## CAT Online Test Series 4 | Free CAT Quiz Series 4 | CAT Free Mock Test

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**CAT Online Test Series 4 | Free CAT Quiz Series 4 | CAT Free Mock Test.** Common Admission Test (CAT) is one of the most challenging and competitive MBA entrance test in our country. Check your level of preparation for *CAT* with free All India *Mock test* 2019. CAT Exam Online Test 2019, CAT Free Mock Test Exam 2019. CAT Exam Free Online Quiz 2019, CAT Full Online Mock Test **Series 4th** in English. CAT Online Test for All Subjects, CAT Free Mock Test Series in English. CAT Free Mock Test **Series 4.** CAT English Language Online Test in English **Series 4th**. Here we are providing** CAT Full Mock Test Paper in English. CAT **Mock Test **Series 4th** 2019. Now Test your self for CAT Exam by using below quiz…

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- Question 1 of 50
##### 1. Question

A batsman in his 17

^{th}innings makes a score of 85 and their by increasing his average by 3. What is his average after the 17^{th}innings?Correct16x + 85 = 17(x + 3)

x = 34 + 3 = 37Incorrect16x + 85 = 17(x + 3)

x = 34 + 3 = 37 - Question 2 of 50
##### 2. Question

The price of a VCR is marked at Rs. 12,000. If successive discounts of 15%, 10% and 5% be allowed, then at what price does a customer buy it?

CorrectActual price = 95% of 90% of 85% of Rs. 12000

= 95/100 * 90/100 * 85/100 * 12000 = Rs. 8721.IncorrectActual price = 95% of 90% of 85% of Rs. 12000

= 95/100 * 90/100 * 85/100 * 12000 = Rs. 8721. - Question 3 of 50
##### 3. Question

The base of a right triangle is 8 and hypotenuse is 10. Its area is?

CorrectIncorrect - Question 4 of 50
##### 4. Question

108 * 107 * 96 = ? (to the nearest hundred)

Correct108 * 107 * 96 ≡ 1109400

Incorrect108 * 107 * 96 ≡ 1109400

- Question 5 of 50
##### 5. Question

Two pipes can separately fill a tank in 20 and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is full, a leak develops in the tank through which one-third of water supplied by both the pipes goes out. What is the total time taken to fill the tank?

Correct1/20 + 1/30 = 1/12

1 + 1/3 = 4/3

1 — 12

4/3 — ?

4/3 * 12 = 16 hrsIncorrect1/20 + 1/30 = 1/12

1 + 1/3 = 4/3

1 — 12

4/3 — ?

4/3 * 12 = 16 hrs - Question 6 of 50
##### 6. Question

A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:

CorrectSpeed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr. x – 5 = 45 ==> x = 50 km/hr.

IncorrectSpeed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr. x – 5 = 45 ==> x = 50 km/hr.

- Question 7 of 50
##### 7. Question

5 * 5 ÷ 5 + 5 ÷ 5 = ?

Correct(5 * 5)/5 + 1/5 * 5 = 5 + 1 = 6

Incorrect(5 * 5)/5 + 1/5 * 5 = 5 + 1 = 6

- Question 8 of 50
##### 8. Question

I. a

^{2}+ 11a + 30 = 0,

II. b^{2}+ 6b + 5 = 0 to solve both the equations to find the values of a and b?CorrectI. (a + 6)(a + 5) = 0

=> a = -6, -5

II. (b + 5)(b + 1) = 0

=> b = -5, -1 => a ≤ bIncorrectI. (a + 6)(a + 5) = 0

=> a = -6, -5

II. (b + 5)(b + 1) = 0

=> b = -5, -1 => a ≤ b - Question 9 of 50
##### 9. Question

9000 + 16 2/3 % of ? = 10500

Correct9000 + 16 2/3 % of ? = 10500 => 9000 + 50/3 % of ? = 10500

50/(3 * 100) of ? = 1500 => ? = 1500 * 6

? = 9000Incorrect9000 + 16 2/3 % of ? = 10500 => 9000 + 50/3 % of ? = 10500

50/(3 * 100) of ? = 1500 => ? = 1500 * 6

? = 9000 - Question 10 of 50
##### 10. Question

9 3/4 + 7 2/17 – 9 1/15 = ?

CorrectGiven sum = 9 + 3/4 + 7 + 2/17 – (9 + 1/15)

= (9 + 7 – 9) + (3/4 + 2/17 – 1/15)

= 7 + (765 + 120 – 68)/1020 = 7 817/1020IncorrectGiven sum = 9 + 3/4 + 7 + 2/17 – (9 + 1/15)

= (9 + 7 – 9) + (3/4 + 2/17 – 1/15)

= 7 + (765 + 120 – 68)/1020 = 7 817/1020 - Question 11 of 50
##### 11. Question

The number of new words that can be formed by rearranging the letters of the word ‘ALIVE’ is -.

CorrectNumber of words which can be formed = 5! – 1 = 120 – 1 = 119.

IncorrectNumber of words which can be formed = 5! – 1 = 120 – 1 = 119.

- Question 12 of 50
##### 12. Question

(84 ÷ 14 ÷ 2 ÷ 9) * ? = 33⅓

Correct(84 ÷ 14 ÷ 2 ÷ 9) * ? = 33⅓

=> (6 ÷ 2 ÷ 9) * ? = 100 / 3

=> 3 / 9 * ? = 100 / 3

=> ? = 100Incorrect(84 ÷ 14 ÷ 2 ÷ 9) * ? = 33⅓

=> (6 ÷ 2 ÷ 9) * ? = 100 / 3

=> 3 / 9 * ? = 100 / 3

=> ? = 100 - Question 13 of 50
##### 13. Question

A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?

CorrectSpeed = 78 * 5/18 = 65/3 m/sec.

Time = 1 min = 60 sec.

Let the length of the train be x meters.

Then, (800 + x)/60 = 65/3

x = 500 m.IncorrectSpeed = 78 * 5/18 = 65/3 m/sec.

Time = 1 min = 60 sec.

Let the length of the train be x meters.

Then, (800 + x)/60 = 65/3

x = 500 m. - Question 14 of 50
##### 14. Question

If 35% of a number is 12 less than 50% of that number, then the number is?

CorrectLet the number be x. Then,

50% of x – 35% of x = 12

50/100 x – 35/100 x = 12

x = (12 * 100)/15 = 80.IncorrectLet the number be x. Then,

50% of x – 35% of x = 12

50/100 x – 35/100 x = 12

x = (12 * 100)/15 = 80. - Question 15 of 50
##### 15. Question

The sum of the digits of a two-digit number is 12. The difference of the digits is 6. Find the number?

CorrectLet the two-digit number be 10a + b

a + b = 12 — (1)

If a>b, a – b = 6

If b>a, b – a = 6

If a – b = 6, adding it to equation (1), we get

2a = 18 => a =9

so b = 12 – a = 3

Number would be 93.

if b – a = 6, adding it to the equation (1), we get

2b = 18 => b = 9

a = 12 – b = 3.

Number would be 39.

There fore, Number would be 39 or 93.IncorrectLet the two-digit number be 10a + b

a + b = 12 — (1)

If a>b, a – b = 6

If b>a, b – a = 6

If a – b = 6, adding it to equation (1), we get

2a = 18 => a =9

so b = 12 – a = 3

Number would be 93.

if b – a = 6, adding it to the equation (1), we get

2b = 18 => b = 9

a = 12 – b = 3.

Number would be 39.

There fore, Number would be 39 or 93. - Question 16 of 50
##### 16. Question

The ratio of the prices of three articles X, Y and Z is 8 : 5 : 3. If the prices of X , Y and Z are increased by 25%, 20% and 33 1/3% respectively, then what would be the ratio of the new prices of X, Y and Z?

CorrectLet the prices of X, Y and Z be 8k, 5k and 3k respectively.

After increase

Price of X = 8k * 125/100 = 10k

Price of Y = 5k * 120/100 = 6k

Price of Z = 3k * (133 1/3)/100 = 4k

Required ratio = 10k : 6k : 4k = 5 : 3 : 2.IncorrectLet the prices of X, Y and Z be 8k, 5k and 3k respectively.

After increase

Price of X = 8k * 125/100 = 10k

Price of Y = 5k * 120/100 = 6k

Price of Z = 3k * (133 1/3)/100 = 4k

Required ratio = 10k : 6k : 4k = 5 : 3 : 2. - Question 17 of 50
##### 17. Question

Find the least multiple of 13 which when divided by 6, 8 and 12 leaves 5, 7 and 11 as remainders respectively?

CorrectIncorrect - Question 18 of 50
##### 18. Question

A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in?

Correct(A + B + C)’s 1 day work = 1/6;

(A + B)’s 1 day work = 1/8

(B + C)’s 1 day work = 1/12

(A + C)’s 1 day work = (2 * 1/6) – (1/8 + 1/12)

= (1/3 – 5/24) = 1/8

So, A and C together will do the work in 8 days.Incorrect(A + B + C)’s 1 day work = 1/6;

(A + B)’s 1 day work = 1/8

(B + C)’s 1 day work = 1/12

(A + C)’s 1 day work = (2 * 1/6) – (1/8 + 1/12)

= (1/3 – 5/24) = 1/8

So, A and C together will do the work in 8 days. - Question 19 of 50
##### 19. Question

If the height of a cone is increased by 100% then its volume is increased by?

CorrectIncorrect - Question 20 of 50
##### 20. Question

The ratio of the radius of two circles is 1: 3, and then the ratio of their areas is?

Correctr

_{1}: r_{2}= 1: 3

Πr_{1}^{2}: Πr_{2}^{2}

r_{1}^{2}: r_{2}^{2 }= 1: 9Incorrectr

_{1}: r_{2}= 1: 3

Πr_{1}^{2}: Πr_{2}^{2}

r_{1}^{2}: r_{2}^{2 }= 1: 9 - Question 21 of 50
##### 21. Question

There is a 30% increase in the price of an article in the first year, a 20% decrease in the second year and a 10% increase in the next year. If the final price of the article is Rs. 2288, then what was the price of the article initially?

CorrectLet the price of the article, four years age be Rs. 100 in the 1st year, price of the article = 100 + 30 = Rs. 130. In the 2nd year, price = 130 – 20% of 130 = 130 – 26 = Rs. 104.

In the 3rd year, price = 104 + 10% of 104 = 104 + 10.4 = Rs. 114.40.

But present price of the article is Rs. 2288

for 114.4 —> 100 ; 2288 —> ?

Required price = (2288 * 100)/114.4 = 20 * 100 = Rs. 2000.IncorrectLet the price of the article, four years age be Rs. 100 in the 1st year, price of the article = 100 + 30 = Rs. 130. In the 2nd year, price = 130 – 20% of 130 = 130 – 26 = Rs. 104.

In the 3rd year, price = 104 + 10% of 104 = 104 + 10.4 = Rs. 114.40.

But present price of the article is Rs. 2288

for 114.4 —> 100 ; 2288 —> ?

Required price = (2288 * 100)/114.4 = 20 * 100 = Rs. 2000. - Question 22 of 50
##### 22. Question

Difference between two numbers is 5, six times of the smaller lacks by 6 from the four times of the greater. Find the numbers?

Correctx – y = 5

4x – 6y = 6

x = 12 y = 7Incorrectx – y = 5

4x – 6y = 6

x = 12 y = 7 - Question 23 of 50
##### 23. Question

The sum of a number and its reciprocal is one-eighth of 34. What is the product of the number and its square root?

CorrectLet the number be x. Then,

x + 1/x = 34/8

8x^{2}– 34x + 8 = 0

4x^{2}– 17x + 4 = 0

(4x – 1)(x – 4) = 0

x = 4

required number = 4 * √4 = 4 * 2 = 8.IncorrectLet the number be x. Then,

x + 1/x = 34/8

8x^{2}– 34x + 8 = 0

4x^{2}– 17x + 4 = 0

(4x – 1)(x – 4) = 0

x = 4

required number = 4 * √4 = 4 * 2 = 8. - Question 24 of 50
##### 24. Question

The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?

CorrectLet the ten’s digit be x and unit’s digit by y

Then, x + y = 15 and x – y = 3 or y – x = 3

Solving x + y = 15 and x – y = 3, we get: x = 9, y = 6

Solving x + y = 15 and y – x = 3, we get: x = 6, y = 9

So, the number is either 96 or 69.

Hence, the number cannot be determined.IncorrectLet the ten’s digit be x and unit’s digit by y

Then, x + y = 15 and x – y = 3 or y – x = 3

Solving x + y = 15 and x – y = 3, we get: x = 9, y = 6

Solving x + y = 15 and y – x = 3, we get: x = 6, y = 9

So, the number is either 96 or 69.

Hence, the number cannot be determined. - Question 25 of 50
##### 25. Question

The length of a rectangle is increased by 25% and its breadth is decreased by 20%. What is the effect on its area?

Correct100 * 100 = 10000

125 * 80 = 10000

No changeIncorrect100 * 100 = 10000

125 * 80 = 10000

No change - Question 26 of 50
##### 26. Question

A and B can finish a work in 16 days while A alone can do the same work in 24 days. In how many days B alone will complete the work?

CorrectB = 1/16 – 1/24 = 1/48 => 48 days

IncorrectB = 1/16 – 1/24 = 1/48 => 48 days

- Question 27 of 50
##### 27. Question

The ratio of the earnings of P and Q is 9 : 10. If the earnings of P increases by one-fourth and the earnings of Q decreases by one-fourth, then find the new ratio of their earnings?

CorrectLet the earnings of P and Q be Rs. 9x and Rs. 10x respectively.

New ration = [9x + 1/4(9x)]/[10x – 1/4(10x)]

=> [9x(1 + 1/4)]/[10x(1 – 1/4)] = 9/10 * (5/4)/(3/4) => 3/2.IncorrectLet the earnings of P and Q be Rs. 9x and Rs. 10x respectively.

New ration = [9x + 1/4(9x)]/[10x – 1/4(10x)]

=> [9x(1 + 1/4)]/[10x(1 – 1/4)] = 9/10 * (5/4)/(3/4) => 3/2. - Question 28 of 50
##### 28. Question

The food in a camp lasts for 30 men for 40 days. If ten more men join, how many days will the food last?

Correctone man can consume the same food in 30*40 = 1200 days.

10 more men join, the total number of men = 40

The number of days the food will last = 1200/40 = 30 days.Incorrectone man can consume the same food in 30*40 = 1200 days.

10 more men join, the total number of men = 40

The number of days the food will last = 1200/40 = 30 days. - Question 29 of 50
##### 29. Question

The roots of the equation 3x

^{2}– 12x + 10 = 0 are?CorrectThe discriminant of the quadratic equation is (-12)

^{2}– 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.IncorrectThe discriminant of the quadratic equation is (-12)

^{2}– 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal. - Question 30 of 50
##### 30. Question

If one third of 3/4 of a number is 21. Then find the number?

Correctx * 1/3 * 3/4 =21 => x = 84

Incorrectx * 1/3 * 3/4 =21 => x = 84

- Question 31 of 50
##### 31. Question

A, B and C can do a work in 6, 8 and 12 days respectively doing the work together and get a payment of Rs.1800. What is B’s share?

CorrectWC = 1/6:1/8:1/12 => 4:3:2

3/9 * 1800 = 600IncorrectWC = 1/6:1/8:1/12 => 4:3:2

3/9 * 1800 = 600 - Question 32 of 50
##### 32. Question

Two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post. If the length of each train be 120 m, in what time will they cross other travelling in opposite direction?

CorrectSpeed of the first train = 120/10 = 12 m/sec.

Speed of the second train = 120/5 = 8 m/sec.

Relative speed = 12 + 8 = 20 m/sec.

Required time = (120 + 120)/20 = 12 sec.IncorrectSpeed of the first train = 120/10 = 12 m/sec.

Speed of the second train = 120/5 = 8 m/sec.

Relative speed = 12 + 8 = 20 m/sec.

Required time = (120 + 120)/20 = 12 sec. - Question 33 of 50
##### 33. Question

The owner of a furniture shop charges his customer 24% more than the cost price. If a customer paid Rs. 8339 for a computer table, then what was the cost price of the computer table?

CorrectCP = SP * (100/(100 + profit%))

= 8339(100/124) = Rs. 6725.IncorrectCP = SP * (100/(100 + profit%))

= 8339(100/124) = Rs. 6725. - Question 34 of 50
##### 34. Question

A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?

CorrectSpeed = [54 * 5/18] m/sec = 15 m/sec.

Length of the train = (15 * 20) m = 300 m.

Let the length of the platform be x meters.

Then, x + 300 / 36 = 15

x + 300 = 540

x = 240 m.IncorrectSpeed = [54 * 5/18] m/sec = 15 m/sec.

Length of the train = (15 * 20) m = 300 m.

Let the length of the platform be x meters.

Then, x + 300 / 36 = 15

x + 300 = 540

x = 240 m. - Question 35 of 50
##### 35. Question

Six points are marked on a straight line and five points are marked on another line which is parallel to the first line. How many straight lines, including the first two, can be formed with these points?

CorrectWe know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear and no other set of three points, except those that can be selected out of these 6 points are collinear.

Hence, the required number of straight lines

= ¹¹C₂ – ⁶C₂ – ⁵C₂ + 1 + 1

= 55 – 15 – 10 + 2 = 32IncorrectWe know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear and no other set of three points, except those that can be selected out of these 6 points are collinear.

Hence, the required number of straight lines

= ¹¹C₂ – ⁶C₂ – ⁵C₂ + 1 + 1

= 55 – 15 – 10 + 2 = 32 - Question 36 of 50
##### 36. Question

The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 10, then their difference is:

CorrectLet the numbers be x and (100 – x).

Then, x(100 – x) = 5 * 495

x^{2}– 100x + 2475 = 0

(x – 55)(x – 45) = 0

x = 55 or 45

The numbers are 45 and 55.

Required difference = 55 – 45 = 10.IncorrectLet the numbers be x and (100 – x).

Then, x(100 – x) = 5 * 495

x^{2}– 100x + 2475 = 0

(x – 55)(x – 45) = 0

x = 55 or 45

The numbers are 45 and 55.

Required difference = 55 – 45 = 10. - Question 37 of 50
##### 37. Question

(743.30)

^{2}= ?Correct(743.30)

^{2}= 552500Incorrect(743.30)

^{2}= 552500 - Question 38 of 50
##### 38. Question

9 men and 12 boys finish a job in 12 days, 12 men and 12 boys finish it in 10 days. 10 men and 10 boys shall finish it in how many days?

Correct9M + 12B —– 12 days

12M + 12B ——- 10 days

10M + 10B ——-?

108M + 144B = 120M +120B

24B = 12M => 1M = 2B

18B + 12B = 30B —- 12 days

20B + 10B = 30B —–? => 12 daysIncorrect9M + 12B —– 12 days

12M + 12B ——- 10 days

10M + 10B ——-?

108M + 144B = 120M +120B

24B = 12M => 1M = 2B

18B + 12B = 30B —- 12 days

20B + 10B = 30B —–? => 12 days - Question 39 of 50
##### 39. Question

The difference between a number and its three-fifth is 50. What is the number?

CorrectLet the number be x. Then,

x – 3/5 x = 50 => 2/5 x = 50

x = (50 * 5)/2 = 125.IncorrectLet the number be x. Then,

x – 3/5 x = 50 => 2/5 x = 50

x = (50 * 5)/2 = 125. - Question 40 of 50
##### 40. Question

If the cost price of 50 articles is equal to the selling price of 40 articles, then the gain or loss percent is?

CorrectGiven that, cost price of 50 article is equal to selling price of 40 articles.

Let cost price of one article = Rs. 1

Selling price of 40 articles = Rs. 50

But Cost price of 40 articles = Rs. 40

Therefore, the trader made profit.\Percentage of profit = 10/40 * 100 = 25%IncorrectGiven that, cost price of 50 article is equal to selling price of 40 articles.

Let cost price of one article = Rs. 1

Selling price of 40 articles = Rs. 50

But Cost price of 40 articles = Rs. 40

Therefore, the trader made profit.\Percentage of profit = 10/40 * 100 = 25% - Question 41 of 50
##### 41. Question

A train running at the speed of 60 km/hr crosses a pole in 9 seconds. Find the length of the train.

CorrectSpeed = 60*(5/18) m/sec = 50/3 m/sec

Length of Train (Distance) = Speed * Time

(50/3) * 9 = 150 meterIncorrectSpeed = 60*(5/18) m/sec = 50/3 m/sec

Length of Train (Distance) = Speed * Time

(50/3) * 9 = 150 meter - Question 42 of 50
##### 42. Question

A watch was sold at a loss of 10%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price?

Correct90%

104%

——–

14% —- 140

100% —- ? => Rs.1000Incorrect90%

104%

——–

14% —- 140

100% —- ? => Rs.1000 - Question 43 of 50
##### 43. Question

Sum of two numbers is 80. Greater number exceeds by 5 from four times of the smaller. Find the numbers?

Correctx + y = 80

x – 4y = 5

x = 65 y = 15Incorrectx + y = 80

x – 4y = 5

x = 65 y = 15 - Question 44 of 50
##### 44. Question

The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?

CorrectLet the rate of interest be R% p.a.

4400{[1 + R/100]^{2}– 1} = 11193.60

[1 + R/100]^{2}= (44000 + 11193.60)/44000

[1 + R/100]^{2}= 1 + 2544/1000 = 1 + 159/625

[1 + R/100]^{2}= 784/625 = (28/25)^{2}

1 + R/100 = 28/25

R/100 = 3/25

herefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100

=Rs.10560IncorrectLet the rate of interest be R% p.a.

4400{[1 + R/100]^{2}– 1} = 11193.60

[1 + R/100]^{2}= (44000 + 11193.60)/44000

[1 + R/100]^{2}= 1 + 2544/1000 = 1 + 159/625

[1 + R/100]^{2}= 784/625 = (28/25)^{2}

1 + R/100 = 28/25

R/100 = 3/25

herefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100

=Rs.10560 - Question 45 of 50
##### 45. Question

If the L.C.M of two numbers is 750 and their product is 18750, find the H.C.F of the numbers.

CorrectH.C.F = (Product of the numbers) / (Their L.C.M) = 18750/750 = 25.

IncorrectH.C.F = (Product of the numbers) / (Their L.C.M) = 18750/750 = 25.

- Question 46 of 50
##### 46. Question

Albert buys 4 horses and 9 cows for Rs. 13,400. If he sells the horses at 10% profit and the cows at 20% profit, then he earns a total profit of Rs. 1880. The cost of a horse is:

CorrectLet C.P. of each horse be Rs. x and C.P. of each cow be Rs. y.

Then, 4x + 9y = 13400 — (i)

And, 10% of 4x + 20% of 9y = 1880

2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 — (ii)

Solving (i) and (ii), we get : x = 2000 and y = 600.

Cost price of each horse = Rs. 2000.IncorrectLet C.P. of each horse be Rs. x and C.P. of each cow be Rs. y.

Then, 4x + 9y = 13400 — (i)

And, 10% of 4x + 20% of 9y = 1880

2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 — (ii)

Solving (i) and (ii), we get : x = 2000 and y = 600.

Cost price of each horse = Rs. 2000. - Question 47 of 50
##### 47. Question

Robert is traveling on his cycle and has calculated to reach point A at 2 p.m. if he travels at 10 km/hr; he will reach there at 12 noon if he travels at 15 km/hr. At what speed must he travel to reach A at 1 p.m.?

CorrectLet the distance traveled be x km.

Then, x/10 – x/15 = 2

3x – 2x = 60 => x = 60 km.

Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.

So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.

Required speed = 60/5 = 12 kmph.IncorrectLet the distance traveled be x km.

Then, x/10 – x/15 = 2

3x – 2x = 60 => x = 60 km.

Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.

So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.

Required speed = 60/5 = 12 kmph. - Question 48 of 50
##### 48. Question

A certain sum is invested at simple interest at 18% p.a. for two years instead of investing at 12% p.a. for the same time period. Therefore the interest received is more by Rs. 840. Find the sum?

CorrectLet the sum be Rs. x.

(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840

=> 12x/100 = 840 => x = 7000.IncorrectLet the sum be Rs. x.

(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840

=> 12x/100 = 840 => x = 7000. - Question 49 of 50
##### 49. Question

A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?

CorrectMilk = 3/5 * 20 = 12 liters, water = 8 liters

If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.

Remaining milk = 12 – 6 = 6 liters

Remaining water = 8 – 4 = 4 liters

10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.

The ratio of milk and water in the new mixture = 16:4 = 4:1

If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.

Amount of water removed = 2 liters.

Remaining milk = (16 – 8) = 8 liters.

Remaining water = (4 -2) = 2 liters.

The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.IncorrectMilk = 3/5 * 20 = 12 liters, water = 8 liters

If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.

Remaining milk = 12 – 6 = 6 liters

Remaining water = 8 – 4 = 4 liters

10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.

The ratio of milk and water in the new mixture = 16:4 = 4:1

If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.

Amount of water removed = 2 liters.

Remaining milk = (16 – 8) = 8 liters.

Remaining water = (4 -2) = 2 liters.

The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1. - Question 50 of 50
##### 50. Question

How much interest can a person get on Rs. 8200 at 17.5% p.a. simple interest for a period of two years and six months?

CorrectI = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50

IncorrectI = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50