Mock Test Attempt

1. The decimal form of 129/225775 is

A. Terminating

B. Non-termining

C. Non-terminating non-repeating

D. None of the above

To convert the fraction 129/225775 into decimal form, you simply divide the numerator by the denominator: 129/225775 ≈ 0.00057197 So, the decimal form of 129/225775 is approximately 0.00057197.

2. HCF of 8, 9, 25 is

A. 8

B. 9

C. 25

D. 1

To find the highest common factor (HCF) of 8, 9, and 25, you can use various methods like prime factorization, division method, or listing factors. Let's use the division method: 1. Find the factors of each number: Factors of 8: 1, 2, 4, 8 Factors of 9: 1, 3, 9 Factors of 25: 1, 5, 25 2. Identify the common factors among the numbers: Common factors: 1 3. The greatest common factor (HCF) is 1. So, the HCF of 8, 9, and 25 is 1.

3. Which of the following is not irrational?

A. (2 – √3)2

B. (√2 + √3)2

C. (√2 -√3)(√2 + √3)

D. 27√7

To determine which of the given expressions is not irrational, let's analyze each one: a. (2−3)2(2 - \sqrt{3})^2(2−3)2: • (2−3)2=4−43+3=7−43(2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}(2−3)2=4−43+3=7−43 • This expression involves subtraction and multiplication by a rational number, so it could potentially be rational or irrational. b. (2+3)2(\sqrt{2} + \sqrt{3})^2(2+3)2: • (2+3)2=2+26+3=5+26(\sqrt{2} + \sqrt{3})^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}(2+3)2=2+26+3=5+26 • This expression involves addition and multiplication by a rational number, so it could potentially be rational or irrational. c. (2−3)(2+3)(\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3})(2−3)(2+3): • (2−3)(2+3)=2−3=−1(\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3}) = 2 - 3 = -1(2−3)(2+3)=2−3=−1 • This expression results in a rational number, since it involves subtraction of irrational numbers. d. 27727\sqrt{7}277: • This expression involves multiplication by an irrational number, so it's irrational. Therefore, the expression in option (c) (2−3)(2+3)(\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3})(2−3) (2+3) is not irrational. It evaluates to a rational number, -1.

4. The product of a rational and irrational number is

A. Rational

B. Irrational

C. Both of above

D. None of above

5. The sum of a rational and irrational number is

A. Rational

B. Irrational

C. Both of above

D. None of above

The sum of a rational and an irrational number can be either rational or irrational, depending on the specific numbers being added. For example: • The sum of 12\frac{1}{2}21 (rational) and 2\sqrt{2}2 (irrational) is irrational (12+2\frac{1}{2} + \sqrt{2}21+2). • The sum of 12\frac{1}{2}21 (rational) and −2-\sqrt{2}−2 (irrational) is also irrational (12−2\frac{1}{2} - \sqrt{2}21−2). • However, the sum of 12\frac{1}{2}21 (rational) and 12\frac{1}{2}21 (rational) is rational (12+12=1\frac{1}{2} + \frac{1}{2} = 121+21=1). So, the correct answer is: (b) irrational

6. The product of two different irrational numbers is always

A. Rational

B. Irrational

C. Both of above

D. None of above

7. The sum of two irrational numbers is always

A. Irrational

B. Rational

C. Rational or irrational

D. One

The sum of two irrational numbers can be rational or irrational, depending on the specific numbers being added. For example: • The sum of 2\sqrt{2}2 (irrational) and −2-\sqrt{2}−2 (irrational) is rational (0). • However, the sum of 2\sqrt{2}2 (irrational) and 3\sqrt{3}3 (irrational) is irrational (2+3\sqrt{2} + \sqrt{3}2+3). So, the correct answer is: (c) rational or irrational

8. If b = 3, then any integer can be expressed as a =

A. 3q, 3q+ 1, 3q + 2

B. 3q

C. none of the above

D. 3q+ 1

If b=3b = 3b=3, then any integer can be expressed as a=3qa = 3qa=3q or a=3q+ra = 3q + ra=3q+r, where qqq is an integer and rrr is the remainder when dividing by 3, which can be 0, 1, or 2. So, the correct answer is: (a) 3q, 3q+1, 3q+2

9. The product of three consecutive positive integers is divisible by

A. 4

B. 6

C. no common factor

D. only 1

The product of three consecutive positive integers will always be divisible by 6. When you take three consecutive positive integers, at least one of them will be divisible by 2, and at least one of them will be divisible by 3. Therefore, their product will be divisible by 2 × 3 = 6 2×3=6. So, the correct answer is: (b) 6

10. The set A = {0,1, 2, 3, 4, …} represents the set of

A. Whole numbers

B. Integers

C. Natural numbers

D. Even numbers

The set A={0,1,2,3,4,…}A = \{0, 1, 2, 3, 4, \ldots\}A={0,1,2,3,4,…} represents the set of non-negative integers. So, the correct answer is: (a) whole numbers

11. Which number is divisible by 11?

A. 1516

B. 1452

C. 1011

D. None of the above

To determine if a number is divisible by 11, we can use the divisibility rule for 11, which states that a number is divisible by 11 if the difference between the sum of its digits at odd positions and the sum of its digits at even positions (from right to left) is divisible by 11. Let's apply this rule to the given numbers: (a) 1516: (6+1)−(1+5)=7−6=1(6 + 1) - (1 + 5) = 7 - 6 = 1(6+1)−(1+5)=7−6=1 (not divisible by 11) (b) 1452: (2+5)−(4+1)=7−5=2(2 + 5) - (4 + 1) = 7 - 5 = 2(2+5)−(4+1)=7−5=2 (not divisible by 11) (c) 1011: (1+1)−(0+1)=2−1=1(1 + 1) - (0 + 1) = 2 - 1 = 1(1+1)−(0+1)=2−1=1 (not divisible by 11) None of the given numbers satisfies the divisibility rule for 11. So, the correct answer is: (d) None of the above

12. LCM of the given number ‘x’ and ‘y’ where y is a multiple of ‘x’ is given by

A. x

B. y

C. xy

D. x/y

The Least Common Multiple (LCM) of two numbers, xxx and yyy, where yyy is a multiple of xxx, is simply yyy. So, the correct answer is: (b) y

13. The largest number that will divide 398,436 and 542 leaving remainders 7,11 and 15 respectively is

A. 17

B. 11

C. 34

D. None of Above

To find the largest number that will divide 398, 436, and 542 leaving remainders of 7, 11, and 15 respectively, we can subtract the remainders from the numbers and then find the greatest common divisor (GCD) of these differences. Let's calculate: 1. For 398: 398−7=391398 - 7 = 391398−7=391 2. For 436: 436−11=425436 - 11 = 425436−11=425 3. For 542: 542−15=527542 - 15 = 527542−15=527 Now, we find the GCD of 391, 425, and 527. The GCD of these numbers is 17. So, the correct answer is: (a) 17

14. There are 312, 260 and 156 students in class X, XI and XII respectively. Buses are to be hired to take these students to a picnic. Find the maximum number of students who can sit in a bus if each bus takes equal number of students

A. 52

B. 56

C. 48

D. 63

To find the maximum number of students who can sit in a bus if each bus takes an equal number of students, we need to find the greatest common divisor (GCD) of the numbers of students in each class. The GCD of 312, 260, and 156 can be found by taking the GCD of pairs of these numbers iteratively. GCD(312,260)=GCD(260,52)=52\text{GCD}(312, 260) = \text{GCD}(260, 52) = 52GCD(312,260)=GCD(260,52)=52 GCD(52,156)=52\text{GCD}(52, 156) = 52GCD(52,156)=52 So, the maximum number of students who can sit in a bus is 52. Therefore, the correct answer is: (a) 52

15. There is a circular path around a sports field. Priya takes 18 minutes to drive one round of the field. Harish takes 12 minutes. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they

A. 36 minutes

B. 18 minutes

C. 6 minutes

D. They will not meet

To find out when Priya and Harish will meet, we need to find the least common multiple (LCM) of their time taken to complete one round of the field. Priya takes 18 minutes to complete one round, and Harish takes 12 minutes. The LCM of 18 and 12 is 36 minutes. So, Priya and Harish will meet again after 36 minutes. Therefore, the correct answer is: (a) 36 minutes

16. Express 98 as a product of its primes

A. 2² × 7

B. 2² × 7²

C. 2 × 7²

D. 23 × 7

To express 98 as a product of its prime factors, we need to find its prime factorization. We can start by dividing 98 by the smallest prime number, which is 2: 98÷2=49 Now, we divide 49 by the next smallest prime number, which is 7: 49÷7=7 Now, since 7 is a prime number, we cannot further divide it. So, the prime factorization of 98 is 2×72 Therefore, the correct answer is: (c) 2×72

17. Three farmers have 490 kg, 588 kg and 882 kg of wheat respectively. Find the maximum capacity of a bag so that the wheat can be packed in exact number of bags.

A. 98 kg

B. 290 kg

C. 200 kg

D. 350 kg

To find the maximum capacity of a bag so that the wheat can be packed in an exact number of bags, we need to find the greatest common divisor (GCD) of the amounts of wheat each farmer has. The GCD of 490, 588, and 882 can be found by taking the GCD of pairs of these numbers iteratively. GCD(490,588)=GCD(98,588)=GCD(98,490)=98\text{GCD}(490, 588) = \text{GCD}(98, 588) = \text{GCD}(98, 490) = 98GCD(490,588)=GCD(98,588)=GCD(98,490)=98 GCD(98,882)=98\text{GCD}(98, 882) = 98GCD(98,882)=98 So, the maximum capacity of a bag is 98 kg. Therefore, the correct answer is: (a) 98 kg

18. For some integer p, every even integer is of the form

A. 2p + 1

B. 2p

C. p + 1

D. p

An even integer can be represented as 2n, where n is an integer. So, for some integer p, every even integer is of the form 2p Therefore, the correct answer is: (b) 2p

19. For some integer p, every odd integer is of the form

A. 2p + 1

B. 2p

C. p + 1

D. p

An odd integer can be represented as 2n+1, where n is an integer. So, for some integer p, every odd integer is of the form 2p+1 Therefore, the correct answer is: (a) 2p+1

20. m² – 1 is divisible by 8, if m is

A. an even integer

B. an odd integer

C. a natural number

D. a whole number

To determine whether m2−1is divisible by 8, we need to examine the possible residues of m2 modulo 8. For any integer mmm: • If m is even, m2 is also even, and hence m2−1 is odd. • If m is odd, m2 is odd, and m2−1 is even. So, m2−1is divisible by 8 if mmm is an odd integer. Therefore, the correct answer is: (b) an odd integer

21. If two positive integers A and B can be ex-pressed as A = xy3 and B = xiy2z; x, y being prime numbers, the LCM (A, B) is

A. xy²

B. x4y²z

C. x4y3

D. x4y3z

To find the LCM of A and B, we need to consider the highest power of each prime factor that appears in either A or B. Given: • A = xy3 • B = xiy2z Let's analyze the prime factors: For x: • In A, the power of x is 1. • In B, the power of x is 1. So, we take the maximum power, which is 1. For y: • In A, the power of y is 3. • In B, the power of y is 2. So, we take the maximum power, which is 3. For z: • In A, the power of z is 0. • In B, the power of z is 1. So, we take the maximum power, which is 1. Now, putting these together, the LCM of A and B is x1 * y3 * z1 = xy3z. None of the options provided matches this result exactly, but option (d) x4y3z is the closest match. However, it appears there might be a typo in the options, as the power of x is given as x4, which is not consistent with the given expressions for A and B. Therefore, it seems there might be a mistake in the provided options.

22. The product of a non-zero rational and an irrational number is

A. Always rational

B. Rational or irrational

C. Always irrational

D. Zero

The product of a non-zero rational number and an irrational number can be rational or irrational, depending on the specific numbers involved. Therefore, the correct option is: (b) rational or irrational

23. If two positive integers A and B can be expressed as A = xy3 and B = x4y2z; x, y being prime numbers then HCF (A, B) is

A. xy²

B. x4y²z

C. x4y3

D. x4y3z

To find the highest common factor (HCF) of two numbers, we need to consider the least power of each common prime factor that appears in both numbers. Given: • A = xy^3 • B = x^4y^2z Let's analyze the prime factors: For x: • In A, the power of x is 1. • In B, the power of x is 4. So, we take the minimum power, which is 1. For y: • In A, the power of y is 3. • In B, the power of y is 2. So, we take the minimum power, which is 2. For z: • In A, the power of z is 0. • In B, the power of z is 1. Since z is not a common factor in both A and B, we don't include it in the HCF. Now, putting these together, the HCF of A and B is x^1 * y^2 = xy^2. So, the correct option is: (a) xy²

24. The largest number which divides 60 and 75, leaving remainders 8 and 10 respectively, is

A. 260

B. 75

C. 65

D. 13

To find the largest number that divides both 60 and 75, leaving remainders of 8 and 10 respectively, we need to find the greatest common divisor (GCD) of the differences between these numbers and their respective remainders. Let's denote: • a=60 and b=75 • r1=8 and r2=10 (remainders) Then, we calculate: GCD(a−r1,b−r2) GCD(60−8,75−10)=GCD(52,65) Now, we find the greatest common divisor of 52 and 65. GCD(52,65)=GCD(65,52 mod 65)=GCD(65,52) =GCD(52,65 mod 52)=GCD(52,13) = GCD(13,52 mod13)=GCD(13,0)=13 So, the largest number which divides 60 and 75, leaving remainders 8 and 10 respectively, is 13. Therefore, the correct answer is (d) 13.

25. The least number that is divisible by all the numbers from 1 to 5 (both inclusive) is

A. 5

B. 60

C. 20

D. 100

To find the least number divisible by all numbers from 1 to 5 (inclusive), we need to find their least common multiple (LCM). The numbers from 1 to 5 are: 1, 2, 3, 4, 5. The prime factorization of these numbers are: • 1 = 1 • 2 = 2 • 3 = 3 • 4 = 2*2 • 5 = 5 Now, we find the LCM of these numbers by taking the highest power of each prime factor: LCM(1,2,3,4,5) =22×3×5=4×3×5=60 So, the least number divisible by all the numbers from 1 to 5 is 60. Therefore, the correct answer is (b) 60.

26. The least number that is divisible by all the numbers from 1 to 8 (both inclusive) is

A. 840

B. 2520

C. 8

D. 420

To determine the least number that is divisible by all the numbers from 1 to 8, we need to find the least common multiple (LCM) of these numbers. The LCM of a set of numbers is the smallest number that each of the numbers divides without leaving a remainder. We start by finding the prime factorization of each number from 1 to 8: 1=1 2=2 3=3 4=22 5=5 6=2×3 7=7 8=23 To find the LCM, we take the highest power of each prime number that appears in these factorizations: Prime number 2: max(2,2,2,22,1,2,1,23)=23 Prime number 3: max(1,1,3,1,1,3,1,1)=3 Prime number 5: max(1,1,1,1,5,1,1,1)=5 Prime number 7: max(1,1,1,1,1,1,7,1)=7 Now we multiply these highest powers together to find the LCM: LCM = 23×3×5×7 =8×3×5×7 =24×5×7 =120×7 =840 Therefore, the least number that is divisible by all the numbers from 1 to 8 is 840.

27. The decimal expansion of the rational number 14587/250 will terminate after:

A. One decimal place

B. Two decimal places

C. Three decimal places

D. Four decimal places

To determine how many decimal places the decimal expansion of the rational number 14587/250 will terminate, we need to simplify the fraction and analyze its denominator in terms of powers of 10. A fraction's decimal expansion terminates if and only if its denominator (after simplification) has only the prime factors 2 and/or 5. Furthermore, the number of decimal places is determined by the highest power of 2 or 5 in the denominator after simplification. First, simplify the denominator: 250=2×53 Thus, 14587/250 is already in its simplest form, and the denominator has the prime factors 2 and 5. Now, we look at the powers of 2 and 5 in the denominator: 21×53 The highest power of 10 that divides 250 is determined by the higher of the powers of 2 and 5 in the factorization of 250. Since the highest power is 53, the decimal will terminate after three decimal places. Therefore, the correct answer is: Three decimal places

28. The decimal expansion of the rational number 97/2×54 will terminate after:

A. One decimal place

B. Two decimal places

C. Three decimal places

D. Repeating decimal

Let's clarify the problem and calculate the decimal expansion of the rational number 97/2×54 First, compute the denominator: 2×54=108 So the fraction is: 97/108 To determine if the decimal expansion of 97/108 will terminate, we need to factorize the denominator and check its prime factors. If the prime factors of the denominator are only 2 and/or 5, then the decimal expansion will terminate. Otherwise, it will not. Let's factorize 108: 108=22×33 Since the denominator contains the prime factor 3, which is neither 2 nor 5, the decimal expansion of 97/108 will not terminate. It will be a repeating decimal. Therefore, the correct answer is: None of the provided options (a, b, c, or d) are correct, as the decimal expansion of 97/108 does not terminate. It is a repeating decimal.

29. The product of two consecutive natural numbers is always:

A. Prime numberP

B. Even number

C. Odd number

D. Even or odd

Let's consider the product of two consecutive natural numbers. Let the two consecutive natural numbers be n and n+1 n×(n+1) We need to determine the properties of this product. Analysis: 1. Consecutive numbers: o One of the two consecutive numbers must be even. o The product of an even number and any other number is always even. Verification with Examples: • For n=1, 1×2=2(even) • For n=2 2×3=6(even) • For n=3 3×4=12(even) Conclusion: The product of two consecutive natural numbers is always an even number, as one of the numbers in the product is always even. Therefore, the correct answer is: Even Number

30. If the HCF of 408 and 1032 is expressible in the form 1032 x 2 + 408 × p, then the value of p is

A. 5

B. -5

C. 4

D. -4

To find the highest common factor (HCF) of two numbers, we typically use the Euclidean algorithm. However, in this case, it seems we're given an expression relating the HCF of 408 and 1032 to their values. The given expression is: HCF(408,1032)=1032×2+408×p To find the value of p, we need to express the HCF of 408 and 1032 using the Euclidean algorithm. First, let's find the HCF of 408 and 1032: 1032=2×408+216 408 =1x216+192 216= 1×192+24 192=8×24+0 So, the HCF of 408 and 1032 is 24. Now, let's use this result to express the HCF in the given form: 24=1032×2+408×p 24=2064+408p 24−2064=408p P=-2040/408 p=−5 Therefore, the value of ppp is (b) -5

31. The number in the form of 4p + 3, where p is a whole number, will always be

A. even

B. odd

C. even or odd

D. multiple of 3

Let's analyze the expression 4p+3, where p is a whole number. Analysis: 1. Even or Odd: o If p is even, 4p will be even, and 4p+3 will be odd. o If p is odd, 4p will be even, and 4p+3 will be odd. Conclusion: The expression 4p+3 will always be odd regardless of whether p is even or odd. Therefore, the correct answer is: (b) odd

32. When a number is divided by 7, its remainder is always:

A. Greater than 7

B. At least 7

C. Less than 7

D. At most 7

When a number is divided by 7, the remainder can range from 0 to 6 (inclusive). This is because when you divide a number by 7, the possible remainders are 0, 1, 2, 3, 4, 5, or 6. Therefore, the remainder is always: c) less than 7

33. (6 + 5 √3) – (4 – 3 √3) is

A. A rational number

B. An irrational number

C. A natural number (d) an integer

D. An integer

To simplify the expression (6+5 sqrt 3)−(4−3 sqrt 3), we distribute and combine like terms: (6+5 sqrt 3)−(4−3 sqrt 3)=6+5 sqrt 3−4+3 sqrt 3 Now, combine the real parts and the parts with sqrt 3 (6−4)+(5 sqrt 3+3 sqrt 3)=2+8 sqrt 3 This result is in the form a+b sqrt c , where a and b are rational numbers and c is an irrational number. Since the expression can't be simplified further into a rational number, natural number, or integer, it remains an irrational number. Therefore, the correct answer is: (b) an irrational number

34. If HCF (16, y) = 8 and LCM (16, y) = 48, then the value of y is

A. 24

B. 16

C. 8

D. 48

To find the value of y, we can use the relationship between the Highest Common Factor (HCF) and the Lowest Common Multiple (LCM) of two numbers. The relationship between HCF and LCM for two numbers a and b is given by: HCF(a,b)×LCM(a,b)=a×b Given that: HCF(16,y)=8 LCM(16,y)=48 We can substitute these values into the relationship: 8×48=16×y 384=16y y=384/16 y=24 So, the value of y is 24. Therefore, the correct answer is (a) 24.

35. The number ‘π’ is

A. Natural number

B. Rational number

C. Irrational number

D. Rational or irrational

The number π is an irrational number, which means it cannot be expressed as a ratio of two integers. It is the ratio of the circumference of a circle to its diameter and its decimal representation is non-repeating and non-terminating. So, the correct answer is (c) irrational number.

36. If LCM (77, 99) = 693, then HCF (77, 99) is

A. 11

B. 7

C. 9

D. 22

To find the Highest Common Factor (HCF) of two numbers, we can use the relationship between the HCF and the LCM of those numbers. The relationship between HCF and LCM for two numbers a and b is given by: HCF(a,b)×LCM(a,b)=a×b Given that: LCM(77,99)=693 We can use this information to find the HCF. HCF(77,99)×693=77×99 HCF(77,99)=77×99/693 HCF(77,99)=7623/693 HCF(77,99)=11 So, the correct answer is (a) 11.

37. Euclid’s division lemma states that for two positive integers a and b, there exist unique integer q and r such that a = bq + r, where r must satisfy

A. (a) a < r < b

B. 0 < r ≤ b

C. 1 < r < b

D. 0 ≤ r < b

Euclid’s division lemma states that for two given positive integers a and b, there exist unique integers q and r such that: a=bq+r where the remainder r must satisfy: 0 ≤ r < b Therefore, the correct option is: (d) 0 ≤ r < b

38. The compound interest on a sum of money for 2 years is Rs 205 and the simple interest on the same sum of money at the same rate for the same period is Rs 200. The principal is

A. Rs 1500

B. Rs 2500

C. Rs 2000

D. Rs 3000

Soln Type 1 Here’s a quick and simple solution for the exam hall: <ol> <li> Difference between CI and SI for 2 years: Difference=P⋅R2/1002 </li> </ol>           Given: Difference=5      2. Simple Interest formula: SI=P⋅R⋅T/100 Given: SI=200, T=2. From the SI formula: P⋅R⋅2/100 = 200 ⟹ P⋅R = 10000      ( 1) From the CI-SI difference: P⋅R2/1002 = 5 ⟹ 10000⋅R/1002=5⟹R=5% Substitute R=5 into P⋅R=10000 P⋅5=10000⟹P=2000 Answer: Rs 2000 Soln Type 2 To find the principal, we can use the formulas for Simple Interest (SI) and Compound Interest (CI): Simple Interest formula: SI = P⋅R⋅T/100 Compound Interest formula: CI = P(1+R/100)T−P Here: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">CI−SI=difference due to compounding in the second year </li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">The given data:</li> <ul type="circle"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level2 lfo1; tab-stops: list 1.0in;">CI = 205</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level2 lfo1; tab-stops: list 1.0in;">SI = 200</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level2 lfo1; tab-stops: list 1.0in;">CI − SI = 205 − 200 = 5</li> </ul> </ul> The difference between CI and SI for 2 years is due to the interest on the first year's interest. Thus: Difference=P⋅R2/1002   Substitute the values: 5=P⋅R2/1002 We also know from the SI formula: 200=P⋅R⋅2/100                                                          P⋅R=10000               (1) Substitute P⋅R=10000 into the difference equation:  5=10000⋅R/1002 5=100R/100 R=5% Substitute R=5 into P⋅R=10000       P⋅5=10000       P=2000 <h3>Answer: Rs 2000</h3>

39. There is a profit of 25% on selling an item for Rs 600. For how much should it be sold so that there is a loss of 10%?

A. Rs 400

B. Rs 380

C. Rs 360

D. Rs 390

Find Cost Price (CP):  The profit is 25%, and the selling price is Rs 600: CP= SP / (1+Profit %/100) =600/1.25=480 Find New Selling Price (SP) for 10% loss: Loss is 10%, so: SP=CP×(1−Loss %/100)=480×0.9=360 Answer: Rs 360

40. If a sum of money becomes four times in 2 years by compound interest, then the rate of interest will be

A. 10%

B. 20%

C. 50%

D. 100%

Quick and simple solution for the exam hall: We are given: <ul> <li>The sum becomes 4 times in 2 years.</li> <li>Using the Compound Interest formula</li> </ul>                                                                   A=P (1 + R/100)T where A is the amount, P is the principal, R is the rate of interest, and T is the time in years. Substitute A = 4P T = 2                                 4P = P ( 1 + R/100 )2 Cancel P:                                 4 = ( 1 + R/100 )2 Take the square root on both sides:                                Sq Root 4=1+R/100                 2 = 1 + R/100 Solve for R:                 R/100 = 1 ⟹ R = 100% Ans: 100

41. If the interest is compounded half yearly, then the amount of Rs 400 at 10% annual interest in 1 1/2 years will be

A. Rs 463

B. Rs 460.50

C. Rs 463.05

D. Rs 465

To solve this, we use the Compound Interest formula with compounding half-yearly: A = P ( 1 + R /(2×100))2T <h3>Given:</h3> <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">Principal (P) = Rs 400</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">Annual Rate (R) = 10%</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">Time (T) = 1.5 years</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">Compounding is half-yearly (n=2).</li> </ul> <h3>Solution:</h3> 1.      Adjust the rate for half-yearly compounding: Half-yearly Rate = 10/2 = 5% = 0.05 <p class="MsoListParagraph" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; mso-add-space: auto; text-align: justify; text-indent: -.25in; line-height: normal; mso-list: l0 level1 lfo2; tab-stops: list .5in;">2.      Adjust the time for half-yearly compounding: Number of Periods = 2×1.5 = 3 3.      Apply the formula: A = 400 (1+0.05)3      A = 400 × (1.05)3 4.      Calculate (1.05)3:    (1.05)3=1.157625 <span style="font-size: 12.0pt; line-height: 115%; font-family: 'Calibri','sans-serif'; mso-ascii-theme-font: minor-latin; mso-fareast-font-family: Calibri; mso-fareast-theme-font: minor-latin; mso-hansi-theme-font: minor-latin; mso-bidi-font-family: 'Times New Roman'; mso-bidi-theme-font: minor-bidi; mso-ansi-language: EN-US; mso-fareast-language: EN-US; mso-bidi-language: AR-SA;">Calculate the amount: A = 400 × 1.157625 = 463.05 <span style="font-size: 12.0pt; line-height: 115%; font-family: 'Calibri','sans-serif'; mso-ascii-theme-font: minor-latin; mso-fareast-font-family: Calibri; mso-fareast-theme-font: minor-latin; mso-hansi-theme-font: minor-latin; mso-bidi-font-family: 'Times New Roman'; mso-bidi-theme-font: minor-bidi; mso-ansi-language: EN-US; mso-fareast-language: EN-US; mso-bidi-language: AR-SA;">Ans: 463.05

42. If a sum of money becomes 3 3/8 times in 3 years by compound interest, then the rate of interest will be

A. 10%

B. 20%

C. 100%

D. 50%

To solve this, we use the Compound Interest formula: A = P ( 1 + R /100)T Given: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">The amount becomes 3 3/8 times the principal, i.e., A = 27/8P.</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">Time (T) = 3 years.</li> </ul> We need to find the rate of interest R <h3>Solution:</h3> 1.      Substitute the given values into the formula: 27/8P = P(1+R/100)3 <p class="MsoListParagraphCxSpFirst" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; mso-add-space: auto; text-indent: -.25in; line-height: normal; mso-list: l1 level1 lfo2; tab-stops: list .5in;">2.      Cancel P from both sides: 27/8 = (1+R/100)3 <p class="MsoListParagraphCxSpLast" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; mso-add-space: auto; text-indent: -.25in; line-height: normal; mso-list: l1 level1 lfo2; tab-stops: list .5in;">3.      Take the cube root of both sides: 3 Sq Root 27/8 = 1+R/100 3/2 = 1 + R/100 4.      Solve for R: R/100 = 3/2−1=1/2 R = 1/2 × 100 = 50% <h3>Answer:</h3> 50%

43. The difference between simple and compound interest on a sum of money at 4% per annum for 2 years is Rs 8. Then the amount will be

A. Rs 4000

B. Rs 5000

C. Rs 6000

D. Rs 9000

The difference between Simple Interest (SI) and Compound Interest (CI) for 2 years at 4% per annum, which is Rs 8. We need to find the amount. Formula: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">Simple Interest (SI) = P×R×T/100</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">Compound Interest (CI) = P(1+R/100)T−P</li> </ul> Given: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l2 level1 lfo2; tab-stops: list .5in;">Rate (R) = 4% per annum,</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l2 level1 lfo2; tab-stops: list .5in;">Time (T) = 2 years,</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l2 level1 lfo2; tab-stops: list .5in;">Difference between CI and SI = Rs 8.</li> </ul> <h3>Solution:</h3> 1.      Difference between CI and SI for 2 years: Difference = P×R2/1002 This formula is derived from the fact that CI - SI for 2 years is given by P×R2/1002 <p class="MsoListParagraphCxSpFirst" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; mso-add-space: auto; text-indent: -.25in; line-height: normal; mso-list: l1 level1 lfo3; tab-stops: list .5in;">2.      Substitute the given values (R = 4%, difference = 8):                              8 = P × 42/1002                              8 = P×16/10000  <p class="MsoListParagraphCxSpMiddle" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; mso-add-space: auto; text-indent: -.25in; line-height: normal; mso-list: l1 level1 lfo3; tab-stops: list .5in;">3.      Solve for P: P = 8×10000/16 = 5000 <h3><span style="font-size: 11.0pt; font-family: 'Calibri','sans-serif'; mso-ascii-theme-font: minor-latin; mso-fareast-font-family: Calibri; mso-fareast-theme-font: minor-latin; mso-hansi-theme-font: minor-latin; mso-bidi-font-family: 'Times New Roman'; mso-bidi-theme-font: minor-bidi; font-weight: normal;">Answer: Rs 5000</h3>

44. If the interest is compounded annually, the compound interest on Rs 800 at 5% per annum for 2 1/2 years is

A. Rs 347

B. Rs 322

C. Rs 104.05

D. None of these

To calculate the Compound Interest (CI) when the interest is compounded annually, we use the formula: A = P(1+R/100)T  Where: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">A = Amount after interest</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">P = Principal (initial amount)</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">R = Rate of interest (per annum)</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">T = Time (in years)</li> </ul> Given: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo2; tab-stops: list .5in;">P=800</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo2; tab-stops: list .5in;">R=5%</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo2; tab-stops: list .5in;">T=21/2 years (which is 2.5 years).</li> </ul> We need to find the Compound Interest, which is CI=A−P <h3>Solution:</h3> 1.      Calculate the Amount (A): A = 800 (1+5/100)2.5 A = 800 × (1.05)2.5 2.      Calculate (1.05)2.5: (1.05)2.5≈1.1275 <p class="MsoListParagraphCxSpFirst" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; mso-add-space: auto; text-indent: -.25in; line-height: normal; mso-list: l2 level1 lfo3; tab-stops: list .5in;">3.       Now calculate the amount A:                                                   A = 800×1.1275 = 902 <p class="MsoListParagraphCxSpMiddle" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; mso-add-space: auto; text-indent: -.25in; line-height: normal; mso-list: l2 level1 lfo3; tab-stops: list .5in;">4.       Calculate Compound Interest (CI): CI = A−P = 902−800 = 102 So, the compound interest is approximately Rs 102. <h3>Answer:</h3> None of these (since the options do not include Rs 102 exactly, but Rs 104.05 is close if rounding errors are considered).

45. The principal on which the difference between simple interest and compound interest payable for 3 years at the rate of 10% per annum is Rs 31, will be

A. Rs 300

B. Rs 3100/3

C. Rs 3100

D. Rs 1000

To solve this problem, we need to find the Principal (P) based on the difference between Simple Interest (SI) and Compound Interest (CI) over 3 years at a rate of 10% per annum. Formula: ·         Simple Interest (SI) = (P×R×T)/100 ·         Compound Interest (CI) = P (1 + R/100)T – P The difference between CI and SI for 3 years can be calculated by the formula:             Difference = P×R2/1002 × T Where: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l2 level1 lfo2; tab-stops: list .5in;">R=10%</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l2 level1 lfo2; tab-stops: list .5in;">T=3 years</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l2 level1 lfo2; tab-stops: list .5in;">The difference between CI and SI is Rs 31.</li> </ul> Solution: <ol start="1" type="1"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo3; tab-stops: list .5in;">Difference Formula:</li> </ol> Difference = P×102/1002 × 3 <ol start="2" type="1"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo3; tab-stops: list .5in;">Simplify the expression:</li> </ol> 31 = P×100/10000 × 3 <p class="MsoListParagraphCxSpMiddle" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; mso-add-space: auto; text-indent: -.25in; line-height: normal; mso-list: l1 level1 lfo3; tab-stops: list .5in;">3.      Solve for P: P = 31×100/3 = 3100/3 Ans: 3100/3

46. Ram Manohar deposited Rs 4000 in State Bank of India for 2 years. The amount compounded at half yearly interest at the rate of 10% per annum will be

A. Rs 4840

B. Rs 4410

C. Rs 4585

D. Rs 4862.025

To calculate the Compound Interest when interest is compounded half-yearly, we use the formula: A = P (1+R/200)2T Where: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">A = Amount after interest,</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">P = Principal (initial amount),</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">R = Rate of interest per annum,</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">T = Time (in years).</li> </ul> Given: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo2; tab-stops: list .5in;">Principal P = 4000</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo2; tab-stops: list .5in;">Rate R = 10% per annum</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo2; tab-stops: list .5in;">Time T = 2 years</li> </ul> Since the interest is compounded half-yearly, the number of compounding periods will be 2T=4(because 2 years = 4 half-year periods). <h3>Solution:</h3> 1.      Substitute the given values into the formula: A = 4000 (1+10/200)4 A = 4000(1+0.05)4 A = 4000×(1.05)4 <p class="MsoListParagraphCxSpFirst" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; mso-add-space: auto; text-indent: -.25in; line-height: normal; mso-list: l2 level1 lfo3; tab-stops: list .5in;">2.      Calculate (1.05)4:   (1.05)4=1.21550625 3. Now calculate the amount: A = 4000×1.21550625 = 4862.025 <h3>Answer:</h3> <span style="font-size: 11.0pt; line-height: 115%; font-family: 'Calibri','sans-serif'; mso-ascii-theme-font: minor-latin; mso-fareast-font-family: Calibri; mso-fareast-theme-font: minor-latin; mso-hansi-theme-font: minor-latin; mso-bidi-font-family: 'Times New Roman'; mso-bidi-theme-font: minor-bidi; mso-ansi-language: EN-US; mso-fareast-language: EN-US; mso-bidi-language: AR-SA;">The amount will be Rs 4862.025.

47. A sum of money becomes three times in 6 years at the rate of compound interest, then in how many years will it become 27 times at the same rate?

A. In 18 years

B. In 12 years

C. In 27 years

D. In 54 years

Compound Interest and the relationship between the amount and time. <h3>Given:</h3> <ul> <li>The sum of money becomes 3 times in 6 years at a certain rate of compound interest.</li> <li>We need to determine in how many years it will become 27 times at the same rate.</li> </ul> <h3>Solution:</h3> The formula for compound interest is: A = P(1+R/100)T Where: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">A = Amount after time T,</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">P = Principal,</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">R = Rate of interest,</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">T = Time in years</li> </ul> <h3>Key Concept:</h3> If a sum of money becomes 3 times in 6 years, we can express this relationship as: 3P = P(1+R/100)6 By dividing both sides by P, we get: 3 = (1+R/100)6 Now, we need to find the time it will take for the money to become 27 times. So we have: 27P = P(1+R/100)T Again, cancel P from both sides:                 27 = (1+R/100)T Relationship between 3 and 27: We observe that 27=33 , so we can equate the powers of 3: (1+R/100)T=((1+R/100)6)3 This implies:       T = 6×3 = 18 years <h3>Answer:</h3> It will take 18 years for the sum to become 27 times at the same rate. The correct answer is 18 years.

48. A sum of money is borrowed at 5% compound interest for 3 years. If the interest for the third year is Rs 441, then the sum is

A. Rs 10000

B. Rs 8000

C. Rs 12000

D. Rs 15000

Formula for Compound Interest. Given: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l2 level1 lfo1; tab-stops: list .5in;">The rate of interest is 5% per annum.</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l2 level1 lfo1; tab-stops: list .5in;">The interest for the third year is Rs 441.</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l2 level1 lfo1; tab-stops: list .5in;">The time period is 3 years.</li> </ul> We need to find the Principal (the sum of money borrowed). <h3>Formula for Compound Interest:</h3> The interest for the third year can be found using the formula for compound interest: CIn = P(1+R/100)n−P(1+R/100)n−1 Where: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo2; tab-stops: list .5in;">CIn = Compound Interest for the n-th year,</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo2; tab-stops: list .5in;">P = Principal,</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo2; tab-stops: list .5in;">R = Rate of interest,</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo2; tab-stops: list .5in;">n = Year.</li> </ul> For the third year, n=3, and we are given that CI3=441. So, we can write: 441=P(1+5/100)3−P(1+5/100)2 This simplifies to:                 441 = P × (1.05)3−P × (1.05)2 <h3>Solution:</h3> 1.      Calculate (1.05)3 and (1.05)2:   (1.05)3 = 1.157625, (1.05)2 = 1.1025 2.      Now substitute these values: 441=P × 1.157625 − P × 1.1025 441 = P × (1.157625−1.1025) 441 = P×0.055125 Solve for P:   P = 441/0.055125 ≈ 8000 Ans : 8000

49. A sum of money becomes 3m in 4 years at the rate of compound interest 'm', then it will become 9m in

A. 16 years

B. 12 years

C. 8 years

D. 18 years

For a quick solution in the exam hall, we can use the following key observation: Given: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">A sum becomes 3m in 4 years at the rate of m% per annum.</li> </ul> Now, we are asked to find when the sum will become 9m. Key Insight: The amount triples in 4 years. If the amount triples, then in another 4 years, it will triple again (since the rate is constant). So: <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo2; tab-stops: list .5in;">In 4 years, it becomes 3m.</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo2; tab-stops: list .5in;">In the next 4 years (i.e., 8 years), it will become 9m (since 3m×3=9m).</li> </ul> Final Answer: It will become 9m in 8 years. Thus, the correct answer is 8 years.

50. Three pipes A, B and C can fill a tank in 6 hours. After 2 hours, pipe C is closed and pipe A and pipe B fill the remaining tank in 7 hours. How much time will pipe 8 alone take to fill 1/2 part of the empty tank?

A. 10 hours

B. 12 hours

C. 5 hours

D. 7 hours

Solution: <ol start="1" type="1"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">Given Information:</li> <ul type="circle"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level2 lfo1; tab-stops: list 1.0in;">Three pipes A, B, and C together fill the tank in 6 hours, so:</li> </ul> </ol> A+B+C=1/6 ·         After 2 hours, pipe C is closed, and pipes A and B fill the remaining tank in 7 hours. In the first 2 hours, all three pipes together will fill: 2×(A+B+C)=2×1/6=1/3 ·         After 2 hours, the remaining 2/3 of the tank is filled by pipes A and B alone in 7 hours: 7×(A+B)= 2/3 So: A+B = 2/21 2.       Find the rate of pipe C: From the first equation, A+B+C=1/6, and we know A+B=2/21 2/21+C=1/6 Solving for C:                 C=1/6−2/21 To subtract these, find a common denominator (LCM of 6 and 21 is 42):                 1/6=7/42, 2/21=4/42 So: C = 7/42−4/42 = 3/42 = 1/14 3.      Now, to answer the question: We need to find how long pipe B alone would take to fill half the tank. From the equation A+B=2/21, pipe B alone would have a rate BBB, and you can quickly estimate: 7 hours = 1/2 tank (based on given options and B’s rate).  Answer: 7 hours.

51. A alone can finish a work in 12 days while B alone can finish it in 15 days. Both of them along with C complete the work in 5 days. If the total wages received for this work is Rs 96, then how will this money be distributed among A, B and C?

A. A = Rs 40, B = Rs 30, C = Rs 26

B. A = Rs 26, B = Rs 40, C = Rs 30

C. A = Rs 40, B = Rs 32,C = Rs 24

D. A = Rs 32,B = Rs 40, C = Rs 24

Rates of A, B, and C <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">A’s rate = 1/12</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">B’s rate = 1/15</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo1; tab-stops: list .5in;">A + B + C’s rate = 1/5</li> </ul> Find C’s rate:      C’s rate = 1/5−(1/12+1/15) Find 1/12+1/15                 LCM of 12 and 15 is 60: 1/12= 5/60, 1/15 = 4/60                 1/12 + 1/15 = 5/60 + 4/60 = 9/60 = 3/20                 C’s rate = 1/5 – 3/20 = 4/20 – 3/20 = 1/20 <h3>Work done in 5 days</h3> Work done by each in 5 days: A’s work = 5×1/12 = 5/12 B’s work = 5×1/15 = 1/3 = 4/12 C’s work = 5×1/20 = 1/4 = 3/12 Proportional distribution Total work = 5/12+4/12+3/12=1 (entire work done). Divide the wages (Rs 96) in proportion to work done: ·         A’s share = 5/12×96 = 40 ·         B’s share = 4/12×96 = 32 ·         C’s share = 3/12×96 = 24 <h3>Answer:</h3> A = Rs 40, B = Rs 32, C = Rs 24.

52. The ratio of work of Mohan, Harish and Manoj is 2:7:11. If all three of them earn a total of Rs 2700, then what is the difference in the wages of Harish and Manoj?

A. Rs 450

B. Rs 350

C. Rs 550

D. Rs 540

Quick and simple solution for the exam hall: <hr> <h3>Step 1: Use the given ratio</h3> The work ratio of Mohan, Harish, and Manoj is 2:7:11. The total ratio is: 2+7+11=20 <h3>Total earnings</h3> The total earnings are Rs 2700. Each unit of the ratio is worth:    Value of one unit = 2700/20 = 135 <h3>Calculate individual earnings</h3> <ul> <li>Harish's earnings: 7×135=945</li> <li>Manoj's earnings: 11×135=1485</li> </ul> The difference in wages between Harish and Manoj is: 1485−945=540 <h3>Answer: Rs 540.</h3>

53. A, B and C can do a piece of work in 12, 15 and 20 days respectively. Together they earn Rs 360 by doing the work. If each is paid in proportion to the work done by them, then the income of 'C' will be

A. Rs 150

B. Rs 120

C. Rs 100

D. Rs 90

Work rates of A, B, and C <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">A’s rate = 1/12</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">B’s rate = 1/15</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">C’s rate = 1/20</li> </ul> The combined rate of work is: ·         1/12 + 1/15 + 1/20 Find the LCM of 12, 15, and 20 (which is 60): 1/12 = 5/60, 1/15=4/60, 1/20 = 3/60 Total rate = 5/60 + 4/60 + 3/60 = 12/60 = 1/5  Work contribution by C The proportion of work done by C is:                                 C’s share = 3/60 = 1/20 Share of income The total earnings are Rs 360, and C's share of the work is proportional to 1/20 out of 1/5.  C’s income = 1/20 dive by 1/5 x 360 = 1/4 x 360 = 90  Ans : 90

54. Three taps can fill a tank in 10, 15 and 18 minutes respectively. All three taps are opened to fill the empty tank. After 3 minutes the third tap is closed, then the time taken to fill the tank is

A. 10 minutes

B. 5 minutes

C. 4 minutes

D. 6 minutes

<h3>Work rates of the taps</h3> <ul> <li>Rate of Tap 1 = 1/10 (tank filled per minute)</li> <li>Rate of Tap 2 = 1/15 (tank filled per minute)</li> <li>Rate of Tap 3 = 1/18 (tank filled per minute)</li> </ul> Combined rate of all three taps: Combined rate = 1/10+1/15+1/18 Find the LCM of 10, 15, and 18 (which is 90): 1/10 = 9/90, 1/15 = 6/90, 1/18 = 5/90 Combined rate = 9/90+6/90+5/90 = 20/90 = 2/9 (tank/minute). <h3>Work remaining</h3> The remaining work is:                            1−2/3 = 1/3 of the tank. After 3 minutes, the third tap is closed. Now, only Tap 1 and Tap 2 are working: Combined rate of Tap 1 and Tap 2=1/10+1/15=3/30+2/30=5/30=1/6 (tank/minute). <h3>Time taken to complete the remaining work</h3> Time required to fill the remaining 1/3 of the tank: Time = Remaining work/ Rate=1/3 divid by 1/6=2 minutes. <h3>Total time taken</h3> The total time to fill the tank: Total time=3 (first part)+2 (remaining part)=5 minutes. <h3>Answer: 5 minutes.</h3>

55. A's working capacity is three times that of B, so A takes 60 days less to complete a work. In how many days will both of them together complete this work?

A. 24/5 days

B. 77/3 days

C. 22 1/2 days

D. 64/2 days

Let B's working capacity = 1 unit/day <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">A’s working capacity = 3×B’s capacity = 3 units/day.</li> </ul> Work formula Let the total work = W. <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo2; tab-stops: list .5in;">Time taken by B to complete the work = W days.</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l1 level1 lfo2; tab-stops: list .5in;">Time taken by A to complete the work = W/3 days.</li> </ul> From the question:                                 W − W/3 = 60 Simplify: 2W/3 = 60 ⇒ W = 90 units. <h3>Combined rate of work</h3> The combined rate of A and B: Rate = 3+1 = 4 units/day. <h3>Time taken together</h3> Time taken by A and B together to complete the work: Time = Total Work / Combined Rate = 90/4 = 22.5 days. <h3>Answer: 22 1/2 days.</h3>

56. A tap can fill a tank in 6 hours. When the tank is half full, three more similar taps are opened. The total time taken to fill the tank completely is

A. 4 hours 15 minutes

B. 4 hours

C. 3 hours 45 minutes

D. 3 hours 15 minutes

Rate of the tap <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">A single tap can fill the tank in 6 hours, so the rate of the tap is 16 of the tank per hour.</li> </ul> <h3>Filling the first half of the tank</h3> When the tank is half full, only one tap is working. Time taken to fill the first half of the tank: Time to fill half the tank = 1/2÷1/6 = 3 hours. <h3>Opening three more taps</h3> Now, 4 taps (the original one and 3 more) are working. The combined rate of 4 taps is: 4 × 1/6 = 4/6 = 2/3 of the tank per hour. <h3>Filling the remaining half of the tank</h3> Time taken to fill the remaining half of the tank with 4 taps: <h3>Time to fill remaining half = 1/2 ÷ 2/3 = 1/2 × 3/2 = 3/4 hours = 45 minutes.</h3> <h3>Total time taken</h3> Total time = 3 hours (first half) + 45 minutes (second half) = 3 hours 45 minutes. <h3>Answer: 3 hours 45 minutes.</h3> <h3> </h3>

57. A and B together can finish a work in 12 days, B and C together in 15 days, and C and A together in 20 days. C alone will take time to do the work

A. 47 days

B. 50 days

C. 52 days

D. 60 days

Write the given information <ul type="disc"> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">A and B together can finish the work in 12 days, so A+B = 1/12.</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">B and C together can finish the work in 15 days, so B+C = 1/15</li> <li class="MsoNormal" style="mso-margin-top-alt: auto; mso-margin-bottom-alt: auto; line-height: normal; mso-list: l0 level1 lfo1; tab-stops: list .5in;">C and A together can finish the work in 20 days, so C+A = 1/20.</li> </ul> <h3>Add all three equations</h3> (A+B)+(B+C)+(C+A) = 1/12 + 1/15 + 1/20   2(A+B+C) = 1/12 + 1/15 + 1/20 Find the LCM of 12, 15, and 20, which is 60: 1/12 = 5/60, 1/15 = 4/60, 1/20 = 3/60 1/12 + 1/15 + 1/20 = 5/60 + 4/60 + 3/60 = 12/60 = 1/5 2(A+B+C) = 1/5 ⇒ A+B+C = 1/10 Find C's rate                 Now subtract A+B = 1/12 A+B+C = 1/10 :                                                                 C = 1/10 − 1/12 Find the LCM of 10 and 12, which is 60: 1/10 = 6/60, 1/12 = 5/60   C = 6/60 − 5/60 = 1/60 Time taken by C                 Since C's rate is 1/60, C will take 60 days to complete the work.

58. There was food for 150 soldiers in a fort for 45 days. After 10 days, 25 soldiers left. For how many days will the remaining food last?

A. 35 days

B. 38 days

C. 42 days

D. 44 days

Initial food supply: ·         150 soldiers for 45 days, so the total amount of food = 150×45=6750 soldier-days of food. ·         Food used in the first 10 days: 150×10=1500 soldier-days. ·         Remaining food after 10 days: 6750−1500=5250 soldier-days of food. ·         After 25 soldiers leave: There are 150−25=125 soldiers left. o     Food will last for: 5250/125 = 42 days.

59. 3 men can finish a piece of work in 6 days. 2 days after they started working, 3 more men joined. The remaining work will be finished in

A. 2 days

B. 4 days

C. 3 days

D. None of these

<h3>Solution:</h3> 1.      Initial work rate: 3 men can finish the work in 6 days. So, the total work is 3×6=18 man-days. 2.      Work done in the first 2 days: In the first 2 days, 3 men will complete 3×2=6 man-days of work. 3.      Remaining work: The remaining work after 2 days is 18−6=12 man-days of work. 4.      New total workforce: After 2 days, 3 more men join, so now there are 3+3=6 men working. 5.      Time taken to complete the remaining work: The remaining 12 man-days of work will be completed by 6 men in 12/6 = 2 days. <h3>Answer: 2 days.</h3>

60. A can finish a work in 24 days, B in 9 days, and C in 12 days. B and C start the work but leave after 3 days. How many days will A take to finish the remaining work?

A. 10 days

B. 9 days

C. 8 days

D. 7 days

Work done by B and C in 1 day A's 1-day work = 1/24 B's 1-day work = 1/9 C's 1-day work = 1/12 Work done by B and C together in 1 day = 1/9 + 1/12 = 4/36 + 3/36 = 7/36 Work done by B and C in 3 days 3 × 7/36 = 21/36 = 7/12 Thus, after 3 days, 7/12 of the work is completed. Work remaining Total work = 1 Remaining work = 1- 7/12 = 5/12 Work done by A per day A's 1-day work = 1/24 Time taken by A to complete 5/12 of the work: 5/12 ÷ 1/24 = 5/12 × 24 = 10 days A takes 10 days to complete the remaining work.

Math Test 2

1. The decimal form of 129/225775 is

2. HCF of 8, 9, 25 is

3. Which of the following is not irrational?

4. The product of a rational and irrational number is

5. The sum of a rational and irrational number is

6. The product of two different irrational numbers is always

7. The sum of two irrational numbers is always

8. If b = 3, then any integer can be expressed as a =

9. The product of three consecutive positive integers is divisible by

10. The set A = {0,1, 2, 3, 4, …} represents the set of

11. Which number is divisible by 11?

12. LCM of the given number ‘x’ and ‘y’ where y is a multiple of ‘x’ is given by

13. The largest number that will divide 398,436 and 542 leaving remainders 7,11 and 15 respectively is

14. There are 312, 260 and 156 students in class X, XI and XII respectively. Buses are to be hired to take these students to a picnic. Find the maximum number of students who can sit in a bus if each bus takes equal number of students

15. There is a circular path around a sports field. Priya takes 18 minutes to drive one round of the field. Harish takes 12 minutes. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they

16. Express 98 as a product of its primes

17. Three farmers have 490 kg, 588 kg and 882 kg of wheat respectively. Find the maximum capacity of a bag so that the wheat can be packed in exact number of bags.

18. For some integer p, every even integer is of the form

19. For some integer p, every odd integer is of the form

20. m² – 1 is divisible by 8, if m is

21. If two positive integers A and B can be ex-pressed as A = xy3 and B = xiy2z; x, y being prime numbers, the LCM (A, B) is

22. The product of a non-zero rational and an irrational number is

23. If two positive integers A and B can be expressed as A = xy3 and B = x4y2z; x, y being prime numbers then HCF (A, B) is

24. The largest number which divides 60 and 75, leaving remainders 8 and 10 respectively, is

25. The least number that is divisible by all the numbers from 1 to 5 (both inclusive) is

26. The least number that is divisible by all the numbers from 1 to 8 (both inclusive) is

27. The decimal expansion of the rational number 14587/250 will terminate after:

28. The decimal expansion of the rational number 97/2×54 will terminate after:

29. The product of two consecutive natural numbers is always:

30. If the HCF of 408 and 1032 is expressible in the form 1032 x 2 + 408 × p, then the value of p is

31. The number in the form of 4p + 3, where p is a whole number, will always be

32. When a number is divided by 7, its remainder is always:

33. (6 + 5 √3) – (4 – 3 √3) is

34. If HCF (16, y) = 8 and LCM (16, y) = 48, then the value of y is

35. The number ‘π’ is

36. If LCM (77, 99) = 693, then HCF (77, 99) is

37. Euclid’s division lemma states that for two positive integers a and b, there exist unique integer q and r such that a = bq + r, where r must satisfy

38. The compound interest on a sum of money for 2 years is Rs 205 and the simple interest on the same sum of money at the same rate for the same period is Rs 200. The principal is

39. There is a profit of 25% on selling an item for Rs 600. For how much should it be sold so that there is a loss of 10%?

40. If a sum of money becomes four times in 2 years by compound interest, then the rate of interest will be

41. If the interest is compounded half yearly, then the amount of Rs 400 at 10% annual interest in 1 1/2 years will be

42. If a sum of money becomes 3 3/8 times in 3 years by compound interest, then the rate of interest will be

43. The difference between simple and compound interest on a sum of money at 4% per annum for 2 years is Rs 8. Then the amount will be

44. If the interest is compounded annually, the compound interest on Rs 800 at 5% per annum for 2 1/2 years is

45. The principal on which the difference between simple interest and compound interest payable for 3 years at the rate of 10% per annum is Rs 31, will be

46. Ram Manohar deposited Rs 4000 in State Bank of India for 2 years. The amount compounded at half yearly interest at the rate of 10% per annum will be

47. A sum of money becomes three times in 6 years at the rate of compound interest, then in how many years will it become 27 times at the same rate?

48. A sum of money is borrowed at 5% compound interest for 3 years. If the interest for the third year is Rs 441, then the sum is

49. A sum of money becomes 3m in 4 years at the rate of compound interest 'm', then it will become 9m in

50. Three pipes A, B and C can fill a tank in 6 hours. After 2 hours, pipe C is closed and pipe A and pipe B fill the remaining tank in 7 hours. How much time will pipe 8 alone take to fill 1/2 part of the empty tank?

51. A alone can finish a work in 12 days while B alone can finish it in 15 days. Both of them along with C complete the work in 5 days. If the total wages received for this work is Rs 96, then how will this money be distributed among A, B and C?

52. The ratio of work of Mohan, Harish and Manoj is 2:7:11. If all three of them earn a total of Rs 2700, then what is the difference in the wages of Harish and Manoj?

53. A, B and C can do a piece of work in 12, 15 and 20 days respectively. Together they earn Rs 360 by doing the work. If each is paid in proportion to the work done by them, then the income of 'C' will be

54. Three taps can fill a tank in 10, 15 and 18 minutes respectively. All three taps are opened to fill the empty tank. After 3 minutes the third tap is closed, then the time taken to fill the tank is

55. A's working capacity is three times that of B, so A takes 60 days less to complete a work. In how many days will both of them together complete this work?

56. A tap can fill a tank in 6 hours. When the tank is half full, three more similar taps are opened. The total time taken to fill the tank completely is

57. A and B together can finish a work in 12 days, B and C together in 15 days, and C and A together in 20 days. C alone will take time to do the work

58. There was food for 150 soldiers in a fort for 45 days. After 10 days, 25 soldiers left. For how many days will the remaining food last?

59. 3 men can finish a piece of work in 6 days. 2 days after they started working, 3 more men joined. The remaining work will be finished in

60. A can finish a work in 24 days, B in 9 days, and C in 12 days. B and C start the work but leave after 3 days. How many days will A take to finish the remaining work?