation:**

Let the face value of the share = Rs. 100

Dividend per share = 9

As per the question, an income of Rs. 180 is obtained from an investment of Rs. 4000.

So, an income of Rs. 9 is obtained from an investment of Rs. = (4000/180) * 9 = 200

So, the market price of the share = Rs. 200

2) What will be the investment in order to obtain an income of 550 from 10% stock at Rs. 96?

- 5280
- 6280
- 7280
- 8280

**Answer:** A

**Explanation:**

At this time, Market value = Rs. 96

Dividend = 10 % = Rs. 10

The face value is not given, so let it be Rs. 100

To obtain Rs. 10 (10 % of face value) the required investment = Rs. 96

To obtain Rs. 550, the required investment = 96/10 * 550 = 5280

##### 28. Problems on Chain Rule:

1) If 15 men can reap the crops of a field in 28 days, in how many days will 5 men reap it?

- 50 days
- 60 days
- 84 days
- 9.333 days

**Answer:** C

**Explanation:**

Let 5 men can reap a field in x days

So, put the same quantities on the same side.

Men: Days

Now, Men and Days are inversely proportional to each other. If we increase the number of men fewer days will be required to complete the work.

Inversely proportional means 15= 1/28 5= 1/x

So, 5/15 = 28/x

i.e., 5: 15 = 28: x

Or, x = (28*15)/ 5

Or, x = 84 days

Hence, 5 men can reap a field in 84 days.

2) If 8 men can reap 80 hectares in 24 days, how many hectares can 36 men reap in 30 days?

- 450
- 350
- 152
- 652

**Answer:** A

**Explanation:**

We can resolve the problem in 2 steps:

i. If 8 men can reap 80 hectares, then how many hectares can reap by 36 men in the same number of days

Now, the same type should be on the same side

Let the required number of hectares = x

Men and hectares are directly proportional to each other.

So, 8: 36 = 80: x

Or, x = (36*80)/8

Or, x = 360 hectares

ii. If 360 hectares in 24 days, then how many hectares can reap in 30 days?

Similarly, 24: 30 = 360: x

So, x = (360*30)/24

Or, x = 450

Hence, 36 men can reap 450 hectares in 30 days.

**Solution 2:**

We know that **8 men work 24 days and reap a field of 80 hectares.Similarly, 36 men work 30 days and reap a field of x (let) hectares**.

Now, we know that

**men * days = total work**

So, 8*24 = 192, that means total work done is equals to 192

Now, 36*30 = 1080, that means the work done by 36 men is 1080 unit.

Now, put the same unit on same side

Or, 192: 1080 = 80: x

Or, x = (1080*80)/ 192 = 450

Hence, 36 men can reap 450 hectares in 30 days.

Try to solve more Problems on chain rule.

##### 29. True and Banker's Discount:

1) The true discount on a bill of Rs. 2700 is Rs. 200. What is the banker’s discount?

- Rs. 210
- Rs. 212
- Rs. 216
- Rs. 218

**Answer:** C

**Explanation:**

Face value = Rs. 2700

TD = Rs. 200

PW (present worth) = FV (face value) – TD (true discount)

= 2700 ? 200 = Rs. 2500

True discount is the simple interest on the present value for the unexpired time.

Now, simple interest on Rs. 2500 for unexpired time = Rs. 200

The rate of simple interest = 200/2500 * 100 = 8%

Banker's discount is the simple interest on the face value of the bill for unexpired time, i.e., simple interest on Rs. 2700 for unexpired time or remaining time.

R = 8%

Banker's discount = 2700 * 8/100 = Rs. 216

2) The banker’s discount and the true discount on a sum of money due 8 months hence are 140 and 130 respectively. What are the sum and the rate percent?

- 1820, 11.5%
- 1920, 12.5%
- 1930, 10.5%
- 1940, 9.5%

**Answer:** A

**Explanation:**

Sum = (B.D. * T.D. / B.D. -T.D.)

= (140 * 130) / (140 – 130)

So, the required sum = 18200/10 = Rs. 1820

As B.D. is the S.I. on sum due, so S.I. on Rs. 1820 for 8 months is Rs. 140.

S.I. = Amount * rate of interest * time

140 = 1820 * r * 8/12

140 = 1820 * (r/100) * (2/3)

r = 140 * 100 * 3 / 1820 * 2

= 1050/91 = 11.5 %

##### 30. Problems on Logarithm:

1) Find the logarithm of 1/256 to the base 2√2.

- 16
- 13/5
- -16/3
- 12

**Answer:** C

**Explanation:**

Let log_{2√2} [1/256] = x

We know that log_{a} y = x is similar to a^{x} = y

So, we can write it as [1/256] = (2√2) ^{x}

Or, (2√2) ^{x} = [1/2^{8}]

Or, [2^{1} * 2^{1/2}]^{x} = 1/2^{8}

Or, 2^{3x/2} = 2-8

Therefore, 3x/2 = -8

Hence, x = (-8 * 2)/ 3 = -16/3

2) If log_{a} [1/36] = -2/3, find the value of a.

- 6
- 8
- 9
- 216

**Answer:** D

**Explanation:**

ATQ, log_{a} [1/36] = -2/3

We know that **log _{a} y = x is similar to a^{x} = y**

So, a ^{(-2/3) }= 1/36

Or, a ^{(-2/3) }= 1/6^{2}

Or, a^{ (-2/3) }= 6^{-2}

Now, multiply and divide by 3 in the power of 6 to make the power equals to power of a.

So, a ^{(-2/3) }= 6^{3(-2/3)}

On comparing both side, we get a = 6^{3}

Therefore, a = 216

##### 31. Square Roots and Cube Roots:

2) Value of

- 5
- 2
- 3
- suare root 2

**Answer:** B

**Explanation:**

As we know, (a+b) (a-b) = a^{2} ?b^{2}

By using above formula:

=

=

= 5-3=2

##### 32. Number Problems:

1) If 30% of a certain number is 12.6, what is the number?

- 24
- 42
- 23
- 32

**Answer:** B

**Explanation:**

Let the required number be x.

So, according to the question:

2) One-fourth of one-third of two-fifth of a number is 15. What will be 40% of that number?

- 120
- 180
- 270
- 350

**Answer:** B

**Explanation:**

Let x be the required number. Then

##### 33. Odd man out and Series:

1)Complete the series 2, 5, 9, 19, 37……

- 76
- 74
- 75
- None of these.

**Answer:** C

**Explanation:**

The second number is one more than twice the first,

2 x 2 +1 = 5

The third number is one less than twice the second,

5 x 2 -1 = 9

The fourth number is one more than twice the third,

9 x 2 +1 = 19

The fifth number is one less than twice the fourth.

19 x 2 – 1 = 37

Therefore, the sixth number is one more than twice the fifth. So, the missing number:

37 x 2 + 1 = 75.

2) The next number of the sequence 0, 7, 26, 63….

- 124
- 126
- 215
- 217

**Answer:** A

**Explanation:**

The terms of the given series are

(1^{3} – 1) = 1 -1 = 0,

(2^{3} – 1) = 8 – 1 = 7,

(3^{3} – 1) = 27 – 1 = 26,

(4^{3} -1) = 64 – 1 = 63…..

So, the next term will be

5^{3} – 1 = 125 – 1 = 124

##### 34. Problems on Algebric Expression:

1) If x2 +1/ x2 = 34,x+1/x is equal to

- 3
- 4
- 5
- None of these

Answer: D

Explanation:

Adding 2 to the L.H.S and R.H.S of the equation:

x2+2+

(x+

(x+

2) The value of x in the equation: 16x +1/x = 8 is

Answer: A

Explanation:

16x+

16x2+1=8x

=>16x2+1-8x=0

=>(4x-1)2=0

=>x =1/4 , 1/4

##### 35. Problems on Permutaton and Combination:

1) In what ways the letters of the word “RUMOUR” can be arranged?

- 180
- 150
- 200
- 230

Answer: D

Answer with the explanation:

The word RUMOUR consists of 6 words in which R and U are repeated twice.

Therefore, the required number of permutations =

Or,

Hence, 180 words can be formed by arranging the word RUMOUR.

2) In what ways the letters of the word “PUZZLE” can be arranged to form the different new words so that the vowels always come together?

- 280
- 450
- 630
- 240

Answer: D

Answer with the explanation:

The word PUZZLE has 6 different letters, but ATQ, the vowels should always come together.

Now, let the vowels UE as a single entity.

Therefore, the number of letters is PZZL = 4, and UE = 1

Since the total number of letters = 4+1 = 5

So the arrangement would be in 5P5 =

#### Note: we know that 0! = 1

Now, the vowels AE can arrange in 2 different ways, i.e., 2P2 = 2! = 2*1 = 2 ways

Hence, the new words that can be formed = 120 *2 = 240.

[/bg_collapse

##### 1. Problems on Pie Charts:

**1.The following pie-chart shows the percentage distribution of the expenditure incurred in publishing a book. Study the pie-chart and the answer the questions based on it.**

**Various Expenditures (in percentage) Incurred in Publishing a Book **

1) If for a certain quantity of books, the publisher has to pay Rs. 30,600 as printing cost, then what will be amount of royalty to be paid for these books?

- Rs. 19,450
- Rs. 21,200
- Rs. 22,950
- Rs. 26,150

Answer: Option C

Explanation:

Let the amount of Royalty to be paid for these books be Rs. *r*.

Then, 20 : 15 = 30600 : *r*

r= Rs. (30600*15)/20

= Rs. 22950

2)What is the central angle of the sector corresponding to the expenditure incurred on Royalty?

- 15°
- 24°
- 54°
- 48°

Answer: Option C

Explanation:

Central angle corresponding to Royalty = (15% of 360)°

=(15/100) *3

= 54°

##### 2. Problems on Table Charts:

**1.Study the following table and answer the questions based on it.**

Expenditures of a Company (in Lakh Rupees) per Annum Over the given Years.

Year | Salary | Fuel & Transport | Bonus | Interest on Loans | Taxes |

1998 | 288 | 98 | 3.00 | 23.4 | 83 |

1999 | 342 | 112 | 2.52 | 32.5 | 108 |

2000 | 324 | 101 | 3.84 | 41.6 | 74 |

2001 | 336 | 133 | 3.68 | 36.4 | 88 |

2002 | 420 | 142 | 3.96 | 49.4 | 98 |

Note : Rotate screen to see responsive table

1) What is the average amount of interest per year which the company had to pay during this period

- Rs. 32.43 lakhs
- Rs. 33.72 lakhs
- Rs. 34.18 lakhs
- Rs. 36.66 lakhs

**Answer:** Option D

**Explanation:**

Average amount of interest paid by the Company during the given period

= Rs.(23.4 + 32.5 + 41.6 + 36.4 + 49.4 )/5 lakhs

=Rs (183.3/5) lakhs

= Rs. 36.66 lakhs.

2)The total amount of bonus paid by the company during the given period is approximately what percent of the total amount of salary paid during this period?

- 0.1%
- 0.5%
- 1%
- 1.25%

**Answer:** Option C

**Explanation:**

Required percentage={(3.00 + 2.52 + 3.84 + 3.68 + 3.96)/(288 + 342 + 324 + 336 + 420)}* 100%

=(17/1710) * 100%

= 1%

##### 3. Problems on Bar Charts:

**1.The bar graph given below shows the sales of books (in thousand number) from six branches of a publishing company during two consecutive years 2000 and 2001.**

Sales of Books (in thousand numbers) from Six Branches – B1, B2, B3, B4, B5 and B6 of a publishing Company in 2000 and 2001.

1)What is the ratio of the total sales of branch B2 for both years to the total sales of branch B4 for both years?

- 2:3
- 3:5
- 4:5
- 7:9

Answer: Option D

Explanation:

Required ratio =(75+65)/(85+95)

=140/180

=7/9

2)Total sales of branch B6 for both the years is what percent of the total sales of branches B3 for both the years?

- 68.54%
- 71.11%
- 73.17%
- 75.55%

Answer: Option C

Explanation:

Required percentage= (70+80)/(95+110) * 100 %

= 73.17 %

##### 4. Problems on Line Charts:

**1.Study the following line graph and answer the questions.**

Exports from Three Companies Over the Years (in Rs. crore)

1) For which of the following pairs of years the total exports from the three Companies together are equal?

- 1995 and 1998
- 1996 and 1998
- 1997 and 1998
- 1995 and 1996

Answer: Option D

Explanation:

Total exports of the three Companies X, Y and Z together, during various years are:

In 1993 = Rs. (30 + 80 + 60) crores = Rs. 170 crores.

In 1994 = Rs. (60 + 40 + 90) crores = Rs. 190 crores.

In 1995 = Rs. (40 + 60 + 120) crores = Rs. 220 crores.

In 1996 = Rs. (70 + 60 + 90) crores = Rs. 220 crores.

In 1997 = Rs. (100 + 80 + 60) crores = Rs. 240 crores.

In 1998 = Rs. (50 + 100 + 80) crores = Rs. 230 crores.

In 1999 = Rs. (120 + 140 + 100) crores = Rs. 360 crores.

Clearly, the total exports of the three Companies X, Y and Z together are same during the years 1995 and 1996.

2) Average annual exports during the given period for Company Y is approximately what percent of the average annual exports for Company Z?

- 87.12%
- 89.64%
- 91.21%
- 93.33%

Answer: Option D

Explanation:

__Analysis of the graph: From the graph it is clear that__

The amount of exports of Company X (in crore Rs.) in the years 1993, 1994, 1995, 1996, 1997, 1998 and 1999 are 30, 60, 40, 70, 100, 50 and 120 respectively.

The amount of exports of Company Y (in crore Rs.) in the years 1993, 1994, 1995, 1996, 1997, 1998 and 1999 are 80, 40, 60, 60, 80, 100 and 140 respectively.

The amount of exports of Company Z (in crore Rs.) in the years 1993, 1994, 1995, 1996, 1997, 1998 and 1999 are 60, 90,, 120, 90, 60, 80 and 100 respectively.

Average annual exports (in Rs. crore) of Company Y during the given period

= 1/7 (80 + 40 + 60 + 60 + 80 + 100 + 140)

= 560/7

Average annual exports (in Rs. crore) of Company Z during the given period

= 1/7 *(60 + 90 + 120 + 90 + 60 + 80 + 100)

= 600/7

Required percentage = (560/7)/(600/7) * 100 %

= 93.33 %