18 + Interview Questions in NUMERICAL ABILITY in NUMBER SYSTEM Page 1 InterviewSolution

1.	How many factors of the number 26 × 38 × 54 × 106 are multiple of 240?1). 5402). 6603). 5944). 792
Answer» The prime factorization of 2⁶ × 3⁸ × 5⁴ × 10⁶ is 2¹² × 3⁸ × 5¹⁰. 240 can be prime-factorized as 2⁴ × 3 × 5. All factors of 2¹² × 3⁸ × 5¹⁰ that can be written as multiples of 240 will be of the form 2⁴ × 3 × 5 × K. 2¹² × 3⁸ × 5¹⁰ = 2⁴ × 3 × 5 × K ⇒ K = 2⁸ × 3⁷ × 5⁹. The number of factors of N that are multiples of 240 is identical to the number of factors of K. Number of factors of K = (8 + 1) (7 + 1) (9 + 1) = 9 × 8 × 10 = 720

1.

How many factors of the number 26 × 38 × 54 × 106 are multiple of 240?1). 5402). 6603). 5944). 792

Answer»

The prime factorization of 2⁶ × 3⁸ × 5⁴ × 10⁶ is 2¹² × 3⁸ × 5¹⁰.

240 can be prime-factorized as 2⁴ × 3 × 5.

All factors of 2¹² × 3⁸ × 5¹⁰ that can be written as multiples of 240 will be of the form 2⁴ × 3 × 5 × K.

2¹² × 3⁸ × 5¹⁰ = 2⁴ × 3 × 5 × K

⇒ K = 2⁸ × 3⁷ × 5⁹.

The number of factors of N that are multiples of 240 is identical to the number of factors of K.

Number of factors of K = (8 + 1) (7 + 1) (9 + 1) = 9 × 8 × 10 = 720

Discussion

2.	The average age of a Husband and his wife is 39 years. Six years ago, the ratio between their ages was 6 ∶5. What is the difference between the present ages of the husband and his wife?1). 12 years2). 6 years3). 3 years4). 4 years
Answer» Let, Present AGES of husband and wife are x years and y years respectively. ? AVERAGE of their ages = 39 ∴ x + y = 2 × 39 ⇒ x + y = 78 ⇒ x = 78 – y----- (1) Six years ago, the ratio between their ages was 6 ? 5 (x – 6) / (y – 6) = 6 : 5 ⇒ 5x – 30 = 6y – 36 Put the value of x from Equation (1) ⇒ 5(78 – y) – 30 = 6y – 36 ⇒ 390 – 5y – 30 = 6y – 36 ⇒ 11y = 360 + 36 ⇒ y = 36 Putting this value in (1) x = 78 – 36 = 42 ∴ REQUIRED difference between their ages = 42 years – 36 years = 6 years

2.

The average age of a Husband and his wife is 39 years. Six years ago, the ratio between their ages was 6 ∶5. What is the difference between the present ages of the husband and his wife?1). 12 years2). 6 years3). 3 years4). 4 years

Answer»

Let,

Present AGES of husband and wife are x years and y years respectively.

? AVERAGE of their ages = 39

∴ x + y = 2 × 39

⇒ x + y = 78

⇒ x = 78 – y----- (1)

Six years ago, the ratio between their ages was 6 ? 5

(x – 6) / (y – 6) = 6 : 5

⇒ 5x – 30 = 6y – 36

Put the value of x from Equation (1)

⇒ 5(78 – y) – 30 = 6y – 36

⇒ 390 – 5y – 30 = 6y – 36

⇒ 11y = 360 + 36

⇒ y = 36

Putting this value in (1)

x = 78 – 36 = 42

∴ REQUIRED difference between their ages = 42 years – 36 years = 6 years

Discussion

3.	Two numbers are in the ratio 5 : 6 and their LCM is 240, the smaller number is1). 162). 203). 244). 36
Answer» LET the number be 5x and 6x Then, their LCM be 30x = 240 ⇒ X = 8 ∴ SMALLER number be 5x = 5 × 8 = 40

3.

Two numbers are in the ratio 5 : 6 and their LCM is 240, the smaller number is1). 162). 203). 244). 36

Answer»

LET the number be 5x and 6x

Then, their LCM be 30x = 240

⇒ X = 8

∴ SMALLER number be 5x = 5 × 8 = 40

Discussion

4.	1). 272). 323). 364). 35
Answer» Let, the bigger digit = x and smaller digit = y According to PROBLEM, ⇒ x + y = 12.… (1) ⇒ x – y = 4.… (2) From (1) and (2) we get, x = 8 and y = 4 ∴ The product of the two DIGITS of thenumber = xy = 8 × 4 = 32

4.

1). 272). 323). 364). 35

Answer»

Let, the bigger digit = x and smaller digit = y

According to PROBLEM,

⇒ x + y = 12.… (1)

⇒ x – y = 4.… (2)

From (1) and (2) we get, x = 8 and y = 4

∴ The product of the two DIGITS of thenumber = xy = 8 × 4 = 32

Discussion

5.	Unit's digits of a 2 digit number is 5 more than the ten's digit, and if we put the digits of the number in reverse order, the new number is 4 less than twice the original number. The number is :-1). 272). 493). 724). 38
Answer» Let the digits of the number be a and b. ∴ the number is 10A + b Given Unit's digits of a 2 digit number is 5 more than the ten's digit ∴ b = 5 + a-----[1] On reversing the digits number OBTAINED = 10b + a Now, this number is 4 less than TWICE the original number ∴ 2(10a + b) – (10b + a) = 4 ⇒ 19a – 8b = 4 -----[2] Putting the value of b = 5 + a from equation [1] in equation [2] ⇒ 19a – 40 – 8a = 4 ⇒ 11a = 44 ⇒ a = 44/11 = 4 ∴ b = 5 + a = 5 + 4 = 9 The original number is 49

5.

Unit's digits of a 2 digit number is 5 more than the ten's digit, and if we put the digits of the number in reverse order, the new number is 4 less than twice the original number. The number is :-1). 272). 493). 724). 38

Answer»

Let the digits of the number be a and b.

∴ the number is 10A + b

Given Unit's digits of a 2 digit number is 5 more than the ten's digit

∴ b = 5 + a-----[1]

On reversing the digits number OBTAINED = 10b + a

Now, this number is 4 less than TWICE the original number

∴ 2(10a + b) – (10b + a) = 4

⇒ 19a – 8b = 4 -----[2]

Putting the value of b = 5 + a from equation [1] in equation [2]

⇒ 19a – 40 – 8a = 4

⇒ 11a = 44

⇒ a = 44/11 = 4

∴ b = 5 + a

= 5 + 4

= 9

The original number is 49

Discussion

6.	1). 72). 363). 494). 6
Answer» Let, the NUMBER is = X According to PROBLEM, ⇒ (3675/x²) × 37 = 2775 ⇒ 3675/x² = 2775/37 ⇒ 3675/x² = 75 ⇒ x² = 49 ⇒ x = 7

6.

1). 72). 363). 494). 6

Answer»

Let, the NUMBER is = X

According to PROBLEM,

⇒ (3675/x²) × 37 = 2775

⇒ 3675/x² = 2775/37

⇒ 3675/x² = 75

⇒ x² = 49

⇒ x = 7

Discussion

7.	Bob reads 157 pages of a 445 page book. He finished the rest in 9 days. How many pages did he read on an average each day while completing the book?1). 322). 333). 424). 43
Answer» Total pages in book = 445 Pages READ = 157 ⇒ REMAINING pages = 445 – 157 = 288 Time taken to read remaining pages = 9 days ∴ Pages read PER day = 288/9 = 32

7.

Bob reads 157 pages of a 445 page book. He finished the rest in 9 days. How many pages did he read on an average each day while completing the book?1). 322). 333). 424). 43

Answer»

Total pages in book = 445

Pages READ = 157

⇒ REMAINING pages = 445 – 157 = 288

Time taken to read remaining pages = 9 days

∴ Pages read PER day = 288/9 = 32

Discussion

8.	1). 6242). 2463). 2644). 426
Answer» Let, TEN’s place DIGIT = x ∴ UNIT’s place digit = 3X ∴ Hundred’s place digit = 2x According to problem, ⇒ x + 3x + 2x = 12 ⇒ x = 2 ∴ Hundred’s place digit = 2x = 4 ∴ Ten’s place digit = x = 2 ∴ Unit’s place digit = 3x = 6 ∴ The 3 digit NUMBER = 426

8.

1). 6242). 2463). 2644). 426

Answer»

Let, TEN’s place DIGIT = x

∴ UNIT’s place digit = 3X

∴ Hundred’s place digit = 2x

According to problem,

⇒ x + 3x + 2x = 12

⇒ x = 2

∴ Hundred’s place digit = 2x = 4

∴ Ten’s place digit = x = 2

∴ Unit’s place digit = 3x = 6

∴ The 3 digit NUMBER = 426

Discussion

9.	Quantity B: Fifth term of a G.P. is 2. Then 75% of the product of its first nine terms:1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B
Answer» First we will find QUANTITY A, Quantity A: Let the number be ‘x’. 85% of the number = 85% of x = 0.85x According to the question, 0.85x + 95 = x – 20 ⇒ 0.15x = 115 ⇒ x = 115/0.15 = 766.6 ∴ The number is 766.6 Now, Quantity B: We know that for a G.P.- n^th term = ar^n-1 Where, a = 1^st term, r = common ratio ⇒ 5^th term = t₅ = ar⁴ = 2 Product of its first NINE terms = (a) × (ar) × (ar²) ×………× (ar⁸) = a⁹r^{1 + 2 + 3 + ….+ 8} = a⁹r³⁶ = (ar⁴)⁹ = 2⁹ = 512 75% of the product of its first nine terms = (75/100) × 512 = 384 ∴ Quantity A > Quantity B

9.

Quantity B: Fifth term of a G.P. is 2. Then 75% of the product of its first nine terms:1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B

Answer»

First we will find QUANTITY A,

Quantity A:

Let the number be ‘x’.

85% of the number = 85% of x = 0.85x

According to the question,

0.85x + 95 = x – 20

⇒ 0.15x = 115

⇒ x = 115/0.15 = 766.6

∴ The number is 766.6

Now,

Quantity B:

We know that for a G.P.-

n^th term = ar^n-1

Where, a = 1^st term, r = common ratio

⇒ 5^th term = t₅ = ar⁴ = 2

Product of its first NINE terms = (a) × (ar) × (ar²) ×………× (ar⁸)

= a⁹r^{1 + 2 + 3 + ….+ 8} = a⁹r³⁶ = (ar⁴)⁹ = 2⁹ = 512

75% of the product of its first nine terms = (75/100) × 512 = 384

∴ Quantity A > Quantity B

Discussion

10.	What is the minimum positive integer that must be added or subtracted with 7806 to make it a perfect square?1). 562). 623). 1084). 115
Answer» √7806 ≈ 88.35 So, NEAREST whole NUMBER are either 88 or 89. In order to get the perfect square of 89, we need to add 89² – 7806 = 115 But in order to get the perfect square of 88, we need to subtract 7806 – 88² = 62. ∴ The MINIMUM POSITIVE integers that must be added or subtracted from 7806 to make it a perfect square is 62.

10.

What is the minimum positive integer that must be added or subtracted with 7806 to make it a perfect square?1). 562). 623). 1084). 115

Answer»

√7806 ≈ 88.35

So, NEAREST whole NUMBER are either 88 or 89.

In order to get the perfect square of 89, we need to add 89² – 7806 = 115

But in order to get the perfect square of 88, we need to subtract 7806 – 88² = 62.

∴ The MINIMUM POSITIVE integers that must be added or subtracted from 7806 to make it a perfect square is 62.

Discussion

11.	How many prime factors are there in the expression 1240 × 1448 × 254?1). 2822). 2373). 1424). 224
Answer» 12⁴⁰ × 14⁴⁸ × 25⁴ ⇒ (2²× 3) ⁴⁰ × (2 × 7) ⁴⁸ × (5²) ⁴ ⇒ (2) ⁸⁰× (3) ⁴⁰ × (2) ⁴⁸ × (7) ⁴⁸ × (5) ⁸ Number of PRIME FACTORS = 80 + 40 + 48 + 48 + 8 = 224

11.

How many prime factors are there in the expression 1240 × 1448 × 254?1). 2822). 2373). 1424). 224

Answer»

12⁴⁰ × 14⁴⁸ × 25⁴

⇒ (2²× 3) ⁴⁰ × (2 × 7) ⁴⁸ × (5²) ⁴

⇒ (2) ⁸⁰× (3) ⁴⁰ × (2) ⁴⁸ × (7) ⁴⁸ × (5) ⁸

Number of PRIME FACTORS = 80 + 40 + 48 + 48 + 8 = 224

Discussion

12.	A milkman produces three kinds of milk. On a particular day, he has 170 litres, 102 litres and 374 litres of the three kinds of milk. He wants to bottle them in bottles of equal sizes so that each of the three varieties of milk would be completely bottled. How many bottle sizes are possible such that the bottle size in terms of litres is an integer?1). 12). 23). 44). 34
Answer» According to QUESTION - ⇒ The no. of bottle SIZES possible = Factors of HCF (170, 102, 374) ⇒ HCF (170, 102, 374) = 34 ⇒ Factors of 34 = 1, 2, 17, 34 ∴ TOTAL no. of bottle sizes possible = 4

12.

A milkman produces three kinds of milk. On a particular day, he has 170 litres, 102 litres and 374 litres of the three kinds of milk. He wants to bottle them in bottles of equal sizes so that each of the three varieties of milk would be completely bottled. How many bottle sizes are possible such that the bottle size in terms of litres is an integer?1). 12). 23). 44). 34

Answer»

According to QUESTION -

⇒ The no. of bottle SIZES possible = Factors of HCF (170, 102, 374)

⇒ HCF (170, 102, 374) = 34

⇒ Factors of 34 = 1, 2, 17, 34

∴ TOTAL no. of bottle sizes possible = 4

Discussion

13.	1). \(\frac{5}{{54}}\)2). \(\frac{7}{{54}}\)3). \(\frac{4}{{45}}\)4). \(\frac{2}{{18}}\)
Answer» Let a and b be two numbers. We know (H.C.F) × (L.C.M) = Product of two numbers Putting H.C.F and L.C.M. as 9 and 54 respectively we get a × b = 9 × 54 Also, a + b = 45 Solving the two EQUATIONS by SUBSTITUTING the value of a from above equation in a × b = 9 × 54 we get a and b as 18 and 27. So, $(\FRAC{1}{a} + \frac{1}{b} = \frac{1}{{18}} + \frac{1}{{27}})$ i.e. 5/54

13.

1). $\frac{5}{{54}}$2). $\frac{7}{{54}}$3). $\frac{4}{{45}}$4). $\frac{2}{{18}}$

Answer»

Let a and b be two numbers.

We know (H.C.F) × (L.C.M) = Product of two numbers

Putting H.C.F and L.C.M. as 9 and 54 respectively we get a × b = 9 × 54

Also, a + b = 45

Solving the two EQUATIONS by SUBSTITUTING the value of a from above equation in a × b = 9 × 54 we get a and b as 18 and 27.

So, $(\FRAC{1}{a} + \frac{1}{b} = \frac{1}{{18}} + \frac{1}{{27}})$

i.e. 5/54

Discussion

14.	An copper wire is sold only in multiple of 7 m and a man requires sever length of wire, each 1 m 70 cm long. To avoid any wastage, he should purchase minimum length of : 1). 119 m2). 11.9 m3). 170 m4). 185 m
Answer» To avoid any wastage MINIMUM length that MUST be PURCHASED would be equal to the least COMMON FACTOR of 7 m and 1 m 70 cm 7 m = 700 cm And 1 m 70 cm = 170 cm LCM of 700 and 170 = 11900 cm 11900 cm = 119 m ∴ To avoid any wastage, he should purchase minimum length of 119 m.

14.

An copper wire is sold only in multiple of 7 m and a man requires sever length of wire, each 1 m 70 cm long. To avoid any wastage, he should purchase minimum length of : 1). 119 m2). 11.9 m3). 170 m4). 185 m

Answer»

To avoid any wastage MINIMUM length that MUST be PURCHASED would be equal to the least COMMON FACTOR of 7 m and 1 m 70 cm

7 m = 700 cm

And 1 m 70 cm = 170 cm

LCM of 700 and 170 = 11900 cm

11900 cm = 119 m

∴ To avoid any wastage, he should purchase minimum length of 119 m.

Discussion

15.	The product of three consecutive even numbers is 2688. The product of the first and the second number is 168. What is five times the third number?1). 802). 1003). 604). 70
Answer» Let, the numbers are = a, a + 2, a + 4 ∴ a (a + 2)(a + 4) = 2688 ……….(1) ∴ a (a + 2) = 168 ………..(2) From (1) and (2) we GET, ⇒ a + 4 = 2688/168 ⇒ a + 4 = 16 ∴ the third number = 16 ∴ Five times the third number = 5 × 16 = 80

15.

The product of three consecutive even numbers is 2688. The product of the first and the second number is 168. What is five times the third number?1). 802). 1003). 604). 70

Answer»

Let, the numbers are = a, a + 2, a + 4

∴ a (a + 2)(a + 4) = 2688 ……….(1)

∴ a (a + 2) = 168 ………..(2)

From (1) and (2) we GET,

⇒ a + 4 = 2688/168

⇒ a + 4 = 16

∴ the third number = 16

∴ Five times the third number = 5 × 16 = 80

Discussion

16.	The smallest number which when diminished by 7, is divisible by 12, 16, 18, 21 and 28 is1). 10152). 10083). 10224). 1032
Answer» If the number is DIVISIBLE by 12, 16, 18, 21 and 28, then it should be divisible by the LCM of 12, 16, 18, 21 and 28. In this case we will evaluate the LCM of 12, 16, 18, 21 and 28 which is 1008. But it is given that the number is only divisible by 12, 16, 18, 21 and 28 after we subtract 7 from it. ∴ The number needs to 1008 + 7 = 1015

16.

The smallest number which when diminished by 7, is divisible by 12, 16, 18, 21 and 28 is1). 10152). 10083). 10224). 1032

Answer»

If the number is DIVISIBLE by 12, 16, 18, 21 and 28, then it should be divisible by the LCM of 12, 16, 18, 21 and 28.

In this case we will evaluate the LCM of 12, 16, 18, 21 and 28 which is 1008.

But it is given that the number is only divisible by 12, 16, 18, 21 and 28 after we subtract 7 from it.

∴ The number needs to 1008 + 7 = 1015

Discussion

17.	1). 182). 153). 104). 12
Answer» MINIMUM NUMBER of rows = Maximum number of balls in each row ∴ HCF of 36, 54 and 90 = 18 ∴ Minimum number of rows = $(\frac{{36}}{{18}} + \;\frac{{54}}{{18}} + \;\frac{{90}}{{18}} = 2 + 3 + 5 = 10)$

17.

1). 182). 153). 104). 12

Answer»

MINIMUM NUMBER of rows = Maximum number of balls in each row

∴ HCF of 36, 54 and 90 = 18

∴ Minimum number of rows = $(\frac{{36}}{{18}} + \;\frac{{54}}{{18}} + \;\frac{{90}}{{18}} = 2 + 3 + 5 = 10)$

Discussion

18.	The difference between two positive numbers is 3. If the sum of their squares is 369, then the sum of the numbers is1). 812). 333). 274). 25
Answer» Let the two numbers be a and b Now as per the given INFORMATION, a – b = 3….(1) a²+ b²= 369…..(2) From (1), a = b + 3 Put the value of a in equation (2) a²+ b²= (b + 3)²+ b²= b²+ 9 + 6b + b²= 369 ⇒ 2b²+ 6b + 9 = 369 ⇒ 2b² + 6b – 360 = 0 ⇒ b² + 3b – 180 =0 ⇒ (b - 12) (b + 15) = 0 ⇒ b = 12(b is not equal to 15 because b is positive) Put the value of b in equation (1) ∴ a = 12 + 3 = 15 Thus sum of two numbers = a + b = 15 + 12 = 27.

18.

The difference between two positive numbers is 3. If the sum of their squares is 369, then the sum of the numbers is1). 812). 333). 274). 25

Answer»

Let the two numbers be a and b

Now as per the given INFORMATION,

a – b = 3….(1)

a²+ b²= 369…..(2)

From (1), a = b + 3

Put the value of a in equation (2)

a²+ b²= (b + 3)²+ b²= b²+ 9 + 6b + b²= 369

⇒ 2b²+ 6b + 9 = 369

⇒ 2b² + 6b – 360 = 0

⇒ b² + 3b – 180 =0

⇒ (b - 12) (b + 15) = 0

⇒ b = 12(b is not equal to 15 because b is positive)

Put the value of b in equation (1)

∴ a = 12 + 3 = 15

Thus sum of two numbers = a + b = 15 + 12 = 27.

Discussion

Explore topic-wise InterviewSolutions in .

How many factors of the number 26 × 38 × 54 × 106 are multiple of 240?1). 5402). 6603). 5944). 792

The average age of a Husband and his wife is 39 years. Six years ago, the ratio between their ages was 6 ∶5. What is the difference between the present ages of the husband and his wife?1). 12 years2). 6 years3). 3 years4). 4 years

Two numbers are in the ratio 5 : 6 and their LCM is 240, the smaller number is1). 162). 203). 244). 36

1). 272). 323). 364). 35

Unit's digits of a 2 digit number is 5 more than the ten's digit, and if we put the digits of the number in reverse order, the new number is 4 less than twice the original number. The number is :-1). 272). 493). 724). 38

1). 72). 363). 494). 6

Bob reads 157 pages of a 445 page book. He finished the rest in 9 days. How many pages did he read on an average each day while completing the book?1). 322). 333). 424). 43

1). 6242). 2463). 2644). 426

Quantity B: Fifth term of a G.P. is 2. Then 75% of the product of its first nine terms:1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B

What is the minimum positive integer that must be added or subtracted with 7806 to make it a perfect square?1). 562). 623). 1084). 115

How many prime factors are there in the expression 1240 × 1448 × 254?1). 2822). 2373). 1424). 224

1). \(\frac{5}{{54}}\)2). \(\frac{7}{{54}}\)3). \(\frac{4}{{45}}\)4). \(\frac{2}{{18}}\)

An copper wire is sold only in multiple of 7 m and a man requires sever length of wire, each 1 m 70 cm long. To avoid any wastage, he should purchase minimum length of : 1). 119 m2). 11.9 m3). 170 m4). 185 m

The product of three consecutive even numbers is 2688. The product of the first and the second number is 168. What is five times the third number?1). 802). 1003). 604). 70

The smallest number which when diminished by 7, is divisible by 12, 16, 18, 21 and 28 is1). 10152). 10083). 10224). 1032

1). 182). 153). 104). 12

The difference between two positive numbers is 3. If the sum of their squares is 369, then the sum of the numbers is1). 812). 333). 274). 25